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LEM, , 1, , MODULE 1, , Systems of Linear, Equations and Matrices, CHAPTER CONTENTS, , INTRODUCTION, , 1.1, , Introduction to Systems of Linear Equations, , 1.2, , Gaussian Elimination, , 1.3, , Matrices and Matrix Operations, , 1.4, 1.5, , Inverses; Algebraic Properties of Matrices, Elementary Matrices and a Method for Finding A−1, , 1.6, , More on Linear Systems and Invertible Matrices, , 1.7, , Diagonal,Triangular, and Symmetric Matrices, , 1.8, , MatrixTransformations, , 2.1, , Determinants by Cofactor expansion, , 2.2, , Evaluation of Determinants by row reduction, , Information in science, business, and mathematics is often organized into rows and, columns to form rectangular arrays called “matrices” (plural of “matrix”). Matrices, often appear as tables of numerical data that arise from physical observations, but they, occur in various mathematical contexts as well. For example, we will see in this chapter, that all of the information required to solve a system of equations such as, 5x + y = 3, 2x − y = 4, is embodied in the matrix, , , , 5, , 1, 2 −1, , , , 3, 4, , and that the solution of the system can be obtained by performing appropriate, operations on this matrix. This is particularly important in developing computer, programs for solving systems of equations because computers are well suited for, manipulating arrays of numerical information. However, matrices are not simply a, notational tool for solving systems of equations; they can be viewed as mathematical, objects in their own right, and there is a rich and important theory associated with, them that has a multitude of practical applications. It is the study of matrices and, related topics that forms the mathematical field that we call “linear algebra.” In this, chapter we will begin our study of matrices., , 1

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2, , Chapter 1 Systems of Linear Equations and Matrices, , 1.1 Introduction to Systems of Linear Equations, Systems of linear equations and their solutions constitute one of the major topics that we, will study in this course. In this first section we will introduce some basic terminology and, discuss a method for solving such systems., , Linear Equations, , Recall that in two dimensions a line in a rectangular xy -coordinate system can be represented by an equation of the form, , ax + by = c (a, b not both 0), and in three dimensions a plane in a rectangular xyz-coordinate system can be represented by an equation of the form, , ax + by + cz = d (a, b, c not all 0), These are examples of “linear equations,” the first being a linear equation in the variables, x and y and the second a linear equation in the variables x , y , and z. More generally, we, define a linear equation in the n variables x1 , x2 , . . . , xn to be one that can be expressed, in the form, (1), a1 x1 + a2 x2 + · · · + an xn = b, where a1 , a2 , . . . , an and b are constants, and the a ’s are not all zero. In the special cases, where n = 2 or n = 3, we will often use variables without subscripts and write linear, equations as, , a1 x + a2 y = b (a1 , a2 not both 0), a1 x + a2 y + a3 z = b (a1 , a2 , a3 not all 0), , (2), (3), , In the special case where b = 0, Equation (1) has the form, , a1 x1 + a2 x2 + · · · + an xn = 0, , (4), , which is called a homogeneous linear equation in the variables x1 , x2 , . . . , xn ., , E X A M P L E 1 Linear Equations, , Observe that a linear equation does not involve any products or roots of variables. All, variables occur only to the first power and do not appear, for example, as arguments of, trigonometric, logarithmic, or exponential functions. The following are linear equations:, , x + 3y = 7, 1, x − y + 3z = −1, 2, , x1 − 2x2 − 3x3 + x4 = 0, x1 + x2 + · · · + xn = 1, , The following are not linear equations:, , x + 3y 2 = 4, sin x + y = 0, , 3x + 2y − xy = 5, , √, , x1 + 2x2 + x3 = 1, , A finite set of linear equations is called a system of linear equations or, more briefly,, a linear system. The variables are called unknowns. For example, system (5) that follows, has unknowns x and y , and system (6) has unknowns x1 , x2 , and x3 ., 5x + y = 3, 2x − y = 4, , 4x1 − x2 + 3x3 = −1, 3x1 + x2 + 9x3 = −4, , (5–6)

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1.1 Introduction to Systems of Linear Equations, , The double subscripting on, the coefficients aij of the unknowns gives their location, in the system—the first subscript indicates the equation, in which the coefficient occurs,, and the second indicates which, unknown it multiplies. Thus,, a12 is in the first equation and, multiplies x2 ., , 3, , A general linear system of m equations in the n unknowns x1 , x2 , . . . , xn can be written, as, , a11 x1 + a12 x2 + · · · + a1n xn = b1, a21 x1 + a22 x2 + · · · + a2n xn = b2, .., .., .., .., ., ., ., ., am1 x1 + am2 x2 + · · · + amn xn = bm, , (7), , A solution of a linear system in n unknowns x1 , x2 , . . . , xn is a sequence of n numbers, s1 , s2 , . . . , sn for which the substitution, , x1 = s1 , x2 = s2 , . . . , xn = sn, makes each equation a true statement. For example, the system in (5) has the solution, , x = 1, y = − 2, and the system in (6) has the solution, , x1 = 1, x2 = 2, x3 = −1, These solutions can be written more succinctly as, , (1, −2) and (1, 2, −1), in which the names of the variables are omitted. This notation allows us to interpret, these solutions geometrically as points in two-dimensional and three-dimensional space., More generally, a solution, , x1 = s1 , x2 = s2 , . . . , xn = sn, of a linear system in n unknowns can be written as, , (s1 , s2 , . . . , sn ), which is called an ordered n-tuple. With this notation it is understood that all variables, appear in the same order in each equation. If n = 2, then the n-tuple is called an ordered, pair, and if n = 3, then it is called an ordered triple., , Linear Systems inTwo and, Three Unknowns, , Linear systems in two unknowns arise in connection with intersections of lines. For, example, consider the linear system, , a1 x + b1 y = c1, a2 x + b2 y = c2, in which the graphs of the equations are lines in the xy-plane. Each solution (x, y) of this, system corresponds to a point of intersection of the lines, so there are three possibilities, (Figure 1.1.1):, 1. The lines may be parallel and distinct, in which case there is no intersection and, consequently no solution., 2. The lines may intersect at only one point, in which case the system has exactly one, solution., 3. The lines may coincide, in which case there are infinitely many points of intersection, (the points on the common line) and consequently infinitely many solutions., In general, we say that a linear system is consistent if it has at least one solution and, inconsistent if it has no solutions. Thus, a consistent linear systemof two equations in

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4, , Chapter 1 Systems of Linear Equations and Matrices, y, , y, , y, , One solution, , No solution, , x, , x, , x, , Figure 1.1.1, , Infinitely many, solutions, (coincident lines), , two unknowns has either one solution or infinitely many solutions—there are no other, possibilities. The same is true for a linear system of three equations in three unknowns, , a1 x + b1 y + c1 z = d1, a2 x + b2 y + c2 z = d2, a3 x + b3 y + c3 z = d3, in which the graphs of the equations are planes. The solutions of the system, if any,, correspond to points where all three planes intersect, so again we see that there are only, three possibilities—no solutions, one solution, or infinitely many solutions (Figure 1.1.2)., , No solutions, (three parallel planes;, no common intersection), , No solutions, (two parallel planes;, no common intersection), , No solutions, (no common intersection), , No solutions, (two coincident planes, parallel to the third;, no common intersection), , One solution, (intersection is a point), , Infinitely many solutions, (intersection is a line), , Infinitely many solutions, (planes are all coincident;, intersection is a plane), , Infinitely many solutions, (two coincident planes;, intersection is a line), , Figure 1.1.2, , We will prove later that our observations about the number of solutions of linear, systems of two equations in two unknowns and linear systems of three equations in, three unknowns actually hold for all linear systems. That is:, Every system of linear equations has zero, one, or infinitely many solutions. There are, no other possibilities.

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1.1 Introduction to Systems of Linear Equations, , 5, , E X A M P L E 2 A Linear System with One Solution, , Solve the linear system, , x−y =1, 2x + y = 6, , x from the second equation by adding −2 times the first, equation to the second. This yields the simplified system, , Solution We can eliminate, , x−y =1, 3y = 4, From the second equation we obtain y = 43 , and on substituting this value in the first, equation we obtain x = 1 + y = 73 . Thus, the system has the unique solution, , x = 73 , y =, , 4, 3, , Geometrically, this means that,  the,  lines represented by the equations in the system, intersect at the single point 73 , 43 . We leave it for you to check this by graphing the, lines., E X A M P L E 3 A Linear System with No Solutions, , Solve the linear system, , x+ y=4, 3x + 3y = 6, , Solution We can eliminate x from the second equation by adding −3 times the first, equation to the second equation. This yields the simplified system, , x+y =, , 4, , 0 = −6, The second equation is contradictory, so the given system has no solution. Geometrically,, this means that the lines corresponding to the equations in the original system are parallel, and distinct. We leave it for you to check this by graphing the lines or by showing that, they have the same slope but different y -intercepts., E X A M P L E 4 A Linear System with Infinitely Many Solutions, , Solve the linear system, , 4x − 2y = 1, 16x − 8y = 4, , Solution We can eliminate x from the second equation by adding −4 times the first, equation to the second. This yields the simplified system, , 4 x − 2y = 1, 0=0, The second equation does not impose any restrictions on x and y and hence can be, omitted. Thus, the solutions of the system are those values of x and y that satisfy the, single equation, 4x − 2y = 1, (8), Geometrically, this means the lines corresponding to the two equations in the original, system coincide. One way to describe the solution set is to solve this equation for x in, terms of y to obtain x = 41 + 21 y and then assign an arbitrary value t (called a parameter)

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6, , Chapter 1 Systems of Linear Equations and Matrices, , In Example 4 we could have, also obtained parametric, equations for the solutions, by solving (8) for y in terms, of x and letting x = t be, the parameter. The resulting, parametric equations would, look different but would, define the same solution set., , to y . This allows us to express the solution by the pair of equations (called parametric, equations), , x=, , 1, 4, , + 21 t, y = t, , We can obtain specific numerical solutions from these equations by substituting,  1  numerical values for the parameter, ., For, example,, 0, yields, the, solution, t, t, =, ,0 , t = 1, 4, , , , yields the solution 43 , 1 , and t = −1 yields the solution − 41 , −1 . You can confirm, that these are solutions by substituting their coordinates into the given equations., , E X A M P L E 5 A Linear System with Infinitely Many Solutions, , Solve the linear system, , x − y + 2z = 5, 2x − 2y + 4z = 10, 3x − 3y + 6z = 15, , Solution This system can be solved by inspection, since the second and third equations, , are multiples of the first. Geometrically, this means that the three planes coincide and, that those values of x , y , and z that satisfy the equation, , x − y + 2z = 5, , (9), , automatically satisfy all three equations. Thus, it suffices to find the solutions of (9)., We can do this by first solving this equation for x in terms of y and z, then assigning, arbitrary values r and s (parameters) to these two variables, and then expressing the, solution by the three parametric equations, , x = 5 + r − 2s, y = r, z = s, Specific solutions can be obtained by choosing numerical values for the parameters r, and s . For example, taking r = 1 and s = 0 yields the solution (6, 1, 0)., Augmented Matrices and, Elementary Row Operations, , As the number of equations and unknowns in a linear system increases, so does the, complexity of the algebra involved in finding solutions. The required computations can, be made more manageable by simplifying notation and standardizing procedures. For, example, by mentally keeping track of the location of the +’s, the x ’s, and the =’s in the, linear system, , a11 x1 + a12 x2 + · · · + a1n xn = b1, a21 x1 + a22 x2 + · · · + a2n xn = b2, .., .., .., .., ., ., ., ., am1 x1 + am2 x2 + · · · + amn xn = bm, , we can abbreviate the system by writing only the rectangular array of numbers, , As noted in the introduction, to this chapter, the term “matrix” is used in mathematics to, denote a rectangular array of, numbers. In a later section, we will study matrices in detail, but for now we will only, be concerned with augmented, matrices for linear systems., , ⎡, a11, ⎢, ⎢a21, ⎢ ., ⎣ .., am1, , a12, a22, .., ., am2, , · · · a1n, · · · a2 n, .., ., · · · amn, , ⎤, b1, ⎥, b2 ⎥, .. ⎥, . ⎦, bm, , This is called the augmented matrix for the system. For example, the augmented matrix, for the system of equations, , ⎡, , x1 + x2 + 2x3 = 9, 2x1 + 4x2 − 3x3 = 1, 3x1 + 6x2 − 5x3 = 0, , is, , 1, , ⎤, , ⎢, ⎣2, , 1, , 2, , 9, , 4, , −3, , 1⎦, , 3, , 6, , −5, , 0, , ⎥

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1.1 Introduction to Systems of Linear Equations, , 7, , The basic method for solving a linear system is to perform algebraic operations on, the system that do not alter the solution set and that produce a succession of increasingly, simpler systems, until a point is reached where it can be ascertained whether the system, is consistent, and if so, what its solutions are. Typically, the algebraic operations are:, 1. Multiply an equation through by a nonzero constant., 2. Interchange two equations., 3. Add a constant times one equation to another., Since the rows (horizontal lines) of an augmented matrix correspond to the equations in, the associated system, these three operations correspond to the following operations on, the rows of the augmented matrix:, 1. Multiply a row through by a nonzero constant., 2. Interchange two rows., 3. Add a constant times one row to another., These are called elementary row operations on a matrix., In the following example we will illustrate how to use elementary row operations and, an augmented matrix to solve a linear system in three unknowns. Since a systematic, procedure for solving linear systems will be developed in the next section, do not worry, about how the steps in the example were chosen. Your objective here should be simply, to understand the computations., E X A M P L E 6 Using Elementary Row Operations, , In the left column we solve a system of linear equations by operating on the equations in, the system, and in the right column we solve the same system by operating on the rows, of the augmented matrix., , ⎡, , x + y + 2z = 9, 2x + 4y − 3z = 1, Add −2 times the first equation to the second, to obtain, , x + y + 2z =, , 9, , 2y − 7z = −17, 3x + 6y − 5z =, , Maxime Bôcher, (1867–1918), , 0, , ⎤, , ⎢, ⎣2, , 1, , 2, , 4, , −3, , 1⎦, , 3, , 6, , −5, , 0, , 3x + 6y − 5z = 0, , 1, , 9, , ⎥, , Add −2 times the first row to the second to, obtain, , ⎡, , 1, , ⎢, ⎣0, , 1, , 2, , 2, , 3, , 6, , −7, −5, , 9, , ⎤, , ⎥, −17⎦, 0, , Historical Note The first known use of augmented matrices appeared, between 200 B.C. and 100 B.C. in a Chinese manuscript entitled Nine, Chapters of Mathematical Art. The coefficients were arranged in, columns rather than in rows, as today, but remarkably the system was, solved by performing a succession of operations on the columns. The, actual use of the term augmented matrix appears to have been introduced by the American mathematician Maxime Bôcher in his book Introduction to Higher Algebra, published in 1907. In addition to being an, outstanding research mathematician and an expert in Latin, chemistry,, philosophy, zoology, geography, meteorology, art, and music, Bôcher, was an outstanding expositor of mathematics whose elementary textbooks were greatly appreciated by students and are still in demand, today., [Image: Courtesy of the American Mathematical Society, www.ams.org]

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8, , Chapter 1 Systems of Linear Equations and Matrices, , Add −3 times the first equation to the third to, obtain, , Add −3 times the first row to the third to obtain, , ⎡, , x + y + 2z =, 9, 2y − 7z = −17, 3y − 11z = −27, Multiply the second equation by, , 1, 2, , x + y + 2z =, , to obtain, , ⎢, ⎣0, , 1, , 2, , 2, , −7, , 0, , 3, , −11, , Multiply the second row by, , ⎡, , 9, , 1, , = − 172, , ⎢, ⎣0, , 3y − 11z = −27, , 0, , y−, , 7, z, 2, , Add −3 times the second equation to the third, to obtain, , x + y + 2z =, , ⎡, , 1, , 9, , − 21 z = − 23, Multiply the third equation by −2 to obtain, , x + y + 2z =, , y−, , 11, z, 2, 7, z, 2, , =, , =, z=, , y, , − 27, , 3, , −11, , ⎢, ⎢0, ⎣, , 2, , 1, , − 27, , 0, , 0, , − 21, , 1, , ⎢, ⎣0, 0, , ⎥, −17⎦, −27, 1, 2, , to obtain, , 9, , ⎤, , ⎥, − 172 ⎦, −27, , ⎤, , 9, , ⎥, − 172 ⎥, ⎦, − 23, , 1, , 2, , 1, , − 27, , 0, , 1, , ⎤, , 9, , ⎥, − 172 ⎦, 3, , Add −1 times the second row to the first to, obtain, , ⎡, , 35, 2, − 172, , ⎢, ⎢0, ⎣, , 0, 1, , 11, 2, − 27, , 0, , 0, , 1, , 1, , 3, , Add −11, times the third equation to the first, 2, and 27 times the third equation to the second to, obtain, , x, , 1, , 1, , ⎡, , 9, , Add −1 times the second equation to the first, to obtain, , +, , 2, , ⎤, , Multiply the third row by −2 to obtain, , y − 27 z = − 172, z= 3, , x, , 1, , 9, , Add −3 times the second row to the third to, obtain, , y − 27 z = − 172, , The solution in this example, can also be expressed as the ordered triple (1, 2, 3) with the, understanding that the numbers in the triple are in the, same order as the variables in, the system, namely, x, y, z., , 1, , ⎤, , 35, 2, ⎥, − 172 ⎥, ⎦, , 3, , Add − 11, times the third row to the first and, 2, times the third row to the second to obtain, , ⎡, , =1, =2, z=3, , 1, , 7, 2, , ⎤, , ⎢, ⎣0, , 0, , 0, , 1, , 1, , 0, , 2⎦, , 0, , 0, , 1, , 3, , ⎥, , The solution x = 1, y = 2, z = 3 is now evident., , Exercise Set 1.1, 1. In each part, determine whether the equation is linear in x1 ,, x2 , and x3 ., (a) x1 + 5x2 −, , √, , 2 x3 = 1, , (c) x1 = −7x2 + 3x3, (e), , 3/5, x1, , − 2x2 + x3 = 4, , 2. In each part, determine whether the equation is linear in x, and y ., , (b) x1 + 3x2 + x1 x3 = 2, , (a) 21/3 x +, , (d) x1−2 + x2 + 8x3 = 5, , (c) cos, , (f ) πx1 −, , (e) xy = 1, , √, , 2 x2 = 7, , 1/3, , π , 7, , √, , 3y = 1, , x − 4y = log 3, , √, , (b) 2x 1/3 + 3 y = 1, (d), , π, 7, , cos x − 4y = 0, , (f ) y + 7 = x

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1.1 Introduction to Systems of Linear Equations, , 3. Using the notation of Formula (7), write down a general linear, system of, , (d), , (a) two equations in two unknowns., (b) three equations in three unknowns., (c) two equations in four unknowns., 4. Write down the augmented matrix for each of the linear systems in Exercise 3., In each part of Exercises 5–6, find a linear system in the unknowns x1 , x2 , x3 , . . . , that corresponds to the given augmented, matrix., , ⎡, , 2, ⎢, 5. (a) ⎣3, 0, , , 6. (a), , ⎤, , 0, −4, 1, , 0, , 3, , −1, , 5, , 2, , 0, , ⎡, , 3, ⎢−4, ⎢, (b) ⎢, ⎣−1, 0, , ⎡, , 0, ⎥, 0⎦, 1, , 0, 0, 3, 0, , 3, ⎢, (b) ⎣7, 0, , −1, −3, −4, , 1, 4, 0, 0, , 1, −2, −1, , −1, −6, , , , 0, 1, −2, , −2, 4, 1, , ⎤, , 5, ⎥, −3⎦, 7, , (c), , x3, , (d), , 2, , , 25 , 2, , , , ,, , 10 2, ,, 7 7, , , , (e), , 5, , , , 7, , , 87 , 0, , 7, , , 227 , 2, , 5, , (c) (5, 8, 1), , , , 11. In each part, solve the linear system, if possible, and use the, result to determine whether the lines represented by the equations in the system have zero, one, or infinitely many points of, intersection. If there is a single point of intersection, give its, coordinates, and if there are infinitely many, find parametric, equations for them., (a) 3x − 2y = 4, 6x − 4 y = 9, , (b) 2x − 4y = 1, 4 x − 8y = 2, , (c) x − 2y = 0, x − 4y = 8, , 12. Under what conditions on a and b will the following linear, system have no solutions, one solution, infinitely many solutions?, 2 x − 3y = a, 4x − 6y = b, , (d) 3v − 8w + 2x − y + 4z = 0, 14. (a) x + 10y = 2, (b) x1 + 3x2 − 12x3 = 3, (c) 4x1 + 2x2 + 3x3 + x4 = 20, (d) v + w + x − 5y + 7z = 0, In Exercises 15–16, each linear system has infinitely many solutions. Use parametric equations to describe its solution set., , (b) 2x1, + 2 x3 = 1, 3x1 − x2 + 4x3 = 7, 6x1 + x2 − x3 = 0, , =1, =2, =3, , 2x1 − 4x2 − x3 = 1, x1 − 3x2 + x3 = 1, 3x1 − 5x2 − 3x3 = 1, ,  13, , 7, , 5, , (b), , (c) −8x1 + 2x2 − 5x3 + 6x4 = 1, , 9. In each part, determine whether the given 3-tuple is a solution, of the linear system, , (a) (3, 1, 1), , , , 87 , 1, , (b) 3x1 − 5x2 + 4x3 = 7, , (b) 6x1 − x2 + 3x3 = 4, 5x2 − x3 = 1, , 8. (a) 3x1 − 2x2 = −1, 4 x1 + 5 x 2 = 3, 7 x1 + 3 x 2 = 2, , x2, , 7, , 13. (a) 7x − 5y = 3, , 2x2, − 3x4 + x5 = 0, −3x1 − x2 + x3, = −1, 6x1 + 2x2 − x3 + 2x4 − 3x5 = 6, , (c) x1, , 5, , In each part of Exercises 13–14, use parametric equations to, describe the solution set of the linear equation., , ⎤, , 3, − 3⎥, ⎥, ⎥, − 9⎦, −2, , In each part of Exercises 7–8, find the augmented matrix for, the linear system., 7. (a) −2x1 = 6, 3x1 = 8, 9x1 = −3, , (a), , 9, , (b) (3, −1, 1), , (c) (13, 5, 2), , (e) (17, 7, 5), , 10. In each part, determine whether the given 3-tuple is a solution, of the linear system, , x + 2y − 2z = 3, 3x − y + z = 1, −x + 5y − 5z = 5, , 15. (a) 2x − 3y = 1, 6 x − 9y = 3, (b), , x1 + 3x2 − x3 = −4, 3x1 + 9x2 − 3x3 = −12, 4, −x1 − 3x2 + x3 =, , 16. (a) 6x1 + 2x2 = −8, 3x1 + x2 = −4, , (b), , 2x − y + 2z = −4, 6x − 3y + 6z = −12, 8, −4 x + 2 y − 4 z =, , In Exercises 17–18, find a single elementary row operation that, will create a 1 in the upper left corner of the given augmented matrix and will not create any fractions in its first row., , ⎡, , −3, , 17. (a) ⎣ 2, 0, , ⎡, , 2, 18. (a) ⎣ 7, −5, , −1, −3, 2, 4, 1, 4, , ⎤, , 2, 3, −3, , 4, 2⎦, 1, , −6, , 8, 3⎦, 7, , 4, 2, , ⎤, , ⎡, , 0, (b) ⎣2, 1, , ⎡, , 7, (b) ⎣ 3, −6, , −1, −9, , ⎤, , −5, , 0, 2⎦, 3, , 3, , −3, , 4, , −4, −1, 3, , −2, 8, , −1, , ⎤, , 2, 1⎦, 4

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10, , Chapter 1 Systems of Linear Equations and Matrices, , In Exercises 19–20, find all values of k for which the given, augmented matrix corresponds to a consistent linear system., 19. (a), , 1, 4, , k, , −4, , 8, , 2, , 20. (a), , 3, −6, , −4, , k, , 8, , 5, , (b), , (b), , 1, 4, , k, , k, , 1, −1, , 4, , 8, , −1, −4, −2, 2, , 21. The curve y = ax 2 + bx + c shown in the accompanying figure passes through the points (x1 , y1 ), (x2 , y2 ), and (x3 , y3 )., Show that the coefficients a , b, and c form a solution of the, system of linear equations whose augmented matrix is, , y, , ⎡ 2, x1, ⎢ 2, ⎣x2, , x1, , 1, , x2, , 1, , ⎥, y2 ⎦, , x32, , x3, , 1, , y3, , y1, , ⎤, , Let x, y, and z denote the number of ounces of the first, second, and third foods that the dieter will consume at the main, meal. Find (but do not solve) a linear system in x, y, and z, whose solution tells how many ounces of each food must be, consumed to meet the diet requirements., 26. Suppose that you want to find values for a, b, and c such that, the parabola y = ax 2 + bx + c passes through the points, (1, 1), (2, 4), and (−1, 1). Find (but do not solve) a system, of linear equations whose solutions provide values for a, b,, and c. How many solutions would you expect this system of, equations to have, and why?, 27. Suppose you are asked to find three real numbers such that the, sum of the numbers is 12, the sum of two times the first plus, the second plus two times the third is 5, and the third number, is one more than the first. Find (but do not solve) a linear, system whose equations describe the three conditions., , True-False Exercises, , y = ax2 + bx + c, , TF. In parts (a)–(h) determine whether the statement is true or, false, and justify your answer., , (x3, y3), (x1, y1), , (a) A linear system whose equations are all homogeneous must, be consistent., , (x2, y2), x, , Figure Ex-21, , 22. Explain why each of the three elementary row operations does, not affect the solution set of a linear system., 23. Show that if the linear equations, , x1 + kx2 = c, , and, , x1 + lx2 = d, , have the same solution set, then the two equations are identical, (i.e., k = l and c = d )., 24. Consider the system of equations, , ax + by = k, cx + dy = l, ex + fy = m, Discuss the relative positions of the lines ax + by = k ,, , cx + dy = l , and ex + fy = m when, (a) the system has no solutions., , (b) the system has exactly one solution., , (b) Multiplying a row of an augmented matrix through by zero is, an acceptable elementary row operation., (c) The linear system, , x− y =3, 2x − 2y = k, cannot have a unique solution, regardless of the value of k ., (d) A single linear equation with two or more unknowns must, have infinitely many solutions., (e) If the number of equations in a linear system exceeds the number of unknowns, then the system must be inconsistent., (f ) If each equation in a consistent linear system is multiplied, through by a constant c, then all solutions to the new system, can be obtained by multiplying solutions from the original, system by c., (g) Elementary row operations permit one row of an augmented, matrix to be subtracted from another., (h) The linear system with corresponding augmented matrix, , (c) the system has infinitely many solutions., 25. Suppose that a certain diet calls for 7 units of fat, 9 units of, protein, and 16 units of carbohydrates for the main meal, and, suppose that an individual has three possible foods to choose, from to meet these requirements:, Food 1: Each ounce contains 2 units of fat, 2 units of, protein, and 4 units of carbohydrates., Food 2: Each ounce contains 3 units of fat, 1 unit of, protein, and 2 units of carbohydrates., Food 3: Each ounce contains 1 unit of fat, 3 units of, protein, and 5 units of carbohydrates., , 2, 0, , −1, 0, , 4, , −1, , is consistent., , Working withTechnology, T1. Solve the linear systems in Examples 2, 3, and 4 to see how, your technology utility handles the three types of systems., T2. Use the result in Exercise 21 to find values of a , b, and c, for which the curve y = ax 2 + bx + c passes through the points, (−1, 1, 4), (0, 0, 8), and (1, 1, 7).

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1.2 Gaussian Elimination, , 11, , 1.2 Gaussian Elimination, In this section we will develop a systematic procedure for solving systems of linear, equations. The procedure is based on the idea of performing certain operations on the rows, of the augmented matrix that simplify it to a form from which the solution of the system, can be ascertained by inspection., , Considerations in Solving, Linear Systems, , When considering methods for solving systems of linear equations, it is important to, distinguish between large systems that must be solved by computer and small systems, that can be solved by hand. For example, there are many applications that lead to, linear systems in thousands or even millions of unknowns. Large systems require special, techniques to deal with issues of memory size, roundoff errors, solution time, and so, forth. Such techniques are studied in the field of numerical analysis and will only be, touched on in this text. However, almost all of the methods that are used for large, systems are based on the ideas that we will develop in this section., , Echelon Forms, , In Example 6 of the last section, we solved a linear system in the unknowns x , y , and z, by reducing the augmented matrix to the form, , ⎡, , 1, ⎢0, ⎣, 0, , 0, 1, 0, , 0, 0, 1, , ⎤, , 1, 2⎥, ⎦, 3, , from which the solution x = 1, y = 2, z = 3 became evident. This is an example of a, matrix that is in reduced row echelon form. To be of this form, a matrix must have the, following properties:, 1. If a row does not consist entirely of zeros, then the first nonzero number in the row, is a 1. We call this a leading 1., 2. If there are any rows that consist entirely of zeros, then they are grouped together at, the bottom of the matrix., 3. In any two successive rows that do not consist entirely of zeros, the leading 1 in the, lower row occurs farther to the right than the leading 1 in the higher row., 4. Each column that contains a leading 1 has zeros everywhere else in that column., A matrix that has the first three properties is said to be in row echelon form. (Thus,, a matrix in reduced row echelon form is of necessity in row echelon form, but not, conversely.), E X A M P L E 1 Row Echelon and Reduced Row Echelon Form, , The following matrices are in reduced row echelon form., , ⎡, , 1, ⎢, ⎣0, 0, , 0, 1, 0, , 0, 0, 1, , ⎤, , ⎡, , 4, 1, ⎥ ⎢, 7⎦ , ⎣0, 0, −1, , 0, 1, 0, , ⎤, , 0, ⎥, 0⎦ ,, 1, , ⎡, , 0, ⎢0, ⎢, ⎢, ⎣0, 0, , 1, 0, 0, 0, , −2, 0, 0, 0, , 0, 1, 0, 0, , ⎤, , 1, 3⎥, ⎥, ⎥,, 0⎦, 0, , 0, 0, , 0, 0, , The following matrices are in row echelon form but not reduced row echelon form., , ⎡, , 1, ⎢, ⎣0, 0, , 4, 1, 0, , −3, 6, 1, , ⎤, , ⎡, , 7, 1, ⎥ ⎢, 2⎦ , ⎣0, 5, 0, , 1, 1, 0, , ⎤, , ⎡, , 0, 0, ⎥ ⎢, 0⎦ , ⎣0, 0, 0, , 1, 0, 0, , 2, 1, 0, , 6, −1, 0, , ⎤, , 0, ⎥, 0⎦, 1

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12, , Chapter 1 Systems of Linear Equations and Matrices, , E X A M P L E 2 More on Row Echelon and Reduced Row Echelon Form, , As Example 1 illustrates, a matrix in row echelon form has zeros below each leading 1,, whereas a matrix in reduced row echelon form has zeros below and above each leading, 1. Thus, with any real numbers substituted for the ∗’s, all matrices of the following types, are in row echelon form:, , ⎡, , 1, ⎢0, ⎢, ⎢, ⎣0, 0, , ⎤, ∗ ∗ ∗, 1 ∗ ∗⎥, ⎥, ⎥,, 0 1 ∗⎦, , ⎡, , 1, ⎢0, ⎢, ⎢, ⎣0, 0, , 0 0 1, , ⎤, ∗ ∗ ∗, 1 ∗ ∗⎥, ⎥, ⎥,, 0 1 ∗⎦, , ⎡, , 1, ⎢0, ⎢, ⎢, ⎣0, 0, , 0 0 0, , ⎤, ∗ ∗ ∗, 1 ∗ ∗⎥, ⎥, ⎥,, 0 0 0⎦, 0 0 0, , ⎡, , 0, ⎢0, ⎢, ⎢, ⎢0, ⎢, ⎣0, 0, , 1, 0, 0, 0, 0, , ⎤, ∗, ∗⎥, ⎥, ⎥, ∗⎥, ⎥, ∗⎦, 0 0 0 0 0 0 1 ∗, , ∗ ∗ ∗ ∗ ∗, 0 1 ∗ ∗ ∗, 0 0 1 ∗ ∗, 0 0 0 1 ∗, , ∗, ∗, ∗, ∗, , All matrices of the following types are in reduced row echelon form:, , ⎡, , ⎤, , ⎡, , ⎤, , ⎡, , 1 0 0 0, 1 0 0 ∗, 1 0, ⎢0 1 0 0⎥ ⎢0 1 0 ∗⎥ ⎢0 1, ⎢, ⎥ ⎢, ⎥ ⎢, ⎢, ⎥, ⎢, ⎥, ⎢, ⎣0 0 1 0⎦ ⎣0 0 1 ∗⎦ ⎣0 0, 0 0 0 1, , 0 0, , 0 0 0 0, , ⎡, , ⎤, , 0, ∗ ∗, ⎢0, ⎢, ∗ ∗⎥, ⎥ ⎢, ⎥ , ⎢0, ⎢, 0 0⎦, ⎣0, 0 0, 0, , 1, 0, 0, 0, 0, , ∗ 0 0 0 ∗, 0 1 0 0 ∗, 0 0 1 0 ∗, 0 0 0 1 ∗, , ∗, ∗, ∗, ∗, , 0, 0, 0, 0, 0 0 0 0 0 0 1, , ∗, ∗, ∗, ∗, , ⎤, ∗, ∗⎥, ⎥, ⎥, ∗⎥, ⎥, ∗⎦, ∗, , If, by a sequence of elementary row operations, the augmented matrix for a system of, linear equations is put in reduced row echelon form, then the solution set can be obtained, either by inspection or by converting certain linear equations to parametric form. Here, are some examples., E X A M P L E 3 Unique Solution, , Suppose that the augmented matrix for a linear system in the unknowns x1 , x2 , x3 , and, x4 has been reduced by elementary row operations to, , ⎡, , 1, ⎢0, ⎢, ⎢, ⎣0, 0, , 0, 1, 0, 0, , 0, 0, 1, 0, , 0, 0, 0, 1, , ⎤, , 3, −1⎥, ⎥, ⎥, 0⎦, 5, , This matrix is in reduced row echelon form and corresponds to the equations, , x1, In Example 3 we could, if, desired, express the solution, more succinctly as the 4-tuple, (3, −1, 0, 5)., , x2, x3, , = 3, = −1, = 0, x4 = 5, , Thus, the system has a unique solution, namely, x1 = 3, x2 = −1, x3 = 0, x4 = 5., E X A M P L E 4 Linear Systems in Three Unknowns, , In each part, suppose that the augmented matrix for a linear system in the unknowns, x , y , and z has been reduced by elementary row operations to the given reduced row, echelon form. Solve the system., , ⎡, , 1, ⎢, (a) ⎣0, 0, , 0, 1, 0, , 0, 2, 0, , ⎤, , 0, ⎥, 0⎦, 1, , ⎡, , 1, ⎢, (b) ⎣0, 0, , 0, 1, 0, , 3, −4, 0, , ⎤, −1, ⎥, 2⎦, 0, , ⎡, , 1, ⎢, (c) ⎣0, 0, , −5, 0, 0, , 1, 0, 0, , ⎤, , 4, ⎥, 0⎦, 0

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1.2 Gaussian Elimination, , 13, , Solution (a) The equation that corresponds to the last row of the augmented matrix is, , 0x + 0y + 0z = 1, Since this equation is not satisfied by any values of x , y , and z, the system is inconsistent., Solution (b) The equation that corresponds to the last row of the augmented matrix is, , 0x + 0y + 0z = 0, This equation can be omitted since it imposes no restrictions on x , y , and z; hence, the, linear system corresponding to the augmented matrix is, , + 3z = −1, y − 4z = 2, , x, , Since x and y correspond to the leading 1’s in the augmented matrix, we call these, the leading variables. The remaining variables (in this case z) are called free variables., Solving for the leading variables in terms of the free variables gives, , x = −1 − 3z, y = 2 + 4z, From these equations we see that the free variable z can be treated as a parameter and, assigned an arbitrary value t , which then determines values for x and y . Thus, the, solution set can be represented by the parametric equations, , x = −1 − 3t, y = 2 + 4t, z = t, By substituting various values for t in these equations we can obtain various solutions, of the system. For example, setting t = 0 yields the solution, , x = −1, y = 2, z = 0, and setting t = 1 yields the solution, , x = −4, y = 6, z = 1, Solution (c) As explained in part (b), we can omit the equations corresponding to the, zero rows, in which case the linear system associated with the augmented matrix consists, of the single equation, x − 5y + z = 4, (1), , We will usually denote parameters in a general solution, by the letters r, s, t, . . . , but, any letters that do not conflict with the names of the, unknowns can be used. For, systems with more than three, unknowns, subscripted letters, such as t1 , t2 , t3 , . . . are convenient., , from which we see that the solution set is a plane in three-dimensional space. Although, (1) is a valid form of the solution set, there are many applications in which it is preferable, to express the solution set in parametric form. We can convert (1) to parametric form, by solving for the leading variable x in terms of the free variables y and z to obtain, , x = 4 + 5y − z, From this equation we see that the free variables can be assigned arbitrary values, say, y = s and z = t , which then determine the value of x . Thus, the solution set can be, expressed parametrically as, , x = 4 + 5s − t, y = s, z = t, , (2), , Formulas, such as (2), that express the solution set of a linear system parametrically, have some associated terminology., DEFINITION 1 If a linear system has infinitely many solutions, then a set of parametric, , equations from which all solutions can be obtained by assigning numerical values to, the parameters is called a general solution of the system.

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14, , Chapter 1 Systems of Linear Equations and Matrices, , Elimination Methods, , We have just seen how easy it is to solve a system of linear equations once its augmented, matrix is in reduced row echelon form. Now we will give a step-by-step elimination, procedure that can be used to reduce any matrix to reduced row echelon form. As we, state each step in the procedure, we illustrate the idea by reducing the following matrix, to reduced row echelon form., , ⎡, , 0, , 0, , ⎢, ⎣2, , 4, , 2, , 4, , −2, −10, −5, , 0, , 7, , 6, , 12, , 6, , 12, , ⎤, ⎥, , 28⎦, , −5 −1, , Step 1. Locate the leftmost column that does not consist entirely of zeros., , ⎡, , 0, ⎢, 2, ⎣, 2, , 0, 4, 4, , 2, 10, 5, , 0, 6, 6, , 7, 12, 5, , ⎤, 12, ⎥, 28⎦, 1, , Leftmost nonzero column, , Step 2. Interchange the top row with another row, if necessary, to bring a nonzero entry, to the top of the column found in Step 1., , ⎡, , 2, , ⎢, ⎣0, 2, , −10, −2, 0, −5, 4, , 4, , 6, , 12, , 0, , 7, , 6, , −5, , ⎤, , 28, , ⎥, , 12⎦, , The first and second rows in the preceding, matrix were interchanged., , −1, , Step 3. If the entry that is now at the top of the column found in Step 1 is a , multiply, the first row by 1/a in order to introduce a leading 1., , ⎡, , 1, , ⎢, ⎣0, 2, , −5, 0 −2, 4 −5, 2, , ⎤, , 3, , 6, , 14, , 0, , 7, , 12⎦, , 6, , ⎥, , The first row of the preceding matrix was, multiplied by 21 ., , −5 −1, , Step 4. Add suitable multiples of the top row to the rows below so that all entries below, the leading 1 become zeros., , ⎡, , 1, , ⎢, ⎣0, 0, , ⎤, , −5, 0 −2, , 3, , 6, , 14, , 0, , 7, , 12⎦, , 0, , 0, , −17, , 2, , 5, , ⎥, , −29, , −2 times the first row of the preceding, matrix was added to the third row., , Step 5. Now cover the top row in the matrix and begin again with Step 1 applied to the, submatrix that remains. Continue in this way until the entire matrix is in row, echelon form., , ⎡, 1, ⎢, ⎣0, 0, , 2, 0, 0, , 5, 2, 5, , 3, 0, 0, , 6, 7, 17, , ⎤, 14, ⎥, 12 ⎦, 29, , Leftmost nonzero column, in the submatrix, , ⎡, 1, ⎢, ⎣0, , 2, , 5, , 3, , 6, , 0, , 1, , 0, , 7, 2, , 0, , 0, , 5, , 0, , 17, , 14, , ⎤, , ⎥, 6⎦, , 29, , The first row in the submatrix was, multiplied by 1 to introduce a, 2, leading 1.

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1.2 Gaussian Elimination, , ⎡, 1, ⎢0, ⎣, , 2, , 5, , 3, , 6, , 0, , 1, , 0, , 0, , 0, , 0, , 7, 2, 1, 2, , 0, ⎡, 1, ⎢, ⎣0, , 2, , 5, , 3, , 6, , 0, , 1, , 0, , 0, , 0, , 0, , 0, , 7, 2, 1, 2, , ⎡, 1, ⎢, ⎣0, 0, , ⎤, , 14, , ⎥, 6⎦, 1, , 14, , 15, , ⎤, , ⎥, 6⎦, 1, , –5 times the first row of the submatrix, was added to the second row of the, submatrix to introduce a zero below, the leading 1., , The top row in the submatrix was, covered, and we returned again to, Step 1., , Leftmost nonzero column, in the new submatrix, , 2, , 5, , 3, , 6, , 0, 0, , 1, 0, , 0, 0, , 7, 2, , 1, , 14, , ⎤, , ⎥, 6⎦, 2, , The first (and only) row in the new, submatrix was multiplied by 2 to, introduce a leading 1., , The entire matrix is now in row echelon form. To find the reduced row echelon form we, need the following additional step., Step 6. Beginning with the last nonzero row and working upward, add suitable multiples, of each row to the rows above to introduce zeros above the leading 1’s., , ⎡, , 1, ⎢, ⎣0, 0, , 2, 0, 0, , −5, , 1, ⎢, ⎣0, 0, , 2, 0, 0, , −5, , 1, ⎢, ⎣0, 0, , 2, 0, 0, , ⎡, , ⎡, , ⎤, , 3, 0, 0, , 6, 0, 1, , 14, ⎥, 1⎦, 2, , 7, times the third row of the preceding, 2, matrix was added to the second row., , 1, 0, , 3, 0, 0, , 0, 0, 1, , 2, ⎥, 1⎦, 2, , −6 times the third row was added to the, first row., , 0, 1, 0, , 3, 0, 0, , 0, 0, 1, , 7, ⎥, 1⎦, 2, , 5 times the second row was added to the, first row., , 1, 0, , ⎤, , ⎤, , The last matrix is in reduced row echelon form., The procedure (or algorithm) we have just described for reducing a matrix to reduced, row echelon form is called Gauss–Jordan elimination. This algorithm consists of two, parts, a forward phase in which zeros are introduced below the leading 1’s and a backward, phase in which zeros are introduced above the leading 1’s. If only theforward phase is, , Carl Friedrich Gauss, (1777–1855), , Wilhelm Jordan, (1842–1899), , Historical Note Although versions of Gaussian elimination were known much, earlier, its importance in scientific computation became clear when the great, German mathematician Carl Friedrich Gauss used it to help compute the orbit, of the asteroid Ceres from limited data. What happened was this: On January 1,, 1801 the Sicilian astronomer and Catholic priest Giuseppe Piazzi (1746–1826), noticed a dim celestial object that he believed might be a “missing planet.” He, named the object Ceres and made a limited number of positional observations, but then lost the object as it neared the Sun. Gauss, then only 24 years old,, undertook the problem of computing the orbit of Ceres from the limited data, using a technique called “least squares,” the equations of which he solved by, the method that we now call “Gaussian elimination.” The work of Gauss created a sensation when Ceres reappeared a year later in the constellation Virgo, at almost the precise position that he predicted! The basic idea of the method, was further popularized by the German engineer Wilhelm Jordan in his book, on geodesy (the science of measuring Earth shapes) entitled Handbuch der Vermessungskunde and published in 1888., [Images: Photo Inc/Photo Researchers/Getty Images (Gauss);, Leemage/Universal Images Group/Getty Images (Jordan)]

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16, , Chapter 1 Systems of Linear Equations and Matrices, , used, then the procedure produces a row echelon form and is called Gaussian elimination., For example, in the preceding computations a row echelon form was obtained at the end, of Step 5., E X A M P L E 5 Gauss–Jordan Elimination, , Solve by Gauss–Jordan elimination., , x1 + 3x2 − 2x3, + 2x5, 2x1 + 6x2 − 5x3 − 2x4 + 4x5 − 3x6, 5x3 + 10x4, + 15x6, 2x1 + 6x2, + 8x4 + 4x5 + 18x6, , = 0, = −1, = 5, = 6, , Solution The augmented matrix for the system is, , ⎡, , 1, ⎢2, ⎢, ⎢, ⎣0, 2, , 3, 6, 0, 6, , −2, 0, −5 − 2, 5, 0, , 10, 8, , 2, 4, 0, 4, , 0, −3, 15, 18, , ⎤, , 0, −1⎥, ⎥, ⎥, 5⎦, 6, , Adding −2 times the first row to the second and fourth rows gives, , ⎡, , 1, ⎢0, ⎢, ⎢, ⎣0, 0, , 3, 0, 0, 0, , −2, 0, −1 −2, 5, 4, , 10, 8, , 2, 0, 0, 0, , 0, −3, 15, 18, , ⎤, , 0, −1⎥, ⎥, ⎥, 5⎦, 6, , Multiplying the second row by −1 and then adding −5 times the new second row to the, third row and −4 times the new second row to the fourth row gives, , ⎡, , 1, ⎢0, ⎢, ⎢, ⎣0, 0, , 3, 0, 0, 0, , −2, 1, 0, 0, , 0, 2, 0, 0, , 2, 0, 0, 0, , ⎤, , 0, 3, 0, 6, , 0, 1⎥, ⎥, ⎥, 0⎦, 2, , Interchanging the third and fourth rows and then multiplying the third row of the resulting matrix by 16 gives the row echelon form, , ⎡, , 1, ⎢0, ⎢, , 3, 0, , −2, , 0, , 0, 0, , ⎢, ⎣0, , ⎤, , 1, , 0, 2, , 2, 0, , 0, 3, , 0, 1⎥, ⎥, , 0, 0, , 0, 0, , 0, 0, , 1, 0, , 0, , 1⎥, ⎦, 3, , This completes the forward phase since, there are zeros below the leading 1’s., , Adding −3 times the third row to the second row and then adding 2 times the second, row of the resulting matrix to the first row yields the reduced row echelon form, , ⎡, , 3, 0, , 0, 1, , 4, 2, , 2, 0, , 0, 0, , 0, 0⎥, ⎥, , 0, , 0, 0, , 0, 0, , 0, 0, , 0, 0, , 1, 0, , 0, , ⎢, ⎣0, Note that in constructing the, linear system in (3) we ignored, the row of zeros in the corresponding augmented matrix., Why is this justified?, , ⎤, , 1, ⎢0, ⎢, , 1⎥, ⎦, 3, , This completes the backward phase since, there are zeros above the leading 1’s., , The corresponding system of equations is, , x1 + 3x2, , + 4x4 + 2x5, x3 + 2x4, , =0, =0, x6 =, , 1, 3, , (3)

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1.2 Gaussian Elimination, , 17, , Solving for the leading variables, we obtain, , x1 = −3x2 − 4x4 − 2x5, x3 = −2x4, x6 =, , 1, 3, , Finally, we express the general solution of the system parametrically by assigning the, free variables x2 , x4 , and x5 arbitrary values r, s , and t , respectively. This yields, , x1 = −3r − 4s − 2t, x2 = r, x3 = −2s, x4 = s, x5 = t, x6 =, , Homogeneous Linear, Systems, , 1, 3, , A system of linear equations is said to be homogeneous if the constant terms are all zero;, that is, the system has the form, , a11 x1 + a12 x2 + · · · + a1n xn = 0, a21 x1 + a22 x2 + · · · + a2n xn = 0, .., .., .., .., ., ., ., ., am1 x1 + am2 x2 + · · · + amn xn = 0, Every homogeneous system of linear equations is consistent because all such systems, have x1 = 0, x2 = 0, . . . , xn = 0 as a solution. This solution is called the trivial solution;, if there are other solutions, they are called nontrivial solutions., Because a homogeneous linear system always has the trivial solution, there are only, two possibilities for its solutions:, • The system has only the trivial solution., • The system has infinitely many solutions in addition to the trivial solution., In the special case of a homogeneous linear system of two equations in two unknowns,, say, a1 x + b1 y = 0 (a1 , b1 not both zero), , a2 x + b2 y = 0 (a2 , b2, , not both zero), , the graphs of the equations are lines through the origin, and the trivial solution corresponds to the point of intersection at the origin (Figure 1.2.1)., , y, , y, a1x + b1y = 0, x, a 2 x + b2 y = 0, , Only the trivial solution, , Figure 1.2.1, , x, a1x + b1y = 0, and, a 2 x + b2 y = 0, Infinitely many, solutions, , There is one case in which a homogeneous system is assured of having nontrivial, solutions—namely, whenever the system involves more unknowns than equations. To, see why, consider the following example of four equations in six unknowns.

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18, , Chapter 1 Systems of Linear Equations and Matrices, , E X A M P L E 6 A Homogeneous System, , Use Gauss–Jordan elimination to solve the homogeneous linear system, , x1 + 3x2 − 2x3, + 2 x5, 2x1 + 6x2 − 5x3 − 2x4 + 4x5 − 3x6, + 15x6, 5x3 + 10x4, + 8x4 + 4x5 + 18x6, 2x1 + 6x2, , =0, =0, =0, =0, , (4), , Solution Observe first that the coefficients of the unknowns in this system are the same, , as those in Example 5; that is, the two systems differ only in the constants on the right, side. The augmented matrix for the given homogeneous system is, , ⎡, , 1, ⎢2, ⎢, ⎢, ⎣0, 2, , 3, 6, 0, 6, , 0, −2, − 5 −2, 5, 0, , 2, 4, 0, 4, , 10, 8, , ⎤, , 0, −3, 15, 18, , 0, 0⎥, ⎥, ⎥, 0⎦, 0, , (5), , which is the same as the augmented matrix for the system in Example 5, except for zeros, in the last column. Thus, the reduced row echelon form of this matrix will be the same, as that of the augmented matrix in Example 5, except for the last column. However,, a moment’s reflection will make it evident that a column of zeros is not changed by an, elementary row operation, so the reduced row echelon form of (5) is, , ⎡, , 1, ⎢0, ⎢, ⎢, ⎣0, 0, , 3, 0, 0, 0, , 0, 1, 0, 0, , 4, 2, 0, 0, , 2, 0, 0, 0, , 0, 0, 1, 0, , ⎤, , 0, 0⎥, ⎥, ⎥, 0⎦, 0, , (6), , The corresponding system of equations is, , x1 + 3x2, , + 4x4 + 2x5, x3 + 2x4, , =0, =0, x6 = 0, , Solving for the leading variables, we obtain, , x1 = −3x2 − 4x4 − 2x5, x3 = −2x4, x6 = 0, , (7), , If we now assign the free variables x2 , x4 , and x5 arbitrary values r , s , and t , respectively,, then we can express the solution set parametrically as, , x1 = −3r − 4s − 2t, x2 = r, x3 = −2s, x4 = s, x5 = t, x6 = 0, Note that the trivial solution results when r = s = t = 0., , Free Variables in, Homogeneous Linear, Systems, , Example 6 illustrates two important points about solving homogeneous linear systems:, 1. Elementary row operations do not alter columns of zeros in a matrix, so the reduced, row echelon form of the augmented matrix for a homogeneous linear system has, a final column of zeros. This implies that the linear system corresponding to the, reduced row echelon form is homogeneous, just like the original system.

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1.2 Gaussian Elimination, , 19, , 2. When we constructed the homogeneous linear system corresponding to augmented, matrix (6), we ignored the row of zeros because the corresponding equation, 0x1 + 0x2 + 0x3 + 0x4 + 0x5 + 0x6 = 0, does not impose any conditions on the unknowns. Thus, depending on whether or, not the reduced row echelon form of the augmented matrix for a homogeneous linear, system has any rows of zero, the linear system corresponding to that reduced row, echelon form will either have the same number of equations as the original system, or it will have fewer., Now consider a general homogeneous linear system with n unknowns, and suppose, that the reduced row echelon form of the augmented matrix has r nonzero rows. Since, each nonzero row has a leading 1, and since each leading 1 corresponds to a leading, variable, the homogeneous system corresponding to the reduced row echelon form of, the augmented matrix must have r leading variables and n − r free variables. Thus, this, system is of the form, , xk1, + ()=0, , +, , xk2, .., , , , , ()=0, .., ., ., , xkr + ( ) = 0, , (8), , where in each equation the expression ( ) denotes a sum that involves the free variables,, if any [see (7), for example]. In summary, we have the following result., , THEOREM 1.2.1 Free Variable Theorem for Homogeneous Systems, , If a homogeneous linear system has n unknowns, and if the reduced row echelon form, of its augmented matrix has r nonzero rows, then the system has n − r free variables., , Note that Theorem 1.2.2 applies only to homogeneous, systems—a nonhomogeneous, system with more unknowns, than equations need not be, consistent. However, we will, prove later that if a nonhomogeneous system with more, unknowns then equations is, consistent, then it has infinitely many solutions., , Theorem 1.2.1 has an important implication for homogeneous linear systems with, more unknowns than equations. Specifically, if a homogeneous linear system has m, equations in n unknowns, and if m < n, then it must also be true that r < n (why?)., This being the case, the theorem implies that there is at least one free variable, and this, implies that the system has infinitely many solutions. Thus, we have the following result., , THEOREM 1.2.2 A homogeneous linear system with more unknowns than equations has, , infinitely many solutions., , In retrospect, we could have anticipated that the homogeneous system in Example 6, would have infinitely many solutions since it has four equations in six unknowns., , Gaussian Elimination and, Back-Substitution, , For small linear systems that are solved by hand (such as most of those in this text),, Gauss–Jordan elimination (reduction to reduced row echelon form) is a good procedure, to use. However, for large linear systems that require a computer solution, it is generally, more efficient to use Gaussian elimination (reduction to row echelon form) followed by, a technique known as back-substitution to complete the process of solving the system., The next example illustrates this technique.

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20, , Chapter 1 Systems of Linear Equations and Matrices, , E X A M P L E 7 Example 5 Solved by Back-Substitution, , From the computations in Example 5, a row echelon form of the augmented matrix is, , ⎡, , 1, ⎢0, ⎢, ⎢, ⎣0, 0, , 3, 0, 0, 0, , −2, , 0, 2, 0, 0, , 1, 0, 0, , 2, 0, 0, 0, , ⎤, , 0, 3, 1, 0, , 0, 1⎥, ⎥, 1⎥, ⎦, 3, , 0, , To solve the corresponding system of equations, , x1 + 3x2 − 2x3, + 2 x5, =0, x3 + 2 x 4, + 3x6 = 1, x6 =, , 1, 3, , we proceed as follows:, Step 1. Solve the equations for the leading variables., , x1 = −3x2 + 2x3 − 2x5, x3 = 1 − 2x4 − 3x6, x6 =, , 1, 3, , Step 2. Beginning with the bottom equation and working upward, successively substitute, each equation into all the equations above it., Substituting x6 = 13 into the second equation yields, , x1 = −3x2 + 2x3 − 2x5, x3 = −2x4, x6 =, , 1, 3, , Substituting x3 = −2x4 into the first equation yields, , x1 = −3x2 − 4x4 − 2x5, x3 = −2x4, x6 =, , 1, 3, , Step 3. Assign arbitrary values to the free variables, if any., If we now assign x2 , x4 , and x5 the arbitrary values r , s , and t , respectively, the, general solution is given by the formulas, , x1 = −3r − 4s − 2t, x2 = r, x3 = −2s, x4 = s, x5 = t, x6 =, , 1, 3, , This agrees with the solution obtained in Example 5., , EXAMPLE 8, , Suppose that the matrices below are augmented matrices for linear systems in the unknowns x1 , x2 , x3 , and x4 . These matrices are all in row echelon form but not reduced row, echelon form. Discuss the existence and uniqueness of solutions to the corresponding, linear systems

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1.2 Gaussian Elimination, , ⎡, , 1, ⎢0, ⎢, (a) ⎢, ⎣0, 0, , −3, 1, 0, 0, , 7, 2, 1, 0, , 2, −4, 6, 0, , ⎤, , 5, 1⎥, ⎥, ⎥, 9⎦, 1, , ⎡, , 1, ⎢0, ⎢, (b) ⎢, ⎣0, 0, , −3, 1, 0, 0, , 7, 2, 1, 0, , 2, −4, 6, 0, , ⎤, , 5, 1⎥, ⎥, ⎥, 9⎦, 0, , ⎡, , 1, ⎢0, ⎢, (c) ⎢, ⎣0, 0, , −3, 1, 0, 0, , 7, 2, 1, 0, , 2, −4, 6, 1, , 21, , ⎤, , 5, 1⎥, ⎥, ⎥, 9⎦, 0, , Solution (a) The last row corresponds to the equation, , 0x1 + 0x2 + 0x3 + 0x4 = 1, from which it is evident that the system is inconsistent., Solution (b) The last row corresponds to the equation, , 0x1 + 0x2 + 0x3 + 0x4 = 0, which has no effect on the solution set. In the remaining three equations the variables, , x1 , x2 , and x3 correspond to leading 1’s and hence are leading variables. The variable x4, is a free variable. With a little algebra, the leading variables can be expressed in terms, of the free variable, and the free variable can be assigned an arbitrary value. Thus, the, system must have infinitely many solutions., Solution (c) The last row corresponds to the equation, , x4 = 0, which gives us a numerical value for x4 . If we substitute this value into the third equation,, namely,, x3 + 6x4 = 9, we obtain x3 = 9. You should now be able to see that if we continue this process and, substitute the known values of x3 and x4 into the equation corresponding to the second, row, we will obtain a unique numerical value for x2 ; and if, finally, we substitute the, known values of x4 , x3 , and x2 into the equation corresponding to the first row, we will, produce a unique numerical value for x1 . Thus, the system has a unique solution., , Some Facts About Echelon, Forms, , There are three facts about row echelon forms and reduced row echelon forms that are, important to know but we will not prove:, 1. Every matrix has a unique reduced row echelon form; that is, regardless of whether, you use Gauss–Jordan elimination or some other sequence of elementary row operations, the same reduced row echelon form will result in the end.*, 2. Row echelon forms are not unique; that is, different sequences of elementary row, operations can result in different row echelon forms., 3. Although row echelon forms are not unique, the reduced row echelon form and all, row echelon forms of a matrix A have the same number of zero rows, and the leading, 1’s always occur in the same positions. Those are called the pivot positions of A. A, column that contains a pivot position is called a pivot column of A., , *, A proof of this result can be found in the article “The Reduced Row Echelon Form of a Matrix Is Unique: A, Simple Proof,” by Thomas Yuster, Mathematics Magazine, Vol. 57, No. 2, 1984, pp. 93–94.

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22, , Chapter 1 Systems of Linear Equations and Matrices, , E X A M P L E 9 Pivot Positions and Columns, , Earlier in this section (immediately after Definition 1) we found a row echelon form of, , ⎡, , 0, ⎢, A = ⎣2, 2, , If A is the augmented matrix for a linear system, then, the pivot columns identify the, leading variables. As an illustration, in Example 5 the pivot, columns are 1, 3, and 6, and, the leading variables are x1 , x3 ,, and x6 ., , ⎡, , to be, , 0, 4, 4, , 1, ⎢, ⎣0, 0, , 2, 0, 0, , ⎤, , −2, −10, −5, , 0, 6, 6, , 7, 12, −5, , 12, ⎥, 28⎦, −1, , −5, , 3, 0, 0, , 6, , 14, ⎥, −6 ⎦, 2, , 1, 0, , − 27, 1, , ⎤, , The leading 1’s occur in positions (row 1, column 1), (row 2, column 3), and (row 3,, column 5). These are the pivot positions. The pivot columns are columns 1, 3, and 5., , Roundoff Error and, Instability, , There is often a gap between mathematical theory and its practical implementation—, Gauss–Jordan elimination and Gaussian elimination being good examples. The problem, is that computers generally approximate numbers, thereby introducing roundoff errors,, so unless precautions are taken, successive calculations may degrade an answer to a, degree that makes it useless. Algorithms (procedures) in which this happens are called, unstable. There are various techniques for minimizing roundoff error and instability., For example, it can be shown that for large linear systems Gauss–Jordan elimination, involves roughly 50% more operations than Gaussian elimination, so most computer, algorithms are based on the latter method. Some of these matters will be considered in, Chapter 9., , Exercise Set 1.2, , ⎡, , In Exercises 1–2, determine whether the matrix is in row echelon form, reduced row echelon form, both, or neither., , ⎡, , 1, ⎢, 1. (a) ⎣0, 0, , , (d), , ⎤, , 0, 1, 0, , ⎡, , 0, ⎥, 0⎦, 1, , 1, ⎢, (b) ⎣0, 0, , 1, , 0, , 3, , 1, , 0, , 1, , 2, , 4, , ⎡, , 0, , ⎢, , (f ) ⎣0, 0, , ⎡, , 0, , 1, , ⎢0, ⎢, (e) ⎢, ⎣0, 0, , , , ⎥, , 0⎦, , (g), , 0, , ⎤, , 1, , 2, , 0, , 1, , 0⎦, , 0, , 0, , ⎡, , 0, ⎥, 0⎦, 0, , ⎡, , , , ⎤, , 2. (a) ⎣0, 0, , ⎢, , 0, 1, 0, , ⎤, , ⎡, , ⎥, ⎤, , 1, ⎢, (d) ⎣0, , 5, 1, , −3, , 0, , 0, , 0, , ⎥, , 1⎦, , 0, ⎢, (c) ⎣0, 0, , ⎤, , 1, 0, 0, 0, , 0, ⎥, 1⎦, 0, , ⎤, , 2, , 0, , 3, , 0, , 1, , 1, , 0, , 0, , 0, , ⎥, 1⎦, , 0, , 0, , 0, , 0, , −7, , 5, , 5, , 0, , 1, , 3, , 2, , ⎤, , 1, , 0, , 0, , 1, , 0⎦, , 2, , 0, , ⎥, , ⎡, , ⎡, , ⎤, , 4, , 0, , 1⎦, , 0, , 0, , ⎤, , 3, , (e) ⎣0, 0, , 0, , 0⎦, , 0, , 1, , ⎥, , ⎥, , ⎤, , 2, , 3, , 4, , 5, , 0, , 7, , 1, , 0, , 0, , 0, , 1⎦, , 0, , 0, , 0, , 0, , , , 3⎥, ⎥, , ⎥, , (g), , 1, , −2, , 0, , 1, , 0, , 0, , 1, , −2, , , , In Exercises 3–4, suppose that the augmented matrix for a linear system has been reduced by row operations to the given row, echelon form. Solve the system., , ⎤, , 1, , −3, , 4, , 7, , 3. (a) ⎣0, 0, , 1, , 2, , 2⎦, , 0, , 1, , 5, , 1, , 0, , 8, , 6, , (b) ⎣0, 0, , 1, , 4, , −5, −9, , 0, , 1, , 1, , 2, , 1, ⎢0, ⎢, (c) ⎢, ⎣0, 0, , 7, 0, 0, 0, , −2, , −8, , 1, 0, 0, , 0, 1, 1, 0, , 1, ⎢, (d) ⎣0, 0, , −3, , 7, 4, 0, , 1, ⎥, 0⎦, 1, , ⎡, ⎢, , 3, , 2, , ⎢, , , , 1, , 1, , 0, , ⎢, , (c) ⎣0, 0, , ⎢, , ⎢1, ⎢, (f ) ⎢, ⎣0, , ⎡, , 0⎥, ⎥, , 1, , (b) ⎣0, 0, , ⎢, , ⎡, , 1, , ⎡, , ⎡, , 1, 0, , ⎥, , ⎤, , ⎤, ⎥, , 3⎦, , 6, 3, 0, , −3, , ⎤, , 5⎥, ⎥, ⎥, 9⎦, 0

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24, , Chapter 1 Systems of Linear Equations and Matrices, , z=2, 26. x + 2y +, 3z = 1, 2x − 2y +, x + 2y − (a 2 − 3)z = a, , 36. Solve the following system for x, y, and z., 1, , x, 2, , In Exercises 27–28, what condition, if any, must a , b, and c, satisfy for the linear system to be consistent?, 27. x + 3y − z = a, x + y + 2z = b, 2 y − 3z = c, , 28., , x + 3y + z = a, −x − 2y + z = b, 3x + 7y − z = c, , In Exercises 29–30, solve the following systems, where a , b,, and c are constants., 29. 2x + y = a, , x, −, , +, , y, 3, , y, 9, , y, , −, , 4, , +, , 8, , +, , 10, , 20, (0, 10), , 2, , 7, , (1, 7), , ⎡, , 2, , (3, –11), –20, , 1, , ⎢, ⎣0, , −2, , 3, , 4, , (4, –14), , Figure Ex-37, , This exercise shows that a matrix can have multiple row echelon forms., 32. Reduce, , =5, , x, , , , 3, , z, , =0, , 6, , 31. Find two different row echelon forms of, 1, , z, , =1, , y, , –2, , , , z, , y = ax 3 + bx 2 + cx + d., , + 2 x3 = b, 3x2 + 3x3 = c, , 2x1, , +, , 2, , 37. Find the coefficients a, b, c, and d so that the curve shown, in the accompanying figure is the graph of the equation, , 30. x1 + x2 + x3 = a, , 3x + 6y = b, , 1, , x, , +, , 3, , ⎤, , ⎥, −29⎦, , 38. Find the coefficients a, b, c, and d so that the circle shown in, the accompanying figure is given by the equation, ax 2 + ay 2 + bx + cy + d = 0., y, (–2, 7), (–4, 5), , 5, , to reduced row echelon form without introducing fractions at, any intermediate stage., , x, , 33. Show that the following nonlinear system has 18 solutions if, 0 ≤ α ≤ 2π , 0 ≤ β ≤ 2π , and 0 ≤ γ ≤ 2π ., sin α + 2 cos β + 3 tan γ = 0, 2 sin α + 5 cos β + 3 tan γ = 0, , − sin α − 5 cos β + 5 tan γ = 0, [Hint: Begin by making the substitutions x = sin α ,, , y = cos β , and z = tan γ .], , 34. Solve the following system of nonlinear equations for the unknown angles α , β , and γ , where 0 ≤ α ≤ 2π , 0 ≤ β ≤ 2π ,, and 0 ≤ γ < π ., 2 sin α − cos β + 3 tan γ = 3, 4 sin α + 2 cos β − 2 tan γ = 2, 6 sin α − 3 cos β + tan γ = 9, 35. Solve the following system of nonlinear equations for x, y,, and z., , x 2 + y 2 + z2 = 6, x 2 − y 2 + 2z 2 = 2, 2x 2 + y 2 − z 2 = 3, [Hint: Begin by making the substitutions X = x 2 , Y = y 2 ,, Z = z 2 .], , (4, –3), , Figure Ex-38, , 39. If the linear system, , a1 x + b1 y + c1 z = 0, a2 x − b2 y + c2 z = 0, a3 x + b3 y − c3 z = 0, has only the trivial solution, what can be said about the solutions of the following system?, , a1 x + b1 y + c1 z = 3, a2 x − b2 y + c2 z = 7, a3 x + b3 y − c3 z = 11, 40. (a) If A is a matrix with three rows and five columns, then, what is the maximum possible number of leading 1’s in its, reduced row echelon form?, (b) If B is a matrix with three rows and six columns, then, what is the maximum possible number of parameters in, the general solution of the linear system with augmented, matrix B ?, (c) If C is a matrix with five rows and three columns, then, what is the minimum possible number of rows of zeros in, any row echelon form of C ?

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1.3 Matrices and Matrix Operations, , 41. Describe all possible reduced row echelon forms of, , ⎡, , a, ⎢, (a) ⎣d, g, , b, e, h, , ⎡, , ⎤, c, ⎥, f⎦, i, , a, ⎢e, ⎢, (b) ⎢, ⎣i, m, , b, f, j, n, , c, g, k, p, , ⎤, , d, h⎥, ⎥, ⎥, l⎦, q, , 42. Consider the system of equations, , ax + by = 0, cx + dy = 0, ex + fy = 0, Discuss the relative positions of the lines ax + by = 0,, cx + dy = 0, and ex + fy = 0 when the system has only the, trivial solution and when it has nontrivial solutions., , Working with Proofs, 43. (a) Prove that if ad − bc  = 0, then the reduced row echelon, form of, , , , , a b, 1 0, is, c d, 0 1, (b) Use the result in part (a) to prove that if ad − bc  = 0, then, the linear system, , ax + by = k, cx + dy = l, , has exactly one solution., , 25, , (d) A homogeneous linear system in n unknowns whose corresponding augmented matrix has a reduced row echelon form, with r leading 1’s has n − r free variables., (e) All leading 1’s in a matrix in row echelon form must occur in, different columns., (f ) If every column of a matrix in row echelon form has a leading, 1, then all entries that are not leading 1’s are zero., (g) If a homogeneous linear system of n equations in n unknowns, has a corresponding augmented matrix with a reduced row, echelon form containing n leading 1’s, then the linear system, has only the trivial solution., (h) If the reduced row echelon form of the augmented matrix for, a linear system has a row of zeros, then the system must have, infinitely many solutions., (i) If a linear system has more unknowns than equations, then it, must have infinitely many solutions., , Working withTechnology, T1. Find the reduced row echelon form of the augmented matrix, for the linear system:, 6x1 + x2, + 4x4 = −3, −9x1 + 2x2 + 3x3 − 8x4 = 1, − 4x3 + 5x4 = 2, 7x1, Use your result to determine whether the system is consistent and,, if so, find its solution., , True-False Exercises, TF. In parts (a)–(i) determine whether the statement is true or, false, and justify your answer., (a) If a matrix is in reduced row echelon form, then it is also in, row echelon form., (b) If an elementary row operation is applied to a matrix that is, in row echelon form, the resulting matrix will still be in row, echelon form., (c) Every matrix has a unique row echelon form., , T2. Find values of the constants A, B , C , and D that make the, following equation an identity (i.e., true for all values of x )., , Ax + B, C, D, 3x 3 + 4x 2 − 6x, = 2, +, +, (x 2 + 2x + 2)(x 2 − 1), x + 2x + 2 x − 1 x + 1, [Hint: Obtain a common denominator on the right, and then, equate corresponding coefficients of the various powers of x in, the two numerators. Students of calculus will recognize this as a, problem in partial fractions.], , 1.3 Matrices and Matrix Operations, Rectangular arrays of real numbers arise in contexts other than as augmented matrices for, linear systems. In this section we will begin to study matrices as objects in their own right, by defining operations of addition, subtraction, and multiplication on them., , Matrix Notation and, Terminology, , In Section 1.2 we used rectangular arrays of numbers, called augmented matrices, to, abbreviate systems of linear equations. However, rectangular arrays of numbers occur, in other contexts as well. For example, the following rectangular array with three rows, and seven columns might describe the number of hours that a student spent studying, three subjects during a certain week:

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26, , Chapter 1 Systems of Linear Equations and Matrices, , Math, History, Language, , Mon., , Tues., , Wed., , Thurs., , Fri., , Sat., , Sun., , 2, 0, 4, , 3, 3, 1, , 2, 1, 3, , 4, 4, 1, , 1, 3, 0, , 4, 2, 0, , 2, 2, 2, , If we suppress the headings, then we are left with the following rectangular array of, numbers with three rows and seven columns, called a “matrix”:, , ⎡, , 2, ⎢, ⎣0, 4, , 3, 3, 1, , 2, 1, 3, , 4, 4, 1, , 1, 3, 0, , ⎤, , 4, 2, 0, , 2, ⎥, 2⎦, 2, , More generally, we make the following definition., DEFINITION 1 A matrix is a rectangular array of numbers. The numbers in the array, are called the entries in the matrix., , E X A M P L E 1 Examples of Matrices, Matrix brackets are often, omitted from 1 × 1 matrices, making it impossible to, tell, for example, whether the, symbol 4 denotes the number “four” or the matrix [4]., This rarely causes problems, because it is usually possible, to tell which is meant from the, context., , Some examples of matrices are, , ⎡, , 1, ⎣ 3, −1, , ⎡, , ⎤, , 2, 0⎦, [2, 4, , 1, , e, , ⎢, − 3], ⎣0, , 0, , 0, , √ ⎤, − 2, ⎥, 1 ⎦,, , π, 1, 2, , 0, , 0, , 1, , [4], 3, , The size of a matrix is described in terms of the number of rows (horizontal lines), and columns (vertical lines) it contains. For example, the first matrix in Example 1 has, three rows and two columns, so its size is 3 by 2 (written 3 × 2). In a size description,, the first number always denotes the number of rows, and the second denotes the number, of columns. The remaining matrices in Example 1 have sizes 1 × 4, 3 × 3, 2 × 1, and, 1 × 1, respectively., A matrix with only one row, such as the second in Example 1, is called a row vector, (or a row matrix), and a matrix with only one column, such as the fourth in that example,, is called a column vector (or a column matrix). The fifth matrix in that example is both, a row vector and a column vector., We will use capital letters to denote matrices and lowercase letters to denote numerical quantities; thus we might write, , A=, , 2, 3, , 1, 4, , 7, 2, , or C =, , a, d, , b, e, , c, f, , When discussing matrices, it is common to refer to numerical quantities as scalars. Unless, stated otherwise, scalars will be real numbers; complex scalars will be considered later in, the text., The entry that occurs in row i and column j of a matrix A will be denoted by aij ., Thus a general 3 × 4 matrix might be written as

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1.3 Matrices and Matrix Operations, , ⎡, , a11, ⎢, A = ⎣a21, a31, and a general m × n matrix as, , ⎡, , a11, ⎢a, ⎢ 21, A=⎢ ., ⎣ .., am1, A matrix with n rows and n, columns is said to be a square, matrix of order n., , a12, a22, a32, , a13, a23, a33, , a12, a22, .., ., , ···, ···, , am2, , 27, , ⎤, a14, ⎥, a24 ⎦, a34, , ···, , ⎤, a1n, a2 n ⎥, ⎥, .. ⎥, . ⎦, amn, , (1), , When a compact notation is desired, the preceding matrix can be written as, , [aij ]m×n or [aij ], the first notation being used when it is important in the discussion to know the size,, and the second when the size need not be emphasized. Usually, we will match the letter, denoting a matrix with the letter denoting its entries; thus, for a matrix B we would, generally use bij for the entry in row i and column j , and for a matrix C we would use, the notation cij ., The entry in row i and column j of a matrix A is also commonly denoted by the, symbol (A)ij . Thus, for matrix (1) above, we have, , (A)ij = aij, and for the matrix, , 2 −3, 7, 0, we have (A)11 = 2, (A)12 = −3, (A)21 = 7, and (A)22 = 0., Row and column vectors are of special importance, and it is common practice to, denote them by boldface lowercase letters rather than capital letters. For such matrices,, double subscripting of the entries is unnecessary. Thus a general 1 × n row vector a and, a general m × 1 column vector b would be written as, , A=, , ⎤, b1, ⎢b ⎥, ⎢ 2⎥, · · · an ] and b = ⎢ .. ⎥, ⎣ . ⎦, bm, ⎡, , a = [a1 a2, , A matrix A with n rows and n columns is called a square matrix of order n, and the, shaded entries a11 , a22 , . . . , ann in (2) are said to be on the main diagonal of A., , ⎡, ⎢, ⎢, ⎢, ⎣, , a11, a21, .., ., an1, , Operations on Matrices, , a12, a22, .., ., an2, , ···, ···, , a1n, a2n, .., ., · · · ann, , ⎤, ⎥, ⎥, ⎥, ⎦, , (2), , So far, we have used matrices to abbreviate the work in solving systems of linear equations. For other applications, however, it is desirable to develop an “arithmetic of matrices” in which matrices can be added, subtracted, and multiplied in a useful way. The, remainder of this section will be devoted to developing this arithmetic., DEFINITION 2 Two matrices are defined to be equal if they have the same size and, , their corresponding entries are equal.

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28, , Chapter 1 Systems of Linear Equations and Matrices, , E X A M P L E 2 Equality of Matrices, The equality of two matrices, , Consider the matrices, , A=, , A = [aij ] and B = [bij ], of the same size can be expressed either by writing, , (A)ij = (B)ij, or by writing, , aij = bij, where it is understood that the, equalities hold for all values of, i and j ., , 2, 3, , 1, , , B=, , x, , 2, 3, , 1, 2, , C=, 5, 3, , 1, 4, , 0, 0, , If x = 5, then A = B , but for all other values of x the matrices A and B are not equal,, since not all of their corresponding entries are equal. There is no value of x for which, A = C since A and C have different sizes., DEFINITION 3 If A and B are matrices of the same size, then the sum A + B is the, matrix obtained by adding the entries of B to the corresponding entries of A, and, the difference A − B is the matrix obtained by subtracting the entries of B from the, corresponding entries of A. Matrices of different sizes cannot be added or subtracted., , In matrix notation, if A = [aij ] and B = [bij ] have the same size, then, , (A + B)ij = (A)ij + (B)ij = aij + bij and (A − B)ij = (A)ij − (B)ij = aij − bij, , E X A M P L E 3 Addition and Subtraction, , Consider the matrices, , ⎡, , Then, , ⎤, , ⎡, , 1, 0, −2, , 0, 2, 7, , −4, 3, ⎥, ⎢, 4⎦, B = ⎣ 2, 0, 3, , −2, , 4, 2, 0, , 5, 2, 3, , 2, ⎢, A = ⎣−1, 4, , ⎡, , ⎢, A+B =⎣ 1, 7, , 3, 2, 2, , ⎤, , 5, 0, −4, , ⎤, , 1, 1, ⎥, −1⎦, C =, 2, 5, , ⎡, , 4, 6, ⎥, ⎢, 3⎦ and A − B = ⎣−3, 5, 1, , −2, −2, −4, , −5, 2, 11, , 1, 2, , ⎤, , 2, ⎥, 5⎦, −5, , The expressions A + C , B + C , A − C , and B − C are undefined., DEFINITION 4 If A is any matrix and c is any scalar, then the product cA is the matrix, obtained by multiplying each entry of the matrix A by c. The matrix cA is said to be, a scalar multiple of A., , In matrix notation, if A = [aij ], then, , (cA)ij = c(A)ij = caij, E X A M P L E 4 Scalar Multiples, , For the matrices, 2, 1, , A=, , 3, 3, , 4, 0, , B=, −1, 1, , 2, 3, , 7, 9, , C=, −5, 3, , −6, , 3, 12, , 0, , we have, 2A =, , 4, 2, , 6, 6, , 8, 0, , (−1)B =, 2, 1, , −2 −7, ,, 5, −3, , It is common practice to denote (−1)B by −B ., , 1, C, 3, , =, , 3, 1, , −2, 0, , 1, 4

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1.3 Matrices and Matrix Operations, , 29, , Thus far we have defined multiplication of a matrix by a scalar but not the multiplication of two matrices. Since matrices are added by adding corresponding entries, and subtracted by subtracting corresponding entries, it would seem natural to define, multiplication of matrices by multiplying corresponding entries. However, it turns out, that such a definition would not be very useful for most problems. Experience has led, mathematicians to the following more useful definition of matrix multiplication., DEFINITION 5 If A is an m × r matrix and B is an r × n matrix, then the product, AB is the m × n matrix whose entries are determined as follows: To find the entry in, row i and column j of AB , single out row i from the matrix A and column j from, the matrix B . Multiply the corresponding entries from the row and column together,, , and then add up the resulting products., , E X A M P L E 5 Multiplying Matrices, , Consider the matrices, , A=, , ⎡, , 1, 2, , 2, 6, , 4, 4, ⎢, , B = ⎣0, 0, 2, , 1, −1, 7, , ⎤, , 4, 3, 5, , 3, ⎥, 1⎦, 2, , Since A is a 2 × 3 matrix and B is a 3 × 4 matrix, the product AB is a 2 × 4 matrix., To determine, for example, the entry in row 2 and column 3 of AB , we single out row 2, from A and column 3 from B . Then, as illustrated below, we multiply corresponding, entries together and add up these products., , , , ⎡, , 4, 1 2 4 ⎢, ⎣0, 2 6 0, 2, , 1, 1, 7, , 4, 3, 5, , ⎤ ⎡, 3, ⎥ ⎢, 1⎦ = ⎣, 2, , ⎤, ⎥, ⎦, , 26, , (2 · 4) + (6 · 3) + (0 · 5) = 26, The entry in row 1 and column 4 of AB is computed as follows:, , ⎡, 4, , 1 2 4 ⎢, ⎣0, 2 6 0, 2, , 1, 1, 7, , 4, 3, 5, , ⎤ ⎡, 3, ⎥ ⎢, 1⎦ = ⎣, 2, , ⎤, 13 ⎥, ⎦, , (1 · 3) + (2 · 1) + (4 · 2) = 13, The computations for the remaining entries are, , (1 · 4) + (2 · 0) + (4 · 2) = 12, (1 · 1) − (2 · 1) + (4 · 7) = 27, (1 · 4) + (2 · 3) + (4 · 5) = 30, (2 · 4) + (6 · 0) + (0 · 2) = 8, (2 · 1) − (6 · 1) + (0 · 7) = −4, (2 · 3) + (6 · 1) + (0 · 2) = 12, , AB =, , 12, 8, , 27, , −4, , 30, 26, , 13, 12, , The definition of matrix multiplication requires that the number of columns of the, first factor A be the same as the number of rows of the second factor B in order to form, the product AB . If this condition is not satisfied, the product is undefined. A convenient

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30, , Chapter 1 Systems of Linear Equations and Matrices, , way to determine whether a product of two matrices is defined is to write down the size, of the first factor and, to the right of it, write down the size of the second factor. If, as in, (3), the inside numbers are the same, then the product is defined. The outside numbers, then give the size of the product., , A, m × r, , B, r × n =, , AB, m × n, (3), , Inside, Outside, , E X A M P L E 6 Determining Whether a Product Is Defined, , Suppose that A, B , and C are matrices with the following sizes:, , A, , B, , C, , 3×4, , 4×7, , 7×3, , Then by (3), AB is defined and is a 3 × 7 matrix; BC is defined and is a 4 × 3 matrix; and, CA is defined and is a 7 × 4 matrix. The products AC , CB , and BA are all undefined., In general, if A = [aij ] is an m × r matrix and B = [bij ] is an r × n matrix, then, as, illustrated by the shading in the following display,, , ⎡, , a11, ⎢a, ⎢ 21, ⎢ ., ⎢ .., AB = ⎢, ⎢ ai 1, ⎢, ⎢ .., ⎣ ., , a12, a22, .., ., ai 2, .., ., , am1, , am2, , ···, ···, ···, ···, , a1r, a2r, .., ., air, .., ., , ⎤, , ⎥ ⎡b, ⎥ 11, ⎥⎢, ⎥ ⎢b21, ⎥⎢ ., ⎥⎣ ., ⎥ ., ⎥, ⎦ br 1, , b12, b22, .., ., , · · · b1 j, · · · b2 j, .., ., , br 2, , br j, , ···, , ⎤, · · · b1n, · · · b2n ⎥, ⎥, .. ⎥, . ⎦, · · · br n, , (4), , amr, , the entry (AB)ij in row i and column j of AB is given by, , (AB)ij = ai 1 b1j + ai 2 b2j + ai 3 b3j + · · · + air brj, , (5), , Formula (5) is called the row-column rule for matrix multiplication., Partitioned Matrices, , A matrix can be subdivided or partitioned into smaller matrices by inserting horizontal, and vertical rules between selected rows and columns. For example, the following are, three possible partitions of a general 3 × 4 matrix A—the first is a partition of A into, , Gotthold Eisenstein, (1823–1852), , Historical Note The concept of matrix multiplication is due to the German mathematician Gotthold, Eisenstein, who introduced the idea around 1844 to, simplify the process of making substitutions in linear systems. The idea was then expanded on and, formalized by Cayley in his Memoir on the Theory, of Matrices that was published in 1858. Eisenstein, was a pupil of Gauss, who ranked him as the equal, of Isaac Newton and Archimedes. However, Eisenstein, suffering from bad health his entire life, died, at age 30, so his potential was never realized., [Image: http://www-history.mcs.st-andrews.ac.uk/, Biographies/Eisenstein.html]

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32, , Chapter 1 Systems of Linear Equations and Matrices, , and from (9) the first row vector of AB can be obtained by the computation, , ⎡, , [1 2, , 4, ⎢, 4 ]⎣0, 2, , 1, 1, 7, , 4, 3, 5, , ⎤, 3, ⎥, 1⎦ =, 2, , [ 12 27 30 13 ], , First row of A, , Matrix Products as Linear, Combinations, Definition 6 is applicable, in, particular, to row and column, vectors. Thus, for example, a, linear combination of column, vectors x1 , x2 , . . . , xr of the, same size is an expression of, the form, , c1 x1 + c2 x2 + · · · + cr xr, , First row of AB, , The following definition provides yet another way of thinking about matrix multiplication., DEFINITION 6 If, , A1 , A2 , . . . , Ar are matrices of the same size, and if c1 , c2 , . . . , cr, are scalars, then an expression of the form, c 1 A 1 + c 2 A2 + · · · + c r A r, is called a linear combination of A1 , A2 , . . . , Ar with coefficients c1 , c2 , . . . , cr ., , To see how matrix products can be viewed as linear combinations, let A be an m × n, matrix and x an n × 1 column vector, say, , ⎡, , Then, , a11, ⎢a, ⎢ 21, A=⎢ ., ⎣ .., , a12, a22, .., ., , am 1, , am2, , ⎤, ⎡ ⎤, a1n, x1, ⎢x ⎥, a2 n ⎥, ⎥, ⎢ 2⎥, .. ⎥ and x = ⎢ .. ⎥, ⎦, ⎣.⎦, ., · · · amn, xn, ···, ···, , ⎡, , ⎤, ⎡, ⎤, ⎡, ⎤, ⎡, ⎤, a11 x1 + a12 x2 + · · · + a1n xn, a11, a12, a1n, ⎢a x + a x +···+ a x ⎥, ⎢, ⎥, ⎢, ⎥, ⎢, ⎥, ⎢ 21 1, ⎢ a21 ⎥, ⎢ a22 ⎥, ⎢ a2 n ⎥, 22 2, 2n n ⎥, Ax = ⎢ ., ⎥ = x1 ⎢ . ⎥ + x2 ⎢ . ⎥ + · · · + xn ⎢ . ⎥, ., ., .., .. ⎦, ⎣ .., ⎣ .. ⎦, ⎣ .. ⎦, ⎣ .. ⎦, am1 x1 + am2 x2 + · · · + amn xn, a m1, am2, amn, (10), , This proves the following theorem., THEOREM 1.3.1 If A is an m × n matrix, and if x is an n × 1 column vector, then the, , product Ax can be expressed as a linear combination of the column vectors of A in which, the coefficients are the entries of x., , E X A M P L E 8 Matrix Products as Linear Combinations, , The matrix product, , ⎡, , −1, , 3, , 2, , 2, 1, , ⎢, ⎣ 1, , 2, , ⎤⎡, , 2, , ⎤, , ⎡, , ⎤, , 1, , ⎥⎢ ⎥ ⎢ ⎥, −3⎦ ⎣−1⎦ = ⎣−9⎦, −2, −3, 3, , can be written as the following linear combination of column vectors:, , ⎡, , ⎤, ⎡ ⎤, ⎡ ⎤ ⎡ ⎤, −1, 3, 2, 1, ⎢ ⎥, ⎢ ⎥, ⎢ ⎥ ⎢ ⎥, −, −, 3, 1, 2, 2 ⎣ ⎦ − 1 ⎣ ⎦ + 3 ⎣ ⎦ = ⎣ 9⎦, −2, −3, 2, 1

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1.3 Matrices and Matrix Operations, , 33, , E X A M P L E 9 Columns of a Product AB as Linear Combinations, , We showed in Example 5 that, , ⎡, AB =, , 1, , 2, , 2, , 6, , 4, , 4 ⎢, ⎣0, 0, 2, , ⎤, , 1, , 4, , 3, , −1, , 3, , 1⎦ =, , 7, , 5, , 2, , ⎥, , 12, , 27, , 30, , 13, , 8, , −4, , 26, , 12, , It follows from Formula (6) and Theorem 1.3.1 that the j th column vector of AB can be, expressed as a linear combination of the column vectors of A in which the coefficients, in the linear combination are the entries from the j th column of B . The computations, are as follows:, ,  , ,  , , 12, , =4, , 8, , , , 27, , , , −4,  , 30, , 1, , =, , 2, , Column-Row Expansion, , 2, 1, 2, , 6, ,  , , +2, ,  , 2, , −, , 6, 2, , +3, , 6, , +7, , 2, 6, , 0, 4, 0, ,  , +5, ,  , +, , 4, ,  , ,  , ,  , =3, , 12, , 1, , 2, , +0, ,  , ,  , 13, , 2, ,  , ,  , , =4, , 26, , 1, , 4, 0, ,  , +2, , 4, 0, , Partitioning provides yet another way to view matrix multiplication. Specifically, suppose that an m × r matrix A is partitioned into its r column vectors c1 , c2 , . . . , cr (each, of size m × 1) and an r × n matrix B is partitioned into its r row vectors r1 , r2 , . . . , rr, (each of size 1 × n). Each term in the sum, c1 r1 + c2 r2 + · · · + cr rr, has size m × n so the sum itself is an m × n matrix. We leave it as an exercise for you to, verify that the entry in row i and column j of the sum is given by the expression on the, right side of Formula (5), from which it follows that, , AB = c1 r1 + c2 r2 + · · · + cr rr, , (11), , We call (11) the column-row expansion of AB ., , E X A M P L E 10 Column-Row Expansion, , Find the column-row expansion of the product, , , AB =, , , , 1, , 3, , 2, , 0, , 4, , 2, , −1, , −3, , 5, , 1, , , (12), , Solution The column vectors of A and the row vectors of B are, respectively,, ,  , c1 =, , 1, , 2, , , , c2 =, , , , 3, , −1, , , ; r1 = 2, , 0, , , , , , 4 , r2 = −3, , 5, , , , 1

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34, , Chapter 1 Systems of Linear Equations and Matrices, , Thus, it follows from (11) that the column-row expansion of AB is, ,  , , 1 , 2, AB =, 2, , , =, The main use of the columnrow expansion is for developing theoretical results rather, than for numerical computations., , Matrix Form of a Linear, System, , 4 +, , 0, , 2, , 0, , 4, , 4, , 0, , 8, , , , , , , , , +, , , , 3 , , −1, , −3 5 1, , , , , , −9, , 15, , 3, , 3, , −5, , −1, , (13), , As a check, we leave it for you to confirm that the product in (12) and the sum in (13), both yield, , , 7, −7 15, , AB =, , 7, , −5, , 7, , Matrix multiplication has an important application to systems of linear equations. Consider a system of m linear equations in n unknowns:, , a11 x1 + a12 x2 + · · · + a1n xn = b1, a21 x1 + a22 x2 + · · · + a2n xn = b2, .., .., .., .., ., ., ., ., am1 x1 + am2 x2 + · · · + amn xn = bm, Since two matrices are equal if and only if their corresponding entries are equal, we can, replace the m equations in this system by the single matrix equation, , ⎡, , ⎤ ⎡ ⎤, a11 x1 + a12 x2 + · · · + a1n xn, b1, ⎢ a x + a x + · · · + a x ⎥ ⎢b ⎥, 22 2, 2n n ⎥, ⎢ 21 1, ⎢ 2⎥, ⎢ .., .., .. ⎥ = ⎢ .. ⎥, ⎣ ., ⎦, ⎣ . ⎦, ., ., bm, am1 x1 + am2 x2 + · · · + amn xn, , The m × 1 matrix on the left side of this equation can be written as a product to give, , ⎡, , a11, ⎢a, ⎢ 21, ⎢ .., ⎣ ., , a12, a22, .., ., , ···, ···, , am1, , am 2, , ···, , ⎤⎡ ⎤ ⎡ ⎤, x1, b1, a1n, ⎥, ⎢, ⎥, ⎢, a 2 n ⎥ ⎢ x2 ⎥ ⎢ b 2 ⎥, ⎥, .. ⎥ ⎢ .. ⎥ = ⎢ .. ⎥, ⎦, ⎣, ⎦, ⎣, ., ., . ⎦, xn, bm, amn, , If we designate these matrices by A, x, and b, respectively, then we can replace the original, system of m equations in n unknowns by the single matrix equation, , Ax = b, , The vertical partition line in, the augmented matrix [A | b], is optional, but is a useful way, of visually separating the coefficient matrix A from the column vector b., , Transpose of a Matrix, , The matrix A in this equation is called the coefficient matrix of the system. The augmented matrix for the system is obtained by adjoining b to A as the last column; thus, the augmented matrix is, , ⎡, , a11, ⎢a, ⎢ 21, [A | b] = ⎢ .., ⎣ ., am1, , a12, a22, .., ., , ···, ···, , a1n, a2 n, .., ., , am2, , ···, , amn, , ⎤, b1, b2 ⎥, ⎥, .. ⎥, . ⎦, bm, , We conclude this section by defining two matrix operations that have no analogs in the, arithmetic of real numbers.

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1.3 Matrices and Matrix Operations, , 35, , If A is any m × n matrix, then the transpose of A, denoted by AT , is, defined to be the n × m matrix that results by interchanging the rows and columns, of A; that is, the first column of AT is the first row of A, the second column of AT is, the second row of A, and so forth., DEFINITION 7, , E X A M P L E 11 Some Transposes, , The following are some examples of matrices and their transposes., , ⎡, , a11, ⎢, A = ⎣a21, a31, ⎡, a11, ⎢a, ⎢ 12, AT = ⎢, ⎣a13, a14, , a12, a22, a32, a21, a22, a23, a24, , a13, a23, a33, , ⎤, ⎡, 2, a14, ⎥, ⎢, a24 ⎦, B = ⎣1, 5, a34, , ⎤, a31, , 2, a32 ⎥, ⎥, ⎥, B T =, 3, a33 ⎦, a34, , 1, 4, , ⎤, , 3, ⎥, 4⎦, C = [1 3 5], D = [4], 6, , ⎡ ⎤, , , , 1, 5, ⎢ ⎥, T, , C = ⎣3⎦, D T = [4], 6, 5, , Observe that not only are the columns of AT the rows of A, but the rows of AT are, the columns of A. Thus the entry in row i and column j of AT is the entry in row j and, column i of A; that is,, , (AT )ij = (A)j i, , (14), , Note the reversal of the subscripts., In the special case where A is a square matrix, the transpose of A can be obtained, by interchanging entries that are symmetrically positioned about the main diagonal. In, (15) we see that AT can also be obtained by “reflecting” A about its main diagonal., , ⎡, , 1, ⎢, A= ⎣ 3, 5, , 2, 7, 8, , ⎤, 4, ⎥, 0⎦, 6, , ⎡, , 1, ⎢, ⎣ 3, 5, , 2, 7, 8, , ⎤, 4, ⎥, 0⎦, 6, , ⎡, AT, , 1, ⎢, 2, ⎣, 4, , 3, 7, 0, , ⎤, 5, ⎥, 8⎦, 6, , (15), , Interchange entries that are, symmetrically positioned, about the main diagonal., , James Sylvester, (1814–1897), , Arthur Cayley, (1821–1895), , Historical Note The term matrix was first used by the English mathematician, James Sylvester, who defined the term in 1850 to be an “oblong arrangement, of terms.” Sylvester communicated his work on matrices to a fellow English, mathematician and lawyer named Arthur Cayley, who then introduced some of, the basic operations on matrices in a book entitled Memoir on the Theory of, Matrices that was published in 1858. As a matter of interest, Sylvester, who was, Jewish, did not get his college degree because he refused to sign a required, oath to the Church of England. He was appointed to a chair at the University of, Virginia in the United States but resigned after swatting a student with a stick, because he was reading a newspaper in class. Sylvester, thinking he had killed, the student, fled back to England on the first available ship. Fortunately, the, student was not dead, just in shock!, [Images: © Bettmann/CORBIS (Sylvester );, Photo Researchers/Getty Images (Cayley )]

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36, , Chapter 1 Systems of Linear Equations and Matrices, , Trace of a Matrix, , DEFINITION 8 If A is a square matrix, then the trace of A, denoted by tr(A), is defined, to be the sum of the entries on the main diagonal of A. The trace of A is undefined, if A is not a square matrix., , E X A M P L E 1 2 Trace, , The following are examples of matrices and their traces., , ⎡, , a11, ⎢, A = ⎣a21, a31, , a12, a22, a32, , ⎡, , ⎤, , −1, , a13, ⎢ 3, ⎥, ⎢, a23 ⎦, B = ⎢, ⎣ 1, a33, 4, , tr(A) = a11 + a22 + a33, , 2, 5, 2, −2, , ⎤, , 7, −8, 7, 1, , 0, 4⎥, ⎥, ⎥, −3⎦, 0, , tr(B) = −1 + 5 + 7 + 0 = 11, , In the exercises you will have some practice working with the transpose and trace, operations., , Exercise Set 1.3, In Exercises 1–2, suppose that A, B , C , D , and E are matrices, with the following sizes:, , A, , B, , C, , D, , E, , (4 × 5), , (4 × 5), , (5 × 2 ), , (4 × 2 ), , (5 × 4 ), , In each part, determine whether the given matrix expression is, defined. For those that are defined, give the size of the resulting, matrix., 1. (a) BA, , T, , 3, 1, , 1, , ⎢, D = ⎣−1, 3, , (i) (CD)E, , ( j) C(BA), , (k) tr(DE T ), , (l) tr(BC), , (b) BA, , (c) (3E)D, , (i) tr(DD T ), , (b) DC, , (c) BC − 3D, , ( j) tr(4E T − D), , (k) tr(C TAT + 2E T ) (l) tr((EC T )TA), , (e) B D + ED, , (f ) BA + D, , , B=, , T, , 4, , , −1, , 0, , 2, , , C=, , ,, , ⎤, , ⎡, , ⎥, , ⎢, , , , 1, , 4, , 2, , 3, , 1, , 5, , ⎤, , 5, , 2, , 6, , 1, , 3, , 0, , 1⎦, E = ⎣−1, 4, 4, , 1, , 2⎦, , 1, , 3, , 2, , (f ) B − B T, , (h) (C TB)AT, , 1, , ⎡, , (h) (2E T − 3D T )T, , 5. (a) AB, , − 41 A, , (g) (DA)T, , 0, , ⎥, 2⎦,, , (g) 2E T − 3D T, , 1 T, C, 2, , (f ) E(5B + A), , ⎤, , ⎢, A = ⎣−1, , (e), , (f ) CC T, , In Exercises 3–6, use the following matrices to compute the, indicated expression if it is defined., , ⎡, , (d) B T + 5C T, , (e) A(BC), , T, , T, , (d) D (BE), , (c) (D − E)T, , (d) (AB)C, , T, , (e) A − 3E, , 2. (a) CD T, , (b) D T − E T, , (c) AC + D, , (b) AB, , (d) E(AC), , 4. (a) 2AT + C, , ⎥, , ,, , 6. (a) (2D T − E)A, , (b) (4B)C + 2B, , (c) (−AC)T + 5D T, , (d) (BAT − 2C)T, , (e) B T(CC T − ATA), , (f ) D T E T − (ED)T, , In Exercises 7–8, use the following matrices and either the row, method or the column method, as appropriate, to find the indicated row or column., , ⎡, , 3, , ⎢, A = ⎣6, , −2, , ⎤, , ⎡, , ⎥, , ⎢, , ⎤, , 6, , −2, , 5, , 4⎦ and B = ⎣0, , 1, , 3⎦, , 4, , 9, , 7, , 5, , 7, , 4, , ⎥, , 3. (a) D + E, , (b) D − E, , (c) 5A, , (d) −7C, , (e) 2B − C, , (f ) 4E − 2D, , (g) −3(D + 2E), , (h) A − A, , (i) tr(D), , (c) the second column of AB, , (d) the first column of BA, , ( j) tr(D − 3E), , (k) 4 tr(7B), , (l) tr(A), , (e) the third row of AA, , (f ) the third column of AA, , 0, , 7. (a) the first row of AB, , 7, , (b) the third row of AB

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1.3 Matrices and Matrix Operations, , 8. (a) the first column of AB, , (b) the third column of BB, , (c) the second row of BB, , (d) the first column of AA, , (e) the third column of AB, , (f ) the first row of BA, , In Exercises 15–16, find all values of k , if any, that satisfy the, equation., , , 15. k, , ⎡, 1, ⎢, 1 ⎣1, , 1, , In Exercises 9–10, use matrices A and B from Exercises 7–8., , (b) Express each column vector of BB as a linear combination, of the column vectors of B ., 10. (a) Express each column vector of AB as a linear combination, of the column vectors of A., (b) Express each column vector of BA as a linear combination, of the column vectors of B ., In each part of Exercises 11–12, find matrices A, x, and b that, express the given linear system as a single matrix equation Ax = b,, and write out this matrix equation., 11. (a) 2x1 − 3x2 + 5x3 = 7, 9x1 − x2 + x3 = −1, x1 + 5x2 + 4x3 = 0, (b) 4x1, − 3x3, 5x1 + x2, 2x1 − 5x2 + 9x3, 3x2 − x3, , + x4, − 8x4, − x4, + 7x4, , 12. (a) x1 − 2x2 + 3x3 = −3, = 0, 2 x1 + x2, − 3x2 + 4x3 = 1, x1, + x3 = 5, , ⎡, , 3, ⎢, 14. (a) ⎣ 4, −2, , ⎡, , 3, , ⎢ 5, ⎢, (b) ⎢, ⎣ 3, −2, , 6, , −1, 3, 1, , −2, 0, 1, 5, , 2, , (b) 3x1 + 3x2 + 3x3 = −3, −x1 − 5x2 − 2x3 = 3, − 4x2 + x3 = 0, , −7, , ⎤⎡ ⎤ ⎡ ⎤, 2, 2, x1, ⎥⎢ ⎥ ⎢ ⎥, 7⎦ ⎣x2 ⎦ = ⎣−1⎦, x3, 5, 4, ⎤⎡ ⎤ ⎡ ⎤, 0, 1, 0, w, ⎢ x ⎥ ⎢0⎥, 2 −2⎥, ⎥⎢ ⎥ ⎢ ⎥, ⎥⎢ ⎥ = ⎢ ⎥, 4, 7⎦ ⎣ y ⎦ ⎣0⎦, 1, 6, 0, z, , ⎤⎡ ⎤, , 2, 0, 3, , 0, , 0, 2, ⎥⎢ ⎥, 3⎦ ⎣ 2 ⎦ = 0, k, 1, , In Exercises 17–20, use the column-row expansion of AB to, express this product as a sum of matrices., , , , 17. A =, , 4, , −3, , 2, , −1, , , 18. A =, , 19. A =, , , , 0, , −2, , 4, , −3, , , , 20. A =, , =1, =3, =0, =2, , ⎤⎡ ⎤ ⎡ ⎤, 2, x1, ⎢, ⎥⎢ ⎥ ⎢ ⎥, 3⎦ ⎣x2 ⎦ = ⎣0⎦, 13. (a) ⎣−1 −2, 3, x3, 0, 4 −1, ⎡, ⎤⎡ ⎤ ⎡ ⎤, 1, 1, 1, 2, x, ⎢, ⎥⎢ ⎥ ⎢ ⎥, 3, 0⎦ ⎣y ⎦ = ⎣ 2⎦, (b) ⎣2, −9, 5 −3 −6, z, 5, , 16. 2, , 1, , ⎢, k ⎣2, , , , , B=, , , 1, , 2, , −2, , , 3, , 1, , 1, , 4, , 1, , −3, , 0, , 2, , ⎡, , , , 2, , 3, , 4, , 5, , 6, , 1, , 2, , ⎢, , B = ⎣3, , 0, , 4, , 2, , 1, , −2, , 5, , , , ⎤, ⎥, , 4⎦, , 5, , , , , , 0, , , B=, , 1, , , , In each part of Exercises 13–14, express the matrix equation, as a system of linear equations., , ⎡, , , , ⎤⎡ ⎤, , 0, k, ⎥⎢ ⎥, 2⎦ ⎣ 1 ⎦ = 0, −3, 1, , 1, 0, 2, , 0, , ⎡, 9. (a) Express each column vector of AA as a linear combination, of the column vectors of A., , 37, , 6, , ⎡, , 2, , ⎢, , B = ⎣4, , 1, , ⎤, −1, ⎥, 0⎦, −1, , 21. For the linear system in Example 5 of Section 1.2, express the, general solution that we obtained in that example as a linear, combination of column vectors that contain only numerical, entries. [Suggestion: Rewrite the general solution as a single, column vector, then write that column vector as a sum of column vectors each of which contains at most one parameter,, and then factor out the parameters.], 22. Follow the directions of Exercise 21 for the linear system in, Example 6 of Section 1.2., In Exercises 23–24, solve the matrix equation for a , b, c,, and d ., 23., , 24., , a, −1, , 3, , 4, , =, , a+b, , d + 2c, , a−b, , b+a, , 3d + c, , 2d − c, , =, , 8, 7, , d − 2c, −2, 1, 6, , 25. (a) Show that if A has a row of zeros and B is any matrix for, which AB is defined, then AB also has a row of zeros., (b) Find a similar result involving a column of zeros., 26. In each part, find a 6 × 6 matrix [aij ] that satisfies the stated, condition. Make your answers as general as possible by using, letters rather than specific numbers for the nonzero entries., (a) aij = 0, , if, , i = j, , (b) aij = 0, , if, , i>j, , (c) aij = 0, , if, , i<j, , (d) aij = 0, , if, , |i − j | > 1

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38, , Chapter 1 Systems of Linear Equations and Matrices, , In Exercises 27–28, how many 3 × 3 matrices A can you find, for which the equation is satisfied for all choices of x , y , and z?, , ⎤, ⎡ ⎤ ⎡, x, x+y, ⎥, ⎢ ⎥ ⎢, 27. A ⎣y ⎦ = ⎣x − y ⎦, 0, z, , ⎡ ⎤ ⎡ ⎤, x, xy, ⎢ ⎥ ⎢ ⎥, 28. A ⎣y ⎦ = ⎣ 0 ⎦, 0, z, , 29. A matrix B is said to be a square root of a matrix A if BB = A., (a) Find two square roots of A =, , 2, 2, , 2, ., 2, , (b) How many different square roots can you find of, 5 0, A=, ?, 0 9, (c) Do you think that every 2 × 2 matrix has at least one, square root? Explain your reasoning., 30. Let 0 denote a 2 × 2 matrix, each of whose entries is zero., (a) Is there a 2 × 2 matrix A such that A  = 0 and AA = 0 ?, Justify your answer., (b) Is there a 2 × 2 matrix A such that A  = 0 and AA = A?, Justify your answer., , 34. The accompanying table shows a record of May and June unit, sales for a clothing store. Let M denote the 4 × 3 matrix of, May sales and J the 4 × 3 matrix of June sales., (a) What does the matrix M + J represent?, (b) What does the matrix M − J represent?, (c) Find a column vector x for which M x provides a list of the, number of shirts, jeans, suits, and raincoats sold in May., (d) Find a row vector y for which yM provides a list of the, number of small, medium, and large items sold in May., (e) Using the matrices x and y that you found in parts (c) and, (d), what does yM x represent?, Table Ex-34, May Sales, Small, , Medium, , Large, , Shirts, , 45, , 60, , 75, , Jeans, , 30, , 30, , 40, , Suits, , 12, , 65, , 45, , Raincoats, , 15, , 40, , 35, , 31. Establish Formula (11) by using Formula (5) to show that, , June Sales, , (AB)ij = (c1 r1 + c2 r2 + · · · + cr rr )ij, 32. Find a 4 × 4 matrix A = [aij ] whose entries satisfy the stated, condition., (b) aij = i j −1, , (a) aij = i + j, , , , (c) aij =, , 1 if, , −1 if, , |i − j | > 1, |i − j | ≤ 1, , 33. Suppose that type I items cost $1 each, type II items cost $2, each, and type III items cost $3 each. Also, suppose that the, accompanying table describes the number of items of each, type purchased during the first four months of the year., Table Ex-33, , Small, , Medium, , Large, , Shirts, , 30, , 33, , 40, , Jeans, , 21, , 23, , 25, , Suits, , 9, , 12, , 11, , Raincoats, , 8, , 10, , 9, , Working with Proofs, 35. Prove: If A and B are n × n matrices, then, tr(A + B) = tr(A) + tr(B), 36. (a) Prove: If AB and BA are both defined, then AB and BA, are square matrices., , Type I, , Type II, , Type III, , Jan., , 3, , 4, , 3, , Feb., , 5, , 6, , 0, , True-False Exercises, , Mar., , 2, , 9, , 4, , Apr., , 1, , 1, , 7, , TF. In parts (a)–(o) determine whether the statement is true or, false, and justify your answer., , What information is represented by the following product?, , ⎡, , 3, ⎢5, ⎢, ⎢, ⎣2, 1, , 4, 6, 9, 1, , ⎤, , 3 ⎡ ⎤, 1, 0⎥, ⎥⎢ ⎥, ⎥ ⎣2⎦, 4⎦, 3, 7, , (b) Prove: If A is an m × n matrix and A(BA) is defined, then, B is an n × m matrix., , (a) The matrix, , 1, 4, , 2, 5, , 3, has no main diagonal., 6, , (b) An m × n matrix has m column vectors and n row vectors., (c) If A and B are 2 × 2 matrices, then AB = BA., (d) The i th row vector of a matrix product AB can be computed, by multiplying A by the i th row vector of B .

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1.4 Inverses; Algebraic Properties of Matrices, , 39, , (e) For every matrix A, it is true that (AT )T = A., , Working withTechnology, , (f ) If A and B are square matrices of the same order, then, , T1. (a) Compute the product AB of the matrices in Example 5,, and compare your answer to that in the text., , tr(AB) = tr(A)tr(B), , (b) Use your technology utility to extract the columns of A, and the rows of B , and then calculate the product AB by, a column-row expansion., , (g) If A and B are square matrices of the same order, then, , (AB)T = ATB T, (h) For every square matrix A, it is true that tr(AT ) = tr(A)., (i) If A is a 6 × 4 matrix and B is an m × n matrix such that B TAT, is a 2 × 6 matrix, then m = 4 and n = 2., ( j) If A is an n × n matrix and c is a scalar, then tr(cA) = c tr(A)., (k) If A, B , and C are matrices of the same size such that, A − C = B − C , then A = B ., , T2. Suppose that a manufacturer uses Type I items at $1.35 each,, Type II items at $2.15 each, and Type III items at $3.95 each. Suppose also that the accompanying table describes the purchases of, those items (in thousands of units) for the first quarter of the year., Write down a matrix product, the computation of which produces, a matrix that lists the manufacturer’s expenditure in each month, of the first quarter. Compute that product., , (l) If A, B , and C are square matrices of the same order such that, AC = BC , then A = B ., (m) If AB + BA is defined, then A and B are square matrices of, the same size., (n) If B has a column of zeros, then so does AB if this product is, defined., (o) If B has a column of zeros, then so does BA if this product is, defined., , Type I, , Type II, , Type III, , Jan., , 3.1, , 4.2, , 3.5, , Feb., , 5.1, , 6.8, , 0, , Mar., , 2.2, , 9.5, , 4.0, , Apr., , 1.0, , 1.0, , 7.4, , 1.4 Inverses; Algebraic Properties of Matrices, In this section we will discuss some of the algebraic properties of matrix operations. We will, see that many of the basic rules of arithmetic for real numbers hold for matrices, but we will, also see that some do not., , Properties of Matrix, Addition and Scalar, Multiplication, , The following theorem lists the basic algebraic properties of the matrix operations., THEOREM 1.4.1 Properties of Matrix Arithmetic, , Assuming that the sizes of the matrices are such that the indicated operations can be, performed, the following rules of matrix arithmetic are valid., (a) A + B = B + A, [Commutative law for matrix addition], , A + (B + C) = (A + B) + C, (c) A(BC) = (AB)C, (d ) A(B + C) = AB + AC, (e) (B + C)A = BA + CA, ( f ) A(B − C) = AB − AC, ( g) (B − C)A = BA − CA, (b), , a(B + C) = aB + aC, (i ) a(B − C) = aB − aC, ( j ) (a + b)C = aC + bC, (k) (a − b)C = aC − bC, (l ) a(bC) = (ab)C, (h), , (m) a(BC) = (aB)C = B(aC), , [Associative law for matrix addition], [Associative law for matrix multiplication], [Left distributive law], [Right distributive law]

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40, , Chapter 1 Systems of Linear Equations and Matrices, , To prove any of the equalities in this theorem we must show that the matrix on the left, side has the same size as that on the right and that the corresponding entries on the two, sides are the same. Most of the proofs follow the same pattern, so we will prove part, (d ) as a sample. The proof of the associative law for multiplication is more complicated, than the rest and is outlined in the exercises., , There are three basic ways, to prove that two matrices, of the same size are equal—, prove that corresponding entries are the same, prove that, corresponding row vectors are, the same, or prove that corresponding column vectors are, the same., , Proof (d) We must show that A(B + C) and AB + AC have the same size and that, corresponding entries are equal. To form A(B + C), the matrices B and C must have, the same size, say m × n, and the matrix A must then have m columns, so its size must, be of the form r × m. This makes A(B + C) an r × n matrix. It follows that AB + AC, is also an r × n matrix and, consequently, A(B + C) and AB + AC have the same size., Suppose that A = [aij ], B = [bij ], and C = [cij ]. We want to show that corresponding entries of A(B + C) and AB + AC are equal; that is,, , , , A(B + C), , , , ij, , = (AB + AC)ij, , for all values of i and j . But from the definitions of matrix addition and matrix multiplication, we have, , , , A(B + C), , , , ij, , = ai 1 (b1j + c1j ) + ai 2 (b2j + c2j ) + · · · + aim (bmj + cmj ), = (ai 1 b1j + ai 2 b2j + · · · + aim bmj ) + (ai 1 c1j + ai 2 c2j + · · · + aim cmj ), = (AB)ij + (AC)ij = (AB + AC)ij, , Remark Although the operations of matrix addition and matrix multiplication were defined for, pairs of matrices, associative laws (b) and (c) enable us to denote sums and products of three, matrices as A + B + C and ABC without inserting any parentheses. This is justified by the fact, that no matter how parentheses are inserted, the associative laws guarantee that the same end, result will be obtained. In general, given any sum or any product of matrices, pairs of parentheses, can be inserted or deleted anywhere within the expression without affecting the end result., , E X A M P L E 1 Associativity of Matrix Multiplication, , As an illustration of the associative law for matrix multiplication, consider, , ⎡, , 1, ⎢, A = ⎣3, 0, Then, , ⎡, , 1, ⎢, AB = ⎣3, 0, Thus, , ⎤, , 2, ⎥ 4, 4⎦, 2, 1, , ⎤, , 2, 4, ⎥, 4 ⎦, B =, 2, 1, , ⎡, , 8, 3, ⎢, = ⎣20, 1, 2, , ⎡, , 8, ⎢, (AB)C = ⎣20, 2, and, , ⎡, , 1, ⎢, A(BC) = ⎣3, 0, , 3, 1, , C=, 1, 2, , 0, 3, , ⎤, , 5, 4, ⎥, 13⎦ and BC =, 2, 1, , ⎤, , 5, ⎥ 1, 13⎦, 2, 1, , ⎤, , 2, ⎥ 10, 4⎦, 4, 1, , ⎡, , 18, 0, ⎢, = ⎣46, 3, 4, , ⎡, , 18, 9, ⎢, = ⎣46, 3, 4, , so (AB)C = A(BC), as guaranteed by Theorem 1.4.1(c)., , 3, 1, , 1, 2, , ⎤, , 15, ⎥, 39⎦, 3, , ⎤, , 15, ⎥, 39⎦, 3, , 0, 10, =, 3, 4, , 9, 3

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1.4 Inverses; Algebraic Properties of Matrices, , Properties of Matrix, Multiplication, , 41, , Do not let Theorem 1.4.1 lull you into believing that all laws of real arithmetic carry over, to matrix arithmetic. For example, you know that in real arithmetic it is always true that, ab = ba, which is called the commutative law for multiplication. In matrix arithmetic,, however, the equality of AB and BA can fail for three possible reasons:, 1. AB may be defined and BA may not (for example, if A is 2 × 3 and B is 3 × 4)., 2. AB and BA may both be defined, but they may have different sizes (for example, if, A is 2 × 3 and B is 3 × 2)., 3. AB and BA may both be defined and have the same size, but the two products may, be different (as illustrated in the next example)., , E X A M P L E 2 Order Matters in Matrix Multiplication, Do not read too much into Example 2—it does not rule out, the possibility that AB and BA, may be equal in certain cases,, just that they are not equal in, all cases. If it so happens that, AB = BA, then we say that, AB and BA commute., , Zero Matrices, , Consider the matrices, , −1 0, , A=, Multiplying gives, , AB =, , 2, , 3, , −1, , −2, , 11, , 4, , and B =, , and BA =, , 1, 3, , 2, 0, 3, −3, , 6, 0, , Thus, AB  = BA., , A matrix whose entries are all zero is called a zero matrix. Some examples are, , ⎡, , 0, 0, , 0, 0, ⎢, , ⎣0, 0, 0, , 0, 0, 0, , ⎤, , 0, ⎥, 0⎦ ,, 0, , ⎡ ⎤, 0, , ⎢0⎥, 0 0 0 0, ⎢ ⎥, , ⎢ ⎥ , [0], 0 0 0 0, ⎣0⎦, 0, , We will denote a zero matrix by 0 unless it is important to specify its size, in which case, we will denote the m × n zero matrix by 0m×n ., It should be evident that if A and 0 are matrices with the same size, then, , A+0=0+A=A, Thus, 0 plays the same role in this matrix equation that the number 0 plays in the, numerical equation a + 0 = 0 + a = a., The following theorem lists the basic properties of zero matrices. Since the results, should be self-evident, we will omit the formal proofs., THEOREM 1.4.2 Properties of Zero Matrices, , If c is a scalar, and if the sizes of the matrices are such that the operations can be, perfomed, then:, (a) A + 0 = 0 + A = A, (b) A − 0 = A, (c), , A − A = A + (−A) = 0, , (d ) 0A = 0, (e), , If cA = 0, then c = 0 or A = 0.

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42, , Chapter 1 Systems of Linear Equations and Matrices, , Since we know that the commutative law of real arithmetic is not valid in matrix, arithmetic, it should not be surprising that there are other rules that fail as well. For, example, consider the following two laws of real arithmetic:, • If ab = ac and a  = 0, then b = c. [The cancellation law], • If ab = 0, then at least one of the factors on the left is 0., The next two examples show that these laws are not true in matrix arithmetic., , E X A M P L E 3 Failure of the Cancellation Law, , Consider the matrices, , , , , , 0, A=, 0, , , , 1, 1, , B=, 2, 3, , We leave it for you to confirm that, , , , , , 1, 2, , C=, 4, 3, , , , 3, AB = AC =, 6, , 4, 8, , 5, 4, , , , , , Although A  = 0, canceling A from both sides of the equation AB = AC would lead, to the incorrect conclusion that B = C . Thus, the cancellation law does not hold, in, general, for matrix multiplication (though there may be particular cases where it is true)., , E X A M P L E 4 A Zero Product with Nonzero Factors, , Here are two matrices for which AB = 0, but A  = 0 and B  = 0:, , , , , , 0, A=, 0, , Identity Matrices, , , , 1, 3, , B=, 2, 0, , 7, 0, , , , A square matrix with 1’s on the main diagonal and zeros elsewhere is called an identity, matrix. Some examples are, , , [1],, , , , ⎡, , 1, 0, ⎢, , ⎣0, 1, 0, , 1, 0, , 0, 1, 0, , ⎤, , ⎡, , 1, 0, ⎢0, ⎥ ⎢, 0⎦ , ⎢, ⎣0, 1, 0, , 0, 1, 0, 0, , 0, 0, 1, 0, , ⎤, , 0, 0⎥, ⎥, ⎥, 0⎦, 1, , An identity matrix is denoted by the letter I . If it is important to emphasize the size, we, will write In for the n × n identity matrix., To explain the role of identity matrices in matrix arithmetic, let us consider the effect, of multiplying a general 2 × 3 matrix A on each side by an identity matrix. Multiplying, on the right by the 3 × 3 identity matrix yields, , ⎡, , AI3 =, , a11, a21, , a12, a22, , a13, a23, , 1, ⎢, ⎣0, 0, , 0, 1, 0, , ⎤, , 0, a11, ⎥, 0⎦ =, a21, 1, , a12, a22, , a13, =A, a23, , and multiplying on the left by the 2 × 2 identity matrix yields, , , , 1, I2 A =, 0, , 0, 1, , , , a11, a21, , a12, a22, , a13, a11, =, a23, a21, , a12, a22, , a13, =A, a23

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1.4 Inverses; Algebraic Properties of Matrices, , 43, , The same result holds in general; that is, if A is any m × n matrix, then, , AIn = A and Im A = A, Thus, the identity matrices play the same role in matrix arithmetic that the number 1, plays in the numerical equation a · 1 = 1 · a = a., As the next theorem shows, identity matrices arise naturally in studying reduced row, echelon forms of square matrices., THEOREM 1.4.3 If R is the reduced row echelon form of an n × n matrix A, then either, , R has a row of zeros or R is the identity matrix In ., , Proof Suppose that the reduced row echelon form of A is, , ⎡, , r11, ⎢r, ⎢ 21, R=⎢ ., ⎣ .., r n1, , r12, r22, .., ., r n2, , ···, ···, ···, , ⎤, r1n, r2n ⎥, ⎥, .. ⎥, . ⎦, rnn, , Either the last row in this matrix consists entirely of zeros or it does not. If not, the, matrix contains no zero rows, and consequently each of the n rows has a leading entry, of 1. Since these leading 1’s occur progressively farther to the right as we move down, the matrix, each of these 1’s must occur on the main diagonal. Since the other entries in, the same column as one of these 1’s are zero, R must be In . Thus, either R has a row of, zeros or R = In ., Inverse of a Matrix, , In real arithmetic every nonzero number a has a reciprocal a −1 (= 1/a) with the property, , a · a −1 = a −1 · a = 1, The number a −1 is sometimes called the multiplicative inverse of a . Our next objective is, to develop an analog of this result for matrix arithmetic. For this purpose we make the, following definition., DEFINITION 1, , If A is a square matrix, and if a matrix B of the same size can be, found such that AB = BA = I , then A is said to be invertible (or nonsingular) and, B is called an inverse of A. If no such matrix B can be found, then A is said to be, singular., , Remark The relationship AB = BA = I is not changed by interchanging A and B , so if A is, invertible and B is an inverse of A, then it is also true that B is invertible, and A is an inverse of, B . Thus, when, AB = BA = I, we say that A and B are inverses of one another., , E X A M P L E 5 An Invertible Matrix, , Let, , A=, , 2, , −5, , −1, , 3, , and B =, , 3, 1, , 5, 2

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44, , Chapter 1 Systems of Linear Equations and Matrices, , Then, , AB =, BA =, , 2, , −5, , −1, , 3, , 3, 1, , 5, 2, , 2, −1, , 3, 1, , 5, 1, =, 2, 0, , −5, 3, , =, , 1, 0, , 0, =I, 1, 0, =I, 1, , Thus, A and B are invertible and each is an inverse of the other., , E X A M P L E 6 A Class of Singular Matrices, , A square matrix with a row or column of zeros is singular. To help understand why this, is so, consider the matrix, ⎡, ⎤, 1 4 0, ⎢, ⎥, A = ⎣2 5 0⎦, 3 6 0, As in Example 6, we will frequently denote a zero matrix, with one row or one column, by a boldface zero., , To prove that A is singular we must show that there is no 3 × 3 matrix B such that, AB = BA = I. For this purpose let c1 , c2 , 0 be the column vectors of A. Thus, for any, 3 × 3 matrix B we can express the product BA as, , BA = B[c1, , c2, , 0] = [B c1, , B c2, , 0] [Formula (6) of Section 1.3], , The column of zeros shows that BA  = I and hence that A is singular., , Properties of Inverses, , It is reasonable to ask whether an invertible matrix can have more than one inverse. The, next theorem shows that the answer is no—an invertible matrix has exactly one inverse., , THEOREM 1.4.4 If B and C are both inverses of the matrix A, then B, , = C., , Proof Since B is an inverse of A, we have BA = I. Multiplying both sides on the right, by C gives (BA)C = I C = C . But it is also true that (BA)C = B(AC) = BI = B , so, C = B., WARNING The symbol, , A−1, , should not be interpreted as, 1/A. Division by matrices will, not be a defined operation in, this text., , As a consequence of this important result, we can now speak of “the” inverse of an, invertible matrix. If A is invertible, then its inverse will be denoted by the symbol A−1 ., Thus,, , AA−1 = I and A−1A = I, , (1), , The inverse of A plays much the same role in matrix arithmetic that the reciprocal a −1, plays in the numerical relationships aa −1 = 1 and a −1 a = 1., In the next section we will develop a method for computing the inverse of an invertible, matrix of any size. For now we give the following theorem that specifies conditions under, which a 2 × 2 matrix is invertible and provides a simple formula for its inverse., , Historical Note The formula for A−1 given in Theorem 1.4.5 first appeared (in a more general form), in Arthur Cayley’s 1858 Memoir on the Theory of Matrices. The more general result that Cayley, discovered will be studied later.

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1.4 Inverses; Algebraic Properties of Matrices, , The quantity ad − bc in Theorem 1.4.5 is called the determinant of the 2 × 2 matrix A, and is denoted by, , THEOREM 1.4.5 The matrix, , A− 1 =, , or alternatively by, , det(A) =, , , b , = ad − bc, d, , a b, = ad – bc, c d, , Figure 1.4.1, , b, d, , is invertible if and only if ad − bc  = 0, in which case the inverse is given by the formula, , det(A) = ad − bc, , , a, , c, , a, c, , A=, , 45, , −b, a, , 1, d, ad − bc −c, , (2), , We will omit the proof, because we will study a more general version of this theorem, later. For now, you should at least confirm the validity of Formula (2) by showing that, AA−1 = A−1A = I ., Remark Figure 1.4.1 illustrates that the determinant of a 2 × 2 matrix A is the product of the, entries on its main diagonal minus the product of the entries off its main diagonal., , E X A M P L E 7 Calculating the Inverse of a 2 × 2 Matrix, , In each part, determine whether the matrix is invertible. If so, find its inverse., (a) A =, , 6, 5, , 1, 2, , (b) A =, , Solution (a) The determinant of A is det(A), , A, , 1, 2, =, 7 −5, , −1, 6, , 2, , 3, , −6, , = (6)(2) − (1)(5) = 7, which is nonzero., , Thus, A is invertible, and its inverse is, −1, , −1, , , , − 17, , 2, 7, − 57, , =, , , , 6, 7, , We leave it for you to confirm that AA−1 = A−1 A = I., Solution (b) The matrix is not invertible since det(A), , = (−1)(−6) − (2)(3) = 0., , E X A M P L E 8 Solution of a Linear System by Matrix Inversion, , A problem that arises in many applications is to solve a pair of equations of the form, , u = ax + by, v = cx + dy, for x and y in terms of u and v. One approach is to treat this as a linear system of, two equations in the unknowns x and y and use Gauss–Jordan elimination to solve, for x and y. However, because the coefficients of the unknowns are literal rather than, numerical, this procedure is a little clumsy. As an alternative approach, let us replace the, two equations by the single matrix equation, , u, ax + by, =, v, cx + dy, which we can rewrite as, , u, a, =, v, c, , b, d, , x, y, If we assume that the 2 × 2 matrix is invertible (i.e., ad − bc  = 0), then we can multiply, through on the left by the inverse and rewrite the equation as, , a, c, , b, d, , −1, , u, a, =, v, c, , b, d, , −1, , a, c, , b, d, , x, y

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46, , Chapter 1 Systems of Linear Equations and Matrices, , which simplifies to, , a, c, , b, d, , −1, , u, x, =, v, y, , Using Theorem 1.4.5, we can rewrite this equation as, , d, ad − bc −c, 1, , from which we obtain, , x=, , −b, a, , u, x, =, v, y, , du − bv, av − cu, , y=, ad − bc, ad − bc, , The next theorem is concerned with inverses of matrix products., THEOREM 1.4.6 If, , A and B are invertible matrices with the same size, then AB is, , invertible and, , (AB)−1 = B −1 A−1, , Proof We can establish the invertibility and obtain the stated formula at the same time, , by showing that, , (AB)(B −1 A−1 ) = (B −1 A−1 )(AB) = I, , But, , (AB)(B −1 A−1 ) = A(BB −1 )A−1 = AIA−1 = AA−1 = I, and similarly, (B −1 A−1 )(AB) = I., Although we will not prove it, this result can be extended to three or more factors:, A product of any number of invertible matrices is invertible, and the inverse of the product, is the product of the inverses in the reverse order., , E X A M P L E 9 The Inverse of a Product, , Consider the matrices, 1, 1, , A=, If a product of matrices is, singular, then at least one of, the factors must be singular., Why?, , We leave it for you to show that, , A, , 3 −2, =, ,, −1 1, , , B, , −1, , =, , 1 −1, , −1, , 3, 2, , 2, 2, , , −3, , , , 4, 6, , (AB)−1 =, 9, 8, −2, , 7, AB =, 9, and also that, , −1, , 2, 3, , B=, 3, 2, , , , , ,, , −1, , −1, , B A, , =, , 7, 2, , 1 −1, , −1, , 3, 2, , , , , , Thus, (AB)−1 = B −1 A−1 as guaranteed by Theorem 1.4.6., Powers of a Matrix, , If A is a square matrix, then we define the nonnegative integer powers of A to be, , A0 = I and An = AA · · · A, , [n factors], , and if A is invertible, then we define the negative integer powers of A to be, , A−n = (A−1 )n = A−1 A−1 · · · A−1, , , , 4 −3, 3 −2, =, 7, 9, −1 1, −2, 2, , [n factors]

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1.4 Inverses; Algebraic Properties of Matrices, , 47, , Because these definitions parallel those for real numbers, the usual laws of nonnegative, exponents hold; for example,, , Ar As = Ar+s and (Ar )s = Ars, In addition, we have the following properties of negative exponents., THEOREM 1.4.7 If A is invertible and n is a nonnegative integer, then:, , (a) A−1 is invertible and (A−1 )−1 = A., (b) An is invertible and (An )−1 = A−n = (A−1 )n ., (c) kA is invertible for any nonzero scalar k, and (kA)−1 = k −1 A−1 ., We will prove part (c) and leave the proofs of parts (a) and (b) as exercises., Proof (c) Properties (m) and (l) of Theorem 1.4.1 imply that, , (kA)(k −1 A−1 ) = k −1 (kA)A−1 = (k −1 k)AA−1 = (1)I = I, and similarly, (k −1 A−1 )(kA) = I. Thus, kA is invertible and (kA)−1 = k −1 A−1 ., E X A M P L E 10 Properties of Exponents, , Let A and A−1 be the matrices in Example 9; that is,, 3, , −2, , −1, , 1, , 1, 1, , 2, 3, , 3, , −2, , 3, , −2, , 3, , −2, , −1, , 1, , −1, , 1, , −1, , 1, , A=, , and A−1 =, , Then, , A−3 = (A−1 )3 =, , =, , 41, , −30, , −15, , 11, , Also,, , A3 =, , 1, 1, , 2, 3, , 1, 1, , 2, 3, , 1, 1, , 2, 11, =, 3, 15, , 30, 41, , so, as expected from Theorem 1.4.7(b),, , (A3 )−1 =, , 41, , −30, , (11)(41) − (30)(15) −15, , 11, , 1, , =, , 41, , −30, , −15, , 11, , = (A−1 )3, , E X A M P L E 11 The Square of a Matrix Sum, , In real arithmetic, where we have a commutative law for multiplication, we can write, , (a + b)2 = a 2 + ab + ba + b2 = a 2 + ab + ab + b2 = a 2 + 2ab + b2, However, in matrix arithmetic, where we have no commutative law for multiplication,, the best we can do is to write, , (A + B)2 = A2 + AB + BA + B 2, , Matrix Polynomials, , It is only in the special case where A and B commute (i.e., AB = BA) that we can go a, step further and write, (A + B)2 = A2 + 2AB + B 2, If A is a square matrix, say n × n, and if, , p(x) = a0 + a1 x + a2 x 2 + · · · + am x m

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48, , Chapter 1 Systems of Linear Equations and Matrices, , is any polynomial, then we define the n × n matrix p(A) to be, , p(A) = a0 I + a1 A + a2 A2 + · · · + am Am, , (3), , where I is the n × n identity matrix; that is, p(A) is obtained by substituting A for x, and replacing the constant term a0 by the matrix a0 I. An expression of form (3) is called, a matrix polynomial in A., E X A M P L E 1 2 A Matrix Polynomial, , Find p(A) for, , p(x) = x 2 − 2x − 3 and A =, Solution, , −1 2, 0, , 3, , p(A) = A2 − 2A − 3I, =, =, , −1 2, 0, 1, 0, , 3, , 2, , −2, , 4, −2, −, 9, 0, , −1 2, 0, , 3, , −3, , 4, 3, −, 6, 0, , 1, 0, , 0, 1, , 0, 0, =, 3, 0, , 0, 0, , or more briefly, p(A) = 0., Remark It follows from the fact that Ar As = Ar+s = As+r = As Ar that powers of a square, matrix commute, and since a matrix polynomial in A is built up from powers of A, any two matrix, polynomials in A also commute; that is, for any polynomials p1 and p2 we have, , p1 (A)p2 (A) = p2 (A)p1 (A), , Properties of theTranspose, , (4), , The following theorem lists the main properties of the transpose., THEOREM 1.4.8 If the sizes of the matrices are such that the stated operations can be, , performed, then:, (a) (AT )T = A, (b) (A + B)T = AT + B T, (c), , (A − B)T = AT − B T, , (d ) (kA)T = kAT, (e), , (AB)T = B TAT, , If you keep in mind that transposing a matrix interchanges its rows and columns, then, you should have little trouble visualizing the results in parts (a)–(d ). For example, part, (a) states the obvious fact that interchanging rows and columns twice leaves a matrix, unchanged; and part (b) states that adding two matrices and then interchanging the, rows and columns produces the same result as interchanging the rows and columns, before adding. We will omit the formal proofs. Part (e) is less obvious, but for brevity, we will omit its proof as well. The result in that part can be extended to three or more, factors and restated as:, The transpose of a product of any number of matrices is the product of the transposes, in the reverse order.

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1.4 Inverses; Algebraic Properties of Matrices, , 49, , The following theorem establishes a relationship between the inverse of a matrix and, the inverse of its transpose., THEOREM 1.4.9 If A is an invertible matrix, then AT is also invertible and, , (AT )−1 = (A−1 )T, Proof We can establish the invertibility and obtain the formula at the same time by, , showing that, , AT(A−1 )T = (A−1 )TAT = I, , But from part (e) of Theorem 1.4.8 and the fact that I T = I, we have, , AT(A−1 )T = (A−1 A)T = I T = I, (A−1 )TAT = (AA−1 )T = I T = I, which completes the proof., E X A M P L E 1 3 Inverse of a Transpose, , Consider a general 2 × 2 invertible matrix and its transpose:, , A=, , a, c, , b, d, , and AT =, , a, b, , c, d, , Since A is invertible, its determinant ad − bc is nonzero. But the determinant of AT is, also ad − bc (verify), so AT is also invertible. It follows from Theorem 1.4.5 that, , ⎡, , (AT )−1, , d, ⎢ ad − bc, =⎢, ⎣, b, −, ad − bc, , which is the same matrix that results if A, , −1, , T −1, , (A ), , −, , ⎤, c, ad − bc ⎥, ⎥, ⎦, a, ad − bc, , is transposed (verify). Thus,, , = (A−1 )T, , as guaranteed by Theorem 1.4.9., , Exercise Set 1.4, In Exercises 1–2, verify that the following matrices and scalars, satisfy the stated properties of Theorem 1.4.1., , , A=, , C=, , 3, , −1, , 2, , 4, , , , , , B=, , 0, , 2, , 1, , −4, , , ,, , , , 4, , 1, , −3, , −2, , , a = 4 , b = −7, , 1. (a) The associative law for matrix addition., (b) The associative law for matrix multiplication., , 2. (a) a(BC) = (aB)C = B(aC), (b) A(B − C) = AB − AC, , In Exercises 3–4, verify that the matrices and scalars in Exercise 1 satisfy the stated properties., 3. (a) (AT )T = A, , (b) (AB)T = B TAT, , 4. (a) (A + B)T = AT + B T, , (b) (aC)T = aC T, , In Exercises 5–8, use Theorem 1.4.5 to compute the inverse of, the matrix., , (c) The left distributive law., (d) (a + b)C = aC + bC, , (c) (B + C)A = BA + CA, , (d) a(bC) = (ab)C, , 5. A =, , 2, 4, , −3, 4, , 6. B =, , 3, 5, , 1, 2

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50, , Chapter 1 Systems of Linear Equations and Matrices, , 7. C =, , 2, 0, , 0, 3, , 8. D =, , 9. Find the inverse of, , ⎡, ⎣, , 1, (ex, 2, , + e−x ), , 1, (ex, 2, , − e−x ), , 6, , 4, , −2, , −1, , In Exercises 25–28, use the method of Example 8 to find the, unique solution of the given linear system., 25. 3x1 − 2x2 = −1, , ⎤, − e−x ), ⎦, 1, (ex + e−x ), 2, 1, (ex, 2, , 10. Find the inverse of, cos θ, − sin θ, , sin θ, cos θ, , 11. (A ), , −1 T, , −1 −1, , = (A ), , 12. (A ), , 13. (ABC)−1 = C −1 B −1 A−1, , −3, , 7, , 1, , −2, −1, , 17. (I + 2A)−1 =, , 4, , 16. (5AT )−1 =, , 18. A−1 =, , −3, , −1, , 5, , 2, , −1, , 2, 3, , −3, , (a) A, , (b) A, , 19. A =, , 3, 2, , (c) A − 2A + I, 20. A =, , 2, 4, , (b) p(x) = 2x 2 − x + 1, (c) p(x) = x 3 − 2x + 1, 1, 1, , 22. A =, , In Exercises 23–24, let, , A=, , , a, , b, , c, , d, , , , , B=, , , , 0, , 1, , 0, , 0, , x1 + 4x2 = 4, , 2, 4, , , , and if A is a square matrix, then it can be proved that, , p(x) = x 2 − 9, p1 (x) = x + 3, p2 (x) = x − 3, 29. The matrix A in Exercise 21., 30. An arbitrary square matrix A., , (b) State a valid formula for multiplying out, , 32. The numerical equation a 2 = 1 has exactly two solutions., Find at least eight solutions of the matrix equation A2 = I3 ., [Hint: Look for solutions in which all entries off the main, diagonal are zero.], 33. (a) Show that if a square matrix A satisfies the equation, A2 + 2A + I = 0, then A must be invertible. What is the, inverse?, , 0, 1, , , , C=, , p(x) = p1 (x)p2 (x), , (c) What condition can you impose on A and B that will allow, you to write (A + B)(A − B) = A2 − B 2 ?, , 0, 1, , (a) p(x) = x − 2, , 3, 2, , 4x1 − 3x2 = −2, , (A + B)(A − B), , In Exercises 21–22, compute p(A) for the given matrix A and, the following polynomials., , 21. A =, , 28. 2x1 − 2x2 = 4, , (A + B)(A − B)  = A2 − B 2, , 2, , 1, 1, , 0, , 31. (a) Give an example of two 2 × 2 matrices such that, , 5, , In Exercises 19–20, compute the following using the given matrix A., 3, , 27. 6x1 + x2 =, , In Exercises 29–30, verify this statement for the stated matrix A, and polynomials, , 14. (ABC)T = C TB TAT, , 2, 5, , −x1 − 3x2 = 1, , p(A) = p1 (A)p2 (A), , =A, , In Exercises 15–18, use the given information to find A., 15. (7A)−1 =, , 3, , If a polynomial p(x) can be factored as a product of lower, degree polynomials, say, , In Exercises 11–14, verify that the equations are valid for the, matrices in Exercises 5–8., T −1, , 26. −x1 + 5x2 = 4, , 4x1 + 5x2 =, , , , 0, , 0, , 1, , 0, , (b) Show that if p(x) is a polynomial with a nonzero constant, term, and if A is a square matrix for which p(A) = 0, then, A is invertible., 34. Is it possible for A3 to be an identity matrix without A being, invertible? Explain., , 23. Find all values of a, b, c, and d (if any) for which the matrices, A and B commute., , 35. Can a matrix with a row of zeros or a column of zeros have an, inverse? Explain., , 24. Find all values of a, b, c, and d (if any) for which the matrices, A and C commute., , 36. Can a matrix with two identical rows or two identical columns, have an inverse? Explain.

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1.4 Inverses; Algebraic Properties of Matrices, , In Exercises 37–38, determine whether A is invertible, and if, so, find the inverse. [Hint: Solve AX = I for X by equating corresponding entries on the two sides.], , ⎡, , 1, , ⎢, 37. A = ⎣1, 0, , 0, , 1, , ⎡, , ⎤, , 1, , ⎥, 0⎦, , 1, , 1, , 1, , ⎢, 38. A = ⎣1, 0, , 1, , 1, , ⎤, , 0, , ⎥, 0⎦, , 1, , 1, , In Exercises 39–40, simplify the expression assuming that A,, B , C , and D are invertible., , 51, , 49. Assuming that all matrices are n × n and invertible, solve, for D ., , C T B −1 A2 BAC −1DA−2 B T C −2 = C T, 50. Assuming that all matrices are n × n and invertible, solve, for D ., , ABC TDBAT C = AB T, , Working with Proofs, In Exercises 51–58, prove the stated result., , 39. (AB)−1 (AC −1 )(D −1 C −1 )−1 D −1, 40. (AC, , −1 −1, , −1, , −1 −1, , ) (AC )(AC ) AD, , 41. Show that if R is a 1 × n matrix and C is an n × 1 matrix,, then RC = tr(CR)., 42. If A is a square matrix and n is a positive integer, is it true that, (An )T = (AT )n ? Justify your answer., 43. (a) Show that if A is invertible and AB = AC , then B = C ., (b) Explain why part (a) and Example 3 do not contradict one, another., 44. Show that if A is invertible and k is any nonzero scalar, then, (kA)n = k nAn for all integer values of n., 45. (a) Show that if A, B , and A + B are invertible matrices with, the same size, then, , A(A−1 + B −1 )B(A + B)−1 = I, (b) What does the result in part (a) tell you about the matrix, A−1 + B −1 ?, 46. A square matrix A is said to be idempotent if A2 = A., (a) Show that if A is idempotent, then so is I − A., (b) Show that if A is idempotent, then 2A − I is invertible, and is its own inverse., 47. Show that if A is a square matrix such that Ak = 0 for some, positive integer k , then the matrix I − A is invertible and, , (I − A)−1 = I + A + A2 + · · · + Ak−1, 48. Show that the matrix, , , A=, , 51. Theorem 1.4.1(a), , 52. Theorem 1.4.1(b), , 53. Theorem 1.4.1( f ), , 54. Theorem 1.4.1(c), , 55. Theorem 1.4.2(c), , 56. Theorem 1.4.2(b), , 57. Theorem 1.4.8(d), , 58. Theorem 1.4.8(e), , −1, , a, , b, , c, , d, , , , satisfies the equation, , A2 − (a + d)A + (ad − bc)I = 0, , True-False Exercises, TF. In parts (a)–(k) determine whether the statement is true or, false, and justify your answer., (a) Two n × n matrices, A and B , are inverses of one another if, and only if AB = BA = 0., (b) For all square matrices A and B of the same size, it is true that, (A + B)2 = A2 + 2AB + B 2 ., (c) For all square matrices A and B of the same size, it is true that, A2 − B 2 = (A − B)(A + B)., (d) If A and B are invertible matrices of the same size, then AB is, invertible and (AB)−1 = A−1 B −1 ., (e) If A and B are matrices such that AB is defined, then it is true, that (AB)T = ATB T ., (f ) The matrix, , A=, , a, c, , b, d, , is invertible if and only if ad − bc  = 0., (g) If A and B are matrices of the same size and k is a constant,, then (kA + B)T = kAT + B T ., (h) If A is an invertible matrix, then so is AT ., (i) If p(x) = a0 + a1 x + a2 x 2 + · · · + am x m and I is an identity, matrix, then p(I ) = a0 + a1 + a2 + · · · + am ., ( j) A square matrix containing a row or column of zeros cannot, be invertible., (k) The sum of two invertible matrices of the same size must be, invertible.

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52, , Chapter 1 Systems of Linear Equations and Matrices, , Working withTechnology, T1. Let A be the matrix, , ⎡, , 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, . . ., 0, , 1, 2, , 1⎤, 3, , ⎢1, A=⎢, ⎣4, , 0, , 1⎥, 5⎦, , 1, 6, , 1, 7, , 0, , the terms of which are commonly denoted as, , ⎥, , F0 , F1 , F2 , F3 , . . . , Fn , . . ., After the initial terms F0 = 0 and F1 = 1, each term is the sum of, the previous two; that is,, , Discuss the behavior of Ak as k increases indefinitely, that is, as, , Fn = Fn−1 + Fn−2, , k → ⬁., T2. In each part use your technology utility to make a conjecture, about the form of An for positive integer powers of n., (a) A =, , , a, , 1, , 0, , a, , , , , , (b) A =, , cos θ, , sin θ, , − sin θ, , cos θ, , Confirm that if, , , Q=, , , , F2, , F1, , F1, , F0, , , then, , T3. The Fibonacci sequence (named for the Italian mathematician, Leonardo Fibonacci 1170–1250) is, , , , Q =, n, , , =, , , , 1, , 1, , 1, , 0, , Fn+1, , Fn, , Fn, , F0, , , , 1.5 Elementary Matrices and a Method for Finding A−1, In this section we will develop an algorithm for finding the inverse of a matrix, and we will, discuss some of the basic properties of invertible matrices., , In Section 1.1 we defined three elementary row operations on a matrix A:, 1. Multiply a row by a nonzero constant c., 2. Interchange two rows., 3. Add a constant c times one row to another., It should be evident that if we let B be the matrix that results from A by performing one, of the operations in this list, then the matrix A can be recovered from B by performing, the corresponding operation in the following list:, 1. Multiply the same row by 1/c., 2. Interchange the same two rows., 3. If B resulted by adding c times row ri of A to row rj , then add −c times rj to ri ., It follows that if B is obtained from A by performing a sequence of elementary row, operations, then there is a second sequence of elementary row operations, which when, applied to B recovers A (Exercise 33). Accordingly, we make the following definition., DEFINITION 1 Matrices, , A and B are said to be row equivalent if either (hence each), can be obtained from the other by a sequence of elementary row operations., , Our next goal is to show how matrix multiplication can be used to carry out an, elementary row operation., DEFINITION 2, , A matrix E is called an elementary matrix if it can be obtained from, an identity matrix by performing a single elementary row operation.

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1.5 Elementary Matrices and a Method for Finding A−1, , 53, , E X A M P L E 1 Elementary Matrices and Row Operations, , Listed below are four elementary matrices and the operations that produce them., , , , 1, 0, , , , 0, , −3, , 1, ⎢0, ⎢, ⎢, ⎣0, 0, , 0, 0, 0, 1, , 0, 0, 1, 0, , ⎤, , 0, 1⎥, ⎥, ⎥, 0⎦, 0, , ⎡, , 1, ⎢, ⎣0, 0, , ⎤, , 0, 1, 0, , 3, ⎥, 0⎦, 1, , Interchange the, second and fourth, rows of I4 ., , ⎡, , 1, ⎢, ⎣0, 0, , 0, 1, 0, , ⎤, , 0, ⎥, 0⎦, 1, , , , , , , , , Multiply the, second row of, I2 by −3., , ⎡, , Add 3 times, the third row of, I3 to the first row., , Multiply the, first row of, I3 by 1., , The following theorem, whose proof is left as an exercise, shows that when a matrix A, is multiplied on the left by an elementary matrix E , the effect is to perform an elementary, row operation on A., THEOREM 1.5.1 Row Operations by Matrix Multiplication, , If the elementary matrix E results from performing a certain row operation on Im and, if A is an m × n matrix, then the product EA is the matrix that results when this same, row operation is performed on A., , E X A M P L E 2 Using Elementary Matrices, , Consider the matrix, , ⎡, , 1, ⎢, A = ⎣2, 1, and consider the elementary matrix, , 0, −1, 4, , ⎡, , 1, ⎢, E = ⎣0, 3, , ⎤, , 2, 3, 4, , 0, 1, 0, , 3, ⎥, 6⎦, 0, , ⎤, , 0, ⎥, 0⎦, 1, , which results from adding 3 times the first row of I3 to the third row. The product EA is, Theorem 1.5.1 will be a useful tool for developing new results about matrices, but as a, practical matter it is usually, preferable to perform row operations directly., , ⎡, , 1, ⎢, EA = ⎣2, 4, , 0, −1, 4, , 2, 3, 10, , ⎤, , 3, ⎥, 6⎦, 9, , which is precisely the matrix that results when we add 3 times the first row of A to the, third row., , We know from the discussion at the beginning of this section that if E is an elementary, matrix that results from performing an elementary row operation on an identity matrix, I , then there is a second elementary row operation, which when applied to E produces, I back again. Table 1 lists these operations. The operations on the right side of the table, are called the inverse operations of the corresponding operations on the left.

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54, , Chapter 1 Systems of Linear Equations and Matrices, Table 1, Row Operation on I, That Produces E, , Row Operation on E, That Reproduces I, , Multiply row i by c  = 0, , Multiply row i by 1/c, , Interchange rows i and j, , Interchange rows i and j, , Add c time row i to row j, , Add −c times row i to row j, , E X A M P L E 3 Row Operations and Inverse Row Operations, , In each of the following, an elementary row operation is applied to the 2 × 2 identity, matrix to obtain an elementary matrix E , then E is restored to the identity matrix by, applying the inverse row operation., 1, 0, , 0, 1, , −→, , 1, 0, , 0, 7, , 0, 1, , −→, , 0, 1, , , , , Multiply the second, row by 7., , 1, 0, , 1, 0, , −→, , 0, 1, , Multiply the second, row by 17 ., , 1, 0, , 1, 0, , −→, , 0, 1, , , , , , Interchange the first Interchange the first, and second rows., and second rows., , 1, 0, , 0, 1, , −→, , 1, 0, , 5, 1, , −→, , 1, 0, , 0, 1, , , , , , Add 5 times the, Add −5 times the, second row to the second row to the, first., first., , The next theorem is a key result about invertibility of elementary matrices. It will be, a building block for many results that follow., THEOREM 1.5.2 Every elementary matrix is invertible, and the inverse is also an ele-, , mentary matrix., Proof If E is an elementary matrix, then E results by performing some row operation, on I . Let E0 be the matrix that results when the inverse of this operation is performed, on I . Applying Theorem 1.5.1 and using the fact that inverse row operations cancel the, effect of each other, it follows that, , E0 E = I and EE0 = I, Thus, the elementary matrix E0 is the inverse of E ., EquivalenceTheorem, , One of our objectives as we progress through this text is to show how seemingly diverse, ideas in linear algebra are related. The following theorem, which relates results we, have obtained about invertibility of matrices, homogeneous linear systems, reduced row

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1.5 Elementary Matrices and a Method for Finding A−1, , 55, , echelon forms, and elementary matrices, is our first step in that direction. As we study, new topics, more statements will be added to this theorem., , THEOREM 1.5.3 Equivalent Statements, , If A is an n × n matrix, then the following statements are equivalent, that is, all true or, all false., (a) A is invertible., (b) Ax = 0 has only the trivial solution., (c), , The reduced row echelon form of A is In ., , (d ) A is expressible as a product of elementary matrices., , The following figure illustrates, visually that from the sequence of implications, , (a) ⇒ (b) ⇒ (c) ⇒ (d ) ⇒ (a), we can conclude that, , Proof We will prove the equivalence by establishing the chain of implications:, , (a) ⇒ (b) ⇒ (c) ⇒ (d ) ⇒ (a)., = 0. Multiplying both, sides of this equation by the matrix A−1 gives A−1 (Ax0 ) = A−1 0, or (A−1 A)x0 = 0, or, I x0 = 0, or x0 = 0. Thus, Ax = 0 has only the trivial solution., (a) ⇒ (b) Assume A is invertible and let x0 be any solution of Ax, , (d ) ⇒ (c) ⇒ (b) ⇒ (a), (b) ⇒ (c) Let Ax, , and hence that, , (a) ⇔ (b) ⇔ (c) ⇔ (d ), (see Appendix A)., (a), , (d), , (b), , = 0 be the matrix form of the system, a11 x1 + a12 x2 + · · · + a1n xn = 0, a21 x1 + a22 x2 + · · · + a2n xn = 0, .., .., .., .., ., ., ., ., an1 x1 + an2 x2 + · · · + ann xn = 0, , (1), , and assume that the system has only the trivial solution. If we solve by Gauss–Jordan, elimination, then the system of equations corresponding to the reduced row echelon, form of the augmented matrix will be, , =0, =0, , x1, , (c), , x2, , .., , ., xn = 0, , Thus the augmented matrix, , ⎡, , a11, ⎢a, ⎢ 21, ⎢ ., ⎣ .., an1, , a12, a22, .., ., an2, , ···, ···, ···, , a1 n, a2n, .., ., ann, , for (1) can be reduced to the augmented matrix, , ⎡, , 1, ⎢, ⎢0, ⎢, ⎢0, ⎢, , 0, 1, 0, , 0, 0, 1, , 0, , 0, , 0, , ⎢ .., ⎣., , .., ., , .., ., , ⎤, , 0, 0⎥, ⎥, , .. ⎥, .⎦, , 0, , ⎤, ··· 0 0, ⎥, · · · 0 0⎥, ⎥, · · · 0 0⎥, .. .. ⎥, ⎥, . .⎦, ··· 1 0, , (2)

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56, , Chapter 1 Systems of Linear Equations and Matrices, , for (2) by a sequence of elementary row operations. If we disregard the last column (all, zeros) in each of these matrices, we can conclude that the reduced row echelon form of, A is In ., (c) ⇒ (d ) Assume that the reduced row echelon form of A is In , so that A can be reduced, , to In by a finite sequence of elementary row operations. By Theorem 1.5.1, each of these, operations can be accomplished by multiplying on the left by an appropriate elementary, matrix. Thus we can find elementary matrices E1 , E2 , . . . , Ek such that, , Ek · · · E2 E1 A = In, , (3), , By Theorem 1.5.2, E1 , E2 , . . . , Ek are invertible. Multiplying both sides of Equation (3), on the left successively by Ek−1 , . . . , E2−1 , E1−1 we obtain, , A = E1−1 E2−1 · · · Ek−1 In = E1−1 E2−1 · · · Ek−1, , (4), , By Theorem 1.5.2, this equation expresses A as a product of elementary matrices., (d ) ⇒ (a) If A is a product of elementary matrices, then from Theorems 1.4.7 and 1.5.2,, , the matrix A is a product of invertible matrices and hence is invertible., , A Method for Inverting, Matrices, , As a first application of Theorem 1.5.3, we will develop a procedure (or algorithm) that, can be used to tell whether a given matrix is invertible, and if so, produce its inverse. To, derive this algorithm, assume for the moment, that A is an invertible n × n matrix. In, Equation (3), the elementary matrices execute a sequence of row operations that reduce, A to In . If we multiply both sides of this equation on the right by A−1 and simplify, we, obtain, , A−1 = Ek · · · E2 E1 In, But this equation tells us that the same sequence of row operations that reduces A to In, will transform In to A−1 . Thus, we have established the following result., Inversion Algorithm To find the inverse of an invertible matrix A, find a sequence of, , elementary row operations that reduces A to the identity and then perform that same, sequence of operations on In to obtain A−1 ., A simple method for carrying out this procedure is given in the following example., , E X A M P L E 4 Using Row Operations to Find A−1, , Find the inverse of, , ⎡, , 1, ⎢, A = ⎣2, 1, , 2, 5, 0, , ⎤, , 3, ⎥, 3⎦, 8, , Solution We want to reduce A to the identity matrix by row operations and simultaneously apply these operations to I to produce A−1 . To accomplish this we will adjoin the, identity matrix to the right side of A, thereby producing a partitioned matrix of the form, , [A | I ]

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1.5 Elementary Matrices and a Method for Finding A−1, , 57, , Then we will apply row operations to this matrix until the left side is reduced to I ; these, operations will convert the right side to A−1 , so the final matrix will have the form, , [I | A−1 ], The computations are as follows:, , ⎡, , 1, , ⎤, , 2, , 3, , 1, , 0, , 0, , 5, , 3, , 0, , 1, , 0⎦, , 1, , 0, , 8, , 0, , 0, , 1, , 1, , 2, , 3, , 1, , 0, , 0, , 1, , −3, , −2, , 1, , 0⎦, , −2, , 5, , −1, , 0, , 1, , 2, , 3, , 1, , 0, , 0, , 1, , −2, −5, , 1, , 0⎦, , 0, , −3, −1, , 2, , 1, , 2, , 3, , 1, , 0, , 0, , 1, , −3, , −2, , 1, , 0⎦, , 0, , 1, , 5, , −2, , −1, , ⎢, ⎣0, , 2, , 0, , −14, , 6, , 3, , 1, , 0, , 13, , 0, , 0, , 1, , ⎢, ⎣0, , 0, , 0, , −40, , 16, , 1, , 0, , 13, , 0, , 0, , 1, , 5, , −5, −2, , ⎥, −3⎦, −1, , 16, −5, −2, , 9, ⎥, −3⎦, −1, , ⎢, ⎣2, ⎡, , ⎢, ⎣0, 0, , ⎡, , 1, , ⎢, ⎣0, 0, , ⎡, , 1, , ⎢, ⎣0, 0, , ⎡, , 1, , ⎡, , 1, , Thus,, , ⎥, , ⎤, ⎥, , ⎤, ⎥, , −40, , ⎢, A−1 = ⎣ 13, 5, , We added 2 times the, second row to the third., , ⎤, ⎥, , We multiplied the, third row by −1., , ⎤, , ⎥, −5 −3⎦, 5 −2 −1, , ⎡, , We added −2 times the first, row to the second and −1 times, the first row to the third., , We added 3 times the third, row to the second and −3 times, the third row to the first., , ⎤, , 9, , We added −2 times the, second row to the first., , ⎤, , Often it will not be known in advance if a given n × n matrix A is invertible. However,, if it is not, then by parts (a) and (c) of Theorem 1.5.3 it will be impossible to reduce A, to In by elementary row operations. This will be signaled by a row of zeros appearing, on the left side of the partition at some stage of the inversion algorithm. If this occurs,, then you can stop the computations and conclude that A is not invertible., E X A M P L E 5 Showing That a Matrix Is Not Invertible, , Consider the matrix ⎡, , 1, ⎢, A=⎣ 2, −1, , 6, 4, 2, , ⎤, , 4, ⎥, −1⎦, 5

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58, , Chapter 1 Systems of Linear Equations and Matrices, , Applying the procedure of Example 4 yields, , ⎡, , ⎤, , 1, ⎢, ⎣ 2, −1, , 6, 4, 2, , 4, −1, 5, , 1, 0, 0, , 0, 1, 0, , 0, ⎥, 0⎦, 1, , 1, ⎢, ⎣ 0, 0, , 6, −8, 8, , 4, −9, 9, , 1, −2, 1, , 0, 1, 0, , 0, ⎥, 0⎦, 1, , 1, ⎢, ⎣ 0, 0, , 6, −8, 0, , 4, −9, 0, , 1, −2, −1, , 0, 1, 1, , 0, ⎥, 0⎦, 1, , ⎡, , ⎡, , ⎤, We added −2 times the first, row to the second and added, the first row to the third., , ⎤, , We added the second, row to the third., , Since we have obtained a row of zeros on the left side, A is not invertible., E X A M P L E 6 Analyzing Homogeneous Systems, , Use Theorem 1.5.3 to determine whether the given homogeneous system has nontrivial, solutions., (a), , x1 + 2x2 + 3x3 = 0, 2x1 + 5x2 + 3x3 = 0, + 8x3 = 0, , x1, , x1 + 6x2 + 4x3 = 0, 2x1 + 4x2 − x3 = 0, , (b), , −x1 + 2x2 + 5x3 = 0, , Solution From parts (a) and (b) of Theorem 1.5.3 a homogeneous linear system has, , only the trivial solution if and only if its coefficient matrix is invertible. From Examples 4, and 5 the coefficient matrix of system (a) is invertible and that of system (b) is not. Thus,, system (a) has only the trivial solution while system (b) has nontrivial solutions., , Exercise Set 1.5, In Exercises 1–2, determine whether the given matrix is elementary., , , , 1. (a), , 1, , 0, , −5, , 1, , ⎡, , 1, , ⎢, (c) ⎣0, 0, , 1, , 1, 0, , ⎡, , (b), , 0, , ⎤, , 0, , ⎥, 1⎦, , 0, , 0, , , 2. (a), , , , 0, , 1, , 1, , 0, , ⎡, , 2, , ⎢0, ⎢, (d) ⎢, ⎣0, 0, , ⎡, , , , 3, , ⎤, , 0, , 0, , (c) ⎣0, 0, , 1, , 9⎦, , 0, , 1, , ⎥, , , , , 2, , 0, , 1, , 0, , 0, , 1, , 0⎦, , 0, , 0, , 1, , ⎥, , ⎤, , 0, , 1, , 1, , 0⎦, , 0, , 0, , 1, , −3, , 0, , 1, , ⎡, , ⎡, , , , 1, , 0, , 0, , (c) ⎣ 0, −5, , 1, , 0⎦, , 0, , 1, , ⎢, , , ⎤, , 0, , 0, , (d) ⎣ 0, 0, , 0, , 1⎦, , 1, , 0, , ⎥, , 4. (a), , 1, , 0, , −3, , 1, , ⎡, , 0, , ⎢0, ⎢, (c) ⎢, ⎣0, 1, , 0, , 0, , (b) ⎣ 0, 0, , 1, , 0⎦, , 0, , 1, , ⎡, , ⎤, , 0, , 1, , 0, , 0, , 0, , 0⎦, , 0, , 0, , 1, , 1, , 0, , 0, , (b) ⎣0, 0, , 1, , 0⎦, , 0, , 3, , ⎡, ⎢, , 1, , ⎤, , 0, , 0, , 1, , 0, , 0, , 1, , ⎥, 0⎦, , 0, , 0, , 0, , 0⎥, ⎥, , ⎡, , 1, , ⎢0, ⎢, ⎣0, , (d) ⎢, , 0, , 0, , ⎤, , 1, , 0, , , , ⎥, , 0, , ⎢0, ⎢, (d) ⎢, ⎣1, , ⎥, , ⎤, , −7, , ⎢, , ⎥, , −1, , ⎢, , 3. (a), , 0⎥, ⎥, , 0, , ⎡, , ⎤, , 0, , (b) ⎣0, 1, , ⎢, , √, , 1, , ⎢, , , −5, , In Exercises 3–4, find a row operation and the corresponding, elementary matrix that will restore the given elementary matrix to, the identity matrix., , 0⎥, ⎥, , ⎥, , ⎤, ⎥, ⎤, , 0, , − 17, , 1, , 0, , 0, , 1, , 0⎦, , 0, , 0, , 1, , 0, , 0⎥, ⎥, , ⎥

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1.5 Elementary Matrices and a Method for Finding A−1, , In Exercises 5–6 an elementary matrix E and a matrix A are, given. Identify the row operation corresponding to E and verify that the product EA results from applying the row operation, to A., , , 5. (a) E =, , 0, , 1, , 1, , 0, , ⎡, , , , , , A=, , 1, , ⎤, , 3, , 0, , 1, , 0 −3, , ⎡, , −2, −6, ⎡, , −6, , 1, , 0, , 4, , (c) E = ⎣0, , 1, , 0⎦ , A = ⎣2, , 5⎦, , 0, , 0, , 1, , 6, , ⎢, , , 6. (a) E =, , −6, , 0, , 0, , 1, , ⎡, , ⎥, , ⎢, , 1, , 4, , 3, , , , , , A=, , −1, 3, , ⎤, , ⎡, , ⎥, , ⎢, , 0, , 0, , (b) E = ⎣−4, , 1, , 0⎦ , A = ⎣ 1, , 0, , 0, , 1, , ⎡, , 1, , ⎢, (c) E = ⎣0, 0, , 0, , 0, , 2, , ⎡, , 4, , 3, , 6, , 5, 0, , 1, , 3, ⎢, A = ⎣2, 8, , 0, , −4, , −1, , 5, , 0, , 1, , 3, , −4, , ⎤, ⎥, , 3⎦, , −1, , ⎡, , 3, ⎢, C = ⎣2, 2, , ⎡, , 8, ⎢, F = ⎣8, 3, , ⎤, , ⎡, , 1, 8, ⎥, ⎢, −1⎦ , B = ⎣2, 5, 3, , 4, −7, −7, , 1, 8, ⎥, ⎢, −1⎦ , D = ⎣−6, 3, 3, , 1, 1, 4, , ⎤, , ⎤, , 1, −7, 4, , ⎡, , 1, 21, 4, , 0, , 8, , ⎡1, , ⎡, , ⎥, , ⎡, , (d) EF = B, , 0, , 1, , 13. ⎣0, 1, , 1, , 1⎦, , 1, , 0, , ⎢, , , , (b) A =, , 6, , 6, , 7, , 6⎦, , 7, , 7, , ⎤, , 0, , 2, , 1 ⎥, 10 ⎦, 1, 10, , (b), , −4, , ⎤, ⎥, , 1⎦, , −9, − 25, , 1, 5, 3, −5, − 45, , ⎤, , ⎥, − 103 ⎦, 1, 10, , , , √, , 2, , 3 2, , √, , ⎢, , 14. ⎣−4 2, 0, , ⎡, , 1, , ⎢1, ⎢, ⎣1, , ⎥, , 16. ⎢, , 1, , 0, , 2, , 12, , 0, , 2, , −1, , −4, , 0, , ⎤, , 0⎥, ⎥, , ⎥, 0⎦, −5, , ⎢0, ⎢, ⎣0, , 0, , 0, , k2, , 0, , 0, , k3, , 0, , 0, , 0, , ⎢0, ⎢, ⎣0, k4, , 8, , −4, , ⎡ √, , −4, , 20. (a) ⎢, , −4, , , , ⎡, , √, , 0, , 0, , 1, , 2, , 0, , 3, , 0, , 3, , 5, , 0⎦, , 3, , 5, , 7, , 0⎥, ⎥, , ⎥, , 0, , 2, , ⎢1, ⎢, ⎣0, , 0, , 0, , −1, , 3, , 2, , 1, , 5, , 18. ⎢, , ⎥, , 0⎦, , ⎤, , 0, , 0, , 0, , ⎤, , 0, , ⎤, , 1⎥, ⎥, , ⎥, , 0⎦, , −3, , In Exercises 19–20, find the inverse of each of the following, 4 × 4 matrices, where k1 , k2 , k3 , k4 , and k are all nonzero., , ⎡, , −4, , 4, , 5, ⎢2, ⎣5, 1, 5, , ⎤, , 2, , ⎢1, ⎢, ⎣0, , 5, ⎥, 3⎦, 1, , 2, , −2, , (b) ⎣ 2, , ⎡1, , ⎥, , 15. ⎣2, 2, , 17. ⎢, , In Exercises 9–10, first use Theorem 1.4.5 and then use the, inversion algorithm to find A−1 , if it exists., , , , −3, , 3, , ⎤, , ⎤, , 1, , 19. (a) ⎢, , (c) EB = F, , 4, , −1, , ⎢, , − 25, , 1, 5, 1, 5, − 45, , 5, ⎢1, ⎣5, 1, 5, , k1, , (b) ED = B, , 7, , 1, , ⎡, , 8. (a) EB = D, , 2, , 3⎦, , ⎤, , (d) EC = A, , 4, , 5, , ⎢, , 2, , 5, ⎥, 1⎦, 1, , (c) EA = C, , 1, , 11. (a) ⎣2, , ⎡, , 5, ⎥, −1⎦, 1, , (b) EB = A, , 9. (a) A =, , 3, , ⎢, , 7. (a) EA = B, , , , ⎤, , 2, , ⎡, , ⎥, 5⎦, , 4, −7, 1, , 6, , In Exercises 13–18, use the inversion algorithm to find the inverse of the matrix (if the inverse exists)., , In Exercises 7–8, use the following matrices and find an elementary matrix E that satisfies the stated equation., , ⎡, , , (b) A =, , ⎤, , 1, , ⎥, ⎢, 0⎦ , A = ⎣2, , −6, , , , −1, −3, , 2, , ⎤, , −1, −6, , 3, , 1, , 12. (a), , 5, , , −5, −16, , 1, , In Exercises 11–12, use the inversion algorithm to find the inverse of the matrix (if the inverse exists)., , ⎡, ⎤, , ⎥, , −2, −6, , 1, , ⎢, , −1, −6, , 5, , 2 −1, 0 −4 −4, ⎥, ⎢, ⎥, 0⎦ , A = ⎣1 −3 −1, 5, 3⎦, 1, 2, 0, 1, 3 −1, ⎤, ⎡, ⎤, , 0, , ⎢, (b) E = ⎣0, , −1, , , , , 10. (a) A =, , 59, , 0, , ⎤, , 0⎥, ⎥, , ⎡, , k, , 0, , 1, , 0, , 0, , k, , 1⎦, , 0, , 0, , 0, , 1, , k, , ⎢1, ⎢, ⎣0, , 0, , 0, , 0, , k, , 0, , 1, , k, , 0⎦, , 0, , 0, , 1, , k, , ⎢0, ⎢, ⎣0, , (b) ⎢, , 0, , ⎥, 0⎦, k4, , 0, , 0, , k1, , ⎤, , ⎡, , 0, , k2, , k3, , 0, , ⎥, 0⎦, , 0, , 0, , 0, , 0⎥, ⎥, , (b) ⎢, , 0, , ⎤, , 1, , 0⎥, ⎥, , ⎥, , ⎤, , 0⎥, ⎥, , ⎥, , In Exercises 21–22, find all values of c, if any, for which the, given matrix is invertible., , ⎡, ⎢, , c, , 21. ⎣1, 1, , c, c, 1, , ⎤, c, ⎥, c⎦, c, , ⎡, , ⎤, , c, , 1, , 22. ⎣1, 0, , c, , 0, ⎥, 1⎦, , 1, , c, , ⎢

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60, , Chapter 1 Systems of Linear Equations and Matrices, , In Exercises 23–26, express the matrix and its inverse as products of elementary matrices., 23., , −3, , 1, 2, , 2, , ⎡, , 1, ⎢, 25. ⎣0, 0, , 1, , 24., , −2, , 0, 4, 0, , ⎤, , 0, 2, , −5, ⎡, , 1, ⎢, 26. ⎣1, 0, , ⎥, , 3⎦, 1, , ⎤, , 1, 1, 1, , True-False Exercises, , 0, ⎥, 1⎦, 1, , TF. In parts (a)–(g) determine whether the statement is true or, false, and justify your answer., , In Exercises 27–28, show that the matrices A and B are row, equivalent by finding a sequence of elementary row operations, that produces B from A, and then use that result to find a matrix, C such that CA = B ., , ⎡, , ⎤, , ⎡, , ⎥, , ⎢, , 1, , 2, , 3, , 1, , 0, , 4, , 1⎦, B = ⎣0, , 2, , ⎥, −2 ⎦, , 2, , 1, , 9, , 1, , 4, , ⎢, ⎡, , 1, , ⎤, , ⎡, , ⎥, , ⎢, , 5, , ⎤, , 27. A = ⎣1, , 2, , 1, , 0, , 6, , 9, , 28. A = ⎣−1, , 1, , 0⎦, B = ⎣−5, , 3, , 0, , −1, −2, , ⎢, , −1, , −1, ⎡, , 29. Show that if, , 1, , ⎢, A = ⎣0, a, , 4, , ⎤, ⎥, , 0⎦, , −1, , ⎤, , 0, , 0, , 1, , 0⎦, , b, , c, , ⎥, , ⎡, , 0, , ⎢, ⎢b, ⎢, A=⎢, ⎢0, ⎢0, ⎣, , a, , 0, , 0, , 0, , c, , 0, , d, , 0, , e, , 0, , f, , 0, , (a) The product of two elementary matrices of the same size must, be an elementary matrix., (b) Every elementary matrix is invertible., (c) If A and B are row equivalent, and if B and C are row equivalent, then A and C are row equivalent., (d) If A is an n × n matrix that is not invertible, then the linear, system Ax = 0 has infinitely many solutions., (e) If A is an n × n matrix that is not invertible, then the matrix, obtained by interchanging two rows of A cannot be invertible., (f ) If A is invertible and a multiple of the first row of A is added, to the second row, then the resulting matrix is invertible., (g) An expression of an invertible matrix A as a product of elementary matrices is unique., , is an elementary matrix, then at least one entry in the third, row must be zero., 30. Show that, , 33. Prove that if B is obtained from A by performing a sequence, of elementary row operations, then there is a second sequence, of elementary row operations, which when applied to B recovers A., , 0, , ⎤, , ⎥, ⎥, 0⎥, ⎥, g⎥, ⎦, , Working withTechnology, T1. It can be proved that if the partitioned matrix, , , , 0⎥, , 0 0 0 h 0, is not invertible for any values of the entries., , Working with Proofs, 31. Prove that if A and B are m × n matrices, then A and B are, row equivalent if and only if A and B have the same reduced, row echelon form., 32. Prove that if A is an invertible matrix and B is row equivalent, to A, then B is also invertible., , A, , B, , C, , D, , , , is invertible, then its inverse is, , , , A−1 + A−1 B(D − CA−1 B)−1 CA−1, , −A−1 B(D − CA−1 B)−1, , −(D − CA−1 B)−1 CA−1, , (D − CA−1 B)−1, , , , provided that all of the inverses on the right side exist. Use this, result to find the inverse of the matrix, , ⎡, , ⎤, , 1, , 2, , 1, , 0, , ⎢0, ⎢, ⎢, ⎣0, , −1, , 0, , 1⎥, ⎥, , 0, , 2, , 0⎦, , 0, , 0, , 3, , 3, , ⎥

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1.6 More on Linear Systems and Invertible Matrices, , 61, , 1.6 More on Linear Systems and Invertible Matrices, In this section we will show how the inverse of a matrix can be used to solve a linear system, and we will develop some more results about invertible matrices., , Number of Solutions of a, Linear System, , In Section 1.1 we made the statement (based on Figures 1.1.1 and 1.1.2) that every linear, system either has no solutions, has exactly one solution, or has infinitely many solutions., We are now in a position to prove this fundamental result., THEOREM 1.6.1 A system of linear equations has zero, one, or infinitely many solutions., , There are no other possibilities., , Ax = b is a system of linear equations, exactly one of the following is true:, (a) the system has no solutions, (b) the system has exactly one solution, or (c) the system, has more than one solution. The proof will be complete if we can show that the system, has infinitely many solutions in case (c)., Assume that Ax = b has more than one solution, and let x0 = x1 − x2 , where x1, and x2 are any two distinct solutions. Because x1 and x2 are distinct, the matrix x0 is, nonzero; moreover,, Proof If, , Ax0 = A(x1 − x2 ) = Ax1 − Ax2 = b − b = 0, If we now let k be any scalar, then, , A(x1 + k x0 ) = Ax1 + A(k x0 ) = Ax1 + k(Ax0 ), = b + k0 = b + 0 = b, But this says that x1 + k x0 is a solution of Ax = b. Since x0 is nonzero and there are, infinitely many choices for k , the system Ax = b has infinitely many solutions., Solving Linear Systems by, Matrix Inversion, , Thus far we have studied two procedures for solving linear systems—Gauss–Jordan, elimination and Gaussian elimination. The following theorem provides an actual formula, for the solution of a linear system of n equations in n unknowns in the case where the, coefficient matrix is invertible., , A is an invertible n × n matrix, then for each n × 1 matrix b, the, system of equations Ax = b has exactly one solution, namely, x = A−1 b., , THEOREM 1.6.2 If, , Proof Since, , A(A−1 b) = b, it follows that x = A−1 b is a solution of Ax = b. To show, , that this is the only solution, we will assume that x0 is an arbitrary solution and then, show that x0 must be the solution A−1 b., If x0 is any solution of Ax = b, then Ax0 = b. Multiplying both sides of this equation by A−1 , we obtain x0 = A−1 b., E X A M P L E 1 Solution of a Linear System Using A−1, , Consider the system of linear equations, , x1 + 2x2 + 3x3 = 5, 2x1 + 5x2 + 3x3 = 3, , x1, , + 8x3 = 17

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62, , Chapter 1 Systems of Linear Equations and Matrices, , In matrix form this system can be written as Ax = b, where, , ⎡, , 1, ⎢, A = ⎣2, 1, , ⎤, , ⎡ ⎤, , ⎡ ⎤, , 3, 5, x1, ⎥, ⎢ ⎥, ⎢ ⎥, 3⎦, x = ⎣x2 ⎦, b = ⎣ 3⎦, 8, 17, x3, , 2, 5, 0, , In Example 4 of the preceding section, we showed that A is invertible and, , ⎡, , −40, , ⎢, A−1 = ⎣ 13, 5, Keep in mind that the method, of Example 1 only applies, when the system has as many, equations as unknowns and, the coefficient matrix is invertible., , Linear Systems with a, Common Coefficient Matrix, , 16, −5, −2, , By Theorem 1.6.2, the solution of the system is, , ⎡, , −40, , ⎢, , x = A−1 b = ⎣ 13, 5, , 16, −5, −2, , ⎤, , 9, ⎥, −3⎦, −1, , ⎤⎡ ⎤, , ⎡, , ⎤, , 9, 5, 1, ⎥⎢ ⎥ ⎢ ⎥, −3⎦ ⎣ 3⎦ = ⎣−1⎦, 17, 2, −1, , or x1 = 1, x2 = −1, x3 = 2., Frequently, one is concerned with solving a sequence of systems, , Ax = b1 , Ax = b2 , Ax = b3 , . . . , Ax = bk, each of which has the same square coefficient matrix A. If A is invertible, then the, solutions, x1 = A−1 b1 , x2 = A−1 b2 , x3 = A−1 b3 , . . . , xk = A−1 bk, can be obtained with one matrix inversion and k matrix multiplications. An efficient, way to do this is to form the partitioned matrix, , [A | b1 | b2 | · · · | bk ], , (1), , in which the coefficient matrix A is “augmented” by all k of the matrices b1 , b2 , . . . , bk ,, and then reduce (1) to reduced row echelon form by Gauss–Jordan elimination. In this, way we can solve all k systems at once. This method has the added advantage that it, applies even when A is not invertible., E X A M P L E 2 Solving Two Linear Systems at Once, , Solve the systems, (a), , x1 + 2x2 + 3x3 = 4, x1, , x1 + 2x2 + 3x3 =, , 1, , 2x1 + 5x2 + 3x3 =, , 6, , (b), , 2x1 + 5x2 + 3x3 = 5, , + 8x3 = 9, , + 8x3 = −6, , x1, , Solution The two systems have the same coefficient matrix. If we augment this co-, , efficient matrix with the columns of constants, obtain, ⎡, 1, 2, 3, ⎢, 2, 5, 3, ⎣, 1, 0, 8, , on the right sides of these systems, we, 4, 5, 9, , ⎤, , 1, ⎥, 6⎦, −6, , Reducing this matrix to reduced row echelon form yields (verify), , ⎡, , 1, ⎢, 0, ⎣, 0, , 0, 1, 0, , 0, 0, 1, , 1, 0, 1, , ⎤, , 2, ⎥, 1⎦, −1

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1.6 More on Linear Systems and Invertible Matrices, , 63, , It follows from the last two columns that the solution of system (a) is x1 = 1, x2 = 0,, x3 = 1 and the solution of system (b) is x1 = 2, x2 = 1, x3 = −1., Properties of Invertible, Matrices, , Up to now, to show that an n × n matrix A is invertible, it has been necessary to find an, n × n matrix B such that, AB = I and BA = I, The next theorem shows that if we produce an n × n matrix B satisfying either condition,, then the other condition will hold automatically., , THEOREM 1.6.3 Let A be a square matrix., , (a) If B is a square matrix satisfying BA = I, then B = A−1 ., (b) If B is a square matrix satisfying AB = I, then B = A−1 ., , We will prove part (a) and leave part (b) as an exercise., , BA = I . If we can show that A is invertible, the proof can be, completed by multiplying BA = I on both sides by A−1 to obtain, , Proof (a) Assume that, , BAA−1 = IA−1 or BI = IA−1 or B = A−1, To show that A is invertible, it suffices to show that the system Ax = 0 has only the trivial, solution (see Theorem 1.5.3). Let x0 be any solution of this system. If we multiply both, sides of Ax0 = 0 on the left by B , we obtain BAx0 = B 0 or I x0 = 0 or x0 = 0. Thus,, the system of equations Ax = 0 has only the trivial solution., EquivalenceTheorem, , We are now in a position to add two more statements to the four given in Theorem 1.5.3., , THEOREM 1.6.4 Equivalent Statements, , If A is an n × n matrix, then the following are equivalent., (a), (b), (c), , A is invertible., Ax = 0 has only the trivial solution., The reduced row echelon form of A is In ., , (e), , A is expressible as a product of elementary matrices., Ax = b is consistent for every n × 1 matrix b., , ( f), , Ax = b has exactly one solution for every n × 1 matrix b., , (d ), , Proof Since we proved in Theorem 1.5.3 that (a), (b), (c), and (d ) are equivalent, it will, , be sufficient to prove that (a) ⇒ ( f ) ⇒ (e) ⇒ (a)., (a) ⇒ (f ) This was already proved in Theorem 1.6.2., (f ) ⇒ (e) This is almost self-evident, for if Ax = b has exactly one solution for every, n × 1 matrix b, then Ax = b is consistent for every n × 1 matrix b.

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64, , Chapter 1 Systems of Linear Equations and Matrices, (e) ⇒ (a) If the system Ax = b is consistent for every n × 1 matrix b, then, in particular,, this is so for the systems, , ⎡ ⎤, , ⎡ ⎤, , ⎡ ⎤, , ⎢ .. ⎥, ⎣.⎦, , ⎢ .. ⎥, ⎣.⎦, , ⎢ .. ⎥, ⎣.⎦, , 0, , 0, , 1, , 1, 0, 0, ⎢ ⎥, ⎢ ⎥, ⎢ ⎥, ⎢0⎥, ⎢1⎥, ⎢0⎥, ⎢ ⎥, ⎢ ⎥, ⎢ ⎥, ⎥, ⎢ ⎥, ⎢ ⎥, Ax = ⎢, ⎢0⎥, Ax = ⎢0⎥, . . . , Ax = ⎢0⎥, , Let x1 , x2 , . . . , xn be solutions of the respective systems, and let us form an n × n matrix C having these solutions as columns. Thus C has the form, , C = [x1 | x2 | · · · | xn ], As discussed in Section 1.3, the successive columns of the product AC will be, , Ax1 , Ax2 , . . . , Axn, [see Formula (8) of Section 1.3]. Thus,, It follows from the equivalency of parts (e) and ( f ) that, if you can show that Ax = b, has at least one solution for every n × 1 matrix b, then you, can conclude that it has exactly one solution for every, n × 1 matrix b., , ⎡, , 1, ⎢, ⎢0, ⎢, AC = [Ax1 | Ax2 | · · · | Axn ] = ⎢, ⎢0, , 0, 1, 0, , 0, , 0, , ⎢ .., ⎣., , .., ., , ⎤, ··· 0, ⎥, · · · 0⎥, ⎥, · · · 0⎥, =I, .. ⎥, ⎥, .⎦, ··· 1, , By part (b) of Theorem 1.6.3, it follows that C = A−1 . Thus, A is invertible., We know from earlier work that invertible matrix factors produce an invertible product. Conversely, the following theorem shows that if the product of square matrices is, invertible, then the factors themselves must be invertible., THEOREM 1.6.5 Let A and B be square matrices of the same size. If AB is invertible,, , then A and B must also be invertible., , B is invertible by showing that the homogeneous system, B x = 0 has only the trivial solution. If we assume that x0 is any solution of this system,, , Proof We will show first that, , then, , (AB)x0 = A(B x0 ) = A0 = 0, so x0 = 0 by parts (a) and (b) of Theorem 1.6.4 applied to the invertible matrix AB ., But the invertibility of B implies the invertibility of B −1 (Theorem 1.4.7), which in turn, implies that, , (AB)B −1 = A(BB −1 ) = AI = A, , is invertible since the left side is a product of invertible matrices. This completes the, proof., In our later work the following fundamental problem will occur frequently in various, contexts., , A be a fixed m × n matrix. Find all m × 1 matrices b, such that the system of equations Ax = b is consistent., , A Fundamental Problem Let

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1.6 More on Linear Systems and Invertible Matrices, , 65, , If A is an invertible matrix, Theorem 1.6.2 completely solves this problem by asserting that for every m × 1 matrix b, the linear system Ax = b has the unique solution, x = A−1 b. If A is not square, or if A is square but not invertible, then Theorem 1.6.2, does not apply., In these cases b must usually satisfy certain conditions in, order for Ax = b to be consistent. The following example illustrates how the methods, of Section 1.2 can be used to determine such conditions., E X A M P L E 3 Determining Consistency by Elimination, , What conditions must b1 , b2 , and b3 satisfy in order for the system of equations, , x1 + x2 + 2x3 = b1, x1, + x3 = b2, 2x1 + x2 + 3x3 = b3, to be consistent?, Solution The augmented matrix is, , ⎡, , 1, ⎢, ⎣1, 2, , 1, 0, 1, , 2, 1, 3, , ⎤, b1, ⎥, b2 ⎦, b3, , which can be reduced to row echelon form as follows:, , ⎡, , 1, ⎢, ⎣0, 0, , 1, −1, −1, , 2, −1, −1, , 1, , 0, , 1, 1, −1, , 2, 1, −1, , 1, ⎢, ⎣0, 0, , 1, 1, 0, , 2, 1, 0, , ⎡, , ⎢, ⎣0, ⎡, , ⎤, b1, ⎥, b1 − b 2 ⎦, b3 − 2b1, ⎤, b1, ⎥, b1 − b 2 ⎦, b3 − 2b1, , −1 times the first row was added, to the second and −2 times the, first row was added to the third., , The second row was, multiplied by −1., , ⎤, b1, ⎥, b1 − b 2 ⎦, b3 − b 2 − b 1, , The second row was added, to the third., , It is now evident from the third row in the matrix that the system has a solution if and, only if b1 , b2 , and b3 satisfy the condition, , b3 − b2 − b1 = 0 or b3 = b1 + b2, To express this condition another way, Ax = b is consistent if and only if b is a matrix, of the form, ⎡, ⎤, , b1, ⎥, b2 ⎦, b1 + b2, , ⎢, , b=⎣, where b1 and b2 are arbitrary., , E X A M P L E 4 Determining Consistency by Elimination, , What conditions must b1 , b2 , and b3 satisfy in order for the system of equations, , x1 + 2x2 + 3x3 = b1, 2x1 + 5x2 + 3x3 = b2, , x1, to be consistent?, , + 8x3 = b3

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1.7 Diagonal, Triangular, and Symmetric Matrices, , ⎡, , −2, ⎢, 20. ⎣ 0, , −1, , 1, , 1, , 0, , 1, , ⎤, , ⎡, , 4, , ⎥, ⎢, −1⎦ X = ⎣6, 1, −4, , ⎤, , 3, , 2, , 1, , 7, , 8, , 9⎦, , 3, , 7, , 9, , 67, , (e) Let A be an n × n matrix and S is an n × n invertible matrix., If x is a solution to the linear system (S −1 AS)x = b, then S x, is a solution to the linear system Ay = S b., , ⎥, , (f ) Let A be an n × n matrix. The linear system Ax = 4x has a, unique solution if and only if A − 4I is an invertible matrix., , Working with Proofs, 21. Let Ax = 0 be a homogeneous system of n linear equations in, n unknowns that has only the trivial solution. Prove that if k, is any positive integer, then the system Ak x = 0 also has only, the trivial solution., , (g) Let A and B be n × n matrices. If A or B (or both) are not, invertible, then neither is AB ., , Working withTechnology, , 22. Let Ax = 0 be a homogeneous system of n linear equations, in n unknowns, and let Q be an invertible n × n matrix., Prove that Ax = 0 has only the trivial solution if and only, if (QA)x = 0 has only the trivial solution., , T1. Colors in print media, on computer monitors, and on television screens are implemented using what are called “color models”. For example, in the RGB model, colors are created by mixing, percentages of red (R), green (G), and blue (B), and in the YIQ, model (used in TV broadcasting), colors are created by mixing, percentages of luminescence (Y) with percentages of a chrominance factor (I) and a chrominance factor (Q). The conversion, from the RGB model to the YIQ model is accomplished by the, matrix equation, , 23. Let Ax = b be any consistent system of linear equations, and, let x1 be a fixed solution. Prove that every solution to the, system can be written in the form x = x1 + x0 , where x0 is a, solution to Ax = 0. Prove also that every matrix of this form, is a solution., 24. Use part (a) of Theorem 1.6.3 to prove part (b)., , ⎡ ⎤, , ⎡, , Y, , ⎢ ⎥ ⎢, ⎣ I ⎦ = ⎣.596, Q, .212, , True-False Exercises, TF. In parts (a)–(g) determine whether the statement is true or, false, and justify your answer., , .587, , .299, , .114, , ⎤⎡ ⎤, R, , ⎥⎢ ⎥, −.321⎦ ⎣G⎦, B, .311, , −.275, −.523, , What matrix would you use to convert the YIQ model to the RGB, model?, , (a) It is impossible for a system of linear equations to have exactly, two solutions., , T2. Let, , (b) If A is a square matrix, and if the linear system Ax = b has a, unique solution, then the linear system Ax = c also must have, a unique solution., , ⎡, , −2, , 1, , ⎢, A = ⎣4, , (c) If A and B are n × n matrices such that AB = In , then, BA = In ., , 2, , 3, , ⎡ ⎤, , ⎡ ⎤, , ⎡, , ⎥, , ⎢ ⎥, , ⎢ ⎥, , ⎢, , 7, , 3, , 0, , 11, , 1, , ⎤, ⎥, , 1⎦ , B1 = ⎣1⎦ , B2 = ⎣ 5⎦ , B3 = ⎣−4⎦, , 5, , 0, , ⎤, , −1, , 2, , Solve the linear systems Ax = B1 , Ax = B2 , Ax = B3 using the, method of Example 2., , (d) If A and B are row equivalent matrices, then the linear systems, Ax = 0 and B x = 0 have the same solution set., , 1.7 Diagonal,Triangular, and Symmetric Matrices, In this section we will discuss matrices that have various special forms. These matrices arise, in a wide variety of applications and will play an important role in our subsequent work., , Diagonal Matrices, , A square matrix in which all the entries off the main diagonal are zero is called a diagonal, matrix. Here are some examples:, , ⎡, , 2, 0, , 1, 0, ⎢, , ⎣0, −5, 0, , 0, 1, 0, , ⎤, , ⎡, , 6, 0, ⎢0, ⎥ ⎢, 0⎦, ⎢, ⎣0, 1, 0, , 0, −4, 0, 0, , 0, 0, 0, 0, , ⎤, , 0, 0⎥, ⎥, ⎥,, 0⎦, 8, , 0, 0, , 0, 0

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1.7 Diagonal, Triangular, and Symmetric Matrices, , Triangular Matrices, , 69, , A square matrix in which all the entries above the main diagonal are zero is called lower, triangular, and a square matrix in which all the entries below the main diagonal are zero, is called upper triangular. A matrix that is either upper triangular or lower triangular is, called triangular., E X A M P L E 2 Upper and Lower Triangular Matrices, , ⎡, a11, ⎢0, ⎢, ⎢, ⎣0, 0, , a12, a22, 0, 0, , a13, a23, a33, 0, , ⎤, a14, a24 ⎥, ⎥, ⎥, a34 ⎦, a44, , A general 4 × 4 upper, triangular matrix, , ⎡, a11, ⎢a, ⎢ 21, ⎢, ⎣a31, a41, , 0, a22, a32, a42, , 0, 0, a33, a43, , ⎤, 0, 0 ⎥, ⎥, ⎥, 0 ⎦, a44, , A general 4 × 4 lower, triangular matrix, , Remark Observe that diagonal matrices are both upper triangular and lower triangular since, they have zeros below and above the main diagonal. Observe also that a square matrix in row, echelon form is upper triangular since it has zeros below the main diagonal., , Properties ofTriangular, Matrices, i<j, i>j, , Figure 1.7.1, , Example 2 illustrates the following four facts about triangular matrices that we will state, without formal proof:, • A square matrix A = [aij ] is upper triangular if and only if all entries to the left of, the main diagonal are zero; that is, aij = 0 if i > j (Figure 1.7.1)., • A square matrix A = [aij ] is lower triangular if and only if all entries to the right of, the main diagonal are zero; that is, aij = 0 if i < j (Figure 1.7.1)., • A square matrix A = [aij ] is upper triangular if and only if the i th row starts with at, least i − 1 zeros for every i., • A square matrix A = [aij ] is lower triangular if and only if the j th column starts with, at least j − 1 zeros for every j., The following theorem lists some of the basic properties of triangular matrices., THEOREM 1.7.1, , (a) The transpose of a lower triangular matrix is upper triangular, and the transpose, of an upper triangular matrix is lower triangular., (b) The product of lower triangular matrices is lower triangular, and the product of, upper triangular matrices is upper triangular., (c), , A triangular matrix is invertible if and only if its diagonal entries are all nonzero., , (d ) The inverse of an invertible lower triangular matrix is lower triangular, and the, inverse of an invertible upper triangular matrix is upper triangular., , Part (a) is evident from the fact that transposing a square matrix can be accomplished by, reflecting the entries about the main diagonal; we omit the formal proof. We will prove, (b), but we will defer the proofs of (c) and (d ) to the next chapter, where we will have the, tools to prove those results more efficiently., Proof (b) We will prove the result for lower triangular matrices; the proof for upper triangular matrices is similar. Let A = [aij ] and B = [bij ] be lower triangular n × n matrices,

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70, , Chapter 1 Systems of Linear Equations and Matrices, , and let C = [cij ] be the product C = AB . We can prove that C is lower triangular by, showing that cij = 0 for i < j . But from the definition of matrix multiplication,, , cij = ai 1 b1j + ai 2 b2j + · · · + ain bnj, If we assume that i < j , then the terms in this expression can be grouped as follows:, , cij = ai 1 b1j + ai 2 b2j + · · · + ai(j −1) b(j −1)j + aij bjj + · · · + ain bnj, , ,  , , , Terms in which the row, number of b is less than, the column number of b, , Terms in which the row, number of a is less than, the column number of a, , In the first grouping all of the b factors are zero since B is lower triangular, and in the, second grouping all of the a factors are zero since A is lower triangular. Thus, cij = 0,, which is what we wanted to prove., E X A M P L E 3 Computations with Triangular Matrices, , Consider the upper triangular matrices, Observe that in Example 3 the, diagonal entries of AB and, BA are the same, and in both, cases they are the products, of the corresponding diagonal, entries of A and B . In the, exercises we will ask you to, prove that this happens whenever two upper triangular matrices or two lower triangular, matrices are multiplied., , ⎡, , 1, ⎢, A = ⎣0, 0, , It is easy to recognize a symmetric matrix by inspection:, The entries on the main diagonal have no restrictions, but, mirror images of entries across, the main diagonal must be, equal. Here is a picture using, the second matrix in Example 4:, , ⎡, , 1, ⎢, ⎣4, 5, , 4, 3, 0, , ⎤, 5, ⎥, 0⎦, 7, , ⎤, ⎡, −1, 3 −2, ⎥, ⎢, 4⎦, B = ⎣0, 0, 0, , 5, , 0, , ⎤, , 2, ⎥, −1⎦, 1, , It follows from part (c) of Theorem 1.7.1 that the matrix A is invertible but the matrix, B is not. Moreover, the theorem also tells us that A−1 , AB , and BA must be upper, triangular. We leave it for you to confirm these three statements by showing that, , ⎡, , 1, , A−1, , ⎢, =⎢, ⎣0, , − 23, 1, 2, , 0, , Symmetric Matrices, , 3, 2, 0, , 0, , ⎤, , 7, 5⎥, 2⎥, −5⎦,, 1, 5, , ⎡, , 3, ⎢, AB = ⎣0, 0, , −2, , −2, , ⎤, , ⎡, , 3, ⎥, ⎢, 2⎦ , BA = ⎣0, 0, 5, , 0, 0, , DEFINITION 1 A square matrix A is said to be symmetric if A, , 5, 0, 0, , ⎤, −1, ⎥, −5⎦, 5, , = AT ., , E X A M P L E 4 Symmetric Matrices, , The following matrices are symmetric, since each is equal to its own transpose (verify)., , , , 7, −3, , −3, 5, , ⎡, , , , 1, , ⎢, , ⎣4, , 5, , 4, −3, 0, , ⎤, , 5, ⎥, 0⎦,, 7, , ⎡, , d1, , ⎢0, ⎢, ⎢, ⎣0, 0, , 0, , d2, 0, 0, , 0, 0, , ⎤, , d3, , 0, 0⎥, ⎥, ⎥, 0⎦, , 0, , d4, , Remark It follows from Formula (14) of Section 1.3 that a square matrix A is symmetric if and, only if, , (A)ij = (A)j i, , (4), , for all values of i and j ., , The following theorem lists the main algebraic properties of symmetric matrices. The, proofs are direct consequences of Theorem 1.4.8 and are omitted.

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1.7 Diagonal, Triangular, and Symmetric Matrices, , 71, , THEOREM 1.7.2 If A and B are symmetric matrices with the same size, and if k is any, , scalar, then:, (a) AT is symmetric., (b) A + B and A − B are symmetric., (c) kA is symmetric., , It is not true, in general, that the product of symmetric matrices is symmetric. To, see why this is so, let A and B be symmetric matrices with the same size. Then it follows, from part (e) of Theorem 1.4.8 and the symmetry of A and B that, , (AB)T = B TAT = BA, Thus, (AB)T = AB if and only if AB = BA, that is, if and only if A and B commute. In, summary, we have the following result., THEOREM 1.7.3 The product of two symmetric matrices is symmetric if and only if the, , matrices commute., , E X A M P L E 5 Products of Symmetric Matrices, , The first of the following equations shows a product of symmetric matrices that is not, symmetric, and the second shows a product of symmetric matrices that is symmetric. We, conclude that the factors in the first equation do not commute, but those in the second, equation do. We leave it for you to verify that this is so., 1, 2, 1, 2, Invertibility of Symmetric, Matrices, , 2, 3, 2, 3, , −4, 1, , −4, 3, , 1, −2, =, −5, 0, 3, 2, =, 1, −1, , 1, 2, 1, 3, , In general, a symmetric matrix need not be invertible. For example, a diagonal matrix, with a zero on the main diagonal is symmetric but not invertible. However, the following, theorem shows that if a symmetric matrix happens to be invertible, then its inverse must, also be symmetric., THEOREM 1.7.4 If A is an invertible symmetric matrix, then A−1 is symmetric., , Proof Assume that A is symmetric and invertible. From Theorem 1.4.9 and the fact, that A = AT , we have, , (A−1 )T = (AT )−1 = A−1, , which proves that A−1 is symmetric., Products AAT and ATA, are Symmetric, , Matrix products of the form AAT and ATA arise in a variety of applications. If A is, an m × n matrix, then AT is an n × m matrix, so the products AAT and ATA are both, square matrices—the matrix AAT has size m × m, and the matrix ATA has size n × n., Such products are always symmetric since, , (AAT )T = (AT )TAT = AAT and (ATA)T = AT(AT )T = ATA

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72, , Chapter 1 Systems of Linear Equations and Matrices, , E X A M P L E 6 The Product of a Matrix and Its Transpose Is Symmetric, , Let A be the 2 × 3 matrix, , ⎡, , Then, , 1, ⎢, ATA = ⎣−2, 4, , AAT =, , 1, 3, , −2, , 1, 3, , A=, , 4, −5, , 0, , ⎤, , −2, 0, , ⎤, , ⎡, , 10 −2 −11, 4, −2, ⎥, ⎢, 4, −8⎦, = ⎣ −2, 0 −5, −11 −8, 41, ⎤, ⎡, 1, 3, 4 ⎢, 21 −17, ⎥, 0⎦ =, ⎣−2, −5, −17, 34, 4 −5, , 3, ⎥ 1, 0⎦, 3, −5, , Observe that ATA and AAT are symmetric as expected., Later in this text, we will obtain general conditions on A under which AAT and ATA, are invertible. However, in the special case where A is square, we have the following, result., THEOREM 1.7.5 If A is an invertible matrix, then AAT and ATA are also invertible., Proof Since A is invertible, so is AT by Theorem 1.4.9. Thus AAT and ATA are invertible,, , since they are the products of invertible matrices., , Exercise Set 1.7, In Exercises 1–2, classify the matrix as upper triangular, lower, triangular, or diagonal, and decide by inspection whether the matrix is invertible. [Note: Recall that a diagonal matrix is both upper and lower triangular, so there may be more than one answer, in some parts.], , , , 1. (a), , , , 2, , 1, , 0, , 3, , ⎡, , , , (b), , ⎤, , −1, , 0, , 0, , (c) ⎣ 0, , 2, , 0⎦, , 0, , 0, , 1, 5, , ⎢, , , 2. (a), , 0, , 1, , 7, , ⎡, , ⎥, , 4, , 0, , −2, , (d) ⎣0, , 0, , 3⎦, , 0, , 0, , 8, , 0, , −3, , 0, , 0, , ⎢, , (b), , ⎤, , 4, , 0, , 0, , 3, 5, , 0⎦, , 0, , 0, , ⎡, , ⎥, , 7, , ⎥, , , , 3, 3. ⎣0, 0, , 0, −1, 0, , ⎤⎡, , 0, 2, 0⎦ ⎣−4, 2, 2, , ⎤, , 1, 1⎦, 5, , 5, 5. ⎣0, 0, , 0, 2, 0, , ⎡, , 2, , 0, , 6. ⎣0, 0, , −1, , 0, , 0, , ⎤⎡, , −3, 0, 0⎦ ⎣ 1, −3, −6, , 0, , 0, 3, 0, , ⎤, , 0, 0⎦, 2, , −4, , −3, , 0, , 0, , 0⎦ ⎣ 0, , 5, , 0⎦, , 0, , 2, , 0, 3, 2, , 4, , −1, , 3, , 0⎦ ⎣ 1, , 2, , −5, , 1, , ⎤⎡, ⎥⎢, , 4, , ⎤, , 4, 0, 2, , 2, −5, 2, , 0, , , , ⎤, , 3, , 0, , 0, , 1, , 0⎦, , 7, , 0, , 0, , ⎥, , In Exercises 3–6, find the product by inspection., , ⎡, , ⎡, , ⎢, , ⎡, −4, −5 ⎣, 0, , 2, −1, , 3⎦, 2, , ⎤⎡, ⎥⎢, , −2, , 0, , ⎤, ⎥, , In Exercises 7–10, find A2 , A−2 , and A−k (where k is any integer) by inspection., , (d) ⎣3, , ⎢, , −2, , ⎤, , 3, , , , (c) ⎣0, , ⎢, , 0, , ⎡, , , , 4, , 0, , , , 1, 4., −3, , 7. A =, , 1, , 0, , 0, , −2, , ⎡1, , ⎡, , , , ⎤, , 0, , 9. A = ⎣ 0, , 1, 3, , 0⎦, , 0, , 0, , 1, 4, , ⎢, , 2, , 0, , 0, , 8. A = ⎣ 0, , 3, , 0⎦, , 0, , 0, , 5, , ⎢, , 0, , ⎥, , ⎤, , −6, , ⎡, , −2, , 0, , ⎥, , 0, , ⎢ 0 −4, 0, ⎢, ⎣ 0, 0 −3, , 10. A = ⎢, , 0, , 0, , 0, , 0, , ⎤, , 0⎥, ⎥, , ⎥, , 0⎦, , 2

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1.7 Diagonal, Triangular, and Symmetric Matrices, , In Exercises 23–24, find the diagonal entries of AB by inspection., , In Exercises 11–12, compute the product by inspection., , ⎡, , ⎤⎡, , ⎤⎡, , ⎤, , 1, , 0, , 0, , 2, , 0, , 0, , 0, , 0, , 0, , 11. ⎣0, , 0, , 0⎦ ⎣ 0, , 5, , 0⎦ ⎣0, , 2, , 0⎦, , 0, , 0, , 3, , 0, , 0, , 0, , 1, , ⎢, , ⎡, , −1, , ⎥⎢, , 0, , ⎤⎡, , 0, , 0, , 3, , ⎢, 12. ⎣ 0, , 2, , ⎥⎢, 0⎦ ⎣0, , 0, , 0, , 4, , 0, , 0, , ⎥⎢, , 0, 0, , ⎤⎡, , 5, , 5, , ⎥⎢, 0⎦ ⎣0, , 0, , 7, , ⎡, , ⎥, , 0, , 0, , 13., , 39, , 1, , 0, , 0, , −1, , , 14., , ⎤⎡, , a, , 0, , 0, , 15. (a) ⎣ 0, , b, , 0⎦ ⎣w, , v, ⎥, x⎦, , 0, , 0, , c, , z, , ⎢, , ⎡, ⎢, , ⎥⎢, , ⎤, v , ⎥ a, x⎦, 0, z, , u, , 16. (a) ⎣w, , y, , u, y, , 0, , , , b, , 0, , −1, , r, , s, , (b) ⎣u, , v, , x, , y, , ⎤⎡, t, a, ⎥⎢, w⎦ ⎣ 0, 0, z, , ⎡, a, ⎢, (b) ⎣ 0, , 0, , 0, , 0, , 2, , −1, , ×, , 3, , 17. (a), , , 18. (a), , ⎡, , , , 0, , 0, , 0⎦, , 0, , c, , r, , s, , t, , b, , 0⎦ ⎣u, , v, , ⎥, w⎦, , 0, , c, , y, , z, , x, , ×, , 3, , 0, , 1, , ×, , −8, , 0, , 2, , −3, , 9, , 0, , 7, , −3, , 2, , ⎡, , , , 0, , ⎢3, ⎢, (b) ⎢, ⎣7, , ⎤, ×, ×⎥, ⎥, ⎥, ×⎦, , 1, , ×, , ×, , 1, , ⎥, , 0, , ⎤, −1, ⎥, −4 ⎦, −2, , 1, , 0, , 0, 0, , 1, , −5, −3, −2, , 0, , 6, , 4, , 5, , ×, , 1, , −7⎥, ⎥, ⎥, −6⎦, , ×, , ×, , ×, , 3, , 19. ⎣0, 0, , 7, , ⎢, , ⎡, , ⎢2, ⎢, 21. ⎢, ⎣4, , ⎡, , 2, , 4, , 20. ⎣ 0, 0, , 3, , 0⎦, , 0, , 5, , ⎢−3, ⎢, ⎣− 4, , 0, , 0, , −1, −6, , 0, , 0, , 3, , 8, , ⎢, , 0, , ⎤, , 0⎥, ⎥, , 4, , ⎥, 0⎦, , 1, , 3, , ⎡, , 2, , 22. ⎢, , 0, , 5, , 3⎦, , 0, , 6, , ⎤, , 6, , 0, , 0, , 5, , 0⎦, , −3, , 0, , 0⎦ , B = ⎣1, 3, 7, , 2, , 6, , ⎢, , ⎥, , ⎥, , , −3, −1, , 4, , a+5, ⎡, , a − 2b + 2c, , 2, , ⎢, , 26. A = ⎣3, , 5, , 0, , −2, , ⎤, , 2a + b + c, , a+c, , ⎤, ⎥, ⎦, , 7, , In Exercises 27–28, find all values of x for which A is invertible., , ⎡, ⎢, , 27. A = ⎣ 0, 0, , ⎡, , x−, , ⎢, , 0, , ⎤, x4, ⎥, x3 ⎦, x−4, , 0, , 0, , x2, x+2, , 1, 2, , 28. A = ⎢, ⎣ x, , x−, , x2, , x3, , 1, 3, , ⎤, ⎥, ⎥, ⎦, , 0, , x+, , 1, 4, , 29. If A is an invertible upper triangular or lower triangular matrix, what can you say about the diagonal entries of A−1 ?, 30. Show that if A is a symmetric n × n matrix and B is any n × m, matrix, then the following products are symmetric:, , B TB, BB T , B TAB, In Exercises 31–32, find a diagonal matrix A that satisfies the, given condition., , ⎡, , ⎤, , −1, , 7, , 0, , ⎥, , ⎤, , 2, , 0, , x−1, , In Exercises 19–22, determine by inspection whether the matrix is invertible., , ⎡, , ⎥, ⎢, −2⎦ , B = ⎣ 0, 0, −1, ⎤, ⎡, , ⎤, , ⎢×, ⎢, ⎣×, , (b) ⎢, , −1, , 0, , ⎤, , b, , ⎥⎢, , ⎡, , 4, , 25. A =, , ⎤⎡, , ⎤, , 24. A = ⎣−2, , , , In Exercises 17–18, create a symmetric matrix by substituting, appropriate numbers for the ×’s., , , , 0, , 6, , In Exercises 25–26, find all values of the unknown constant(s), for which A is symmetric., , 1000, , 0, , ⎢, , 0, , 3, , 1, , ⎡, , ⎤, , 1, , ⎢, , In Exercises 15–16, use what you have learned in this section, about multiplying by diagonal matrices to compute the product, by inspection., , ⎡, , 23. A = ⎣0, , ⎤, , In Exercises 13–14, compute the indicated quantity., , , , 2, , ⎡, , ⎥, 0⎦, , −2, , 0, , 3, , ⎢, , 0, , 73, , 1, 31. A = ⎣0, 0, 5, , 0, −1, 0, , ⎤, , ⎡, , 0, 0⎦, −1, , 32. A, , −2, , 9, = ⎣0, 0, , 0, 4, 0, , ⎤, , 0, 0⎦, 1, , 33. Verify Theorem 1.7.1(b) for the matrix product AB and Theorem 1.7.1(d) for the matrix A, where, , ⎥, , ⎡, , −1, , ⎢, A=⎣ 0, 0, , ⎤, , 0⎥, ⎥, , ⎥, 0⎦, −5, , 0, , ⎤, , ⎡, , ⎥, , ⎢, , ⎤, , 2, , 5, , 2, , −8, , 1, , 3 ⎦, B = ⎣ 0, , 2, , 1⎦, , 0, , 3, , 0, , −4, , 0, , 34. Let A be an n × n symmetric matrix., (a) Show that A2 is symmetric., (b) Show that 2A2 − 3A + I is symmetric., , 0, , ⎥

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74, , Chapter 1 Systems of Linear Equations and Matrices, , 35. Verify Theorem 1.7.4 for the given matrix A., , , (a) A =, , ⎡, , , , 2, , −1, , −1, , 3, , 1, , −2, , (b) A = ⎣−2, , 1, , 3, , −7, , ⎢, , 3, , ⎤, , ⎥, −7 ⎦, 4, , Working with Proofs, , 36. Find all 3 × 3 diagonal matrices A that satisfy, A2 − 3A − 4I = 0., 37. Let A = [aij ] be an n × n matrix. Determine whether A is, symmetric., (a) aij = i 2 + j 2, , (b) aij = i 2 − j 2, , (c) aij = 2i + 2j, , (d) aij = 2i + 2j, 2, , 39. Find an upper triangular matrix that satisfies, 1, 0, , Step 1. Let U x = y, so that LU x = b can be expressed as, Ly = b. Solve this system., Step 2. Solve the system U x = y for x., In each part, use this two-step method to solve the given, system., 1, , ⎡, , ⎤⎡, , 0, 3, 4, , ⎢, (a) ⎣−2, 2, , 2, ⎢, (b) ⎣ 4, −3, , ⎤⎡, , 0, 1, −2, , ⎤⎡ ⎤, , −1, , 0, 2, ⎥⎢, 0⎦ ⎣0, 0, 1, , 0, 3, ⎥⎢, 0⎦ ⎣0, 0, 3, , ⎡, , ⎤, , ⎤⎡ ⎤ ⎡ ⎤, 2, 4, x1, ⎥⎢ ⎥ ⎢ ⎥, 1⎦ ⎣x2 ⎦ = ⎣−5⎦, 2, 2, x3, , −5, 4, 0, , In the text we defined a matrix A to be symmetric if AT = A., Analogously, a matrix A is said to be skew-symmetric if AT = −A., Exercises 41–45 are concerned with matrices of this type., 41. Fill in the missing entries (marked with ×) so the matrix A is, skew-symmetric., , ⎡, ⎢, , ×, , (a) A = ⎣ 0, , ×, , ×, ×, −1, , 4, , ⎡, , ⎤, , ⎥, ×⎦, ×, , ×, , 0, , (b) A = ⎣×, 8, , ×, ×, , ⎢, , ⎤, ×, ⎥, −4 ⎦, ×, , 42. Find all values of a , b, c, and d for which A is skew-symmetric., , ⎡, , 0, ⎢, A = ⎣−2, −3, , 2a − 3b + c, 0, −5, , (b) If A and B are skew-symmetric matrices, then so are AT ,, A + B , A − B , and kA for any scalar k ., 46. Prove: If the matrices A and B are both upper triangular or, both lower triangular, then the diagonal entries of both AB, and BA are the products of the diagonal entries of A and B ., , ⎤, , 3 a − 5b + 5 c, ⎥, 5 a − 8b + 6 c ⎦, , d, , True-False Exercises, TF. In parts (a)–(m) determine whether the statement is true or, false, and justify your answer., (a) The transpose of a diagonal matrix is a diagonal matrix., (b) The transpose of an upper triangular matrix is an upper triangular matrix., (c) The sum of an upper triangular matrix and a lower triangular, matrix is a diagonal matrix., (d) All entries of a symmetric matrix are determined by the entries, occurring on and above the main diagonal., , 3, 1, x1, ⎥⎢ ⎥ ⎢ ⎥, 2⎦ ⎣x2 ⎦ = ⎣−2⎦, 4, 0, x3, , 1, 0, , 45. Prove the following facts about skew-symmetric matrices., , 47. Prove: If ATA = A, then A is symmetric and A = A2 ., , 30, −8, , 40. If the n × n matrix A can be expressed as A = LU , where L is, a lower triangular matrix and U is an upper triangular matrix,, then the linear system Ax = b can be expressed as LU x = b, and can be solved in two steps:, , ⎡, , 44. Prove that every square matrix A can be expressed as the sum, of a symmetric matrix and a skew-symmetric matrix. [Hint:, Note the identity A = 21 (A + AT ) + 21 (A − AT ).], (a) If A is an invertible skew-symmetric matrix, then A−1 is, skew-symmetric., , 3, , 38. On the basis of your experience with Exercise 37, devise a general test that can be applied to a formula for aij to determine, whether A = [aij ] is symmetric., , A3 =, , 43. We showed in the text that the product of symmetric matrices, is symmetric if and only if the matrices commute. Is the product of commuting skew-symmetric matrices skew-symmetric?, Explain., , (e) All entries of an upper triangular matrix are determined by, the entries occurring on and above the main diagonal., (f ) The inverse of an invertible lower triangular matrix is an upper, triangular matrix., (g) A diagonal matrix is invertible if and only if all of its diagonal, entries are positive., (h) The sum of a diagonal matrix and a lower triangular matrix is, a lower triangular matrix., (i) A matrix that is both symmetric and upper triangular must be, a diagonal matrix., ( j) If A and B are n × n matrices such that A + B is symmetric,, then A and B are symmetric., (k) If A and B are n × n matrices such that A + B is upper triangular, then A and B are upper triangular., (l) If A2 is a symmetric matrix, then A is a symmetric matrix., (m) If kA is a symmetric matrix for some k  = 0, then A is a symmetric matrix.

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1.8 Matrix Transformations, , Working withTechnology, T1. Starting with the formula stated in Exercise T1 of Section 1.5,, derive a formula for the inverse of the “block diagonal” matrix, , , , in which D1 and D2 are invertible, and use your result to compute, the inverse of the matrix, , ⎡, , 1.24, , ⎢3.08, ⎢, M=⎢, ⎣ 0, , , , D1, , 0, , 0, , D2, , 75, , 0, , ⎤, , 2.37, , 0, , 0, , −1.01, , 0, , 0 ⎥, ⎥, , ⎥, , 0, , 2.76, , 4.92⎦, , 0, , 3.23, , 5.54, , 1.8 MatrixTransformations, In this section we will introduce a special class of functions that arise from matrix, multiplication. Such functions, called “matrix transformations,” are fundamental in the, study of linear algebra and have important applications in physics, engineering, social, sciences, and various branches of mathematics., , Recall that in Section 1.1 we defined an “ordered n-tuple” to be a sequence of n real, numbers, and we observed that a solution of a linear system in n unknowns, say, , x1 = s1 , x2 = s2 , . . . , xn = sn, can be expressed as the ordered n-tuple, , (s1 , s2 , . . . , sn ), , The term “vector” is used in, various ways in mathematics, physics, engineering, and, other applications. The idea, of viewing n-tuples as vectors, will be discussed in more detail, in Chapter 3, at which point we, will also explain how this idea, relates to more familiar notion, of a vector., , (1), , Recall also that if n = 2, then the n-tuple is called an “ordered pair,” and if n = 3, it is, called an “ordered triple.” For two ordered n-tuples to be regarded as the same, they, must list the same numbers in the same order. Thus, for example, (1, 2) and (2, 1) are, different ordered pairs., The set of all ordered n-tuples of real numbers is denoted by the symbol R n . The, elements of R n are called vectors and are denoted in boldface type, such as a, b, v, w,, and x. When convenient, ordered n-tuples can be denoted in matrix notation as column, vectors. For example, the matrix, , ⎡ ⎤, s1, ⎢s ⎥, ⎢ 2⎥, ⎢ ⎥, ⎢ .. ⎥, ⎣.⎦, , (2), , sn, can be used as an alternative to (1). We call (1) the comma-delimited form of a vector, and (2) the column-vector form. For each i = 1, 2, . . . , n, let ei denote the vector in R n, with a 1 in the i th position and zeros elsewhere. In column form these vectors are, , ⎡ ⎤, 1, , ⎡ ⎤, 0, , ⎡ ⎤, 0, , ⎢ ⎥, ⎢ ⎥, ⎢ ⎥, ⎢0⎥, ⎢1⎥, ⎢0⎥, ⎢ ⎥, ⎢ ⎥, ⎢ ⎥, ⎢0⎥, ⎢0⎥, ⎢ ⎥, e1 = ⎢ ⎥ , e2 = ⎢ ⎥ , . . . , en = ⎢0⎥, ⎢.⎥, ⎢.⎥, ⎢.⎥, ⎢.⎥, ⎢.⎥, ⎢.⎥, ⎣.⎦, ⎣.⎦, ⎣.⎦, 0, , 0, , 1, , We call the vectors e1 , e2 , . . . , en the standard basis vectors for R n . For example, the, vectors, ⎡ ⎤, ⎡ ⎤, ⎡ ⎤, 1, 0, 0, , ⎢ ⎥, , ⎢ ⎥, , ⎢ ⎥, , e1 = ⎣0⎦ , e2 = ⎣1⎦ , e3 = ⎣0⎦, 0, 0, 1, are the standard basis vectors for R 3 .

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76, , Chapter 1 Systems of Linear Equations and Matrices, , The vectors e1 , e2 , . . . , en in R n are termed “basis vectors” because all other vectors, in R n are expressible in exactly one way as a linear combination of them. For example,, if, ⎡ ⎤, , x1, ⎢x ⎥, ⎢ 2⎥, ⎥, x=⎢, ⎢ .. ⎥, ⎣.⎦, xn, , then we can express x as, x = x1 e1 + x2 e2 + · · · + xn en, Functions and, Transformations, , Recall that a function is a rule that associates with each element of a set A one and only, one element in a set B . If f associates the element b with the element a , then we write, , b = f(a), f, a, b = f (a), , Domain, A, , Codomain, B, , Figure 1.8.1, , and we say that b is the image of a under f or that f(a) is the value of f at a . The set, A is called the domain of f and the set B the codomain of f (Figure 1.8.1). The subset, of the codomain that consists of all images of elements in the domain is called the range, of f ., In many applications the domain and codomain of a function are sets of real numbers,, but in this text we will be concerned with functions for which the domain is R n and the, codomain is R m for some positive integers m and n., DEFINITION 1 If f is a function with domain R n and codomain R m , then we say that, , f is a transformation from R n to R m or that f maps from R n to R m , which we denote, by writing, , f : R n →R m, In the special case where m = n, a transformation is sometimes called an operator on, , Rn., , MatrixTransformations, It is common in linear algebra, to use the letter T to denote, a transformation. In keeping, with this usage, we will usually, denote a transformation from, R n to R m by writing, , T : R n →R m, , In this section we will be concerned with the class of transformations from R n to R m, that arise from linear systems. Specifically, suppose that we have the system of linear, equations, , w1 = a11 x1 + a12 x2 + · · · + a1n xn, w2 = a21 x1 + a22 x2 + · · · + a2n xn, .., .., .., .., ., ., ., ., wm = am1 x1 + am2 x2 + · · · + amn xn, which we can write in matrix notation as, , ⎡, , ⎤ ⎡, w1, a11, ⎢w ⎥ ⎢a, ⎢ 2 ⎥ ⎢ 21, ⎢ .. ⎥ = ⎢ .., ⎣ . ⎦ ⎣ ., wm, a m1, , a12, a22, .., ., am 2, , ⎤⎡ ⎤, a1n, x1, ⎥, ⎢, a 2 n ⎥ ⎢ x2 ⎥, ⎥, .. ⎥ ⎢ .. ⎥, . ⎦⎣ . ⎦, · · · amn, xn, ···, ···, , (3), , (4), , or more briefly as, w = Ax, , (5), , Although we could view (5) as a compact way of writing linear system (3), we will view, it instead as a transformation that maps a vector x in R n into thevector w in R m by

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1.8 Matrix Transformations, , 77, , multiplying x on the left by A. We call this a matrix transformation (or matrix operator, in the special case where m = n). We denote it by, , TA : R n → R m, TA, x, , TA(x), , Rn, , (see Figure 1.8.2). This notation is useful when it is important to make the domain, and codomain clear. The subscript on TA serves as a reminder that the transformation, results from multiplying vectors in R n by the matrix A. In situations where specifying, the domain and codomain is not essential, we will express (4) as, w = TA (x), , Rm, TA : R n → R m, , (6), , We call the transformation TA multiplication by A. On occasion we will find it convenient, to express (6) in the schematic form, TA, , x −→ w, , Figure 1.8.2, , (7), , which is read “TA maps x into w.”, E X A M P L E 1 A Matrix Transformation from R 4 to R 3, , The transformation from R 4 to R 3 defined by the equations, , w1 = 2x1 − 3x2 + x3 − 5x4, w2 = 4x1 + x2 − 2x3 + x4, w3 = 5x1 − x2 + 4x3, can be expressed in matrix form as, , ⎡, , ⎤, , ⎡, , 2 −3, w1, ⎢ ⎥ ⎢, 1, ⎣w2 ⎦ = ⎣4, 5 −1, w3, , (8), , ⎡ ⎤, ⎤ x1, −5 ⎢ ⎥, ⎥ ⎢x2 ⎥, 1⎦ ⎢ ⎥, ⎣ x3 ⎦, 0, x4, , 1, , −2, 4, , from which we see that the transformation can be interpreted as multiplication by, , ⎡, , −3, , 2, ⎢, A = ⎣4, 5, , 1, , −1, , ⎤, −5, ⎥, 1⎦, , 1, −2, 4, , (9), , 0, , Although the image under the transformation TA of any vector, , ⎡ ⎤, x1, ⎢x ⎥, ⎢ 2⎥, x=⎢ ⎥, ⎣x3 ⎦, x4, , in R 4 could be computed directly from the defining equations in (8), we will find it, preferable to use the matrix in (9). For example, if, , ⎡, , ⎤, , 1, ⎢−3⎥, ⎢ ⎥, x=⎢ ⎥, ⎣ 0⎦, 2, then it follows from (9) that, , ⎡, , ⎤, , ⎡, , 2 −3, w1, ⎢, ⎢ ⎥, 1, ⎣w2 ⎦ = TA (x) = Ax = ⎣4, 5 −1, w3, , 1, , −2, 4, , ⎡, , ⎤, , ⎡ ⎤, 1, −5 ⎢ ⎥, 1, ⎥ ⎢−3⎥ ⎢ ⎥, 1⎦ ⎢ ⎥ = ⎣3⎦, ⎣ 0⎦, 8, 0, 2, ⎤

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78, , Chapter 1 Systems of Linear Equations and Matrices, , E X A M P L E 2 Zero Transformations, , If 0 is the m × n zero matrix, then, , T0 (x) = 0x = 0, so multiplication by zero maps every vector in R n into the zero vector in R m . We call T0, the zero transformation from R n to R m ., , E X A M P L E 3 Identity Operators, , If I is the n × n identity matrix, then, , T I (x ) = I x = x, so multiplication by I maps every vector in R n to itself. We call TI the identity operator, on R n ., , Properties of Matrix, Transformations, , The following theorem lists four basic properties of matrix transformations that follow, from properties of matrix multiplication., , A the matrix transformation TA : R n →R m has the, following properties for all vectors u and v and for every scalar k :, , THEOREM 1.8.1 For every matrix, , (a) TA (0) = 0, (b) TA (k u) = kTA (u), , [Homogeneity property], , (c) TA (u + v) = TA (u) + TA (v), , [Additivity property], , (d ) TA (u − v) = TA (u) − TA (v), , Proof All four parts are restatements of the following properties of matrix arithmetic, , given in Theorem 1.4.1:, , A0 = 0, A(k u) = k(Au), A(u + v) = Au + Av, A(u − v) = Au − Av, It follows from parts (b) and (c) of Theorem 1.8.1 that a matrix transformation maps, a linear combination of vectors in R n into the corresponding linear combination of, vectors in R m in the sense that, , TA (k1 u1 + k2 u2 + · · · + kr ur ) = k1 TA (u1 ) + k2 TA (u2 ) + · · · + kr TA (ur ), , (10), , Matrix transformations are not the only kinds of transformations. For example, if, , w1 = x12 + x22, w2 = x1 x2, then there are no constants a , b, c, and d for which, ,  , w1, w2, , , , =, , a, , b, , c, , d, ,  , x1, x2, , =, , (11), , , , x12 + x22, , , , x1 x2, , so that the equations in (11) do not define a matrix transformation from R 2 to R 2 .

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1.8 Matrix Transformations, , 79, , This leads us to the following two questions., Question 1. Are there algebraic properties of a transformation T : R n →R m that can, , be used to determine whether T is a matrix transformation?, Question 2. If we discover that a transformation, , T : R n →R m is a matrix transfor-, , mation, how can we find a matrix for it?, , The following theorem and its proof will provide the answers., , T : R n →R m is a matrix transformation if and only if the following, relationships hold for all vectors u and v in R n and for every scalar k :, [Additivity property], (i) T (u + v) = T (u) + T (v), , THEOREM 1.8.2, , (ii) T (k u) = kT (u), , [Homogeneity property], , T is a matrix transformation, then properties (i) and (ii) follow respectively, from parts (c) and (b) of Theorem 1.8.1., Conversely, assume that properties (i) and (ii) hold. We must show that there exists, an m × n matrix A such that, T (x) = Ax, , Proof If, , for every vector x in R n . Recall that the derivation of Formula (10) used only the, additivity and homogeneity properties of TA . Since we are assuming that T has those, properties, it must be true that, , T (k1 u1 + k2 u2 + · · · + kr ur ) = k1 T (u1 ) + k2 T (u2 ) + · · · + kr T (ur ), , (12), , for all scalars k1 , k2 , . . . , kr and all vectors u1 , u2 , . . . , ur in R n . Let A be the matrix, , A = [T (e1 ) | T (e2 ) | · · · | T (en )], , (13), , where e1 , e2 , . . . , en are the standard basis vectors for R n . It follows from Theorem 1.3.1, that Ax is a linear combination of the columns of A in which the successive coefficients, are the entries x1 , x2 , . . . , xn of x. That is,, , Ax = x1 T (e1 ) + x2 T (e2 ) + · · · + xn T (en ), Using Formula (10) we can rewrite this as, , Ax = T (x1 e1 + x2 e2 + · · · + xn en ) = T (x), which completes the proof., , Theorem 1.8.3 tells us that, for transformations from R n to, R m , the terms “matrix transformation” and “linear transformation” are synonymous., , The additivity and homogeneity properties in Theorem 1.8.2 are called linearity, conditions, and a transformation that satisfies these conditions is called a linear transformation. Using this terminology Theorem 1.8.2 can be restated as follows., , THEOREM 1.8.3 Every linear transformation from R n to R m is a matrix transformation,, , and conversely, every matrix transformation from R n to R m is a linear transformation.

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80, , Chapter 1 Systems of Linear Equations and Matrices, , Depending on whether n-tuples and m-tuples are regarded as vectors or points, the, geometric effect of a matrix transformation TA : R n →R m is to map each vector (point), in R n into a vector (point) in R m (Figure 1.8.3)., Rn, x, , 0, , 0, , Figure 1.8.3, , Rn, , Rm, TA(x), , x, , 0, , 0, , TA maps vectors to vectors., , Rm, TA(x), , TA maps points to points., , The following theorem states that if two matrix transformations from R n to R m have, the same image at each point of R n , then the matrices themselves must be the same., THEOREM 1.8.4 If TA : R n →R m and TB : R n →R m are matrix transformations, and if, , TA (x) = TB (x) for every vector x in R n , then A = B ., , Proof To say that TA (x), , = TB (x) for every vector in R n is the same as saying that, Ax = B x, , for every vector x in R n . This will be true, in particular, if x is any of the standard basis, vectors e1 , e2 , . . . , en for R n ; that is,, , Aej = B ej (j = 1, 2, . . . , n), , (14), , Since every entry of ej is 0 except for the j th, which is 1, it follows from Theorem 1.3.1, that Aej is the j th column of A and B ej is the j th column of B . Thus, (14) implies that, corresponding columns of A and B are the same, and hence that A = B ., Theorem 1.8.4 is significant because it tells us that there is a one-to-one correspondence, between m × n matrices and matrix transformations from R n to R m in the sense that, every m × n matrix A produces exactly one matrix transformation (multiplication by A), and every matrix transformation from R n to R m arises from exactly one m × n matrix;, we call that matrix the standard matrix for the transformation., A Procedure for Finding, Standard Matrices, , In the course of proving Theorem 1.8.2 we showed in Formula (13) that if e1 , e2 , . . . , en, are the standard basis vectors for R n (in column form), then the standard matrix for a, linear transformation T : R n →R m is given by the formula, , A = [T (e1 ) | T (e2 ) | · · · | T (en )], , (15), , This suggests the following procedure for finding standard matrices., Finding the Standard Matrix for a Matrix Transformation, Step 1. Find the images of the standard basis vectors e1 , e2 , . . . , en for R n ., Step 2. Construct the matrix that has the images obtained in Step 1 as its successive, columns. This matrix is the standard matrix for the transformation.

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1.8 Matrix Transformations, , 81, , E X A M P L E 4 Finding a Standard Matrix, , Find the standard matrix A for the linear transformation T : R 2 →R 2 defined by the, formula, ⎡, ⎤,  , 2 x1 + x 2, , ⎢, ⎥, = ⎣ x1 − 3x2 ⎦, −x1 + x2, , x1, , T, , x2, , (16), , Solution We leave it for you to verify that, ,  , T (e1 ) = T, , 1, 0, , ⎡, , 2, , ⎤, , ⎢ ⎥, = ⎣ 1⎦ and T (e2 ) = T, −1, ,  , 0, 1, , ⎡, , ⎤, , 1, , ⎢ ⎥, = ⎣−3⎦, 1, , Thus, it follows from Formulas (15) and (16) that the standard matrix is, , ⎡, , 2, , 1, , ⎤, , ⎥, ⎢, A = [T (e1 ) | T (e2 )] = ⎣ 1 −3⎦, −1, 1, , E X A M P L E 5 Computing with Standard Matrices, , For the linear transformation in Example 4, use the standard matrix A obtained in that, example to find,  , 1, , T, , Although we could have obtained the result in Example 5, by substituting values for the, variables in (13), the method, used in Example 5 is preferable, for large-scale problems in that, matrix multiplication is better, suited for computer computations., , 4, , Solution The transformation is multiplication by A, so, ,  , T, , 1, , 4, , ⎡, , 2, , ⎢, =⎣ 1, −1, , ⎤, , ⎡, , ⎤, , 1  , 6, ⎥ 1, ⎥, ⎢, −3⎦, = ⎣−11⎦, 4, 1, 3, , For transformation problems posed in comma-delimited form, a good procedure is, to rewrite the problem in column-vector form and use the methods previously illustrated., E X A M P L E 6 Finding a Standard Matrix, , Rewrite the transformation T (x1 , x2 ) = (3x1 + x2 , 2x1 − 4x2 ) in column-vector form, and find its standard matrix., Solution, ,  ,  ,   , 3x1 + x2, 3, 1 x1, x1, =, =, T, 2x1 − 4x2, x2, 2 −2 x2, , Thus, the standard matrix is, , , , 3, , 1, , 2, , −2, , , , Remark This section is but a first step in the study of linear transformations, which is one of the, major themes in this text. We will delve deeper into this topic in Chapter 4, at which point we will, have more background and a richer source of examples to work with.

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1.8 Matrix Transformations, , ⎡, −2, ⎢, 20. (a) A = ⎣ 3, 6, , ⎡, −1, ⎢, (b) A = ⎣ 2, 7, , ⎤, , ⎡ ⎤, , 30. We proved in the text that if T : R n →R m is a matrix transformation, then T (0) = 0. Show that the converse of this result, is false by finding a mapping T : R n →R m that is not a matrix, transformation but for which T (0) = 0., , 4, x1, ⎥, ⎢ ⎥, 7⎦; x = ⎣x2 ⎦, −1, x3, , 1, 5, 0, , ⎤, , 1, x1, ⎥, 4⎦; x =, x2, 8, , 31. Let TA : R 3 →R 3 be multiplication by, , ⎡, , −1, , In Exercises 21–22, use Theorem 1.8.2 to show that T is a, matrix transformation., 21. (a) T (x, y) = (2x + y, x − y), (b) T (x1 , x2 , x3 ) = (x1 , x3 , x1 + x2 ), , In Exercises 23–24, use Theorem 1.8.2 to show that T is not a, matrix transformation., 23. (a) T (x, y) = (x 2 , y), (b) T (x, y, z) = (x, y, xz), , , , √ , (b) T (x1 , x2 , x3 ) = x1 , x2 , x3, , 25. A function of the form f(x) = mx + b is commonly called a, “linear function” because the graph of y = mx + b is a line., Is f a matrix transformation on R ?, 26. Show that T (x, y) = (0, 0) defines a matrix operator on R 2, but T (x, y) = (1, 1) does not., In Exercises 27–28, the images of the standard basis vectors for R 3 are given for a linear transformation T : R 3 →R 3 ., Find the standard matrix for the transformation, and find, T (x)., 0, , ⎡, , 4, , ⎤, , ⎡ ⎤, 2, , ⎢ ⎥, ⎢ ⎥, ⎢ ⎥, ⎢ ⎥, 27. T (e1 ) = ⎣3⎦ , T (e2 ) = ⎣0⎦ , T (e3 ) = ⎣−3⎦ ; x = ⎣1⎦, 0, 1, 0, −1, ⎡ ⎤, ⎡ ⎤, ⎡ ⎤, ⎡ ⎤, 2, 1, 3, −3, ⎢ ⎥, ⎢ ⎥, ⎢ ⎥, ⎢ ⎥, 28. T (e1 ) = ⎣1⎦ , T (e2 ) = ⎣−1⎦ , T (e3 ) = ⎣0⎦ ; x = ⎣2⎦, 3, , 0, , 1, , 2⎦, , 4, , 5, , ⎥, , −3, , and let e1 , e2 , and e3 be the standard basis vectors for R 3 . Find, the following vectors by inspection., , (c) TA (7e3 ), , Working with Proofs, 32. (a) Prove: If T : R n →R m is a matrix transformation, then, T (0) = 0; that is, T maps the zero vector in R n into the, zero vector in R m ., (b) The converse of this is not true. Find an example of a, function T for which T (0) = 0 but which is not a matrix, transformation., , 24. (a) T (x, y) = (x, y + 1), , 1, , ⎢, A=⎣ 2, , 0, , (b) TA (e1 + e2 + e3 ), , (b) T (x1 , x2 ) = (x2 , x1 ), , ⎡ ⎤, , ⎤, , 3, , (a) TA (e1 ), TA (e2 ), and TA (e3 ), , 22. (a) T (x, y, z) = (x + y, y + z, x), , ⎡ ⎤, , 83, , 2, , 1, , 29. Let T : R 2 →R 2 be a linear operator for which the images, of the standard basis vectors for R 2 are T (e1 ) = (a, b) and, T (e2 ) = (c, d). Find T (1, 1)., , True-False Exercises, TF. In parts (a)–(g) determine whether the statement is true or, false, and justify your answer., (a) If A is a 2 × 3 matrix, then the domain of the transformation, TA is R 2 ., (b) If A is an m × n matrix, then the codomain of the transformation TA is R n ., (c) There is at least one linear transformation T : R n →R m for, which T (2x) = 4T (x) for some vector x in R n ., (d) There are linear transformations from R n to R m that are not, matrix transformations., (e) If TA : R n →R n and if TA (x) = 0 for every vector x in R n , then, A is the n × n zero matrix., (f ) There is only one matrix transformation T : R n →R m such that, T (−x) = −T (x) for every vector x in R n ., (g) If b is a nonzero vector in R n , then T (x) = x + b is a matrix, operator on R n .

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84, , Chapter 1 Systems of Linear Equations and Matrices, , 1.9 Applications of Linear Systems, In this section we will discuss some brief applications of linear systems. These are but a, small sample of the wide variety of real-world problems to which our study of linear, systems is applicable., , Network Analysis, , The concept of a network appears in a variety of applications. Loosely stated, a network, is a set of branches through which something “flows.” For example, the branches might, be electrical wires through which electricity flows, pipes through which water or oil flows,, traffic lanes through which vehicular traffic flows, or economic linkages through which, money flows, to name a few possibilities., In most networks, the branches meet at points, called nodes or junctions, where the, flow divides. For example, in an electrical network, nodes occur where three or more wires, join, in a traffic network they occur at street intersections, and in a financial network, they occur at banking centers where incoming money is distributed to individuals or, other institutions., In the study of networks, there is generally some numerical measure of the rate at, which the medium flows through a branch. For example, the flow rate of electricity is, often measured in amperes, the flow rate of water or oil in gallons per minute, the flow rate, of traffic in vehicles per hour, and the flow rate of European currency in millions of Euros, per day. We will restrict our attention to networks in which there is flow conservation at, each node, by which we mean that the rate of flow into any node is equal to the rate of flow, out of that node. This ensures that the flow medium does not build up at the nodes and, block the free movement of the medium through the network., A common problem in network analysis is to use known flow rates in certain branches, to find the flow rates in all of the branches. Here is an example., , E X A M P L E 1 Network Analysis Using Linear Systems, 30, , Figure 1.9.1 shows a network with four nodes in which the flow rate and direction of, flow in certain branches are known. Find the flow rates and directions of flow in the, remaining branches., , 35, , 55, 15, , 60, , Solution As illustrated in Figure 1.9.2, we have assigned arbitrary directions to the, unknown flow rates x1 , x2 , and x3 . We need not be concerned if some of the directions, are incorrect, since an incorrect direction will be signaled by a negative value for the flow, rate when we solve for the unknowns., It follows from the conservation of flow at node A that, , Figure 1.9.1, , x1 + x2 = 30, Similarly, at the other nodes we have, , 30, x2, 35, , A, , B, x3, , C, 60, , x2 + x3 = 35, , (node B ), , x1, , x3 + 15 = 60, , (node C ), , D, , x1 + 15 = 55, , (node D ), , 15, , 55, , These four conditions produce the linear system, , x1 + x2, = 30, x2 + x3 = 35, , Figure 1.9.2, , x1, , x3 = 45, = 40

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1.9 Applications of Linear Systems, , 85, , which we can now try to solve for the unknown flow rates. In this particular case the, system is sufficiently simple that it can be solved by inspection (work from the bottom, up). We leave it for you to confirm that the solution is, , x1 = 40, x2 = −10, x3 = 45, The fact that x2 is negative tells us that the direction assigned to that flow in Figure 1.9.2, is incorrect; that is, the flow in that branch is into node A., , E X A M P L E 2 Design of Traffic Patterns, , The network in Figure 1.9.3 shows a proposed plan for the traffic flow around a new, park that will house the Liberty Bell in Philadelphia, Pennsylvania. The plan calls for a, computerized traffic light at the north exit on Fifth Street, and the diagram indicates the, average number of vehicles per hour that are expected to flow in and out of the streets, that border the complex. All streets are one-way., (a) How many vehicles per hour should the traffic light let through to ensure that the, average number of vehicles per hour flowing into the complex is the same as the, average number of vehicles flowing out?, (b) Assuming that the traffic light has been set to balance the total flow in and out of, the complex, what can you say about the average number of vehicles per hour that, will flow along the streets that border the complex?, N, W, , E, , Traffic, light, , 200, , x, , 200, , Market St., , 700, , Liberty, Park, , Fifth St., , 500, , Sixth St., , S, , 400, , 500, , C, , x3, , 400, , 700, , Chestnut St., , D, , x1, , 600, , 400, , A, , 400, , 600, , (a), , Figure 1.9.3, , B, x2, , x4, , (b), , Solution (a) If, as indicated in Figure 1.9.3b, we let x denote the number of vehicles per, hour that the traffic light must let through, then the total number of vehicles per hour, that flow in and out of the complex will be, , Flowing in: 500 + 400 + 600 + 200 = 1700, Flowing out: x + 700 + 400, Equating the flows in and out shows that the traffic light should let x = 600 vehicles per, hour pass through., Solution (b) To avoid traffic congestion, the flow in must equal the flow out at each, intersection. For this to happen, the following conditions must be satisfied:, , Intersection, , Flow In, , A, B, C, D, , 400 + 600, , x2 + x 3, 500 + 200, x1 + x4, , Flow Out, , =, =, =, =, , x1 + x 2, 400 + x, x3 + x4, 700

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86, , Chapter 1 Systems of Linear Equations and Matrices, , Thus, with x = 600, as computed in part (a), we obtain the following linear system:, , x1 + x2, x2 + x3, x3 + x4, x1, + x4, , = 1000, = 1000, = 700, = 700, , We leave it for you to show that the system has infinitely many solutions and that these, are given by the parametric equations, , x1 = 700 − t, x2 = 300 + t, x3 = 700 − t, x4 = t, , (1), , However, the parameter t is not completely arbitrary here, since there are physical constraints to be considered. For example, the average flow rates must be nonnegative since, we have assumed the streets to be one-way, and a negative flow rate would indicate a flow, in the wrong direction. This being the case, we see from (1) that t can be any real number, that satisfies 0 ≤ t ≤ 700, which implies that the average flow rates along the streets will, fall in the ranges, 0 ≤ x1 ≤ 700, 300 ≤ x2 ≤ 1000, 0 ≤ x3 ≤ 700, 0 ≤ x4 ≤ 700, Electrical Circuits, + –, , Switch, , Figure 1.9.4, , Next we will show how network analysis can be used to analyze electrical circuits consisting of batteries and resistors. A battery is a source of electric energy, and a resistor,, such as a lightbulb, is an element that dissipates electric energy. Figure 1.9.4 shows a, schematic diagram of a circuit with one battery (represented by the symbol, ), one, resistor (represented by the symbol, ), and a switch. The battery has a positive pole, (+) and a negative pole (−). When the switch is closed, electrical current is considered to, flow from the positive pole of the battery, through the resistor, and back to the negative, pole (indicated by the arrowhead in the figure)., Electrical current, which is a flow of electrons through wires, behaves much like the, flow of water through pipes. A battery acts like a pump that creates “electrical pressure”, to increase the flow rate of electrons, and a resistor acts like a restriction in a pipe that, reduces the flow rate of electrons. The technical term for electrical pressure is electrical, potential; it is commonly measured in volts (V). The degree to which a resistor reduces the, electrical potential is called its resistance and is commonly measured in ohms (). The, rate of flow of electrons in a wire is called current and is commonly measured in amperes, (also called amps) (A). The precise effect of a resistor is given by the following law:, , I amperes passes through a resistor with a resistance of, R ohms, then there is a resulting drop of E volts in electrical potential that is the, , Ohm’s Law If a current of, , product of the current and resistance; that is,, , E = IR, , + –, , Figure 1.9.5, , + –, , A typical electrical network will have multiple batteries and resistors joined by some, configuration of wires. A point at which three or more wires in a network are joined is, called a node (or junction point). A branch is a wire connecting two nodes, and a closed, loop is a succession of connected branches that begin and end at the same node. For, example, the electrical network in Figure 1.9.5 has two nodes and three closed loops—, two inner loops and one outer loop. As current flows through an electrical network, it, undergoes increases and decreases in electrical potential, called voltage rises and voltage, drops, respectively. The behavior of the current at the nodes and around closed loops is, governed by two fundamental laws:

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1.9 Applications of Linear Systems, , 87, , Kirchhoff’s Current Law The sum of the currents flowing into any node is equal to the, sum of the currents flowing out., , Kirchhoff’s Voltage Law In one traversal of any closed loop, the sum of the voltage, rises equals the sum of the voltage drops., , I2, I1, , Kirchhoff’s current law is a restatement of the principle of flow conservation at a node, that was stated for general networks. Thus, for example, the currents at the top node in, Figure 1.9.6 satisfy the equation I1 = I2 + I3 ., In circuits with multiple loops and batteries there is usually no way to tell in advance, which way the currents are flowing, so the usual procedure in circuit analysis is to assign arbitrary directions to the current flows in the branches and let the mathematical, computations determine whether the assignments are correct. In addition to assigning, directions to the current flows, Kirchhoff’s voltage law requires a direction of travel for, each closed loop. The choice is arbitrary, but for consistency we will always take this, direction to be clockwise (Figure 1.9.7).We also make the following conventions:, , I3, , Figure 1.9.6, , + –, , + –, , • A voltage drop occurs at a resistor if the direction assigned to the current through the, resistor is the same as the direction assigned to the loop, and a voltage rise occurs at, a resistor if the direction assigned to the current through the resistor is the opposite, to that assigned to the loop., , Clockwise closed-loop, convention with arbitrary, direction assignments to, currents in the branches, , Figure 1.9.7, , • A voltage rise occurs at a battery if the direction assigned to the loop is from − to +, through the battery, and a voltage drop occurs at a battery if the direction assigned, to the loop is from + to − through the battery., If you follow these conventions when calculating currents, then those currents whose, directions were assigned correctly will have positive values and those whose directions, were assigned incorrectly will have negative values., E X A M P L E 3 A Circuit with One Closed Loop, , I, , Determine the current I in the circuit shown in Figure 1.9.8., , +, 6 V–, , 3, , Figure 1.9.8, , Solution Since the direction assigned to the current through the resistor is the same, as the direction of the loop, there is a voltage drop at the resistor. By Ohm’s law this, voltage drop is E = IR = 3I . Also, since the direction assigned to the loop is from −, to + through the battery, there is a voltage rise of 6 volts at the battery. Thus, it follows, from Kirchhoff’s voltage law that, 3I = 6, , from which we conclude that the current is I = 2 A. Since I is positive, the direction, assigned to the current flow is correct., , I1, , E X A M P L E 4 A Circuit with Three Closed Loops, , I2, , A, , Determine the currents I1 , I2 , and I3 in the circuit shown in Figure 1.9.9., , I3, 5, , + –, 50 V, , 20 , , B, , Figure 1.9.9, , + –, 30 V, , 10 , , Solution Using the assigned directions for the currents, Kirchhoff’s current law provides, one equation for each node:, , Node, , Current In, , A, B, , I1 + I 2, I3, , Current Out, , =, =, , I3, I1 + I2

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88, , Chapter 1 Systems of Linear Equations and Matrices, , However, these equations are really the same, since both can be expressed as, , I1 + I 2 − I 3 = 0, , (2), , To find unique values for the currents we will need two more equations, which we will, obtain from Kirchhoff’s voltage law. We can see from the network diagram that there, are three closed loops, a left inner loop containing the 50 V battery, a right inner loop, containing the 30 V battery, and an outer loop that contains both batteries. Thus,, Kirchhoff’s voltage law will actually produce three equations. With a clockwise traversal, of the loops, the voltage rises and drops in these loops are as follows:, , Left Inside Loop, , Voltage Rises, , Voltage Drops, , 50, , 5I1 + 20I3, , Right Inside Loop 30 + 10I2 + 20I3, Outside Loop, , 30 + 50 + 10I2, , 0, 5I1, , These conditions can be rewritten as, 5I1, , + 20I3 = 50, 10I2 + 20I3 = −30, , 5I1 − 10I2, , =, , (3), , 80, , However, the last equation is superfluous, since it is the difference of the first two. Thus,, if we combine (2) and the first two equations in (3), we obtain the following linear system, of three equations in the three unknown currents:, , I1 +, 5 I1, , 0, I2 − I3 =, + 20I3 = 50, 10I2 + 20I3 = −30, , We leave it for you to show that the solution of this system in amps is I1 = 6, I2 = −5,, and I3 = 1. The fact that I2 is negative tells us that the direction of this current is opposite, to that indicated in Figure 1.9.9., Balancing Chemical, Equations, , Chemical compounds are represented by chemical formulas that describe the atomic, makeup of their molecules. For example, water is composed of two hydrogen atoms and, one oxygen atom, so its chemical formula is H2 O; and stable oxygen is composed of two, oxygen atoms, so its chemical formula is O2 ., When chemical compounds are combined under the right conditions, the atoms in, their molecules rearrange to form new compounds. For example, when methane burns,, , Gustav Kirchhoff, (1824–1887), , Historical Note The German physicist Gustav Kirchhoff was a student of Gauss., His work on Kirchhoff’s laws,, announced in 1854, was a, major advance in the calculation of currents, voltages,, and resistances of electrical circuits. Kirchhoff was, severely disabled and spent, most of his life on crutches, or in a wheelchair., [Image: ullstein bild histopics/akg-im]

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1.9 Applications of Linear Systems, , 89, , the methane (CH4 ) and stable oxygen (O2 ) react to form carbon dioxide (CO2 ) and water, (H2 O). This is indicated by the chemical equation, CH4 + O2 −→ CO2 + H2 O, , (4), , The molecules to the left of the arrow are called the reactants and those to the right, the products. In this equation the plus signs serve to separate the molecules and are, not intended as algebraic operations. However, this equation does not tell the whole, story, since it fails to account for the proportions of molecules required for a complete, reaction (no reactants left over). For example, we can see from the right side of (4) that, to produce one molecule of carbon dioxide and one molecule of water, one needs three, oxygen atoms for each carbon atom. However, from the left side of (4) we see that one, molecule of methane and one molecule of stable oxygen have only two oxygen atoms, for each carbon atom. Thus, on the reactant side the ratio of methane to stable oxygen, cannot be one-to-one in a complete reaction., A chemical equation is said to be balanced if for each type of atom in the reaction,, the same number of atoms appears on each side of the arrow. For example, the balanced, version of Equation (4) is, CH4 + 2O2 −→ CO2 + 2H2 O, , (5), , by which we mean that one methane molecule combines with two stable oxygen molecules, to produce one carbon dioxide molecule and two water molecules. In theory, one could, multiply this equation through by any positive integer. For example, multiplying through, by 2 yields the balanced chemical equation, 2CH4 + 4O2 −→ 2CO2 + 4H2 O, However, the standard convention is to use the smallest positive integers that will balance, the equation., Equation (4) is sufficiently simple that it could have been balanced by trial and error,, but for more complicated chemical equations we will need a systematic method. There, are various methods that can be used, but we will give one that uses systems of linear, equations. To illustrate the method let us reexamine Equation (4). To balance this, equation we must find positive integers, x1 , x2 , x3 , and x4 such that, , x1 (CH4 ) + x2 (O2 ) −→ x3 (CO2 ) + x4 (H2 O), , (6), , For each of the atoms in the equation, the number of atoms on the left must be equal to, the number of atoms on the right. Expressing this in tabular form we have, Left Side, Carbon, Hydrogen, Oxygen, , x1, 4x1, 2x2, , Right Side, , =, =, =, , x3, 2x4, 2 x3 + x 4, , from which we obtain the homogeneous linear system, , x1, 4 x1, , − x3, , =0, − 2 x4 = 0, 2x2 − 2x3 − x4 = 0, , The augmented matrix for this system is, , ⎡, , 1, ⎢, 4, ⎣, 0, , 0, 0, 2, , −1, 0, −2, , 0, −2, −1, , ⎤, , 0, ⎥, 0⎦, 0

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90, , Chapter 1 Systems of Linear Equations and Matrices, , We leave it for you to show that the reduced row echelon form of this matrix is, , ⎡, , ⎤, , ⎢, ⎣0, , 0, , 0, , − 21, , 0, , 1, , 0, , −1, , 0⎦, , 0, , 0, , 1, , − 21, , 0, , 1, , ⎥, , from which we conclude that the general solution of the system is, , x1 = t/2, x2 = t, x3 = t/2, x4 = t, where t is arbitrary. The smallest positive integer values for the unknowns occur when, we let t = 2, so the equation can be balanced by letting x1 = 1, x2 = 2, x3 = 1, x4 = 2., This agrees with our earlier conclusions, since substituting these values into Equation (6), yields Equation (5)., , E X A M P L E 5 Balancing Chemical Equations Using Linear Systems, , Balance the chemical equation, , + Na3 PO4, −→ H3 PO4, + NaCl, [hydrochloric acid] + [sodium phosphate] −→ [phosphoric acid] + [sodium chloride], HCl, , Solution Let x1 , x2 , x3 , and x4 be positive integers that balance the equation, , x1 (HCl) + x2 (Na3 PO4 ) −→ x3 (H3 PO4 ) + x4 (NaCl), , (7), , Equating the number of atoms of each type on the two sides yields, 1x1 = 3x3 Hydrogen (H), 1x1 = 1x4 Chlorine (Cl), 3x2 = 1x4 Sodium (Na), 1x2 = 1x3 Phosphorus (P), 4x2 = 4x3 Oxygen (O), from which we obtain the homogeneous linear system, , x1, x1, , − 3x3, , =0, − x4 = 0, − x4 = 0, 3x2, x2 − x3, =0, =0, 4x2 − 4x3, , We leave it for you to show that the reduced row echelon form of the augmented matrix, for this system is, , ⎡, , 1, , ⎢, ⎢0, ⎢, ⎢, ⎢0, ⎢, ⎢, ⎣0, 0, , 0, , 0, , −1, , 1, , 0, , − 13, , 0, , 1, , − 13, , 0, 0, , 0, 0, , 0, 0, , 0, , ⎤, , ⎥, ⎥, ⎥, 0⎥, ⎥, ⎥, 0⎦, 0⎥, , 0

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1.9 Applications of Linear Systems, , 91, , from which we conclude that the general solution of the system is, , x1 = t, x2 = t/3, x3 = t/3, x4 = t, where t is arbitrary. To obtain the smallest positive integers that balance the equation,, we let t = 3, in which case we obtain x1 = 3, x2 = 1, x3 = 1, and x4 = 3. Substituting, these values in (7) produces the balanced equation, 3HCl + Na3 PO4 −→ H3 PO4 + 3NaCl, Polynomial Interpolation, , An important problem in various applications is to find a polynomial whose graph passes, through a specified set of points in the plane; this is called an interpolating polynomial, for the points. The simplest example of such a problem is to find a linear polynomial, , p(x) = ax + b, y, , whose graph passes through two known distinct points, (x1 , y1 ) and (x2 , y2 ), in the, xy-plane (Figure 1.9.10). You have probably encountered various methods in analytic, geometry for finding the equation of a line through two points, but here we will give a, method based on linear systems that can be adapted to general polynomial interpolation., The graph of (8) is the line y = ax + b, and for this line to pass through the points, (x1 , y1 ) and (x2 , y2 ), we must have, , y = ax + b, (x2, y2), (x1, y1), , (8), , x, , y1 = ax1 + b and y2 = ax2 + b, , Figure 1.9.10, , Therefore, the unknown coefficients a and b can be obtained by solving the linear system, , ax1 + b = y1, ax2 + b = y2, We don’t need any fancy methods to solve this system—the value of a can be obtained, by subtracting the equations to eliminate b, and then the value of a can be substituted, into either equation to find b. We leave it as an exercise for you to find a and b and then, show that they can be expressed in the form, , a=, , y2 − y1, y1 x2 − y2 x1, and b =, x2 − x 1, x2 − x1, , (9), , provided x1  = x2 . Thus, for example, the line y = ax + b that passes through the points, , (2, 1) and (5, 4), , y, , can be obtained by taking (x1 , y1 ) = (2, 1) and (x2 , y2 ) = (5, 4), in which case (9) yields, , y=x–1, (5, 4), , (2, 1), , a=, x, , 4−1, (1)(5) − (4)(2), = 1 and b =, = −1, 5−2, 5−2, , Therefore, the equation of the line is, , y =x−1, Figure 1.9.11, , (Figure 1.9.11)., Now let us consider the more general problem of finding a polynomial whose graph, passes through n points with distinct x -coordinates, , (x1 , y1 ), (x2 , y2 ), (x3 , y3 ), . . . , (xn , yn ), , (10), , Since there are n conditions to be satisfied, intuition suggests that we should begin by, looking for a polynomial of the form, , p(x) = a0 + a1 x + a2 x 2 + · · · + an−1 x n−1, , (11)

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92, , Chapter 1 Systems of Linear Equations and Matrices, , since a polynomial of this form has n coefficients that are at our disposal to satisfy the, n conditions. However, we want to allow for cases where the points may lie on a line or, have some other configuration that would make it possible to use a polynomial whose, degree is less than n − 1; thus, we allow for the possibility that an−1 and other coefficients, in (11) may be zero., The following theorem, which we will prove later in the text, is the basic result on, polynomial interpolation., , THEOREM 1.9.1 Polynomial Interpolation, , Given any n points in the xy-plane that have distinct x-coordinates, there is a unique, polynomial of degree n − 1 or less whose graph passes through those points., , Let us now consider how we might go about finding the interpolating polynomial, (11) whose graph passes through the points in (10). Since the graph of this polynomial, is the graph of the equation, , y = a0 + a1 x + a2 x 2 + · · · + an−1 x n−1, , (12), , it follows that the coordinates of the points must satisfy, , a0 + a1 x1 + a2 x12 + · · · + an−1 x1n−1 = y1, a0 + a1 x2 + a2 x22 + · · · + an−1 x2n−1 = y2, .., .., .., .., .., ., ., ., ., ., a0 + a1 xn + a2 xn2 + · · · + an−1 xnn−1 = yn, , (13), , In these equations the values of x ’s and y ’s are assumed to be known, so we can view, this as a linear system in the unknowns a0 , a1 , . . . , an−1 . From this point of view the, augmented matrix for the system is, , ⎡, , 1, , ⎢, ⎢, ⎢1, ⎢, ⎢ .., ⎢., ⎣, 1, , x1, , x12, , · · · x1n−1, , x2, .., ., , x22, .., ., , · · · x2n−1, .., ., , xn, , xn2, , · · · xnn−1, , ⎤, , y1, , ⎥, ⎥, y2 ⎥, ⎥, .. ⎥, .⎥, ⎦, yn, , (14), , and hence the interpolating polynomial can be found by reducing this matrix to reduced, row echelon form (Gauss–Jordan elimination)., , E X A M P L E 6 Polynomial Interpolation by Gauss–Jordan Elimination, , Find a cubic polynomial whose graph passes through the points, , (1, 3), (2, −2), (3, −5), (4, 0), Solution Since there are four points, we will use an interpolating polynomial of degree, , n = 3. Denote this polynomial by, p(x) = a0 + a1 x + a2 x 2 + a3 x 3, and denote the x - and y -coordinates of the given points by, , x1 = 1, x2 = 2, x3 = 3, x4 = 4 and y1 = 3, y2 = −2, y3 = −5, y4 = 0

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1.9 Applications of Linear Systems, , 93, , Thus, it follows from (14) that the augmented matrix for the linear system in the unknowns a0 , a1 , a2 , and a3 is, , ⎡, , 1, , ⎢, ⎢, ⎢1, ⎢, ⎢, ⎢1, ⎣, , y, , 1, , x1, , x12, , x13, , x2, , x22, , x23, , x3, , x32, , x33, , x4, , x42, , x43, , ⎤, , ⎡, 1, ⎥, ⎥ ⎢, y2 ⎥ ⎢1, ⎥=⎢, ⎥, y3 ⎥ ⎣1, ⎦, 1, y4, y1, , 1, 2, 3, 4, , 1, 4, 9, 16, , 1, 8, 27, 64, , ⎤, , 3, −2 ⎥, ⎥, ⎥, −5⎦, 0, , 4, , We leave it for you to confirm that the reduced row echelon form of this matrix is, , 3, , ⎡, , 2, 1, –1, –1, , 1, ⎢, ⎢0, ⎢, ⎢0, ⎣, 0, , x, 1, , 2, , 3, , 4, , –2, –3, , 0, 1, 0, 0, , 0, 0, 1, 0, , ⎤, , 0, 0, 0, 1, , 4, ⎥, 3⎥, ⎥, −5⎥, ⎦, 1, , –5, , from which it follows that a0 = 4, a1 = 3, a2 = −5, a3 = 1. Thus, the interpolating, polynomial is, p(x) = 4 + 3x − 5x 2 + x 3, , Figure 1.9.12, , The graph of this polynomial and the given points are shown in Figure 1.9.12., , –4, , Remark Later we will give a more efficient method for finding interpolating polynomials that is, better suited for problems in which the number of data points is large., , CA L C U L U S A N D, C A L C U L AT I N G UT I L ITY, REQUIRED, , E X A M P L E 7 Approximate Integration, , There is no way to evaluate the integral, , , , 1, , sin, , π x2, , 0, , !, dx, , 2, , directly since there is no way to express an antiderivative of the integrand in terms of, elementary functions. This integral could be approximated by Simpson’s rule or some, comparable method, but an alternative approach is to approximate the integrand by an, interpolating polynomial and integrate the approximating polynomial. For example, let, us consider the five points, , x0 = 0, x1 = 0.25, x2 = 0.5, x3 = 0.75, x4 = 1, that divide the interval [0, 1] into four equally spaced subintervals (Figure 1.9.13). The, values of, !, , y, 1, , f(x) = sin, , π x2, 2, , at these points are approximately, , 0.5, , f(0) = 0, f(0.25) = 0.098017, f(0.5) = 0.382683,, f(0.75) = 0.77301, f(1) = 1, , x, 0, , 0.25 0.5 0.75 1 1.25, p(x), sin (πx 2/2), , Figure 1.9.13, , The interpolating polynomial is (verify), , p(x) = 0.098796x + 0.762356x 2 + 2.14429x 3 − 2.00544x 4, and, , , , 1, , p(x) dx ≈ 0.438501, , (15), , (16), , 0, , As shown in Figure 1.9.13, the graphs of f and p match very closely over the interval, [0, 1], so the approximation is quite good.

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94, , Chapter 1 Systems of Linear Equations and Matrices, , Exercise Set 1.9, 1. The accompanying figure shows a network in which the flow, rate and direction of flow in certain branches are known. Find, the flow rates and directions of flow in the remaining branches., , (b) Solve the system for the unknown flow rates., (c) Is it possible to close the road from A to B for construction, and keep traffic flowing on the other streets? Explain., , 50, 300, 500, 30, , 200, x1, , A, , 60, , 450, , 2. The accompanying figure shows known flow rates of hydrocarbons into and out of a network of pipes at an oil refinery., (a) Set up a linear system whose solution provides the unknown flow rates., , 600, , 5., , (c) Find the flow rates and directions of flow if x4 = 50 and, x6 = 0., , Figure Ex-4, , 8V, + –, , 2, , I1 2 , , I2 I3, , 4, , 150, x4, , 25, , 400, , In Exercises 5–8, analyze the given electrical circuits by finding, the unknown currents., , (b) Solve the system for the unknown flow rates., , x1, , x7, , 350, , Figure Ex-1, , 40, , x3, , x5, , 400, x6, , 200, , 600, , x4, , x3, , 50, , 100, x2, , B, , x5, , – +, 6V, , x6, , x2, , 200, , Figure Ex-2, , 175, , 6., , +2V, –, , 6, , 3. The accompanying figure shows a network of one-way streets, with traffic flowing in the directions indicated. The flow rates, along the streets are measured as the average number of vehicles per hour., , I2, 4, , (a) Set up a linear system whose solution provides the unknown flow rates., , I1, I3, , –, 1V+, , 2, , (b) Solve the system for the unknown flow rates., (c) If the flow along the road from A to B must be reduced for, construction, what is the minimum flow that is required to, keep traffic flowing on all roads?, 400, 300, , 250, , I1 20 , , x2, , I2 20 , , x4, x1, 300, , I3, , 20 , , 200, , B, , I5, , I6, , A, , 400, 100, , I4, +, 10 V –, , 750, x3, , 20 , , 7., , Figure Ex-3, , 8., , 5V, + –, , 3, I1, 4, , 4. The accompanying figure shows a network of one-way streets, with traffic flowing in the directions indicated. The flow rates, along the streets are measured as the average number of vehicles per hour., , – +, 4V, , (a) Set up a linear system whose solution provides the unknown flow rates., , – +, 3V, , I2, 5, I3, , –, 10 V, +

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1.9 Applications of Linear Systems, , In Exercises 9–12, write a balanced equation for the given, chemical reaction., 9. C3 H8 + O2 → CO2 + H2 O, , (propane combustion), , 10. C6 H12 O6 → CO2 + C2 H5 OH, , (fermentation of sugar), , 11. CH3 COF + H2 O → CH3 COOH + HF, 12. CO2 + H2 O → C6 H12 O6 + O2, , (photosynthesis), , 13. Find the quadratic polynomial whose graph passes through, the points (1, 1), (2, 2), and (3, 5)., 14. Find the quadratic polynomial whose graph passes through, the points (0, 0), (−1, 1), and (1, 1)., 15. Find the cubic polynomial whose graph passes through the, points (−1, −1), (0, 1), (1, 3), (4, −1)., 16. The accompanying figure shows the graph of a cubic polynomial. Find the polynomial., 10, 9, 8, 7, 6, 5, 4, 3, 2, 1, , 95, , (b) When a current passes through a resistor, there is an increase, in the electrical potential in a circuit., (c) Kirchhoff’s current law states that the sum of the currents, flowing into a node equals the sum of the currents flowing out, of the node., (d) A chemical equation is called balanced if the total number of, atoms on each side of the equation is the same., (e) Given any n points in the xy-plane, there is a unique polynomial of degree n − 1 or less whose graph passes through those, points., , Working withTechnology, T1. The following table shows the lifting force on an aircraft wing, measured in a wind tunnel at various wind velocities. Model the, data with an interpolating polynomial of degree 5, and use that, polynomial to estimate the lifting force at 2000 ft/s., Velocity, (100 ft/s), Lifting Force, (100 lb), , 1, , 2, , 4, , 8, , 16, , 32, , 0, , 3.12, , 15.86, , 33.7, , 81.5, , 123.0, , T2. (Calculus required ) Use the method of Example 7 to approximate the integral, , , , 1, , 2, , ex dx, 1, , 2, , 3, , 4, , 5, , 6, , 7, , 8, , 0, , Figure Ex-16, , 17. (a) Find an equation that represents the family of all seconddegree polynomials that pass through the points (0, 1), and (1, 2). [Hint: The equation will involve one arbitrary parameter that produces the members of the family, when varied.], (b) By hand, or with the help of a graphing utility, sketch, four curves in the family., 18. In this section we have selected only a few applications of linear systems. Using the Internet as a search tool, try to find, some more real-world applications of such systems. Select one, that is of interest to you, and write a paragraph about it., , by subdividing the interval of integration into five equal parts and, using an interpolating polynomial to approximate the integrand., Compare your answer to that obtained using the numerical integration capability of your technology utility., T3. Use the method of Example 5 to balance the chemical equation, Fe2 O3 + Al → Al2 O3 + Fe, , (Fe = iron, Al = aluminum, O = oxygen), T4. Determine the currents in the accompanying circuit., 20 V, + –, I2, , True-False Exercises, TF. In parts (a)–(e) determine whether the statement is true or, false, and justify your answer., (a) In any network, the sum of the flows out of a node must equal, the sum of the flows into a node., , I3, , 3, , 470 , , I3, , I2, , I1, , I1, + –, 12 V, , 2

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96, , Chapter 1 Systems of Linear Equations and Matrices, , 1.10 Leontief Input-Output Models, In 1973 the economist Wassily Leontief was awarded the Nobel prize for his work on, economic modeling in which he used matrix methods to study the relationships among, different sectors in an economy. In this section we will discuss some of the ideas developed, by Leontief., , Inputs and Outputs in an, Economy, , Manufacturing, , Agriculture, , Open, Sector, , Utilities, , Figure 1.10.1, , Leontief Model of an Open, Economy, , One way to analyze an economy is to divide it into sectors and study how the sectors, interact with one another. For example, a simple economy might be divided into three, sectors—manufacturing, agriculture, and utilities. Typically, a sector will produce certain outputs but will require inputs from the other sectors and itself. For example, the, agricultural sector may produce wheat as an output but will require inputs of farm machinery from the manufacturing sector, electrical power from the utilities sector, and food, from its own sector to feed its workers. Thus, we can imagine an economy to be a network in which inputs and outputs flow in and out of the sectors; the study of such flows, is called input-output analysis. Inputs and outputs are commonly measured in monetary, units (dollars or millions of dollars, for example) but other units of measurement are, also possible., The flows between sectors of a real economy are not always obvious. For example,, in World War II the United States had a demand for 50,000 new airplanes that required, the construction of many new aluminum manufacturing plants. This produced an unexpectedly large demand for certain copper electrical components, which in turn produced, a copper shortage. The problem was eventually resolved by using silver borrowed from, Fort Knox as a copper substitute. In all likelihood modern input-output analysis would, have anticipated the copper shortage., Most sectors of an economy will produce outputs, but there may exist sectors that, consume outputs without producing anything themselves (the consumer market, for, example). Those sectors that do not produce outputs are called open sectors. Economies, with no open sectors are called closed economies, and economies with one or more open, sectors are called open economies (Figure 1.10.1). In this section we will be concerned with, economies with one open sector, and our primary goal will be to determine the output, levels that are required for the productive sectors to sustain themselves and satisfy the, demand of the open sector., Let us consider a simple open economy with one open sector and three product-producing, sectors: manufacturing, agriculture, and utilities. Assume that inputs and outputs are, measured in dollars and that the inputs required by the productive sectors to produce, one dollar’s worth of output are in accordance with Table 1., , Wassily Leontief, (1906–1999), , Historical Note It is somewhat ironic that it was, the Russian-born Wassily Leontief who won the Nobel prize in 1973 for pioneering the modern methods for analyzing free-market economies. Leontief, was a precocious student who entered the University, of Leningrad at age 15. Bothered by the intellectual, restrictions of the Soviet system, he was put in jail, for anti-Communist activities, after which he headed, for the University of Berlin, receiving his Ph.D. there, in 1928. He came to the United States in 1931, where, he held professorships at Harvard and then New York, University., [Image: © Bettmann/CORBIS]

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1.10 Leontief Input-Output Models, , 97, , Table 1, , Provider, , Input Required per Dollar Output, Manufacturing, , Agriculture, , Utilities, , Manufacturing, , $ 0.50, , $ 0.10, , $ 0.10, , Agriculture, , $ 0.20, , $ 0.50, , $ 0.30, , Utilities, , $ 0.10, , $ 0.30, , $ 0.40, , Usually, one would suppress the labeling and express this matrix as, , ⎡, , 0.5, ⎣, C = 0 .2, 0 .1, , ⎤, , 0.1, 0.5, 0.3, , 0.1, 0.3⎦, 0.4, , (1), , This is called the consumption matrix (or sometimes the technology matrix) for the economy. The column vectors, , ⎡, , ⎤, , ⎡, , ⎡, , ⎤, , ⎤, , 0.5, 0.1, 0.1, ⎣, ⎣, ⎣, ⎦, ⎦, c1 = 0.2 , c2 = 0.5 , c3 = 0.3⎦, 0 .1, 0.3, 0 .4, , What is the economic significance of the row sums of the, consumption matrix?, , in C list the inputs required by the manufacturing, agricultural, and utilities sectors,, respectively, to produce $1.00 worth of output. These are called the consumption vectors, of the sectors. For example, c1 tells us that to produce $1.00 worth of output the manufacturing sector needs $0.50 worth of manufacturing output, $0.20 worth of agricultural, output, and $0.10 worth of utilities output., Continuing with the above example, suppose that the open sector wants the economy, to supply it manufactured goods, agricultural products, and utilities with dollar values:, , d1 dollars of manufactured goods, d2 dollars of agricultural products, d3 dollars of utilities, The column vector d that has these numbers as successive components is called the outside, demand vector. Since the product-producing sectors consume some of their own output,, the dollar value of their output must cover their own needs plus the outside demand., Suppose that the dollar values required to do this are, , x1 dollars of manufactured goods, x2 dollars of agricultural products, x3 dollars of utilities, The column vector x that has these numbers as successive components is called the, production vector for the economy. For the economy with consumption matrix (1), that, portion of the production vector x that will be consumed by the three productive sectors is, , ⎡, , ⎤, , 0. 5, ⎢ ⎥, x1 ⎣0.2⎦, 0 .1, Fractions, consumed by, manufacturing, , ⎡, , ⎤, , 0. 1, ⎢ ⎥, + x2 ⎣0.5⎦, 0 .3, Fractions, consumed by, agriculture, , ⎡, , ⎤, , ⎡, , 0.1, 0.5, ⎢ ⎥ ⎢, + x3 ⎣0.3⎦ = ⎣0.2, 0.4, 0.1, Fractions, consumed, by utilities, , 0.1, 0.5, 0.3, , ⎤⎡ ⎤, , x1, 0.1, ⎥⎢ ⎥, 0.3⎦ ⎣x2 ⎦ = C x, 0.4, x3

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98, , Chapter 1 Systems of Linear Equations and Matrices, , The vector C x is called the intermediate demand vector for the economy. Once the, intermediate demand is met, the portion of the production that is left to satisfy the, outside demand is x − C x. Thus, if the outside demand vector is d, then x must satisfy, the equation, x, −, Cx, =, d, Amount, produced, , Intermediate, demand, , Outside, demand, , which we will find convenient to rewrite as, , (I − C)x = d, , (2), , The matrix I − C is called the Leontief matrix and (2) is called the Leontief equation., E X A M P L E 1 Satisfying Outside Demand, , Consider the economy described in Table 1. Suppose that the open sector has a demand, for $7900 worth of manufacturing products, $3950 worth of agricultural products, and, $1975 worth of utilities., (a) Can the economy meet this demand?, (b) If so, find a production vector x that will meet it exactly., Solution The consumption matrix, production vector, and outside demand vector are, , ⎡, , 0.5, ⎢, C = ⎣0.2, 0 .1, , ⎤, , 0.1, 0.5, 0.3, , ⎡ ⎤, , ⎡, , ⎤, , 0.1, 7900, x1, ⎥, ⎢ ⎥, ⎢, ⎥, 0.3⎦ , x = ⎣x2 ⎦ , d = ⎣3950⎦, 1975, x3, 0.4, , (3), , To meet the outside demand, the vector x must satisfy the Leontief equation (2), so the, problem reduces to solving the linear system, , ⎡, , 0.5, ⎢, ⎣−0.2, −0.1, , ⎤⎡ ⎤ ⎡, ⎤, x1, 7900, −0.1 −0.1, ⎥⎢ ⎥ ⎢, ⎥, 0.5 −0.3⎦ ⎣x2 ⎦ = ⎣3950⎦, 1975, −0.3, x3, 0.6, , I −C, , x, , (4), , d, , (if consistent). We leave it for you to show that the reduced row echelon form of the, augmented matrix for this system is, , ⎡, , 1, ⎢, 0, ⎣, 0, , 0, 1, 0, , 0, 0, 1, , ⎤, , 27,500, ⎥, 33,750⎦, 24,750, , This tells us that (4) is consistent, and the economy can satisfy the demand of the open, sector exactly by producing $27,500 worth of manufacturing output, $33,750 worth of, agricultural output, and $24,750 worth of utilities output., Productive Open, Economies, , In the preceding discussion we considered an open economy with three product-producing, sectors; the same ideas apply to an open economy with n product-producing sectors. In, this case, the consumption matrix, production vector, and outside demand vector have, the form, ⎡, ⎤, ⎡ ⎤, ⎡ ⎤, , c11, ⎢c21, ⎢, C = ⎢ .., ⎣ ., cn1, , c12, c22, .., ., , ···, ···, , c n2, , · · · cnn, , x1, d1, c1n, ⎢ x2 ⎥, ⎢d2 ⎥, c2n ⎥, ⎢ ⎥, ⎢ ⎥, ⎥, .. ⎥ , x = ⎢ .. ⎥ , d = ⎢ .. ⎥, ⎣, ⎣.⎦, ⎦, ⎦, ., ., xn, , dn

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1.10 Leontief Input-Output Models, , 99, , where all entries are nonnegative and, , cij = the monetary value of the output of the i th sector that is needed by the j th, sector to produce one unit of output, , xi = the monetary value of the output of the i th sector, di = the monetary value of the output of the i th sector that is required to meet, the demand of the open sector, , Remark Note that the j th column vector of C contains the monetary values that the j th sector, requires of the other sectors to produce one monetary unit of output, and the i th row vector of C, contains the monetary values required of the i th sector by the other sectors for each of them to, produce one monetary unit of output., , As discussed in our example above, a production vector x that meets the demand d, of the outside sector must satisfy the Leontief equation, , (I − C)x = d, If the matrix I − C is invertible, then this equation has the unique solution, x = (I − C)−1 d, , (5), , for every demand vector d. However, for x to be a valid production vector it must, have nonnegative entries, so the problem of importance in economics is to determine, conditions under which the Leontief equation has a solution with nonnegative entries., It is evident from the form of (5) that if I − C is invertible, and if (I − C)−1 has nonnegative entries, then for every demand vector d the corresponding x will also have nonnegative entries, and hence will be a valid production vector for the economy. Economies, for which (I − C)−1 has nonnegative entries are said to be productive. Such economies, are desirable because demand can always be met by some level of production. The following theorem, whose proof can be found in many books on economics, gives conditions, under which open economies are productive., , THEOREM 1.10.1 If, , C is the consumption matrix for an open economy, and if all of, the column sums are less than 1, then the matrix I − C is invertible, the entries of, (I − C)−1 are nonnegative, and the economy is productive., , Remark The j th column sum of C represents the total dollar value of input that the j th sector, requires to produce $1 of output, so if the j th column sum is less than 1, then the j th sector, requires less than $1 of input to produce $1 of output; in this case we say that the j th sector is, profitable. Thus, Theorem 1.10.1 states that if all product-producing sectors of an open economy, are profitable, then the economy is productive. In the exercises we will ask you to show that an, open economy is productive if all of the row sums of C are less than 1 (Exercise 11). Thus, an open, economy is productive if either all of the column sums or all of the row sums of C are less than 1., , E X A M P L E 2 An Open Economy Whose Sectors Are All Profitable, , The column sums of the consumption matrix C in (1) are less than 1, so (I − C)−1 exists, and has nonnegative entries. Use a calculating utility to confirm this, and use this inverse, to solve Equation (4) in Example 1.

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100, , Chapter 1 Systems of Linear Equations and Matrices, Solution We leave it for you to show that, , ⎡, , (I − C)−1, , 2.65823, ⎣, ≈ 1.89873, 1.39241, , This matrix has nonnegative entries, and, , ⎡, , 2.65823, x = (I − C)−1 d ≈ ⎣1.89873, 1.39241, , 1.13924, 3.67089, 2.02532, , ⎤, , 1.13924, 3.67089, 2.02532, , 1.01266, 2.15190⎦, 2.91139, , ⎤⎡, , ⎤, , ⎡, , ⎤, , 1.01266, 7900, 27,500, 2.15190⎦ ⎣3950⎦ ≈ ⎣33,750⎦, 2.91139, 1975, 24,750, , which is consistent with the solution in Example 1., , Exercise Set 1.10, , (a) Construct a consumption matrix for this economy., (b) How much must M and B each produce to provide customers with $7000 worth of mechanical work and $14,000, worth of body work?, 2. A simple economy produces food (F ) and housing (H ). The, production of $1.00 worth of food requires $0.30 worth of, food and $0.10 worth of housing, and the production of $1.00, worth of housing requires $0.20 worth of food and $0.60 worth, of housing., (a) Construct a consumption matrix for this economy., (b) What dollar value of food and housing must be produced, for the economy to provide consumers $130,000 worth of, food and $130,000 worth of housing?, , 3. Consider the open economy described by the accompanying, table, where the input is in dollars needed for $1.00 of output., , 4. A company produces Web design, software, and networking, services. View the company as an open economy described by, the accompanying table, where input is in dollars needed for, $1.00 of output., (a) Find the consumption matrix for the company., (b) Suppose that the customers (the open sector) have a demand for $5400 worth of Web design, $2700 worth of software, and $900 worth of networking. Use row reduction, to find a production vector that will meet this demand, exactly., Table Ex-4, Input Required per Dollar Output, , Provider, , 1. An automobile mechanic (M ) and a body shop (B ) use each, other’s services. For each $1.00 of business that M does, it, uses $0.50 of its own services and $0.25 of B ’s services, and, for each $1.00 of business that B does it uses $0.10 of its own, services and $0.25 of M ’s services., , Web Design, , Software, , Networking, , Web Design, , $ 0.40, , $ 0.20, , $ 0.45, , Software, , $ 0.30, , $ 0.35, , $ 0.30, , Networking, , $ 0.15, , $ 0.10, , $ 0.20, , In Exercises 5–6, use matrix inversion to find the production, vector x that meets the demand d for the consumption matrix C., , (a) Find the consumption matrix for the economy., , 5. C =, , (b) Suppose that the open sector has a demand for $1930, worth of housing, $3860 worth of food, and $5790 worth, of utilities. Use row reduction to find a production vector, that will meet this demand exactly., , 0.1, 0.5, , 0.3, 50, ; d=, 0. 4, 60, , 6. C =, , 0.3, 0.3, , 0. 1, 22, ; d=, 14, 0. 7, , 1, , Table Ex-3, Input Required per Dollar Output, Housing, , Provider, , 7. Consider an open economy with consumption matrix, , Food, , Utilities, , Housing, , $ 0.10, , $ 0.60, , $ 0.40, , Food, , $ 0.30, , $ 0.20, , $ 0.30, , Utilities, , $ 0.40, , $ 0.10, , $ 0.20, , C=, , , , 2, , 0, , 0, , 1, , (a) Show that the economy can meet a demand of d1 = 2 units, from the first sector and d2 = 0 units from the second sector, but it cannot meet a demand of d1 = 2 units from the, first sector and d2 = 1 unit from the second sector., (b) Give both a mathematical and an economic explanation, of the result in part (a).

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Chapter 1 Supplementary Exercises, , 8. Consider an open economy with consumption matrix, , ⎡1, , ⎤, 1, , 1, 4, , 2, , ⎢1, C=⎢, ⎣2, 1, 2, , 1, 8, 1, 4, , 1, 8, , True-False Exercises, TF. In parts (a)–(e) determine whether the statement is true or, false, and justify your answer., , 4, , ⎥, 1⎥, 4⎦, , 101, , (a) Sectors of an economy that produce outputs are called open, sectors., , If the open sector demands the same dollar value from each, product-producing sector, which such sector must produce the, greatest dollar value to meet the demand? Is the economy productive?, , (b) A closed economy is an economy that has no open sectors., (c) The rows of a consumption matrix represent the outputs in a, sector of an economy., , 9. Consider an open economy with consumption matrix, , c11, C=, c21, , (d) If the column sums of the consumption matrix are all less than, 1, then the Leontief matrix is invertible., , c12, 0, , Show that the Leontief equation x − C x = d has a unique, solution for every demand vector d if c21 c12 < 1 − c11 ., , Working with Proofs, 10. (a) Consider an open economy with a consumption matrix, C whose column sums are less than 1, and let x be the, production vector that satisfies an outside demand d; that, is, (I − C)−1 d = x. Let dj be the demand vector that is, obtained by increasing the j th entry of d by 1 and leaving, the other entries fixed. Prove that the production vector, xj that meets this demand is, , (e) The Leontief equation relates the production vector for an, economy to the outside demand vector., , Working withTechnology, T1. The following table describes an open economy with three sectors in which the table entries are the dollar inputs required to produce one dollar of output. The outside demand during a 1-week, period if $50,000 of coal, $75,000 of electricity, and $1,250,000, of manufacturing. Determine whether the economy can meet the, demand., Input Required per Dollar Output, Electricity, , (b) In words, what is the economic significance of the j th column vector of (I − C)−1 ? [Hint: Look at xj − x.], 11. Prove: If C is an n × n matrix whose entries are nonnegative, and whose row sums are less than 1, then I − C is invertible, and has nonnegative entries. [Hint: (AT )−1 = (A−1 )T for any, invertible matrix A.], , Provider, , xj = x + j th column vector of (I − C)−1, , Coal, , Manufacturing, , Electricity, , $ 0.1, , $ 0.25, , $ 0.2, , Coal, , $ 0.3, , $ 0.4, , $ 0.5, , Manufacturing, , $ 0.1, , $ 0.15, , $ 0.1, , Chapter 1 Supplementary Exercises, In Exercises 1–4 the given matrix represents an augmented, matrix for a linear system. Write the corresponding set of linear, equations for the system, and use Gaussian elimination to solve, the linear system. Introduce free parameters as necessary., , , 1., , 3, , −1, , 0, , 4, , 1, , 2, , 0, , 3, , 3, , −1, , ⎡, , 2, , −4, , 1, , 3. ⎣−4, 0, , 0, , 3, , 1, , −1, , ⎢, , 6, , ⎤, , ⎥, −1⎦, 3, , , , ⎡, , 1, , 4, , −1, , ⎤, , 5. Use Gauss–Jordan elimination to solve for x and y in terms, of x and y ., , x = 35 x − 45 y, y = 45 x + 35 y, , ⎢−2, ⎢, 2. ⎢, ⎣ 3, , −8, 12, , ⎥, −3⎦, , 6. Use Gauss–Jordan elimination to solve for x and y in terms, of x and y ., x = x cos θ − y sin θ, , 0, , 0, , 0, , y = x sin θ + y cos θ, , 3, , 1, , −2, , 4. ⎣−9, 6, , −3, , ⎡, ⎢, , 2, , 2⎥, ⎥, , ⎤, ⎥, , 6⎦, 1, , 7. Find positive integers that satisfy, , x+ y+, , z= 9, , x + 5y + 10z = 44

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102, , Chapter 1 Systems of Linear Equations and Matrices, , 8. A box containing pennies, nickels, and dimes has 13 coins with, a total value of 83 cents. How many coins of each type are in, the box? Is the economy productive?, , ⎡, , 9. Let, , a, ⎢, ⎣a, 0, , ⎤, , 0, , b, , 2, , a, a, , 4, , 4⎦, , 2, , b, , 15. Find values of a , b, and c such that the graph of the polynomial p(x) = ax 2 + bx + c passes through the points (1, 2),, (−1, 6), and (2, 3)., 16. (Calculus required ) Find values of a , b, and c such that, the graph of p(x) = ax 2 + bx + c passes through the point, (−1, 0) and has a horizontal tangent at (2, −9)., , ⎥, , be the augmented matrix for a linear system. Find for what, values of a and b the system has, , 17. Let Jn be the n × n matrix each of whose entries is 1. Show, that if n > 1, then, , (I − Jn )−1 = I −, , (a) a unique solution., (b) a one-parameter solution., (c) a two-parameter solution., , x1 + x 2 + x3 = 4, (a 2 − 4)x3 = a − 2, ⎤, , , 2, ⎥, 3⎦, B =, , ⎢, A = ⎣−2, , 0, , −2, ⎡, , 1, , 8, , −6, , 6, , ⎢, C=⎣ 6, −4, , A3 + 4A2 − 2A + 7I = 0, then so does AT ., , 20. Prove: If A is invertible, then A + B and I + BA−1 are both, invertible or both not invertible., , 11. Find a matrix K such that AKB = C given that, 4, , −1, , , , 0, , 0, , 1, , −1, , ,, , 21. Prove: If A is an m × n matrix and B is the n × 1 matrix each, of whose entries is 1/n, then, , ⎡, , ⎤, r1, ⎢r2 ⎥, ⎢ ⎥, AB = ⎢ . ⎥, ⎣ .. ⎦, , ⎤, ⎥, , 1⎦, , 0, , rm, , 0, , 12. How should the coefficients a , b, and c be chosen so that the, system, ax + by − 3z = −3, , where r i is the average of the entries in the i th row of A., 22. (Calculus required ) If the entries of the matrix, , ⎡, , −2x − by + cz = −1, ax + 3y − cz = −3, has the solution x = 1, y = −1, and z = 2?, 13. In each part, solve the matrix equation for X ., , ⎡, , ⎤, , , , −1, , 0, , 1, , (a) X ⎣ 1, 3, , 1, , 0⎦ =, , ⎢, , , (b) X, , , (c), , 1, , −1, , , 1, , −1, , 2, , 3, , 0, , 1, , 3, , 1, , −1, , 2, , , , ⎥, , X−X, , , =, , , , 1, , 2, , 0, , −3, , 1, , 5, , −5, , −1, −3, , , 6, , 1, , 4, , 2, , 0, , , , =, , , , c11 (x), ⎢ c21 (x), ⎢, C=⎢ ., ⎣ .., , c12 (x), c22 (x), .., ., , ···, ···, , cm1 (x), , cm2 (x), , ···, , , , 0, 7, , 2, , −2, , 5, , 4, , , , (a) Show that (I − A)−1 = I + A + A2 + A3 if A4 = 0., (b) Show that, , (I − A)−1 = I + A + A2 + · · · + An, , ⎤, c1n (x), c2n (x) ⎥, ⎥, .. ⎥, . ⎦, cmn (x), , are differentiable functions of x , then we define, , ⎡, , 14. Let A be a square matrix., , if An+1 = 0., , Jn, , 19. Prove: If B is invertible, then AB −1 = B −1 A if and only if, AB = BA., , x3 = 2, , 1, , n−1, , 18. Show that if a square matrix A satisfies, , (d) no solution., , 10. For which value(s) of a does the following system have zero, solutions? One solution? Infinitely many solutions?, , ⎡, , 1, , c11 (x), ⎢ c (x), dC, ⎢ 21, =⎢ ., ⎣ .., dx, , c12 (x), c22 (x), .., ., , ···, ···, , cm1 (x), , cm2 (x), , ···, , ⎤, c1n (x), c2n (x) ⎥, ⎥, .. ⎥, . ⎦, cmn (x), , Show that if the entries in A and B are differentiable functions of x and the sizes of the matrices are such that the stated, operations can be performed, then, , dA, d, (kA) = k, dx, dx, dA dB, d, (A + B) =, +, (b), dx, dx, dx, dA, dB, d, (AB) =, B +A, (c), dx, dx, dx, (a)

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Chapter 1 Supplementary Exercises, , 23. (Calculus required ) Use part (c) of Exercise 22 to show that, −1, , dA, dx, , = −A−1, , dA −1, A, dx, , State all the assumptions you make in obtaining this formula., 24. Assuming that the stated inverses exist, prove the following, equalities., (a) (C, , −1, , −1 −1, , +D ), , (a) Confirm that the sizes of all matrices are such that the, product AB can be obtained using Formula (∗)., (b) Confirm that the result obtained using Formula (∗) agrees, with that obtained using ordinary matrix multiplication., 26. Suppose that an invertible matrix A is partitioned as, , A=, , −1, , = C(C + D) D, , −1, , , A11, , A12, , A21, , A22, , −1, , (b) (I + CD) C = C(I + DC), , Show that, , (c) (C + DD T )−1 D = C −1 D(I + D T C −1 D)−1, , −1, , A, Partitioned matrices can be multiplied by the row-column rule, just as if the matrix entries were numbers provided that the sizes, of all matrices are such that the necessary operations can be performed. Thus, for example, if A is partitioned into a 2 × 2 matrix, and B into a 2 × 1 matrix, then, , , , AB =, , A11, , A12, , A21, , A22, , , , B1, , , , , , =, , B2, , A11 B1 + A12 B2, , , , A21 B1 + A22 B2, , 1, , ⎢, 4, A=⎢, ⎣, , 0, , ⎡, , 3, , ⎢, ⎢2, ⎢, ⎢, B = ⎢4, ⎢, ⎢, ⎣0, 2, , 0, , 2, , 1, , 1, , 0, , 3, , −3, ⎤, , 4, , 2, , 0, , ⎥,  , ⎥, B1, −1⎥, ⎥=, ⎥, B2, ⎥, 3⎦, 1⎥, , 5, , 4, , ⎤, , , , ⎥, A11, −1⎥ =, ⎦, A21, −2, , A12, A22, , =, , , , , B11, , B12, , B21, , B22, , , , where, −1, B11 = (A11 − A12 A22, A21 )−1 ,, −1, B21 = −A22, A21 B11 ,, , 1, B12 = −B11 A12 A−, 22, , −1, B22 = (A22 − A21 A11, A12 )−1, , provided all the inverses in these formulas exist., (*), , provided that the sizes are such that AB , the two sums, and the, four products are all defined., 25. Let A and B be the following partitioned matrices., , ⎡, , 103, , , , 27. In the special case where matrix A21 in Exercise 26 is zero, the, matrix A simplifies to, , A=, , , A11, 0, , A12, , , , A22, , which is said to be in block upper triangular form. Use the, result of Exercise 26 to show that in this case, , , , A−1 =, , −1, A11, , −1, 1, −A−, 11 A12 A22, , 0, , 1, A−, 22, , , , 28. A linear system whose coefficient matrix has a pivot position, in every row must be consistent. Explain why this must be so., 29. What can you say about the consistency or inconsistency of, a linear system of three equations in five unknowns whose, coefficient matrix has three pivot columns?

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CHAPTER, , 2, , Determinants, CHAPTER CONTENTS, , 2.1 Determinants by Cofactor Expansion, , 105, , 2.2 Evaluating Determinants by Row Reduction, 2.3 Properties of Determinants; Cramer’s Rule, INTRODUCTION, , 113, 118, , In this chapter we will study “determinants” or, more precisely, “determinant, functions.” Unlike real-valued functions, such as f(x) = x 2 , that assign a real number, to a real variable x , determinant functions assign a real number f(A) to a matrix, variable A. Although determinants first arose in the context of solving systems of, linear equations, they are rarely used for that purpose in real-world applications. While, they can be useful for solving very small linear systems (say two or three unknowns),, our main interest in them stems from the fact that they link together various concepts, in linear algebra and provide a useful formula for the inverse of a matrix., , 2.1 Determinants by Cofactor Expansion, In this section we will define the notion of a “determinant.” This will enable us to develop a, specific formula for the inverse of an invertible matrix, whereas up to now we have had only, a computational procedure for finding it. This, in turn, will eventually provide us with a, formula for solutions of certain kinds of linear systems., , Recall from Theorem 1.4.5 that the 2 × 2 matrix, , A=, WARNING It is important to, , keep in mind that det(A) is a, number, whereas A is a matrix., , a, c, , b, d, , is invertible if and only if ad − bc  = 0 and that the expression ad − bc is called the, determinant of the matrix A. Recall also that this determinant is denoted by writing, , , a, det(A) = ad − bc or , c, , , b , = ad − bc, d, , (1), , and that the inverse of A can be expressed in terms of the determinant as, , A −1 =, Minors and Cofactors, , d, 1, det(A) −c, , −b, a, , (2), , One of our main goals in this chapter is to obtain an analog of Formula (2) that is, applicable to square matrices of all orders. For this purpose we will find it convenient, to use subscripted entries when writing matrices or determinants. Thus, if we denote a, 2 × 2 matrix as, , A=, , a11, a21, , a12, a22, , 105

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106, , Chapter 2 Determinants, , then the two equations in (1) take the form, , , a11, a21, , , a12 , = a11 a22 − a12 a21, a22 , , det(A) = , , (3), , In situations where it is inconvenient to assign a name to the matrix, we can express this, formula as, det, , a11, a21, , a12, = a11 a22 − a12 a21, a22, , (4), , There are various methods for defining determinants of higher-order square matrices., In this text, we will us an “inductive definition” by which we mean that the determinant, of a square matrix of a given order will be defined in terms of determinants of square, matrices of the next lower order. To start the process, let us define the determinant of a, 1 × 1 matrix [a11 ] as, (5), det [a11 ] = a11, from which it follows that Formula (4) can be expressed as, det, , a11, a21, , a12, = det[a11 ] det[a22 ] − det[a12 ] det[a21 ], a22, , Now that we have established a starting point, we can define determinants of 3 × 3, matrices in terms of determinants of 2 × 2 matrices, then determinants of 4 × 4 matrices, in terms of determinants of 3 × 3 matrices, and so forth, ad infinitum. The following, terminology and notation will help to make this inductive process more efficient., , A is a square matrix, then the minor of entry aij is denoted by Mij, and is defined to be the determinant of the submatrix that remains after the i th row, and j th column are deleted from A. The number (−1)i+j Mij is denoted by Cij and, is called the cofactor of entry aij ., DEFINITION 1 If, , E X A M P L E 1 Finding Minors and Cofactors, , ⎡, , Let, , 3, ⎢, A = ⎣2, 1, WARNING We have followed, , 1, 5, 4, , −4, , ⎤, ⎥, , 6⎦, 8, , The minor of entry a11 is, , the standard convention of using capital letters to denote, minors and cofactors even, though they are numbers, not, matrices., , M11 =, , The cofactor of a11 is, , 3, 2, 1, , 1, 5, 4, , 4, 6, 8, , =, , 5, 4, , 6, = 16, 8, , C11 = (−1)1+1 M11 = M11 = 16, , Historical Note The term determinant was first introduced by the German mathematician Carl, Friedrich Gauss in 1801 (see p. 15), who used them to “determine” properties of certain kinds of, functions. Interestingly, the term matrix is derived from a Latin word for “womb” because it was, viewed as a container of determinants.

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2.1 Determinants by Cofactor Expansion, , 107, , Similarly, the minor of entry a32 is, , 3, M32 = 2, 1, The cofactor of a32 is, , 1, 5, 4, , 4, 3, 6 =, 2, 8, , 4, = 26, 6, , C32 = (−1)3+2 M32 = −M32 = −26, , Remark Note that a minor Mij and its corresponding cofactor Cij are either the same or negatives, of each other and that the relating sign (−1)i+j is either +1 or −1 in accordance with the pattern, in the “checkerboard” array, ⎤, ⎡, +, ⎢−, ⎢, ⎢, ⎢+, ⎢, ⎢−, ⎣, .., ., , −, +, −, +, .., ., , +, −, +, −, .., ., , −, +, −, +, .., ., , +, −, +, −, .., ., , ···, · · ·⎥, ⎥, ⎥, · · ·⎥, ⎥, · · ·⎥, ⎦, , For example,, , C11 = M11 , C21 = −M21 , C22 = M22, and so forth. Thus, it is never really necessary to calculate (−1)i+j to calculate Cij —you can simply, compute the minor Mij and then adjust the sign in accordance with the checkerboard pattern. Try, this in Example 1., , E X A M P L E 2 Cofactor Expansions of a 2 × 2 Matrix, , The checkerboard pattern for a 2 × 2 matrix A = [aij ] is, , +, −, so that, , −, +, , C11 = M11 = a22, C21 = −M21 = −a12, , C12 = −M12 = −a21, C22 = M22 = a11, We leave it for you to use Formula (3) to verify that det(A) can be expressed in terms of, cofactors in the following four ways:, , , a11, , det(A) = , , a21, , , a12 , a22 , , = a11 C11 + a12 C12, = a21 C21 + a22 C22, = a11 C11 + a21 C21, = a12 C12 + a22 C22, , (6), , Each of the last four equations is called a cofactor expansion of det(A). In each cofactor, expansion the entries and cofactors all come from the same row or same column of A., Historical Note The term minor is apparently due to the English mathematician James Sylvester (see, p. 35), who wrote the following in a paper published in 1850: “Now conceive any one line and any one, column be struck out, we get…a square, one term less in breadth and depth than the original square;, and by varying in every possible selection of the line and column excluded, we obtain, supposing, the original square to consist of n lines and n columns, n2 such minor squares, each of which will, represent what I term a “First Minor Determinant” relative to the principal or complete determinant.”

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108, , Chapter 2 Determinants, , For example, in the first equation the entries and cofactors all come from the first row of, A, in the second they all come from the second row of A, in the third they all come from, the first column of A, and in the fourth they all come from the second column of A., Definition of a General, Determinant, , Formula (6) is a special case of the following general result, which we will state without, proof., THEOREM 2.1.1 If A is an n × n matrix, then regardless of which row or column of A, , is chosen, the number obtained by multiplying the entries in that row or column by the, corresponding cofactors and adding the resulting products is always the same., , This result allows us to make the following definition., DEFINITION 2 If A is an n × n matrix, then the number obtained by multiplying the, , entries in any row or column of A by the corresponding cofactors and adding the, resulting products is called the determinant of A, and the sums themselves are called, cofactor expansions of A. That is,, det(A) = a1j C1j + a2j C2j + · · · + anj Cnj, , (7), , [cofactor expansion along the jth column], , and, , det(A) = ai 1 Ci 1 + ai 2 Ci 2 + · · · + ain Cin, [cofactor expansion along the ith row], , E X A M P L E 3 Cofactor Expansion Along the First Row, , Find the determinant of the matrix, , ⎡, , 3, A = ⎣−2, 5, , 1, −4, 4, , ⎤, , 0, 3⎦, −2, , by cofactor expansion along the first row., , Historical Note Cofactor expansion is not, the only method for expressing the determinant of a matrix in terms of determinants, of lower order. For example, although it is, not well known, the English mathematician, Charles Dodgson, who was the author of Alice’s Adventures in Wonderland and Through, the Looking Glass under the pen name of, Lewis Carroll, invented such a method, called, condensation. That method has recently been, resurrected from obscurity because of its suitability for parallel processing on computers., [Image: Oscar G. Rejlander/, Time & Life Pictures/Getty Images], , Charles Lutwidge Dodgson, (Lewis Carroll), (1832–1898), , (8)

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2.1 Determinants by Cofactor Expansion, Solution, , ,  3, , det(A) =  −2,  5, , 1, −4, 4, , , , , 0, ,  −4, , 3  = 3 , 4, −2 , , , , , ,  −2, 3 , − 1 , , 5, −2, , , , , ,  −2, 3 , + 0 , , 5, −2, , 109, , , −4 , 4, , = 3(−4) − (1)(−11) + 0 = −1, E X A M P L E 4 Cofactor Expansion Along the First Column, Note that in Example 4 we had, to compute three cofactors,, whereas in Example 3 only two, were needed because the third, was multiplied by zero. As a, rule, the best strategy for cofactor expansion is to expand, along a row or column with the, most zeros., , Let A be the matrix in Example 3, and evaluate det(A) by cofactor expansion along the, first column of A., Solution, , ,  3, 1, , det(A) =  −2 −4,  5, 4, , , , , 0, ,  −4, 3  = 3 , 4, −2 , , , , , , 1, 3 , − (−2) , , 4, −2, , , , , ,  1, 0 , + 5 , , −2, −4, , , , 0 , 3, , = 3(−4) − (−2)(−2) + 5(3) = −1, This agrees with the result obtained in Example 3., , E X A M P L E 5 Smart Choice of Row or Column, , If A is the 4 × 4 matrix, , ⎤, , ⎡, , 1, 0, 0 −1, ⎢3, 1, 2, 2⎥, ⎥, ⎢, A=⎢, ⎥, 0 −2, 1⎦, ⎣1, 2, 0, 0, 1, then to find det(A) it will be easiest to use cofactor expansion along the second column,, since it has the most zeros:, , , 1, 0, , , det(A) = 1 · 1 −2, , 2, 0, , , −1, , 1, , 1, , For the 3 × 3 determinant, it will be easiest to use cofactor expansion along its second, column, since it has the most zeros:, , , , 1, det(A) = 1 · −2 · , 2, , , −1, 1, , = −2(1 + 2), = −6, E X A M P L E 6 Determinant of a Lower Triangular Matrix, , The following computation shows that the determinant of a 4 × 4 lower triangular matrix, is the product of its diagonal entries. Each part of the computation uses a cofactor, expansion along the first row., , , a11, , a,  21, , a31, , a41, , 0, , a22, a32, a42, , 0, 0, , a33, a43, , , , , , a22 0, , 0 , , , , , ,  = a11 a32 a33 0 , , , , a42 a43 a44 , , a44 , , , a33 0 , , = a11 a22 , a, a , 0, 0, 0, , 43, , 44, , = a11 a22 a33 |a44 | = a11 a22 a33 a44

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2.1 Determinants by Cofactor Expansion, , 111, , Exercise Set 2.1, In Exercises 1–2, find all the minors and cofactors of the matrix A., , ⎡, , ⎤, , −2, , 1, ⎢, 1. A = ⎣ 6, −3, , 3, ⎥, −1 ⎦, 4, , 7, 1, , 3. Let, , ⎡, , 4, ⎢0, ⎢, A=⎢, ⎣4, 4, , ⎡, , 1, ⎢, 2. A = ⎣3, 0, , −1, 0, 1, 1, , 1, −3, 0, 3, , ⎤, , 1, 3, 1, , 2, ⎥, 6⎦, 4, , ⎤, , 6, 3⎥, ⎥, ⎥, 14⎦, 2, , (d) M21 and C21 ., , 4. Let, , ⎡, , 2, ⎢−3, ⎢, A=⎢, ⎣ 3, 3, , 3, 2, −2, −2, , ⎤, , −1, , (d) M24 and C24 ., , √, , 7, −2, , 8., , 2, , 4, , √ , 6, √, 3, , In Exercises 9–14, use the arrow technique to evaluate the determinant., , , a − 3, 9. , −3, , −2, , , 11.  3, ,  1, , 3, , , 13. 2, , 1, , , 5 , a − 2, , , 4, , −7, , 2, , 1, 5, 6, 0, −1, 9, , , , , −2, , , 10.  5, ,  3, , 7, 1, 8, , 6, , −2, , 4, , , −1, , , 12.  3, ,  1, , 1, 0, 7, , 2, , −5, , 2, , , c, , , 14. 2, , 4, , 0, , 5, , −4, , , , 1, , λ+4, , 1, , c−1, , 0, , (f ) the third column., , (a) the first row., , (b) the first column., , (c) the second row., , (d) the second column., , (e) the third row., , (f ) the third column., , , , c, , 2, , 3, , 3, ⎢2, ⎢, 25. A = ⎢, ⎣4, 2, , ⎡, , 1, ⎢, 27. ⎣0, 0, , 2, , λ, , ⎡, , ⎡, , 3, , 0, , 1, ⎢, 23. A = ⎣1, 1, , ⎤, , 0, 5, 0, , ⎡, , 7, ⎥, 1⎦, 5, , k, k, k, , ⎤, k2, ⎥, k2 ⎦, k2, , 3, 2, 1, 10, , 0, 0, −3, 3, , 0, 3, 2, 4, 2, , 0, 3, 4, 6, 4, , 3, ⎢, 22. A = ⎣1, 1, , ⎤, , 3, 0, −3, , ⎡, , 1, ⎥, −4 ⎦, 5, , k−1, k−3, k+1, , k+1, , ⎢, , 24. A = ⎣ 2, 5, , ⎤, , 7, ⎥, 4⎦, , k, , ⎤, , 5, −2⎥, ⎥, ⎥, 0⎦, 2, 1, −1, 2, 2, 2, , ⎤, , 0, 0⎥, ⎥, ⎥, 3⎥, ⎥, 3⎦, 3, , In Exercises 27–32, evaluate the determinant of the given matrix by inspection., , In Exercises 15–18, find all values of λ for which det(A) = 0., , λ−2, 15. A =, −5, , (e) the third row., , 4, ⎢3, ⎢, ⎢, 26. A = ⎢1, ⎢, ⎣9, 2, , , , −4, , ⎡, λ−4, ⎢, 16. A = ⎣ 0, , (d) the second column., , ⎡, , In Exercises 5–8, evaluate the determinant of the given matrix., If the matrix is invertible, use Equation (2) to find its inverse., , −5, 7., −7, , (b) the first column., , ⎢, , (c) M41 and C41 ., , 1, 2, , 0, , (c) the second row., , 21. A = ⎣ 2, −1, , (b) M44 and C44 ., , 4, 6., 8, , λ, , 0, , (a) the first row., , −3, , (a) M32 and C32 ., , 5, 4, , λ+1, , ⎡, , Find, , 3, 5., −2, , 0, , 2, , ⎤, , 0, ⎥, 0 ⎦, λ−5, , In Exercises 21–26, evaluate det(A) by a cofactor expansion, along a row or column of your choice., , 1, 3⎥, ⎥, ⎥, 0⎦, 4, , 0, 1, 1, , λ−1, , 4, , 20. Evaluate the determinant in Exercise 12 by a cofactor expansion along, , (b) M23 and C23 ., , (c) M22 and C22 ., , 17. A =, , λ−4, ⎢, 18. A = ⎣ −1, , 19. Evaluate the determinant in Exercise 13 by a cofactor expansion along, , Find, (a) M13 and C13 ., , ⎡, , ⎤, , 0, ⎥, 2 ⎦, λ−1, , ⎡, , 0, , ⎢1, ⎢, ⎣0, , 29. ⎢, , 1, , ⎤, , 0, −1, 0, , 0, ⎥, 0⎦, 1, 0, , ⎤, , 0, , 0, , 2, , 0, , 4, , 3, , ⎥, 0⎦, , 2, , 3, , 8, , 0⎥, ⎥, , ⎡, , 2, ⎢, 28. ⎣0, 0, , ⎤, , 0, 2, 0, , 0, ⎥, 0⎦, 2, , ⎢0, ⎢, ⎣0, , 1, , 1, , 2, , 2, , 0, , 3, , 3⎦, , 0, , 0, , 0, , 4, , ⎡, , 1, , 30. ⎢, , 1, , ⎤, , 2⎥, ⎥, , ⎥

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112, , Chapter 2 Determinants, , ⎡, , 1, , ⎢0, ⎢, 31. ⎢, ⎣0, 0, , ⎤, , ⎡, , 1⎥, ⎥, , −3, , 2, , 7, , 1, , −4, , 0, , 2, , 7⎦, , 0, , 0, , 3, , ⎥, , −3, , ⎢ 1, ⎢, 32. ⎢, ⎣ 40, , 0, , 0, , 2, 10, , 0, −1, , 100, , 200, , −23, , 0, , ⎤, , 0⎥, ⎥, ⎥, 0⎦, , , , 0, , A=, , a, , b, c, , 0, , , and B =, , d, , e, f, , 0, , , , 1, 0, , 42. Prove that if A is upper triangular and Bij is the matrix that, results when the i th row and j th column of A are deleted, then, Bij is upper triangular if i < j ., , a, c, , b, is ad + bc., d, , (b) Two square matrices that have the same determinant must have, the same size., (c) The minor Mij is the same as the cofactor Cij if i + j is even., (d) If A is a 3 × 3 symmetric matrix, then Cij = Cj i for all i and j ., , , a − c , =0, d − f, , (e) The number obtained by a cofactor expansion of a matrix A is, independent of the row or column chosen for the expansion., , 35. By inspection, what is the relationship between the following, determinants?, , b, , , , 1, , 1 = 0, , (a) The determinant of the 2 × 2 matrix, , commute if and only if, , , a, , , d1 = d, , g, , , , 1 , , TF. In parts (a)–( j) determine whether the statement is true or, false, and justify your answer., , , , , b, , , e, , y, b1, b2, , True-False Exercises, , , , 1, 0, , 34. Show that the matrices, , , , , x, , ,  a1, ,  a2, , 3, , 33. In each part, show that the value of the determinant is independent of θ ., , , ,  sin θ cos θ , , , (a) , , − cos θ sin θ , , , sin θ, cos θ, , , sin θ, (b)  − cos θ, , sin θ − cos θ sin θ + cos θ, , 41. Prove that the equation of the line through the distinct points, (a1 , b1 ) and (a2 , b2 ) can be written as, , , , a + λ, c, , , , f  and d2 =  d, , ,  g, 1, , 36. Show that, 1, det(A) =, 2, , ,  tr(A), , tr(A2 ), , b, 1, 0, , , c , , f, , 1, , (f ) If A is a square matrix whose minors are all zero, then, det(A) = 0., (g) The determinant of a lower triangular matrix is the sum of the, entries along the main diagonal., (h) For every square matrix A and every scalar c, it is true that, det(cA) = c det(A)., , , , (i) For all square matrices A and B , it is true that, , tr(A), , det(A + B) = det(A) + det(B), , 1 , , ( j) For every 2 × 2 matrix A it is true that det(A2 ) = (det(A))2 ., , for every 2 × 2 matrix A., 37. What can you say about an nth-order determinant all of whose, entries are 1? Explain., 38. What is the maximum number of zeros that a 3 × 3 matrix can, have without having a zero determinant? Explain., 39. Explain why the determinant of a matrix with integer entries, must be an integer., , Working withTechnology, T1. (a) Use the determinant capability of your technology utility, to find the determinant of the matrix, , ⎡, , 4.2, , ⎢0.0, ⎢, A=⎢, ⎣4.5, 4.7, , ⎤, , −1.3, , 1.1, , 0 .0, , −3.2, , 1 .3, , 0.0, , 14.8⎦, , 1 .0, , 3.4, , 2.3, , 6.0, , 3.4⎥, ⎥, , ⎥, , Working with Proofs, 40. Prove that (x1 , y1 ), (x2 , y2 ), and (x3 , y3 ) are collinear points, if and only if, , , x1, , , x2, , x3, , y1, y2, y3, , , , 1, , 1 = 0, , 1, , (b) Compare the result obtained in part (a) to that obtained by a, cofactor expansion along the second row of A., T2. Let An be the n × n matrix with 2’s along the main diagonal,, 1’s along the diagonal lines immediately above and below the main, diagonal, and zeros everywhere else. Make a conjecture about the, relationship between n and det(An ).

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2.2 Evaluating Determinants by Row Reduction, , 113, , 2.2 Evaluating Determinants by Row Reduction, In this section we will show how to evaluate a determinant by reducing the associated, matrix to row echelon form. In general, this method requires less computation than, cofactor expansion and hence is the method of choice for large matrices., , A BasicTheorem, , We begin with a fundamental theorem that will lead us to an efficient procedure for, evaluating the determinant of a square matrix of any size., THEOREM 2.2.1 Let, , A be a square matrix. If A has a row of zeros or a column of, , zeros, then det(A) = 0., Proof Since the determinant of A can be found by a cofactor expansion along any row, , or column, we can use the row or column of zeros. Thus, if we let C1 , C2 , . . . , Cn denote, the cofactors of A along that row or column, then it follows from Formula (7) or (8) in, Section 2.1 that, det(A) = 0 · C1 + 0 · C2 + · · · + 0 · Cn = 0, The following useful theorem relates the determinant of a matrix and the determinant, of its transpose., Because transposing a matrix, changes its columns to rows, and its rows to columns, almost every theorem about the, rows of a determinant has, a companion version about, columns, and vice versa., , Elementary Row, Operations, , THEOREM 2.2.2 Let A be a square matrix. Then det(A), , = det(AT )., , Proof Since transposing a matrix changes its columns to rows and its rows to columns,, the cofactor expansion of A along any row is the same as the cofactor expansion of AT, along the corresponding column. Thus, both have the same determinant., , The next theorem shows how an elementary row operation on a square matrix affects the, value of its determinant. In place of a formal proof we have provided a table to illustrate, the ideas in the 3 × 3 case (see Table 1)., Table 1, , The first panel of Table 1, shows that you can bring a, common factor from any row, (column) of a determinant, through the determinant sign., This is a slightly different way, of thinking about part (a) of, Theorem 2.2.3., , Relationship, , , ka11, , ,  a21, ,  a31, , Operation, , , , a11, ka13 , , , , a23  = k a21, , , a31, a33 , , ka12, a22, a32, , , a13 , , a23 , , a33 , , a12, a22, a32, , In the matrix B the first, row of A was multiplied, by k ., , det(B) = k det(A), , , a21, , , a11, , a31, , , , a11, a23 , , , , a13  = − a21, , , a31, a33 , , a22, a12, a32, , a12, a22, a32, , , a13 , , a23 , , a33 , , In the matrix B the first and, second rows of A were, interchanged., , det(B) = − det(A), , , a11 + ka21, , ,  a21, ,  a31, , a12 + ka22, a22, a32, ,  , a13 + ka23  a11,  , a23,  = a21,  ,  a31, a33, , det(B) = det(A), , a12, a22, a32, , , a13 , , a23 , , a33 , , In the matrix B a multiple of, the second row of A was, added to the first row.

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114, , Chapter 2 Determinants, THEOREM 2.2.3 Let A be an n × n matrix., , (a) If B is the matrix that results when a single row or single column of A is multiplied, by a scalar k, then det(B) = k det(A)., (b) If B is the matrix that results when two rows or two columns of A are interchanged,, then det(B) = − det(A)., (c) If B is the matrix that results when a multiple of one row of A is added to another, or when a multiple of one column is added to another, then det(B) = det(A)., We will verify the first equation in Table 1 and leave the other two for you. To start,, note that the determinants on the two sides of the equation differ only in the first row, so, these determinants have the same cofactors, C11 , C12 , C13 , along that row (since those, cofactors depend only on the entries in the second two rows). Thus, expanding the left, side by cofactors along the first row yields, , , ka11, ,  a21, , a, 31, , Elementary Matrices, , ka12, a22, a32, , , ka13 , , a23  = ka11 C11 + ka12 C12 + ka13 C13, a33 , = k(a11 C11 + a12 C12 + a13 C13 ), , , a11 a12 a13 , , , = k a21 a22 a23 , a, a32 a33 , 31, , It will be useful to consider the special case of Theorem 2.2.3 in which A = In is the, n × n identity matrix and E (rather than B ) denotes the elementary matrix that results, when the row operation is performed on In . In this special case Theorem 2.2.3 implies, the following result., THEOREM 2.2.4 Let E be an n × n elementary matrix., , (a) If E results from multiplying a row of In by a nonzero number k, then det(E) = k ., (b) If E results from interchanging two rows of In , then det(E) = −1., (c) If E results from adding a multiple of one row of In to another, then det(E) = 1., , E X A M P L E 1 Determinants of Elementary Matrices, Observe that the determinant, of an elementary matrix cannot be zero., , The following determinants of elementary matrices, which are evaluated by inspection,, illustrate Theorem 2.2.4., , , 1, , 0, , , 0, , 0, , 0, , 0, , 3, , 0, , , , 0 , 0 , ,  = 3,, 0 1 0, , 0 0 1, , The second row of I4, was multiplied by 3., , Matrices with Proportional, Rows or Columns, , , 0, , 0, , , 0, , 1, , 0, , 0, , 1, , 0, , 0, , 1, , , , 1 , 0 , ,  = −1,, , 0 0 0, 0, , The first and last rows of, , I4 were interchanged., , , 1, , 0, , , 0, , 0, , , , 1 0 0 , =1, 0 1 0, , 0 0 1, 0, , 0, , 7, , 7 times the last row of I4, was added to the first row., , If a square matrix A has two proportional rows, then a row of zeros can be introduced, by adding a suitable multiple of one of the rows to the other. Similarly for columns. But, adding a multiple of one row or column to another does not change the determinant, so, from Theorem 2.2.1, we must have det(A) = 0. This proves the following theorem.

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2.2 Evaluating Determinants by Row Reduction, , 115, , THEOREM 2.2.5 If A is a square matrix with two proportional rows or two proportional, , columns, then det(A) = 0., , E X A M P L E 2 Proportional Rows or Columns, , Each of the following matrices has two proportional rows or columns; thus, each has a, determinant of zero., , ⎡, −1 4, ,, −2 8, , Evaluating Determinants, by Row Reduction, , 1, , ⎢, ⎣−4, , 2, , −2, , 7, , ⎡, , ⎤, , ⎢ 6, ⎢, ⎢, ⎣ 5, −9, , ⎥, , 5⎦,, , 8, , −4, , 3, , 3, , −1, −2, , 4, , 8, , 1, , 3, , −12, , 5, , ⎤, −5, 2⎥, ⎥, ⎥, 4⎦, 15, , We will now give a method for evaluating determinants that involves substantially less, computation than cofactor expansion. The idea of the method is to reduce the given, matrix to upper triangular form by elementary row operations, then compute the determinant of the upper triangular matrix (an easy computation), and then relate that, determinant to that of the original matrix. Here is an example., , E X A M P L E 3 Using Row Reduction to Evaluate a Determinant, , Evaluate det(A) where, , ⎡, , 0, ⎢, A = ⎣3, 2, , ⎤, , 1, −6, 6, , 5, ⎥, 9⎦, 1, , Solution We will reduce A to row echelon form (which is upper triangular) and then, apply Theorem 2.1.2., , , 0, , , det(A) =  3, , 2, , Even with today’s fastest computers it would take millions of, years to calculate a 25 × 25 determinant by cofactor expansion, so methods based on row, reduction are often used for, large determinants. For determinants of small size (such as, those in this text), cofactor expansion is often a reasonable, choice., , 1, −6, 6, , , , , , 3, 5 , , , , 9 = − 0, , , 2, 1, , −6, 1, 6, , , , 9 , , 5, , 1, , , ,  1 −2 3 , , , , , 1, 5, = −3  0, , , 2, 6, 1, , ,  1 −2, 3 , , , , 1, 5, = −3  0, , ,  0 10 −5 , , ,  1 −2, 3 , , , , 1, 5, = −3  0, , , 0, 0 −55 , , ,  1 −2, 3 , , , , 1, 5, = (−3)(−55)  0, , , 0, 0, 1, = (−3)(−55)(1) = 165, , The first and second rows of, A were interchanged., , A common factor of 3 from, the first row was taken, through the determinant sign., , −2 times the first row was, added to the third row., , −10 times the second row, was added to the third row., , A common factor of −55, from the last row was taken, through the determinant sign.

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116, , Chapter 2 Determinants, , E X A M P L E 4 Using Column Operations to Evaluate a Determinant, , Compute the determinant of, , ⎡, , 1, ⎢2, ⎢, A=⎢, ⎣0, 7, , 0, 7, 6, 3, , 0, 0, 3, 1, , ⎤, , 3, 6⎥, ⎥, ⎥, 0⎦, −5, , Solution This determinant could be computed as above by using elementary row oper-, , ations to reduce A to row echelon form, but we can put A in lower triangular form in, one step by adding −3 times the first column to the fourth to obtain, , ⎡, , 1, ⎢2, ⎢, det(A) = det ⎢, ⎣0, 7, , Example 4 points out that it, is always wise to keep an eye, open for column operations, that can shorten computations., , 0, 7, 6, 3, , 0, 0, 3, 1, , ⎤, , 0, 0⎥, ⎥, ⎥ = (1)(7)(3)(−26) = −546, 0⎦, −26, , Cofactor expansion and row or column operations can sometimes be used in combination to provide an effective method for evaluating determinants. The following, example illustrates this idea., E X A M P L E 5 Row Operations and Cofactor Expansion, , Evaluate det(A) where, , ⎡, , 3, ⎢1, ⎢, A=⎢, ⎣2, 3, , 5, 2, 4, 7, , −2, −1, 1, 5, , ⎤, , 6, 1⎥, ⎥, ⎥, 5⎦, 3, , Solution By adding suitable multiples of the second row to the remaining rows, we, , obtain, , ,  0 −1, 1, , 1, 2, 1, −, , det(A) = , 0, 0, 3, , 0, 1, 8, ,  −1, 1, , , 3, = − 0, ,  1, 8, ,  −1, 1, , , 3, = − 0, ,  0, 9, , 3, = −(−1) , , , , 3 , 1 , , 3, , 0, , , , 3 , , 3, , 0, , Cofactor expansion along, the first column, , , , 3 , , 3, , 3, , We added the first row to the, third row., , , , 3 , 9 3, = −18, , Cofactor expansion along, the first column

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118, , Chapter 2 Determinants, , ⎡, a, ⎢b, ⎢, ⎢, ⎣b, b, , In Exercises 29–30, show that det(A) = 0 without directly evaluating the determinant., , ⎡, , −2, , 8, 2, 10, −6, , ⎢ 3, , 29. A = ⎢, ⎣, , 1, 4, , ⎡, , −4, ⎢ 1, ⎢, ⎢, 30. A = ⎢ 1, ⎢, ⎣ 1, , 1, −4, 1, 1, 1, , 1, , ⎤, , 1, 5, 6, 4, , 4, 1⎥, ⎥, 5⎦, −3, , 1, 1, −4, 1, 1, , 1, 1, 1, −4, 1, , A, M=, C, , 0, , TF. In parts (a)–(f ) determine whether the statement is true or, false, and justify your answer., , 1, 1⎥, ⎥, ⎥, 1⎥, ⎥, 1⎦, −4, , , , (a) If A is a 4 × 4 matrix and B is obtained from A by interchanging the first two rows and then interchanging the last two rows,, then det(B) = det(A)., , , , or M =, , B, , A, 0, , C, B, , , , in which A and B are square, then det(M) = det(A) det(B). Use, this result to compute the determinants of the matrices in Exercises 31 and 32., , ⎡, , −9, , ⎤, , 1, ⎢ 2, ⎢, ⎢, ⎢−1, , 2, 5, 3, , 0, 0, 2, , 8, 4, 6, , 6, 7, 9, , 5⎥, ⎥, ⎥, −2⎥, , ⎢ 0, ⎢, ⎣ 0, , 0, 0, 0, , 0, 0, 0, , 3, 2, −3, , 0, 1, 8, , 0⎥, ⎥, 0⎦, −4, , 31. M = ⎢, ⎢, , 0, , ⎡, , 1, ⎢0, ⎢, ⎢, 0, 32. M = ⎢, ⎢, , ⎢, ⎣0, 2, , 2, 1, 0, , 0, 2, 1, , 0, 0, 0, , 0, 0, , 0, 0, , 1, 0, , ⎤, b, b⎥, ⎥, ⎥, b⎦, a, , b, b, a, b, , True-False Exercises, ⎤, , It can be proved that if a square matrix M is partitioned into, block triangular form as, , , , b, a, b, b, , (b) If A is a 3 × 3 matrix and B is obtained from A by multiplying, the first column by 4 and multiplying the third column by 43 ,, then det(B) = 3 det(A)., (c) If A is a 3 × 3 matrix and B is obtained from A by adding 5, times the first row to each of the second and third rows, then, det(B) = 25 det(A)., (d) If A is an n × n matrix and B is obtained from A by multiplying each row of A by its row number, then, det(B) =, , ⎥, ⎥, , n(n + 1), 2, , det(A), , (e) If A is a square matrix with two identical columns, then, det(A) = 0., (f ) If the sum of the second and fourth row vectors of a 6 × 6, matrix A is equal to the last row vector, then det(A) = 0., , ⎤, , 0, 0⎥, ⎥, ⎥, 0⎥, , Working withTechnology, T1. Find the determinant of, , ⎡, , ⎥, ⎥, 2⎦, , 4.2, , 1, , 33. Let A be an n × n matrix, and let B be the matrix that results when the rows of A are written in reverse order. State a, theorem that describes how det(A) and det(B) are related., 34. Find the determinant of the following matrix., , ⎢0.0, ⎢, A=⎢, ⎣4.5, 4.7, , ⎤, , −1.3, , 1.1, , 0 .0, , −3.2, , 1 .3, , 0 .0, , 14.8⎦, , 1 .0, , 3.4, , 2.3, , 6.0, , 3.4⎥, ⎥, , ⎥, , by reducing the matrix to reduced row echelon form, and compare, the result obtained in this way to that obtained in Exercise T1 of, Section 2.1., , 2.3 Properties of Determinants; Cramer’s Rule, In this section we will develop some fundamental properties of matrices, and we will use, these results to derive a formula for the inverse of an invertible matrix and formulas for the, solutions of certain kinds of linear systems., , Basic Properties of, Determinants, , Suppose that A and B are n × n matrices and k is any scalar. We begin by considering, possible relationships among det(A), det(B), and, det(kA), det(A + B), and det(AB), Since a common factor of any row of a matrix can be moved through the determinant, sign, and since each of the n rows in kA has a common factor of k , it follows that

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CHAPTER, , 4, , General Vector Spaces, CHAPTER CONTENTS, , 4.1, , Real Vector Spaces, , 4.2, , Subspaces, , 4.3, , Linear Independence, , 4.4, , Coordinates and Basis, , 4.5, , Dimension, , 4.6, , Change of Basis, , 4.7, , Row Space, Column Space, and Null Space, , 4.8, , Rank, Nullity, and the Fundamental Matrix Spaces, , 4.9, , Basic Matrix Transformations in R 2 and R 3, , 183, , 191, 202, 212, , 221, 229, , 4.10 Properties of Matrix Transformations, , 270, , 2, , 280, , 4.11 Geometry of Matrix Operators on R, INTRODUCTION, , 237, 248, , 259, , Recall that we began our study of vectors by viewing them as directed line segments, (arrows). We then extended this idea by introducing rectangular coordinate systems,, which enabled us to view vectors as ordered pairs and ordered triples of real numbers., As we developed properties of these vectors we noticed patterns in various formulas, that enabled us to extend the notion of a vector to an n-tuple of real numbers., Although n-tuples took us outside the realm of our “visual experience,” it gave us a, valuable tool for understanding and studying systems of linear equations. In this, chapter we will extend the concept of a vector yet again by using the most important, algebraic properties of vectors in R n as axioms. These axioms, if satisfied by a set of, objects, will enable us to think of those objects as vectors., , 4.1 Real Vector Spaces, In this section we will extend the concept of a vector by using the basic properties of vectors, in R n as axioms, which if satisfied by a set of objects, guarantee that those objects behave, like familiar vectors., , Vector Space Axioms, , The following definition consists of ten axioms, eight of which are properties of vectors, in R n that were stated in Theorem 3.1.1. It is important to keep in mind that one does, not prove axioms; rather, they are assumptions that serve as the starting point for proving, theorems., 183

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184, , Chapter 4 General Vector Spaces, DEFINITION 1 Let V be an arbitrary nonempty set of objects on which two operations, , In this text scalars will be either real numbers or complex, numbers. Vector spaces with, real scalars will be called real, vector spaces and those with, complex scalars will be called, complex vector spaces. There, is a more general notion of a, vector space in which scalars, can come from a mathematical, structure known as a “field,”, but we will not be concerned, with that level of generality., For now, we will focus exclusively on real vector spaces,, which we will refer to simply as “vector spaces.” We, will consider complex vector, spaces later., , are defined: addition, and multiplication by numbers called scalars. By addition we, mean a rule for associating with each pair of objects u and v in V an object u + v,, called the sum of u and v; by scalar multiplication we mean a rule for associating with, each scalar k and each object u in V an object k u, called the scalar multiple of u by k ., If the following axioms are satisfied by all objects u, v, w in V and all scalars k and, m, then we call V a vector space and we call the objects in V vectors., 1. If u and v are objects in V, then u + v is in V., 2. u + v = v + u, 3. u + (v + w) = (u + v) + w, 4. There is an object 0 in V, called a zero vector for V, such that 0 + u = u + 0 = u, for all u in V., 5. For each u in V, there is an object −u in V, called a negative of u, such that, u + (−u) = (−u) + u = 0., 6. If k is any scalar and u is any object in V, then k u is in V., 7. k(u + v) = k u + k v, 8. (k + m)u = k u + mu, 9. k(mu) = (km)(u), 10. 1u = u, Observe that the definition of a vector space does not specify the nature of the vectors, or the operations. Any kind of object can be a vector, and the operations of addition, and scalar multiplication need not have any relationship to those on R n . The only, requirement is that the ten vector space axioms be satisfied. In the examples that follow, we will use four basic steps to show that a set with two operations is a vector space., To Show That a Set with Two Operations Is a Vector Space, Step 1. Identify the set V of objects that will become vectors., Step 2. Identify the addition and scalar multiplication operations on V ., Step 3. Verify Axioms 1 and 6; that is, adding two vectors in V produces a vector, in V , and multiplying a vector in V by a scalar also produces a vector in V ., Axiom 1 is called closure under addition, and Axiom 6 is called closure under, scalar multiplication., Step 4. Confirm that Axioms 2, 3, 4, 5, 7, 8, 9, and 10 hold., , Hermann Günther, Grassmann, (1809–1877), , Historical Note The notion of an “abstract vector, space” evolved over many years and had many, contributors. The idea crystallized with the work, of the German mathematician H. G. Grassmann,, who published a paper in 1862 in which he considered abstract systems of unspecified elements, on which he defined formal operations of addition and scalar multiplication. Grassmann’s work, was controversial, and others, including Augustin, Cauchy (p. 121), laid reasonable claim to the idea., [Image: © Sueddeutsche Zeitung Photo/The, Image Works]

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4.1 Real Vector Spaces, , 185, , Our first example is the simplest of all vector spaces in that it contains only one, object. Since Axiom 4 requires that every vector space contain a zero vector, the object, will have to be that vector., , E X A M P L E 1 The Zero Vector Space, , Let V consist of a single object, which we denote by 0, and define, 0 + 0 = 0 and k 0 = 0, for all scalars k . It is easy to check that all the vector space axioms are satisfied. We call, this the zero vector space., , Our second example is one of the most important of all vector spaces—the familiar, space R n . It should not be surprising that the operations on R n satisfy the vector space, axioms because those axioms were based on known properties of operations on R n ., E X A M P L E 2 R n Is a Vector Space, , Let V = R n , and define the vector space operations on V to be the usual operations of, addition and scalar multiplication of n-tuples; that is,, u + v = (u1 , u2 , . . . , un ) + (v1 , v2 , . . . , vn ) = (u1 + v1 , u2 + v2 , . . . , un + vn ), k u = (ku1 , ku2 , . . . , kun ), The set V = R n is closed under addition and scalar multiplication because the foregoing, operations produce n-tuples as their end result, and these operations satisfy Axioms 2,, 3, 4, 5, 7, 8, 9, and 10 by virtue of Theorem 3.1.1., Our next example is a generalization of R n in which we allow vectors to have infinitely, many components., , E X A M P L E 3 The Vector Space of Infinite Sequences of Real Numbers, , Let V consist of objects of the form, u = (u1 , u2 , . . . , un , . . .), in which u1 , u2 , . . . , un , . . . is an infinite sequence of real numbers. We define two infinite sequences to be equal if their corresponding components are equal, and we define, addition and scalar multiplication componentwise by, u + v = (u1 , u2 , . . . , un , . . .) + (v1 , v2 , . . . , vn , . . .), , = (u1 + v1 , u2 + v2 , . . . , un + vn , . . .), k u = (ku1 , ku2 , . . . , kun , . . .), E(t), Voltage, , In the exercises we ask you to confirm that V with these operations is a vector space. We, will denote this vector space by the symbol R ⬁ ., , 1, t, Time, –1, , Figure 4.1.1, , Vector spaces of the type in Example 3 arise when a transmitted signal of indefinite, duration is digitized by sampling its values at discrete time intervals (Figure 4.1.1)., In the next example our vectors will be matrices. This may be a little confusing at, first because matrices are composed of rows and columns, which are themselves vectors, (row vectors and column vectors). However, from the vector space viewpoint we are not

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186, , Chapter 4 General Vector Spaces, , concerned with the individual rows and columns but rather with the properties of the, matrix operations as they relate to the matrix as a whole., E X A M P L E 4 The Vector Space of 2 × 2 Matrices, Note that Equation (1) involves three different addition, operations: the addition operation on vectors, the addition operation on matrices,, and the addition operation on, real numbers., , Let V be the set of 2 × 2 matrices with real entries, and take the vector space operations, on V to be the usual operations of matrix addition and scalar multiplication; that is,, u+v=, , ku = k, , u11, u21, u11, u21, , v12, u11 + v11, =, v22, u21 + v21, , u12, v11, +, u22, v21, u12, ku11, =, u22, ku21, , u12 + v12, u22 + v22, , (1), , ku12, ku22, , The set V is closed under addition and scalar multiplication because the foregoing operations produce 2 × 2 matrices as the end result. Thus, it remains to confirm that Axioms, 2, 3, 4, 5, 7, 8, 9, and 10 hold. Some of these are standard properties of matrix operations., For example, Axiom 2 follows from Theorem 1.4.1(a) since, u+v=, , u11, u21, , u12, v11, +, u22, v21, , v12, v11, =, v22, v21, , v12, u11, +, v22, u21, , u12, =v+u, u22, , Similarly, Axioms 3, 7, 8, and 9 follow from parts (b), (h), ( j), and (e), respectively, of, that theorem (verify). This leaves Axioms 4, 5, and 10 that remain to be verified., To confirm that Axiom 4 is satisfied, we must find a 2 × 2 matrix 0 in V for which, u + 0 = 0 + u for all 2 × 2 matrices in V . We can do this by taking, 0=, , 0, 0, , 0, 0, , With this definition,, 0+u=, , 0, 0, , u11, 0, +, 0, u21, , u12, u11, =, u22, u21, , u12, =u, u22, , and similarly u + 0 = u. To verify that Axiom 5 holds we must show that each object, u in V has a negative −u in V such that u + (−u) = 0 and (−u) + u = 0. This can be, done by defining the negative of u to be, , −u =, , −u11, −u21, , −u12, −u22, , With this definition,, u + (−u) =, , u11, u21, , u12, −u11, +, u22, −u21, , 0 0, −u12, =, =0, 0 0, −u22, , and similarly (−u) + u = 0. Finally, Axiom 10 holds because, 1u = 1, , u11, u21, , u12, u11, =, u22, u21, , u12, =u, u22, , E X A M P L E 5 The Vector Space of m × n Matrices, , Example 4 is a special case of a more general class of vector spaces. You should have, no trouble adapting the argument used in that example to show that the set V of all, m × n matrices with the usual matrix operations of addition and scalar multiplication is, a vector space. We will denote this vector space by the symbol Mmn . Thus, for example,, the vector space in Example 4 is denoted as M22 .

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4.1 Real Vector Spaces, , 187, , E X A M P L E 6 The Vector Space of Real-Valued Functions, , Let V be the set of real-valued functions that are defined at each x in the interval (−⬁, ⬁)., If f = f(x) and g = g(x) are two functions in V and if k is any scalar, then define the, operations of addition and scalar multiplication by, , (f + g)(x) = f(x) + g(x), , (2), , (k f)(x) = kf(x), , (3), , One way to think about these operations is to view the numbers f(x) and g(x) as “components” of f and g at the point x , in which case Equations (2) and (3) state that two, functions are added by adding corresponding components, and a function is multiplied, by a scalar by multiplying each component by that scalar—exactly as in R n and R ⬁ . This, idea is illustrated in parts (a) and (b) of Figure 4.1.2. The set V with these operations is, denoted by the symbol F(−⬁, ⬁). We can prove that this is a vector space as follows:, Axioms 1 and 6: These closure axioms require that if we add two functions that are, defined at each x in the interval (−⬁, ⬁), then sums and scalar multiples of those functions must also be defined at each x in the interval (−⬁, ⬁). This follows from Formulas, (2) and (3)., Axiom 4: This axiom requires that there exists a function 0 in F (−⬁, ⬁), which when, added to any other function f in F(−⬁, ⬁) produces f back again as the result. The, function whose value at every point x in the interval (−⬁, ⬁) is zero has this property., Geometrically, the graph of the function 0 is the line that coincides with the x -axis., In Example 6 the functions, were defined on the entire interval (−⬁, ⬁). However, the, arguments used in that example apply as well on all subintervals of (−⬁, ⬁), such as, a closed interval [a, b] or an, open interval (a, b). We will, denote the vector spaces of, functions on these intervals by, F [a, b] and F(a, b), respectively., , Axiom 5: This axiom requires that for each function f in F(−⬁, ⬁) there exists a function, −f in F(−⬁, ⬁), which when added to f produces the function 0. The function defined, by −f(x) = −f(x) has this property. The graph of −f can be obtained by reflecting the, graph of f about the x -axis (Figure 4.1.2c)., Axioms 2, 3, 7, 8, 9, 10: The validity of each of these axioms follows from properties of, real numbers. For example, if f and g are functions in F(−⬁, ⬁), then Axiom 2 requires, that f + g = g + f. This follows from the computation, , (f + g)(x) = f(x) + g(x) = g(x) + f(x) = (g + f)(x), in which the first and last equalities follow from (2), and the middle equality is a property, of real numbers. We will leave the proofs of the remaining parts as exercises., , y, , y, , y, , f+g, g, , g(x), f, , f(x), , f(x) + g(x), x, , kf, f, , f, , k f (x), , f (x), , x, , 0, , f(x), , x, , x, , x, , (a), , (b), , –f, , –f(x), , (c), , Figure 4.1.2, , It is important to recognize that you cannot impose any two operations on any set, , V and expect the vector space axioms to hold. For example, if V is the set of n-tuples, with positive components, and if the standard operations from R n are used, then V is not, closed under scalar multiplication, because if u is a nonzero n-tuple in V , then (−1)u has

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188, , Chapter 4 General Vector Spaces, , at least one negative component and hence is not in V . The following is a less obvious, example in which only one of the ten vector space axioms fails to hold., E X A M P L E 7 A Set That Is Not a Vector Space, , Let V = R 2 and define addition and scalar multiplication operations as follows: If, u = (u1 , u2 ) and v = (v1 , v2 ), then define, u + v = (u1 + v1 , u2 + v2 ), and if k is any real number, then define, , k u = (ku1 , 0), For example, if u = (2, 4), v = (−3, 5), and k = 7, then, u + v = (2 + (−3), 4 + 5) = (−1, 9), , k u = 7u = (7 · 2, 0) = (14, 0), The addition operation is the standard one from R 2 , but the scalar multiplication is not., In the exercises we will ask you to show that the first nine vector space axioms are satisfied., However, Axiom 10 fails to hold for certain vectors. For example, if u = (u1 , u2 ) is such, that u2  = 0, then, 1u = 1(u1 , u2 ) = (1 · u1 , 0) = (u1 , 0)  = u, Thus, V is not a vector space with the stated operations., Our final example will be an unusual vector space that we have included to illustrate, how varied vector spaces can be. Since the vectors in this space will be real numbers,, it will be important for you to keep track of which operations are intended as vector, operations and which ones as ordinary operations on real numbers., E X A M P L E 8 An Unusual Vector Space, , Let V be the set of positive real numbers, let u = u and v = v be any vectors (i.e., positive, real numbers) in V , and let k be any scalar. Define the operations on V to be, , u + v = uv, ku = uk, , [ Vector addition is numerical multiplication. ], [ Scalar multiplication is numerical exponentiation. ], , Thus, for example, 1 + 1 = 1 and (2)(1) = 12 = 1—strange indeed, but nevertheless, the set V with these operations satisfies the ten vector space axioms and hence is a vector, space. We will confirm Axioms 4, 5, and 7, and leave the others as exercises., • Axiom 4—The zero vector in this space is the number 1 (i.e., 0 = 1) since, , u+1=u·1=u, • Axiom 5—The negative of a vector u is its reciprocal (i.e., −u = 1/u) since, , u+, , 1, , u, , =u, , 1, , u, , !, , = 1 (= 0), , • Axiom 7—k(u + v) = (uv)k = uk v k = (ku) + (kv)., Some Properties of Vectors, , The following is our first theorem about vector spaces. The proof is very formal with, each step being justified by a vector space axiom or a known property of real numbers., There will not be many rigidly formal proofs of this type in the text, but we have included, this one to reinforce the idea that the familiar properties of vectors can all be derived, from the vector space axioms.

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4.1 Real Vector Spaces, , 189, , THEOREM 4.1.1 Let V be a vector space, u a vector in V, and k a scalar; then:, , (a) 0u = 0, (b) k 0 = 0, (c), , (−1)u = −u, , (d ) If k u = 0, then k = 0 or u = 0., , We will prove parts (a) and (c) and leave proofs of the remaining parts as exercises., Proof (a) We can write, , 0u + 0u = (0 + 0)u [ Axiom 8 ], , = 0u, , [ Property of the number 0 ], , By Axiom 5 the vector 0u has a negative, −0u. Adding this negative to both sides above, yields, [0u + 0u] + (−0u) = 0u + (−0u), or, , 0u + [0u + (−0u)] = 0u + (−0u) [ Axiom 3 ], , Proof (c) To prove that, , as follows:, , 0u + 0 = 0, , [ Axiom 5 ], , 0u = 0, , [ Axiom 4 ], , (−1)u = −u, we must show that u + (−1)u = 0. The proof is, , u + (−1)u = 1u + (−1)u, , = (1 + (−1))u, = 0u, =0, , A Closing Observation, , [ Axiom 10 ], [ Axiom 8 ], [ Property of numbers ], [ Part (a) of this theorem ], , This section of the text is important to the overall plan of linear algebra in that it establishes a common thread among such diverse mathematical objects as geometric vectors,, vectors in R n , infinite sequences, matrices, and real-valued functions, to name a few., As a result, whenever we discover a new theorem about general vector spaces, we will, at the same time be discovering a theorem about geometric vectors, vectors in R n , sequences, matrices, real-valued functions, and about any new kinds of vectors that we, might discover., To illustrate this idea, consider what the rather innocent-looking result in part (a), of Theorem 4.1.1 says about the vector space in Example 8. Keeping in mind that the, vectors in that space are positive real numbers, that scalar multiplication means numerical, exponentiation, and that the zero vector is the number 1, the equation, 0u = 0, is really a statement of the familiar fact that if u is a positive real number, then, , u0 = 1

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190, , Chapter 4 General Vector Spaces, , Exercise Set 4.1, 1. Let V be the set of all ordered pairs of real numbers, and, consider the following addition and scalar multiplication operations on u = (u1 , u2 ) and v = (v1 , v2 ):, u + v = (u1 + v1 , u2 + v2 ), k u = (0, ku2 ), (a) Compute u + v and k u for u = (−1, 2), v = (3, 4), and, k = 3., (b) In words, explain why V is closed under addition and, scalar multiplication., (c) Since addition on V is the standard addition operation on, R 2 , certain vector space axioms hold for V because they, are known to hold for R 2 . Which axioms are they?, (d) Show that Axioms 7, 8, and 9 hold., (e) Show that Axiom 10 fails and hence that V is not a vector, space under the given operations., 2. Let V be the set of all ordered pairs of real numbers, and, consider the following addition and scalar multiplication operations on u = (u1 , u2 ) and v = (v1 , v2 ):, u + v = (u1 + v1 + 1, u2 + v2 + 1), k u = (ku1 , ku2 ), (a) Compute u + v and k u for u = (0, 4), v = (1, −3), and, k = 2., (b) Show that (0, 0)  = 0., , 9. The set of all 2 × 2 matrices of the form, , a, , 0, , 0, , b, , with the standard matrix addition and scalar multiplication., 10. The set of all real-valued functions f defined everywhere on, the real line and such that f(1) = 0 with the operations used, in Example 6., 11. The set of all pairs of real numbers of the form (1, x) with the, operations, , (1, y) + (1, y ) = (1, y + y ) and k(1, y) = (1, ky), 12. The set of polynomials of the form a0 + a1 x with the operations, , (a0 + a1 x) + (b0 + b1 x) = (a0 + b0 ) + (a1 + b1 )x, and, , k(a0 + a1 x) = (ka0 ) + (ka1 )x, 13. Verify Axioms 3, 7, 8, and 9 for the vector space given in Example 4., 14. Verify Axioms 1, 2, 3, 7, 8, 9, and 10 for the vector space given, in Example 6., 15. With the addition and scalar multiplication operations defined, in Example 7, show that V = R 2 satisfies Axioms 1–9., , (c) Show that (−1, −1) = 0., (d) Show that Axiom 5 holds by producing an ordered pair, −u such that u + (−u) = 0 for u = (u1 , u2 )., (e) Find two vector space axioms that fail to hold., In Exercises 3–12, determine whether each set equipped with, the given operations is a vector space. For those that are not vector, spaces identify the vector space axioms that fail., 3. The set of all real numbers with the standard operations of, addition and multiplication., 4. The set of all pairs of real numbers of the form (x, 0) with the, standard operations on R 2 ., 5. The set of all pairs of real numbers of the form (x, y), where, x ≥ 0, with the standard operations on R 2 ., 6. The set of all n-tuples of real numbers that have the form, (x, x, . . . , x) with the standard operations on R n ., 7. The set of all triples of real numbers with the standard vector, addition but with scalar multiplication defined by, , k(x, y, z) = (k x, k y, k z), 2, , 2, , 2, , 8. The set of all 2 × 2 invertible matrices with the standard matrix addition and scalar multiplication., , 16. Verify Axioms 1, 2, 3, 6, 8, 9, and 10 for the vector space given, in Example 8., 17. Show that the set of all points in R 2 lying on a line is a vector, space with respect to the standard operations of vector addition and scalar multiplication if and only if the line passes, through the origin., 18. Show that the set of all points in R 3 lying in a plane is a vector, space with respect to the standard operations of vector addition and scalar multiplication if and only if the plane passes, through the origin., In Exercises 19–20, let V be the vector space of positive real, numbers with the vector space operations given in Example 8. Let, u = u be any vector in V , and rewrite the vector statement as a, statement about real numbers., 19. −u = (−1)u, 20. k u = 0 if and only if k = 0 or u = 0., , Working with Proofs, 21. The argument that follows proves that if u, v, and w are vectors, in a vector space V such that u + w = v + w, then u = v (the, cancellation law for vector addition). As illustrated, justify the, steps by filling in the blanks.

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4.2 Subspaces, , u+w=v+w, Hypothesis, (u + w) + (−w) = (v + w) + (−w) Add −w to both sides., u + [w + (−w)] = v + [w + (−w)], u+0=v+0, u=v, , 191, , 25. The set V = {0} with the operations of addition and scalar, multiplication given in Example 1., 26. The set R ⬁ of all infinite sequences of real numbers with the, operations of addition and scalar multiplication given in Example 3., , 22. Below is a seven-step proof of part (b) of Theorem 4.1.1., Justify each step either by stating that it is true by hypothesis, or by specifying which of the ten vector space axioms applies., , 27. The set Mmn of all m × n matrices with the usual operations, of addition and scalar multiplication., , Hypothesis: Let u be any vector in a vector space V, let 0 be, the zero vector in V, and let k be a scalar., Conclusion: Then k 0 = 0., , 28. Prove: If u is a vector in a vector space V and k a scalar such, that k u = 0, then either k = 0 or u = 0. [Suggestion: Show, that if k u = 0 and k  = 0, then u = 0. The result then follows, as a logical consequence of this.], , Proof: (1) k 0 + k u = k(0 + u), , True-False Exercises, , (2), , = ku, , (3) Since k u is in V, −k u is in V., , TF. In parts (a)–(f) determine whether the statement is true or, false, and justify your answer., , (4) Therefore, (k 0 + k u) + (−k u) = k u + (−k u)., , (a) A vector is any element of a vector space., , (5), , k 0 + (k u + (−k u)) = k u + (−k u), , (6), , k0 + 0 = 0, , (7), , k0 = 0, , (b) A vector space must contain at least two vectors., (c) If u is a vector and k is a scalar such that k u = 0, then it must, be true that k = 0., , In Exercises 23–24, let u be any vector in a vector space V ., Give a step-by-step proof of the stated result using Exercises 21, and 22 as models for your presentation., , (d) The set of positive real numbers is a vector space if vector, addition and scalar multiplication are the usual operations of, addition and multiplication of real numbers., , 23. 0u = 0, , (e) In every vector space the vectors (−1)u and −u are the same., , 24. −u = (−1)u, , In Exercises 25–27, prove that the given set with the stated, operations is a vector space., , (f ) In the vector space F (−⬁, ⬁) any function whose graph passes, through the origin is a zero vector., , 4.2 Subspaces, It is often the case that some vector space of interest is contained within a larger vector space, whose properties are known. In this section we will show how to recognize when this is the, case, we will explain how the properties of the larger vector space can be used to obtain, properties of the smaller vector space, and we will give a variety of important examples., , We begin with some terminology., DEFINITION 1 A subset W of a vector space V is called a subspace of V if W is itself, a vector space under the addition and scalar multiplication defined on V., , In general, to show that a nonempty set W with two operations is a vector space one, must verify the ten vector space axioms. However, if W is a subspace of a known vector, space V , then certain axioms need not be verified because they are “inherited” from V ., For example, it is not necessary to verify that u + v = v + u holds in W because it holds, for all vectors in V including those in W . On the other hand, it is necessary to verify

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192, , Chapter 4 General Vector Spaces, , that W is closed under addition and scalar multiplication since it is possible that adding, two vectors in W or multiplying a vector in W by a scalar produces a vector in V that is, outside of W (Figure 4.2.1). Those axioms that are not inherited by W are, Axiom 1—Closure of W under addition, Axiom 4—Existence of a zero vector in W, Axiom 5—Existence of a negative in W for every vector in W, Axiom 6—Closure of W under scalar multiplication, so these must be verified to prove that it is a subspace of V . However, the next theorem, shows that if Axiom 1 and Axiom 6 hold in W , then Axioms 4 and 5 hold in W as a, consequence and hence need not be verified., , u+v, , ku, u, v, , Figure 4.2.1 The vectors u, and v are in W , but the vectors, u + v and k u are not., , W, V, , THEOREM 4.2.1 If, , W is a set of one or more vectors in a vector space V, then W is a, subspace of V if and only if the following conditions are satisfied., (a) If u and v are vectors in W, then u + v is in W ., (b) If k is a scalar and u is a vector in W, then k u is in W ., , Theorem 4.2.1 states that W is, a subspace of V if and only if, it is closed under addition and, scalar multiplication., , Proof If W is a subspace of V, then all the vector space axioms hold in W , including, Axioms 1 and 6, which are precisely conditions (a) and (b)., Conversely, assume that conditions (a) and (b) hold. Since these are Axioms 1 and, 6, and since Axioms 2, 3, 7, 8, 9, and 10 are inherited from V , we only need to show, that Axioms 4 and 5 hold in W . For this purpose, let u be any vector in W . It follows, from condition (b) that k u is a vector in W for every scalar k . In particular, 0u = 0 and, (−1)u = −u are in W , which shows that Axioms 4 and 5 hold in W ., , E X A M P L E 1 The Zero Subspace, Note that every vector space, has at least two subspaces, itself and its zero subspace., , If V is any vector space, and if W = {0} is the subset of V that consists of the zero vector, only, then W is closed under addition and scalar multiplication since, 0 + 0 = 0 and k 0 = 0, for any scalar k . We call W the zero subspace of V ., E X A M P L E 2 Lines Through the Origin Are Subspaces of R 2 and of R 3, , If W is a line through the origin of either R 2 or R 3 , then adding two vectors on the line, or multiplying a vector on the line by a scalar produces another vector on the line, so, W is closed under addition and scalar multiplication (see Figure 4.2.2 for an illustration, in R 3 ).

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4.2 Subspaces, , 193, , W, , W, u+v, ku, , v, u, , u, , (a) W is closed under addition., , (b) W is closed under scalar, multiplication., , Figure 4.2.2, , E X A M P L E 3 Planes Through the Origin Are Subspaces of R 3, , If u and v are vectors in a plane W through the origin of R 3 , then it is evident geometrically, that u + v and k u also lie in the same plane W for any scalar k (Figure 4.2.3). Thus W, is closed under addition and scalar multiplication., , u+v, v, ku, , u, , Table 1 below gives a list of subspaces of R 2 and of R 3 that we have encountered thus, far. We will see later that these are the only subspaces of R 2 and of R 3 ., , W, , Figure 4.2.3 The vectors, u + v and k u both lie in the same, plane as u and v., , Table 1, Subspaces of R 2, , Subspaces of R 3, , •, •, •, , •, •, •, •, , {0}, Lines through the origin, , R2, , {0}, Lines through the origin, Planes through the origin, , R3, , E X A M P L E 4 A Subset of R 2 That Is Not a Subspace, y, W, , (1, 1), , Let W be the set of all points (x, y) in R 2 for which x ≥ 0 and y ≥ 0 (the shaded region, in Figure 4.2.4). This set is not a subspace of R 2 because it is not closed under scalar, multiplication. For example, v = (1, 1) is a vector in W , but (−1)v = (−1, −1) is not., , x, , E X A M P L E 5 Subspaces of M nn, (–1, –1), , Figure 4.2.4 W is not closed, under scalar multiplication., , We know from Theorem 1.7.2 that the sum of two symmetric n × n matrices is symmetric, and that a scalar multiple of a symmetric n × n matrix is symmetric. Thus, the set of, symmetric n × n matrices is closed under addition and scalar multiplication and hence, is a subspace of Mnn . Similarly, the sets of upper triangular matrices, lower triangular, matrices, and diagonal matrices are subspaces of Mnn ., E X A M P L E 6 A Subset of M nn That Is Not a Subspace, , The set W of invertible n × n matrices is not a subspace of Mnn , failing on two counts—it, is not closed under addition and not closed under scalar multiplication. We will illustrate, this with an example in M22 that you can readily adapt to Mnn . Consider the matrices, , U=, , 1, 2, , 2, 5, , and V =, , −1 2, −2 5, , The matrix 0U is the 2 × 2 zero matrix and hence is not invertible, and the matrix U + V, has a column of zeros so it also is not invertible.

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194, , Chapter 4 General Vector Spaces, , CA L C U L U S R E Q U I R E D, , E X A M P L E 7 The Subspace C (−ⴥ, ⴥ), , There is a theorem in calculus which states that a sum of continuous functions is continuous and that a constant times a continuous function is continuous. Rephrased in, vector language, the set of continuous functions on (−⬁, ⬁) is a subspace of F (−⬁, ⬁)., We will denote this subspace by C(−⬁, ⬁)., CA L C U L U S R E Q U I R E D, , E X A M P L E 8 Functions with Continuous Derivatives, , A function with a continuous derivative is said to be continuously differentiable. There, is a theorem in calculus which states that the sum of two continuously differentiable, functions is continuously differentiable and that a constant times a continuously differentiable function is continuously differentiable. Thus, the functions that are continuously, differentiable on (−⬁, ⬁) form a subspace of F (−⬁, ⬁). We will denote this subspace, by C 1 (−⬁, ⬁), where the superscript emphasizes that the first derivatives are continuous., To take this a step further, the set of functions with m continuous derivatives on (−⬁, ⬁), is a subspace of F (−⬁, ⬁) as is the set of functions with derivatives of all orders on, (−⬁, ⬁). We will denote these subspaces by C m (−⬁, ⬁) and C ⬁ (−⬁, ⬁), respectively., E X A M P L E 9 The Subspace of All Polynomials, , Recall that a polynomial is a function that can be expressed in the form, , p(x) = a0 + a1 x + · · · + an x n, , In this text we regard all constants to be polynomials of degree zero. Be aware, however,, that some authors do not assign a degree to the constant 0., , (1), , where a0 , a1 , . . . , an are constants. It is evident that the sum of two polynomials is a, polynomial and that a constant times a polynomial is a polynomial. Thus, the set W of all, polynomials is closed under addition and scalar multiplication and hence is a subspace, of F (−⬁, ⬁). We will denote this space by P⬁ ., E X A M P L E 10 The Subspace of Polynomials of Degree ≤ n, , Recall that the degree of a polynomial is the highest power of the variable that occurs with, a nonzero coefficient. Thus, for example, if an  = 0 in Formula (1), then that polynomial, has degree n. It is not true that the set W of polynomials with positive degree n is a, subspace of F (−⬁, ⬁) because that set is not closed under addition. For example, the, polynomials, 1 + 2x + 3x 2 and 5 + 7x − 3x 2, both have degree 2, but their sum has degree 1. What is true, however, is that for each, nonnegative integer n the polynomials of degree n or less form a subspace of F (−⬁, ⬁)., We will denote this space by Pn ., , The Hierarchy of Function, Spaces, , It is proved in calculus that polynomials are continuous functions and have continuous, derivatives of all orders on (−⬁, ⬁). Thus, it follows that P⬁ is not only a subspace of, F (−⬁, ⬁), as previously observed, but is also a subspace of C ⬁ (−⬁, ⬁). We leave it, for you to convince yourself that the vector spaces discussed in Examples 7 to 10 are, “nested” one inside the other as illustrated in Figure 4.2.5., Remark In our previous examples we considered functions that were defined at all points of the, interval (−⬁, ⬁). Sometimes we will want to consider functions that are only defined on some, subinterval of (−⬁, ⬁), say the closed interval [a, b] or the open interval (a, b). In such cases, we will make an appropriate notation change. For example, C[a, b] is the space of continuous, functions on [a, b] and C(a, b) is the space of continuous functions on (a, b).

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4.2 Subspaces, , 195, , Pn, , Figure 4.2.5, , Building Subspaces, , C ∞(–∞, ∞), C m(–∞, ∞), C 1(–∞, ∞), C(–∞, ∞), F(–∞, ∞), , The following theorem provides a useful way of creating a new subspace from known, subspaces., THEOREM 4.2.2 If W1 , W2 , . . . , Wr are subspaces of a vector space V, then the inter-, , section of these subspaces is also a subspace of V ., , Note that the first step in, proving Theorem 4.2.2 was, to establish that W contained, at least one vector. This is important, for otherwise the subsequent argument might be, logically correct but meaningless., If k = 1, then Equation (2) has, the form w = k1 v1 , in which, case the linear combination is, just a scalar multiple of v1 ., , Proof Let W be the intersection of the subspaces W1 , W2 , . . . , Wr . This set is not, empty because each of these subspaces contains the zero vector of V , and hence so does, their intersection. Thus, it remains to show that W is closed under addition and scalar, multiplication., To prove closure under addition, let u and v be vectors in W . Since W is the intersection of W1 , W2 , . . . , Wr , it follows that u and v also lie in each of these subspaces., Moreover, since these subspaces are closed under addition and scalar multiplication, they, also all contain the vectors u + v and k u for every scalar k , and hence so does their intersection W . This proves that W is closed under addition and scalar multiplication., , Sometimes we will want to find the “smallest” subspace of a vector space V that contains all of the vectors in some set of interest. The following definition, which generalizes, Definition 4 of Section 3.1, will help us to do that., DEFINITION 2 If w is a vector in a vector space, , V, then w is said to be a linear, combination of the vectors v1 , v2 , . . . , vr in V if w can be expressed in the form, w = k1 v1 + k2 v2 + · · · + kr vr, , (2), , where k1 , k2 , . . . , kr are scalars. These scalars are called the coefficients of the linear, combination., , THEOREM 4.2.3 If S, , V, then:, , = {w1 , w2 , . . . , wr } is a nonempty set of vectors in a vector space, , (a) The set W of all possible linear combinations of the vectors in S is a subspace of V ., (b) The set W in part (a) is the “smallest” subspace of V that contains all of the vectors, in S in the sense that any other subspace that contains those vectors contains W ., , W be the set of all possible linear combinations of the vectors in S . We, must show that W is closed under addition and scalar multiplication. To prove closure, under addition, let, Proof (a) Let, , u = c1 w1 + c2 w2 + · · · + cr wr and v = k1 w1 + k2 w2 + · · · + kr wr, be two vectors in W . It follows that their sum can be written as, u + v = (c1 + k1 )w1 + (c2 + k2 )w2 + · · · + (cr + kr )wr

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196, , Chapter 4 General Vector Spaces, , which is a linear combination of the vectors in S . Thus, W is closed under addition. We, leave it for you to prove that W is also closed under scalar multiplication and hence is a, subspace of V ., Proof (b) Let W be any subspace of V that contains all of the vectors in S . Since W, is closed under addition and scalar multiplication, it contains all linear combinations of, the vectors in S and hence contains W ., , The following definition gives some important notation and terminology related to, Theorem 4.2.3., In the case where S is the, empty set, it will be convenient, to agree that span(Ø) = {0}., , = {w1 , w2 , . . . , wr } is a nonempty set of vectors in a vector space, V , then the subspace W of V that consists of all possible linear combinations of the, vectors in S is called the subspace of V generated by S , and we say that the vectors, w1 , w2 , . . . , wr span W . We denote this subspace as, , DEFINITION 3 If S, , W = span{w1 , w2 , . . . , wr } or W = span(S), , E X A M P L E 11 The Standard Unit Vectors Span R n, , Recall that the standard unit vectors in R n are, e1 = (1, 0, 0, . . . , 0), e2 = (0, 1, 0, . . . , 0), . . . , en = (0, 0, 0, . . . , 1), These vectors span R n since every vector v = (v1 , v2 , . . . , vn ) in R n can be expressed as, v = v1 e1 + v2 e2 + · · · + vn en, which is a linear combination of e1 , e2 , . . . , en . Thus, for example, the vectors, i = (1, 0, 0), j = (0, 1, 0), k = (0, 0, 1), span R since every vector v = (a, b, c) in this space can be expressed as, 3, , v = (a, b, c) = a(1, 0, 0) + b(0, 1, 0) + c(0, 0, 1) = a i + bj + ck, E X A M P L E 1 2 A Geometric View of Spanning in R 2 and R 3, , (a) If v is a nonzero vector in R 2 or R 3 that has its initial point at the origin, then span{v},, which is the set of all scalar multiples of v, is the line through the origin determined, by v. You should be able to visualize this from Figure 4.2.6a by observing that the, tip of the vector k v can be made to fall at any point on the line by choosing the, value of k to lengthen, shorten, or reverse the direction of v appropriately., , George William Hill, (1838–1914), , Historical Note The term linear combination is due to the American, mathematician G.W. Hill, who introduced it in a research paper on planetary motion published in 1900. Hill was a “loner” who preferred to, work out of his home in West Nyack, NewYork, rather than in academia,, though he did try lecturing at Columbia University for a few years. Interestingly, he apparently returned the teaching salary, indicating that, he did not need the money and did not want to be bothered looking, after it. Although technically a mathematician, Hill had little interest in, modern developments of mathematics and worked almost entirely on, the theory of planetary orbits., [Image: Courtesy of the American Mathematical Society, www.ams.org]

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4.2 Subspaces, , 197, , (b) If v1 and v2 are nonzero vectors in R 3 that have their initial points at the origin,, then span{v1 , v2 }, which consists of all linear combinations of v1 and v2 , is the plane, through the origin determined by these two vectors. You should be able to visualize, this from Figure 4.2.6b by observing that the tip of the vector k1 v1 + k2 v2 can be, made to fall at any point in the plane by adjusting the scalars k1 and k2 to lengthen,, shorten, or reverse the directions of the vectors k1 v1 and k2 v2 appropriately., z, , z, , span{v}, kv, , span{v1, v2}, , v2, , v, y, , v1, , x, , Figure 4.2.6, , k1v1 + k2v2, , k 2v 2, k1v1, , y, , x, , (a) Span{v} is the line through the, origin determined by v., , (b) Span{v1, v2} is the plane through the, origin determined by v1 and v2., , E X A M P L E 1 3 A Spanning Set for Pn, , The polynomials 1, x, x 2 , . . . , x n span the vector space Pn defined in Example 10 since, each polynomial p in Pn can be written as, p = a0 + a1 x + · · · + an x n, which is a linear combination of 1, x, x 2 , . . . , x n . We can denote this by writing, , Pn = span{1, x, x 2 , . . . , x n }, The next two examples are concerned with two important types of problems:, • Given a nonempty set S of vectors in R n and a vector v in R n , determine whether v is, a linear combination of the vectors in S ., • Given a nonempty set S of vectors in R n , determine whether the vectors span R n ., E X A M P L E 1 4 Linear Combinations, , Consider the vectors u = (1, 2, −1) and v = (6, 4, 2) in R 3 . Show that w = (9, 2, 7) is, a linear combination of u and v and that w = (4, −1, 8) is not a linear combination of, u and v., Solution In order for w to be a linear combination of u and v, there must be scalars k1, and k2 such that w = k1 u + k2 v; that is,, , (9, 2, 7) = k1 (1, 2, −1) + k2 (6, 4, 2) = (k1 + 6k2 , 2k1 + 4k2 , −k1 + 2k2 ), Equating corresponding components gives, , k1 + 6k2 = 9, 2k1 + 4k2 = 2, −k1 + 2k2 = 7, Solving this system using Gaussian elimination yields k1 = −3, k2 = 2, so, w = −3u + 2v

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198, , Chapter 4 General Vector Spaces, , Similarly, for w to be a linear combination of u and v, there must be scalars k1 and, , k2 such that w = k1 u + k2 v; that is,, , (4, −1, 8) = k1 (1, 2, −1) + k2 (6, 4, 2) = (k1 + 6k2 , 2k1 + 4k2 , −k1 + 2k2 ), Equating corresponding components gives, , k1 + 6k2 =, , 4, , 2k1 + 4k2 = −1, , −k1 + 2k2 =, , 8, , This system of equations is inconsistent (verify), so no such scalars k1 and k2 exist., Consequently, w is not a linear combination of u and v., , E X A M P L E 1 5 Testing for Spanning, , Determine whether the vectors v1 = (1, 1, 2), v2 = (1, 0, 1), and v3 = (2, 1, 3) span the, vector space R 3 ., Solution We must determine whether an arbitrary vector b, , = (b1 , b2 , b3 ) in R 3 can be, , expressed as a linear combination, b = k1 v1 + k2 v2 + k3 v3, of the vectors v1 , v2 , and v3 . Expressing this equation in terms of components gives, , (b1 , b2 , b3 ) = k1 (1, 1, 2) + k2 (1, 0, 1) + k3 (2, 1, 3), or, , (b1 , b2 , b3 ) = (k1 + k2 + 2k3 , k1 + k3 , 2k1 + k2 + 3k3 ), or, , k1 + k2 + 2k3 = b1, k1, + k3 = b2, 2k1 + k2 + 3k3 = b3, , Thus, our problem reduces to ascertaining whether this system is consistent for all values, of b1 , b2 , and b3 . One way of doing this is to use parts (e) and (g) of Theorem 2.3.8,, which state that the system is consistent if and only if its coefficient matrix, , ⎡, , 1, ⎢, A = ⎣1, 2, , 1, 0, 1, , ⎤, , 2, ⎥, 1⎦, 3, , has a nonzero determinant. But this is not the case here since det(A) = 0 (verify), so v1 ,, v2 , and v3 do not span R 3 ., Solution Spaces of, Homogeneous Systems, , The solutions of a homogeneous linear system Ax = 0 of m equations in n unknowns, can be viewed as vectors in R n . The following theorem provides a useful insight into the, geometric structure of the solution set., THEOREM 4.2.4 The solution set of a homogeneous linear system, , tions in n unknowns is a subspace of R n ., , Ax = 0 of m equa-, , W be the solution set of the system. The set W is not empty because it, contains at least the trivial solution x = 0., , Proof Let

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4.2 Subspaces, , 199, , To show that W is a subspace of R n , we must show that it is closed under addition, and scalar multiplication. To do this, let x1 and x2 be vectors in W . Since these vectors, are solutions of Ax = 0, we have, , Ax1 = 0 and Ax2 = 0, It follows from these equations and the distributive property of matrix multiplication, that, A(x1 + x2 ) = Ax1 + Ax2 = 0 + 0 = 0, so W is closed under addition. Similarly, if k is any scalar then, , A(k x1 ) = kAx1 = k 0 = 0, so W is also closed under scalar multiplication., Because the solution set of a homogeneous system in n unknowns is actually a, subspace of R n , we will generally refer to it as the solution space of the system., E X A M P L E 1 6 Solution Spaces of Homogeneous Systems, , In each part, solve the system by any method and then give a geometric description of, the solution set., , ⎡, , 1, ⎢, (a) ⎣2, 3, , ⎡, , 1, ⎢, (c) ⎣−3, 4, , ⎤⎡ ⎤, , −2, −4, −6, −2, 7, 1, , ⎡ ⎤, , ⎡, , 3, 0, x, ⎥⎢ ⎥ ⎢ ⎥, 6⎦ ⎣y ⎦ = ⎣0⎦, 0, z, 9, , 1, ⎢, (b) ⎣−3, −2, , x, 3, 0, ⎥⎢ ⎥ ⎢ ⎥, −8⎦ ⎣y ⎦ = ⎣0⎦, 0, 2, z, , 0, ⎢, (d) ⎣0, 0, , ⎤⎡ ⎤, , ⎡ ⎤, , ⎡, , −2, 7, 4, 0, 0, 0, , ⎤⎡ ⎤, , ⎡ ⎤, , 3, 0, x, ⎥⎢ ⎥ ⎢ ⎥, −8⎦ ⎣y ⎦ = ⎣0⎦, 0, −6, z, , ⎤⎡ ⎤, , ⎡ ⎤, , 0, x, 0, ⎥⎣ ⎦ ⎣ ⎦, 0⎦ y = 0, 0, z, 0, , Solution, , (a) The solutions are, , x = 2s − 3t, y = s, z = t, from which it follows that, , x = 2y − 3z or x − 2y + 3z = 0, This is the equation of a plane through the origin that has n = (1, −2, 3) as a, normal., (b) The solutions are, , x = −5t, y = −t, z = t, which are parametric equations for the line through the origin that is parallel to the, vector v = (−5, −1, 1)., (c) The only solution is x = 0, y = 0, z = 0, so the solution space consists of the single, point {0}., (d) This linear system is satisfied by all real values of x , y , and z, so the solution space, is all of R 3 ., Remark Whereas the solution set of every homogeneous system of m equations in n unknowns is, a subspace of R n , it is never true that the solution set of a nonhomogeneous system of m equations, in n unknowns is a subspace of R n . There are two possible scenarios: first, the system may not, have any solutions at all, and second, if there are solutions, then the solution set will not be closed, either under addition or under scalar multiplication (Exercise 18).

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200, , Chapter 4 General Vector Spaces, , The LinearTransformation, Viewpoint, , Theorem 4.2.4 can be viewed as a statement about matrix transformations by letting, TA : R n →R m be multiplication by the coefficient matrix A. From this point of view, the solution space of Ax = 0 is the set of vectors in R n that TA maps into the zero, vector in R m . This set is sometimes called the kernel of the transformation, so with this, terminology Theorem 4.2.4 can be rephrased as follows., THEOREM 4.2.5 If A is an m × n matrix, then the kernel of the matrix transformation, , TA : R n →R m is a subspace of R n ., , A Concluding Observation, , It is important to recognize that spanning sets are not unique. For example, any nonzero, vector on the line in Figure 4.2.6a will span that line, and any two noncollinear vectors, in the plane in Figure 4.2.6b will span that plane. The following theorem, whose proof, is left as an exercise, states conditions under which two sets of vectors will span the same, space., , = {v1 , v2 , . . . , vr } and S = {w1 , w2 , . . . , wk } are nonempty sets, of vectors in a vector space V, then, , THEOREM 4.2.6 If S, , span{v1 , v2 , . . . , vr } = span{w1 , w2 , . . . , wk }, if and only if each vector in S is a linear combination of those in S , and each vector in, S is a linear combination of those in S ., , Exercise Set 4.2, 1. Use Theorem 4.2.1 to determine which of the following are, subspaces of R 3 ., (a) All vectors of the form (a, 0, 0)., (b) All vectors of the form (a, 1, 1)., (c) All vectors of the form (a, b, c), where b = a + c., (d) All vectors of the form (a, b, c), where b = a + c + 1., (e) All vectors of the form (a, b, 0)., 2. Use Theorem 4.2.1 to determine which of the following are, subspaces of Mnn ., (a) The set of all diagonal n × n matrices., , 3. Use Theorem 4.2.1 to determine which of the following are, subspaces of P3 ., (a) All polynomials a0 + a1 x + a2 x 2 + a3 x 3 for which, a0 = 0., (b) All polynomials a0 + a1 x + a2 x 2 + a3 x 3 for which, a0 + a1 + a2 + a3 = 0., (c) All polynomials of the form a0 + a1 x + a2 x 2 + a3 x 3 in, which a0 , a1 , a2 , and a3 are rational numbers., (d) All polynomials of the form a0 + a1 x , where a0 and a1 are, real numbers., 4. Which of the following are subspaces of F (−⬁, ⬁)?, , (b) The set of all n × n matrices A such that det(A) = 0., , (a) All functions f in F (−⬁, ⬁) for which f(0) = 0., , (c) The set of all n × n matrices A such that tr(A) = 0., , (b) All functions f in F (−⬁, ⬁) for which f(0) = 1., , (d) The set of all symmetric n × n matrices., , (c) All functions f in F (−⬁, ⬁) for which f(−x) = f(x)., , (e) The set of all n × n matrices A such that AT = −A., , (d) All polynomials of degree 2., , (f ) The set of all n × n matrices A for which Ax = 0 has only, the trivial solution., (g) The set of all n × n matrices A such that AB = BA for, some fixed n × n matrix B ., , 5. Which of the following are subspaces of R ⬁ ?, (a) All sequences v in R ⬁ of the form, v = (v, 0, v, 0, v, 0, . . .).

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4.2 Subspaces, , (b) All sequences v in R ⬁ of the form, v = (v, 1, v, 1, v, 1, . . .)., , origin only. If it is a plane, find an equation for it. If it is a, line, find parametric equations for it., , (c) All sequences v in R ⬁ of the form, v = (v, 2v, 4v, 8v, 16v, . . .)., (d) All sequences in R ⬁ whose components are 0 from some, point on., 6. A line L through the origin in R 3 can be represented by parametric equations of the form x = at , y = bt , and z = ct . Use, these equations to show that L is a subspace of R 3 by showing, that if v1 = (x1 , y1 , z1 ) and v2 = (x2 , y2 , z2 ) are points on L, and k is any real number, then k v1 and v1 + v2 are also points, on L., 7. Which of the following are linear combinations of, u = (0, −2, 2) and v = (1, 3, −1)?, (a) (2, 2, 2), , (b) (0, 4, 5), , (c) (0, 0, 0), , (b) (6, 11, 6), , ⎡, −1, ⎢, (a) A = ⎣ 3, 2, , ⎡, , 1, ⎢, (c) A = ⎣2, 3, , 1, −1, −4, , −3, −6, −9, , ⎤, , 1, ⎥, 0⎦, −5, , ⎤, , 1, ⎥, 2⎦, 3, , (a), , 4, , 0, , −2, , −2, , −8, −8, , 6, −1, , , B=, (b), , 0, 0, , 1, 2, , 3, , , C=, , 0, 0, , (c), , 0, 1, , 2, ?, 4, , −1, , 5, 1, , 7, , (a) −9 − 7x − 15x 2, , (b) 6 + 11x + 6x 2, , (c) 0, , (d) 7 + 8x + 9x 2, , (a) v1 = (2, 2, 2), v2 = (0, 0, 3), v3 = (0, 1, 1), (b) v1 = (2, −1, 3), v2 = (4, 1, 2), v3 = (8, −1, 8), 12. Suppose that v1 = (2, 1, 0, 3), v2 = (3, −1, 5, 2), and, v3 = (−1, 0, 2, 1). Which of the following vectors are in, span{v1 , v2 , v3 }?, (a) (2, 3, −7, 3), , (b) (0, 0, 0, 0), , (c) (1, 1, 1, 1), , (d) (−4, 6, −13, 4), , 13. Determine whether the following polynomials span P2 ., p1 = 1 − x + 2x 2 , p2 = 3 + x,, p3 = 5 − x + 4x 2 , p4 = −2 − 2x + 2x 2, 14. Let f = cos2 x and g = sin2 x . Which of the following lie in, the space spanned by f and g?, (d) sin x, , 1, ⎥, 4⎦, 11, , 1, , 17. (Calculus required ) Show that the set of continuous functions, f = f(x) on [a, b] such that, , , , b, , f(x) dx = 0, , is a subspace of C [a, b]., 18. Show that the solution vectors of a consistent nonhomogeneous system of m linear equations in n unknowns do not, form a subspace of R n ., 19. In each part, let TA : R 2 →R 2 be multiplication by A, and, let u1 = (1, 2) and u2 = (−1, 1). Determine whether the set, {TA (u1 ), TA (u2 )} spans R 2 ., (a) A =, , 11. In each part, determine whether the vectors span R ., , (c) 1, , ⎤, , −1, −1, , 1, ⎢, (d) A = ⎣2, 3, , (c) All differentiable functions on (−⬁, ⬁) that satisfy, f + 2f = 0., , , , 3, , (b) 3 + x 2, , ⎡, , 3, ⎥, 3⎦, 8, , a, , 10. In each part express the vector as a linear combination of, p1 = 2 + x + 4x 2 , p2 = 1 − x + 3x 2 , and, p3 = 3 + 2x + 5x 2 ., , (a) cos 2x, , ⎤, , 2, 5, 0, , (a) All continuous functions on (−⬁, ⬁)., , (c) (0, 0, 0), , −1, , 1, ⎢, (b) A = ⎣2, 1, , 16. (Calculus required ) Show that the following sets of functions, are subspaces of F (−⬁, ⬁)., , 9. Which of the following are linear combinations of, , A=, , ⎡, , (b) All differentiable functions on (−⬁, ⬁)., , 8. Express the following as linear combinations of u = (2, 1, 4),, v = (1, −1, 3), and w = (3, 2, 5)., (a) (−9, −7, −15), , 201, , (e) 0, , 15. Determine whether the solution space of the system Ax = 0, is a line through the origin, a plane through the origin, or the, , 1, , −1, , 0, , 2, , , , , , (b) A =, , 1, , −1, , −2, , 2, , , , 20. In each part, let TA : R 3 →R 2 be multiplication by A, and let, u1 = (0, 1, 1) and u2 = (2, −1, 1) and u3 = (1, 1, −2). Determine whether the set {TA (u1 ), TA (u2 ), TA (u3 )} spans R 2 ., , , , (a) A =, , 1, , 1, , 0, , 0, , 1, , −1, , , , , , (b) A =, , 0, , 1, , 0, , 1, , 1, , −3, , , , 21. If TA is multiplication by a matrix A with three columns, then, the kernel of TA is one of four possible geometric objects. What, are they? Explain how you reached your conclusion., 22. Let v1 = (1, 6, 4), v2 = (2, 4, −1), v3 = (−1, 2, 5), and, w1 = (1, −2, −5), w2 = (0, 8, 9). Use Theorem 4.2.6 to show, that span{v1 , v2 , v3 } = span{w1 , w2 }., 23. The accompanying figure shows a mass-spring system in which, a block of mass m is set into vibratory motion by pulling the, block beyond its natural position at x = 0 and releasing it at, time t = 0. If friction and air resistance are ignored, then the, x -coordinate x(t) of the block at time t is given by a function, of the form, x(t) = c1 cos ωt + c2 sin ωt

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202, , Chapter 4 General Vector Spaces, , where ω is a fixed constant that depends on the mass of the, block and the stiffness of the spring and c1 and c2 are arbitrary. Show that this set of functions forms a subspace of, C ⬁ (−⬁, ⬁)., Natural position, , m, , (g) The intersection of any two subspaces of a vector space V is a, subspace of V., (h) The union of any two subspaces of a vector space V is a subspace of V., (i) Two subsets of a vector space V that span the same subspace, of V must be equal., , x, , ( j) The set of upper triangular n × n matrices is a subspace of the, vector space of all n × n matrices., , 0, Stretched, , m, , (k) The polynomials x − 1, (x − 1)2 , and (x − 1)3 span P3 ., , x, , 0, , Working withTechnology, , Released, , m, , T1. Recall from Theorem 1.3.1 that a product Ax can be expressed, as a linear combination of the column vectors of the matrix A in, which the coefficients are the entries of x. Use matrix multiplication to compute, , x, 0, , Figure Ex-23, , v = 6(8, −2, 1, −4) + 17(−3, 9, 11, 6) − 9(13, −1, 2, 4), , Working with Proofs, , T2. Use the idea in Exercise T1 and matrix multiplication to determine whether the polynomial, , 24. Prove Theorem 4.2.6., , p = 1 + x + x2 + x3, , True-False Exercises, TF. In parts (a)–(k) determine whether the statement is true or, false, and justify your answer., (a) Every subspace of a vector space is itself a vector space., , is in the span of, p1 = 8 − 2x + x 2 − 4x 3 , p2 = −3 + 9x + 11x 2 + 6x 3 ,, p3 = 13 − x + 2x 2 + 4x 3, , (b) Every vector space is a subspace of itself., (c) Every subset of a vector space V that contains the zero vector, in V is a subspace of V., , T3. For the vectors that follow, determine whether, , (d) The kernel of a matrix transformation TA : R →R is a subspace of R m ., n, , span{v1 , v2 , v3 } = span{w1 , w2 , w3 }, , m, , (e) The solution set of a consistent linear system Ax = b of m, equations in n unknowns is a subspace of R n ., (f ) The span of any finite set of vectors in a vector space is closed, under addition and scalar multiplication., , v1 = (−1, 2, 0, 1, 3), v2 = (7, 4, 6, −3, 1),, v3 = (−5, 3, 1, 2, 4), w1 = (−6, 5, 1, 3, 7), w2 = (6, 6, 6, −2, 4),, w3 = (2, 7, 7, −1, 5), , 4.3 Linear Independence, In this section we will consider the question of whether the vectors in a given set are, interrelated in the sense that one or more of them can be expressed as a linear combination, of the others. This is important to know in applications because the existence of such, relationships often signals that some kind of complication is likely to occur., , Linear Independence and, Dependence, , In a rectangular xy -coordinate system every vector in the plane can be expressed in, exactly one way as a linear combination of the standard unit vectors. For example, the, only way to express the vector (3, 2) as a linear combination of i = (1, 0) and j = (0, 1), is, (1), (3, 2) = 3(1, 0) + 2(0, 1) = 3i + 2j

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4.3 Linear Independence, y, 2, , (3, 2), 3i, , +, , (Figure 4.3.1). Suppose, however, that we were to introduce a third coordinate axis that, makes an angle of 45◦ with the x -axis. Call it the w -axis. As illustrated in Figure 4.3.2,, the unit vector along the w -axis is, , 2j, , j, 3, , 1, , √ ,√, 2, , !, , 2, , Whereas Formula (1) shows the only way to express the vector (3, 2) as a linear combination of i and j, there are infinitely many ways to express this vector as a linear combination, of i, j, and w. Three possibilities are, , Figure 4.3.1, y, , 1, , w=, , x, i, , w, , 1, , w, , ( √2, , 45°, , ,, , 1, , ), √2, , x, , Figure 4.3.2, , (3, 2) = 2(1, 0) + (0, 1) +, , √, , (3, 2) = 4(1, 0) + 3(0, 1) −, , 2, , 2, , 1, , 1, , = 3i + 2j + 0w, !, = 3i + j +, , √ ,√, , 2, , √, , !, , 1, , (3, 2) = 3(1, 0) + 2(0, 1) + 0 √ , √, 1, , 203, , 2, , 2, , 2, , 1, , 1, , √ ,√, 2, , 2, , √, , !, = 4i + 3j −, , 2w, , √, , 2w, , In short, by introducing a superfluous axis we created the complication of having multiple ways of assigning coordinates to points in the plane. What makes the vector w, superfluous is the fact that it can be expressed as a linear combination of the vectors i, and j, namely,, !, 1, 1, 1, 1, w= √ ,√, = √ i+ √ j, 2, 2, 2, 2, This leads to the following definition., DEFINITION 1 If S = {v1 , v2 , . . . , vr } is a set of two or more vectors in a vector space, V , then S is said to be a linearly independent set if no vector in S can be expressed as, , a linear combination of the others. A set that is not linearly independent is said to be, linearly dependent., , In the case where the set S in, Definition 1 has only one vector, we will agree that S is linearly independent if and only, if that vector is nonzero., , In general, the most efficient way to determine whether a set is linearly independent, or not is to use the following theorem whose proof is given at the end of this section., , A nonempty set S = {v1 , v2 , . . . , vr } in a vector space V is linearly, independent if and only if the only coefficients satisfying the vector equation, , THEOREM 4.3.1, , k1 v1 + k2 v2 + · · · + kr vr = 0, are k1 = 0, k2 = 0, . . . , kr = 0., , E X A M P L E 1 Linear Independence of the Standard Unit Vectors in R n, , The most basic linearly independent set in R n is the set of standard unit vectors, e1 = (1, 0, 0, . . . , 0), e2 = (0, 1, 0, . . . , 0), . . . , en = (0, 0, 0, . . . , 1), To illustrate this in R 3 , consider the standard unit vectors, i = (1, 0, 0), j = (0, 1, 0), k = (0, 0, 1)

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204, , Chapter 4 General Vector Spaces, , To prove linear independence we must show that the only coefficients satisfying the vector, equation, k1 i + k 2 j + k 3 k = 0, are k1 = 0, k2 = 0, k3 = 0. But this becomes evident by writing this equation in its, component form, (k1 , k2 , k3 ) = (0, 0, 0), You should have no trouble adapting this argument to establish the linear independence, of the standard unit vectors in R n ., , E X A M P L E 2 Linear Independence in R 3, , Determine whether the vectors, v1 = (1, −2, 3), v2 = (5, 6, −1), v3 = (3, 2, 1), , (2), , are linearly independent or linearly dependent in R 3 ., Solution The linear independence or dependence of these vectors is determined by, , whether the vector equation, , k1 v1 + k2 v2 + k3 v3 = 0, , (3), , can be satisfied with coefficients that are not all zero. To see whether this is so, let us, rewrite (3) in the component form, , k1 (1, −2, 3) + k2 (5, 6, −1) + k3 (3, 2, 1) = (0, 0, 0), Equating corresponding components on the two sides yields the homogeneous linear, system, k1 + 5k2 + 3k3 = 0, , −2k1 + 6k2 + 2k3 = 0, 3 k1 − k 2 + k 3 = 0, , (4), , Thus, our problem reduces to determining whether this system has nontrivial solutions., There are various ways to do this; one possibility is to simply solve the system, which, yields, , k1 = − 21 t, k2 = − 21 t, k3 = t, (we omit the details). This shows that the system has nontrivial solutions and hence, that the vectors are linearly dependent. A second method for establishing the linear, dependence is to take advantage of the fact that the coefficient matrix, , ⎡, , 1, A = ⎣−2, 3, , 5, 6, −1, , ⎤, , 3, 2⎦, 1, , is square and compute its determinant. We leave it for you to show that det(A) = 0 from, which it follows that (4) has nontrivial solutions by parts (b) and (g) of Theorem 2.3.8., Because we have established that the vectors v1 , v2 , and v3 in (2) are linearly dependent, we know that at least one of them is a linear combination of the others. We leave, it for you to confirm, for example, that, v3 = 21 v1 + 21 v2

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4.3 Linear Independence, , 205, , E X A M P L E 3 Linear Independence in R 4, , Determine whether the vectors, v1 = (1, 2, 2, −1), v2 = (4, 9, 9, −4), v3 = (5, 8, 9, −5), in R 4 are linearly dependent or linearly independent., Solution The linear independence or linear dependence of these vectors is determined, , by whether there exist nontrivial solutions of the vector equation, , k1 v1 + k2 v2 + k3 v3 = 0, or, equivalently, of, , k1 (1, 2, 2, −1) + k2 (4, 9, 9, −4) + k3 (5, 8, 9, −5) = (0, 0, 0, 0), Equating corresponding components on the two, system, k1 + 4k2 + 5k3, 2k1 + 9k2 + 8k3, 2k1 + 9k2 + 9k3, −k1 − 4k2 − 5k3, , sides yields the homogeneous linear, , =0, =0, =0, =0, , We leave it for you to show that this system has only the trivial solution, , k1 = 0, k2 = 0, k3 = 0, from which you can conclude that v1 , v2 , and v3 are linearly independent., E X A M P L E 4 An Important Linearly Independent Set in Pn, , Show that the polynomials, 1, x, x 2 , . . . , x n, form a linearly independent set in Pn ., Solution For convenience, let us denote the polynomials as, , p0 = 1, p1 = x, p2 = x 2 , . . . , pn = x n, We must show that the only coefficients satisfying the vector equation, , a0 p0 + a1 p1 + a2 p2 + · · · + an pn = 0, , (5), , are, , a0 = a1 = a2 = · · · = an = 0, But (5) is equivalent to the statement that, , a0 + a1 x + a2 x 2 + · · · + an x n = 0, , (6), , for all x in (−⬁, ⬁), so we must show that this is true if and only if each coefficient in, (6) is zero. To see that this is so, recall from algebra that a nonzero polynomial of degree, n has at most n distinct roots. That being the case, each coefficient in (6) must be zero,, for otherwise the left side of the equation would be a nonzero polynomial with infinitely, many roots. Thus, (5) has only the trivial solution., , The following example shows that the problem of determining whether a given set of, vectors in Pn is linearly independent or linearly dependent can be reduced to determining, whether a certain set of vectors in R n is linearly dependent or independent.

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206, , Chapter 4 General Vector Spaces, , E X A M P L E 5 Linear Independence of Polynomials, , Determine whether the polynomials, p1 = 1 − x, p2 = 5 + 3x − 2x 2 , p3 = 1 + 3x − x 2, are linearly dependent or linearly independent in P2 ., Solution The linear independence or dependence of these vectors is determined by, , whether the vector equation, , k1 p1 + k2 p2 + k3 p3 = 0, , (7), , can be satisfied with coefficients that are not all zero. To see whether this is so, let us, rewrite (7) in its polynomial form, , k1 (1 − x) + k2 (5 + 3x − 2x 2 ) + k3 (1 + 3x − x 2 ) = 0, , (8), , or, equivalently, as, , (k1 + 5k2 + k3 ) + (−k1 + 3k2 + 3k3 )x + (−2k2 − k3 )x 2 = 0, Since this equation must be satisfied by all x in (−⬁, ⬁), each coefficient must be zero, , In Example 5, what relationship do you see between, the coefficients of the given, polynomials and the column, vectors of the coefficient matrix of system (9)?, , Sets with One orTwo, Vectors, , (as explained in the previous example). Thus, the linear dependence or independence, of the given polynomials hinges on whether the following linear system has a nontrivial, solution:, k1 + 5 k 2 + k 3 = 0, , −k1 + 3k2 + 3k3 = 0, − 2 k2 − k 3 = 0, , (9), , We leave it for you to show that this linear system has nontrivial solutions either by, solving it directly or by showing that the coefficient matrix has determinant zero. Thus,, the set {p1 , p2 , p3 } is linearly dependent., The following useful theorem is concerned with the linear independence and linear dependence of sets with one or two vectors and sets that contain the zero vector., THEOREM 4.3.2, , (a) A finite set that contains 0 is linearly dependent., (b) A set with exactly one vector is linearly independent if and only if that vector is, not 0., (c) A set with exactly two vectors is linearly independent if and only if neither vector, is a scalar multiple of the other., , We will prove part (a) and leave the rest as exercises., Proof (a) For any vectors v1 , v2 , . . . , vr , the set S, , = {v1 , v2 , . . . , vr , 0} is linearly depen-, , dent since the equation, 0v1 + 0v2 + · · · + 0vr + 1(0) = 0, expresses 0 as a linear combination of the vectors in S with coefficients that are not, all zero., E X A M P L E 6 Linear Independence of Two Functions, , The functions f1 = x and f2 = sin x are linearly independent vectors in F (−⬁, ⬁) since, neither function is a scalar multiple of the other. On the other hand, the two functions, g1 = sin 2x and g2 = sin x cos x are linearly dependent because the trigonometric identity sin 2x = 2 sin x cos x reveals that g1 and g2 are scalar multiples of each other.

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4.3 Linear Independence, , A Geometric Interpretation, of Linear Independence, , 207, , Linear independence has the following useful geometric interpretations in R 2 and R 3 :, • Two vectors in R 2 or R 3 are linearly independent if and only if they do not lie on the, same line when they have their initial points at the origin. Otherwise one would be a, scalar multiple of the other (Figure 4.3.3)., z, , z, , z, , v2, , v1, , v1, , v1, y, , x, , v2, x, , (a) Linearly dependent, , Figure 4.3.3, , v2, y, , y, , x, , (b) Linearly dependent, , (c) Linearly independent, , • Three vectors in R 3 are linearly independent if and only if they do not lie in the same, plane when they have their initial points at the origin. Otherwise at least one would, be a linear combination of the other two (Figure 4.3.4)., z, , z, , z, , v1, v3, , v3, v2, , x, , Figure 4.3.4, , v2, y, , v2, , y, v1, , v1, , x, , x, , (a) Linearly dependent, , (b) Linearly dependent, , y, , v3, , (c) Linearly independent, , At the beginning of this section we observed that a third coordinate axis in R 2 is, superfluous by showing that a unit vector along such an axis would have to be expressible, as a linear combination of unit vectors along the positive x - and y -axis. That result is, a consequence of the next theorem, which shows that there can be at most n vectors in, any linearly independent set R n ., THEOREM 4.3.3 Let S, , = {v1 , v2 , . . . , vr } be a set of vectors in R n . If r > n, then S is, , linearly dependent., Proof Suppose that, , v1 = (v11 , v12 , . . . , v1n ), v2 = (v21 , v22 , . . . , v2n ), , .., ., , .., ., vr = (vr 1 , vr 2 , . . . , vrn ), , and consider the equation, , k1 v1 + k2 v2 + · · · + kr vr = 0

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208, , Chapter 4 General Vector Spaces, , It follows from Theorem 4.3.3, that a set in R 2 with more than, two vectors is linearly dependent and a set in R 3 with more, than three vectors is linearly, dependent., , If we express both sides of this equation in terms of components and then equate the, corresponding components, we obtain the system, , v11 k1 + v21 k2 + · · · + vr 1 kr = 0, v12 k1 + v22 k2 + · · · + vr 2 kr = 0, .., .., .., .., ., ., ., ., v1n k1 + v2n k2 + · · · + vrn kr = 0, This is a homogeneous system of n equations in the r unknowns k1 , . . . , kr . Since, r > n, it follows from Theorem 1.2.2 that the system has nontrivial solutions. Therefore,, S = {v1 , v2 , . . . , vr } is a linearly dependent set., , CA L C U L U S R E Q U I R E D, , Linear Independence of, Functions, , Sometimes linear dependence of functions can be deduced from known identities. For, example, the functions, f1 = sin2 x, f2 = cos2 x, and f3 = 5, form a linearly dependent set in F (−⬁, ⬁), since the equation, 5f1 + 5f2 − f3 = 5 sin2 x + 5 cos2 x − 5, , = 5(sin2 x + cos2 x) − 5 = 0, expresses 0 as a linear combination of f1 , f2 , and f3 with coefficients that are not all zero., However, it is relatively rare that linear independence or dependence of functions can, be ascertained by algebraic or trigonometric methods. To make matters worse, there is, no general method for doing that either. That said, there does exist a theorem that can, be useful for that purpose in certain cases. The following definition is needed for that, theorem., , = f1 (x), f2 = f2 (x), . . . , fn = fn (x) are functions that are, n − 1 times differentiable on the interval (−⬁, ⬁), then the determinant, , ,  f1 (x), , f2 (x), · · · fn (x), , , , , f2 (x), · · · fn (x),  f1 (x), , , W (x) =  .., .., .., , ., .,  ., ,  (n−1), ,  f1, (x) f2(n−1) (x) · · · fn(n−1) (x) , , DEFINITION 2 If f1, , is called the Wronskian of f1 , f2 , . . . , fn ., , Józef Hoëné de Wroński, (1778–1853), , Historical Note The Polish-French mathematician Józef Hoëné de, Wroński was born Józef Hoëné and adopted the name Wroński after, he married. Wroński’s life was fraught with controversy and conflict,, which some say was due to psychopathic tendencies and his exaggeration of the importance of his own work. Although Wroński’s work, was dismissed as rubbish for many years, and much of it was indeed, erroneous, some of his ideas contained hidden brilliance and have survived. Among other things, Wroński designed a caterpillar vehicle to, compete with trains (though it was never manufactured) and did research on the famous problem of determining the longitude of a ship, at sea. His final years were spent in poverty., [Image: © TopFoto/The Image Works]

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4.3 Linear Independence, , 209, , Suppose for the moment that f1 = f1 (x), f2 = f2 (x), . . . , fn = fn (x) are linearly, dependent vectors in C (n−1) (−⬁, ⬁). This implies that the vector equation, , k1 f1 + k2 f2 + · · · + kn fn = 0, is satisfied by values of the coefficients k1 , k2 , . . . , kn that are not all zero, and for these, coefficients the equation, , k1 f1 (x) + k2 f2 (x) + · · · + kn fn (x) = 0, is satisfied for all x in (−⬁, ⬁). Using this equation together with those that result by, differentiating it n − 1 times we obtain the linear system, , k1 f1 (x), , + k2 f2 (x), , + · · · + kn fn (x), , =0, , k1 f1 (x), .., ., , + k2 f2 (x), .., ., , + · · · + kn fn (x), .., ., , =0, .., ., , k1 f1(n−1) (x) + k2 f2(n−1) (x) + · · · + kn fn(n−1) (x) = 0, Thus, the linear dependence of f1 , f2 , . . . , fn implies that the linear system, , ⎡, , f1 (x), ⎢, ⎢f1 (x), ⎢ ., ⎢ ., ⎣ ., f1(n−1) (x), , ⎤⎡ ⎤ ⎡ ⎤, 0, k1, ⎥⎢ ⎥ ⎢ ⎥, ⎥ ⎢k2 ⎥ ⎢0⎥, ⎥ ⎢ . ⎥ = ⎢.⎥, ⎥ ⎢ . ⎥ ⎢.⎥, ⎦ ⎣ . ⎦ ⎣.⎦, (n−1), 0, kn, · · · fn, (x), , · · · fn (x), · · · fn (x), .., ., , f2 (x), f2 (x), .., ., f2(n−1) (x), , (10), , has a nontrivial solution for every x in the interval (−⬁, ⬁), and this in turn implies, that the determinant of the coefficient matrix of (10) is zero for every such x . Since this, determinant is the Wronskian of f1 , f2 , . . . , fn , we have established the following result., WARNING The, , converse of, Theorem 4.3.4 is false. If the, Wronskian of f1 , f2 , . . . , fn is, identically zero on (−⬁, ⬁),, then no conclusion can be, reached about the linear independence of {f1 , f2 , . . . , fn }—, this set of vectors may be linearly independent or linearly, dependent., , n − 1 continuous derivatives, on the interval (−⬁, ⬁), and if the Wronskian of these functions is not identically, zero on (−⬁, ⬁), then these functions form a linearly independent set of vectors in, C (n−1) (−⬁, ⬁)., , THEOREM 4.3.4 If the functions f1 , f2 , . . . , fn have, , In Example 6 we showed that x and sin x are linearly independent functions by, observing that neither is a scalar multiple of the other. The following example illustrates, how to obtain the same result using the Wronskian (though it is a more complicated, procedure in this particular case)., E X A M P L E 7 Linear Independence Using the Wronskian, , Use the Wronskian to show that f1 = x and f2 = sin x are linearly independent vectors, in C ⬁ (−⬁, ⬁)., Solution The Wronskian is, , , x, W (x) = , 1, , , , sin x , = x cos x − sin x, cos x , , This function is not identically zero on the interval (−⬁, ⬁) since, for example,, , W, , 'π (, , =, , π, , cos, , 'π (, , 2, 2, 2, Thus, the functions are linearly independent., , − sin, , 'π (, 2, , =, , π, 2

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210, , Chapter 4 General Vector Spaces, , E X A M P L E 8 Linear Independence Using the Wronskian, , Use the Wronskian to show that f1 = 1, f2 = ex , and f3 = e2x are linearly independent, vectors in C ⬁ (−⬁, ⬁)., Solution The Wronskian is, , ,  1 ex, , , W(x) =  0 ex, ,  0 ex, , , e2x , , 2 e 2 x  = 2 e 3x, , 4 e 2x , , This function is obviously not identically zero on (−⬁, ⬁), so f1 , f2 , and f3 form a linearly, independent set., We will close this section by proving Theorem 4.3.1., , O PT I O N A L, , Proof of Theorem 4.3.1 We will prove this theorem in the case where the set S has two, or more vectors, and leave the case where S has only one vector as an exercise. Assume, first that S is linearly independent. We will show that if the equation, , k1 v1 + k2 v2 + · · · + kr vr = 0, , (11), , can be satisfied with coefficients that are not all zero, then at least one of the vectors in, , S must be expressible as a linear combination of the others, thereby contradicting the, assumption of linear independence. To be specific, suppose that k1  = 0. Then we can, rewrite (11) as, v1 =, , −, , k2, k1, , !, v2 + · · · + −, , kr, k1, , !, vr, , which expresses v1 as a linear combination of the other vectors in S ., Conversely, we must show that if the only coefficients satisfying (11) are, , k1 = 0, k2 = 0, . . . , kr = 0, then the vectors in S must be linearly independent. But if this were true of the coefficients and the vectors were not linearly independent, then at least one of them would be, expressible as a linear combination of the others, say, v1 = c2 v2 + · · · + cr vr, which we can rewrite as, v1 + (−c2 )v2 + · · · + (−cr )vr = 0, But this contradicts our assumption that (11) can only be satisfied by coefficients that, are all zero. Thus, the vectors in S must be linearly independent., , Exercise Set 4.3, 1. Explain why the following form linearly dependent sets of vectors. (Solve this problem by inspection.), (a) u1 = (−1, 2, 4) and u2 = (5, −10, −20) in R 3, (b) u1 = (3, −1), u2 = (4, 5), u3 = (−4, 7) in R 2, (c) p1 = 3 − 2x + x 2 and p2 = 6 − 4x + 2x 2 in P2, , , , (d) A =, , −3, , 4, , 2, , 0, , , , , , and B =, , 3, , −4, , −2, , 0, , , , 2. In each part, determine whether the vectors are linearly independent or are linearly dependent in R 3 ., (a) (−3, 0, 4), (5, −1, 2), (1, 1, 3), (b) (−2, 0, 1), (3, 2, 5), (6, −1, 1), (7, 0, −2), 3. In each part, determine whether the vectors are linearly independent or are linearly dependent in R 4 ., (a) (3, 8, 7, −3), (1, 5, 3, −1), (2, −1, 2, 6), (4, 2, 6, 4), , in M22, , (b) (3, 0, −3, 6), (0, 2, 3, 1), (0, −2, −2, 0), (−2, 1, 2, 1)

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4.3 Linear Independence, , whether the set {TA (u1 ), TA (u2 ), TA (u3 )} is linearly independent in R 3 ., , 4. In each part, determine whether the vectors are linearly independent or are linearly dependent in P2 ., , ⎡, , (a) 2 − x + 4x 2 , 3 + 6x + 2x 2 , 2 + 10x − 4x 2, , , , (a), , , (b), , , , , , 1, , 0, , 1, , 2, , 1, , 0, , 0, , 0, , 0, , 0, , ,, , , 1, , 2, , 2, , 1, , , , ,, , , , , , ,, , , , 0, , 1, , 2, , 1, , 0, , 0, , 1, , 0, , 0, , 0, , , , in M22, , , , ,, , 0, , 0, , 0, , 0, , 1, , 0, , , , , −1, 0, ,, k, k, , 1, 1, , 0, , , , 1, , , , ,, , 2, , 0, , 1, , 3, , 1, , (a) A = ⎣1, , 0, , ⎥, −3⎦, , 2, , 2, , 0, , ⎡, , ⎤, , 1, , 1, , (b) A = ⎣1, , 1, , ⎥, −3⎦, , 2, , 2, , 0, , ⎢, , 1, , 15. Are the vectors v1 , v2 , and v3 in part (a) of the accompanying figure linearly independent? What about those in part (b)?, Explain., , , , z, , in M23, , z, v3, , 6. Determine all values of k for which the following matrices are, linearly independent in M22 ., , , , 2, , ⎤, , 1, , ⎢, , (b) 1 + 3x + 3x 2 , x + 4x 2 , 5 + 6x + 3x 2 , 7 + 2x − x 2, 5. In each part, determine whether the matrices are linearly independent or dependent., , 211, , v3, v2, , , , v2, , v1, , y, v1, , 7. In each part, determine whether the three vectors lie in a plane, in R 3 ., (a) v1 = (2, −2, 0), v2 = (6, 1, 4), v3 = (2, 0, −4), , x, , x, , (a), , (b), , Figure Ex-15, , (b) v1 = (−6, 7, 2), v2 = (3, 2, 4), v3 = (4, −1, 2), 8. In each part, determine whether the three vectors lie on the, same line in R 3 ., (a) v1 = (−1, 2, 3), v2 = (2, −4, −6), v3 = (−3, 6, 0), , 16. By using appropriate identities, where required, determine, which of the following sets of vectors in F (−⬁, ⬁) are linearly dependent., , (b) v1 = (2, −1, 4), v2 = (4, 2, 3), v3 = (2, 7, −6), , (a) 6, 3 sin2 x, 2 cos2 x, , (b) x, cos x, , (c) v1 = (4, 6, 8), v2 = (2, 3, 4), v3 = (−2, −3, −4), , (c) 1, sin x, sin 2x, , (d) cos 2x, sin2 x, cos2 x, , (e) (3 − x)2 , x 2 − 6x, 5, , (f ) 0, cos3 πx, sin5 3πx, , 9. (a) Show that the three vectors v1 = (0, 3, 1, −1),, v2 = (6, 0, 5, 1), and v3 = (4, −7, 1, 3) form a linearly, dependent set in R 4 ., (b) Express each vector in part (a) as a linear combination of, the other two., 10. (a) Show that the vectors v1 = (1, 2, 3, 4), v2 = (0, 1, 0, −1),, and v3 = (1, 3, 3, 3) form a linearly dependent set in R 4 ., (b) Express each vector in part (a) as a linear combination of, the other two., 11. For which real values of λ do the following vectors form a, linearly dependent set in R 3 ?, v1 =, , , , λ, − 21 , − 21, , , , , v2 =, , , , − 21 , λ, − 21, , , , , v3 =, , , , − 21 , − 21 , λ, , , , 12. Under what conditions is a set with one vector linearly independent?, 13. In each part, let TA : R 2 →R 2 be multiplication by A, and, let u1 = (1, 2) and u2 = (−1, 1). Determine whether the set, {TA (u1 ), TA (u2 )} is linearly independent in R 2 ., , , , (a) A =, , 1, , −1, , 0, , 2, , , , , , (b) A =, , y, , 1, , −1, , −2, , 2, , , , 14. In each part, let TA : R 3 →R 3 be multiplication by A, and let, u1 = (1, 0, 0), u2 = (2, −1, 1), and u3 = (0, 1, 1). Determine, , 17. (Calculus required ) The functions, , f1 (x) = x and f2 (x) = cos x, are linearly independent in F (−⬁, ⬁) because neither function, is a scalar multiple of the other. Confirm the linear independence using the Wronskian., 18. (Calculus required ) The functions, , f1 (x) = sin x and f2 (x) = cos x, are linearly independent in F (−⬁, ⬁) because neither function, is a scalar multiple of the other. Confirm the linear independence using the Wronskian., 19. (Calculus required ) Use the Wronskian to show that the following sets of vectors are linearly independent., (a) 1, x, ex, , (b) 1, x, x 2, , 20. (Calculus required ) Use the Wronskian to show that the functions f1 (x) = ex , f2 (x) = xex , and f3 (x) = x 2 ex are linearly, independent vectors in C ⬁ (−⬁, ⬁)., 21. (Calculus required ) Use the Wronskian to show that the functions f1 (x) = sin x, f2 (x) = cos x , and f3 (x) = x cos x are, linearly independent vectors in C ⬁ (−⬁, ⬁).

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212, , Chapter 4 General Vector Spaces, , 22. Show that for any vectors u, v, and w in a vector space V , the, vectors u − v, v − w, and w − u form a linearly dependent set., 23. (a) In Example 1 we showed that the mutually orthogonal vectors i, j, and k form a linearly independent set of vectors in, R 3 . Do you think that every set of three nonzero mutually, orthogonal vectors in R 3 is linearly independent? Justify, your conclusion with a geometric argument., (b) Justify your conclusion with an algebraic argument. [Hint:, Use dot products.], , Working with Proofs, 24. Prove that if {v1 , v2 , v3 } is a linearly independent set of vectors,, then so are {v1 , v2 }, {v1 , v3 }, {v2 , v3 }, {v1 }, {v2 }, and {v3 }., 25. Prove that if S = {v1 , v2 , . . . , vr } is a linearly independent set, of vectors, then so is every nonempty subset of S ., 26. Prove that if S = {v1 , v2 , v3 } is a linearly dependent set of vectors in a vector space V, and v4 is any vector in V that is not, in S , then {v1 , v2 , v3 , v4 } is also linearly dependent., 27. Prove that if S = {v1 , v2 , . . . , vr } is a linearly dependent set of, vectors in a vector space V, and if vr+1 , . . . , vn are any vectors, in V that are not in S , then {v1 , v2 , . . . , vr , vr+1 , . . . , vn } is also, linearly dependent., 28. Prove that in P2 every set with more than three vectors is linearly dependent., 29. Prove that if {v1 , v2 } is linearly independent and v3 does not lie, in span{v1 , v2 }, then {v1 , v2 , v3 } is linearly independent., , (a) A set containing a single vector is linearly independent., (b) The set of vectors {v, k v} is linearly dependent for every, scalar k ., (c) Every linearly dependent set contains the zero vector., (d) If the set of vectors {v1 , v2 , v3 } is linearly independent, then, {k v1 , k v2 , k v3 } is also linearly independent for every nonzero, scalar k ., (e) If v1 , . . . , vn are linearly dependent nonzero vectors, then, at least one vector vk is a unique linear combination of, v1 , . . . , vk−1 ., (f ) The set of 2 × 2 matrices that contain exactly two 1’s and two, 0’s is a linearly independent set in M22 ., (g) The three polynomials (x − 1)(x + 2), x(x + 2), and, x(x − 1) are linearly independent., (h) The functions f1 and f2 are linearly dependent if there is a real, number x such that k1 f1 (x) + k2 f2 (x) = 0 for some scalars k1, and k2 ., , Working withTechnology, T1. Devise three different methods for using your technology utility to determine whether a set of vectors in R n is linearly independent, and then use each of those methods to determine whether, the following vectors are linearly independent., v1 = (4, −5, 2, 6), v2 = (2, −2, 1, 3),, , 30. Use part (a) of Theorem 4.3.1 to prove part (b)., 31. Prove part (b) of Theorem 4.3.2., , v3 = (6, −3, 3, 9), v4 = (4, −1, 5, 6), T2. Show that S = {cos t, sin t, cos 2t, sin 2t} is a linearly independent set in C(−⬁, ⬁) by evaluating the left side of the equation, , 32. Prove part (c) of Theorem 4.3.2., , c1 cos t + c2 sin t + c3 cos 2t + c4 sin 2t = 0, , True-False Exercises, TF. In parts (a)–(h) determine whether the statement is true or, false, and justify your answer., , at sufficiently many values of t to obtain a linear system whose, only solution is c1 = c2 = c3 = c4 = 0., , 4.4 Coordinates and Basis, We usually think of a line as being one-dimensional, a plane as two-dimensional, and the, space around us as three-dimensional. It is the primary goal of this section and the next to, make this intuitive notion of dimension precise. In this section we will discuss coordinate, systems in general vector spaces and lay the groundwork for a precise definition of, dimension in the next section., , Coordinate Systems in, Linear Algebra, , In analytic geometry one uses rectangular coordinate systems to create a one-to-one correspondence between points in 2-space and ordered pairs of real numbers and between, points in 3-space and ordered triples of real numbers (Figure 4.4.1). Although rectangular coordinate systems are common, they are not essential. For example, Figure 4.4.2, shows coordinate systems in 2-space and 3-space in which the coordinate axes are not, mutually perpendicular.

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4.4 Coordinates and Basis, , 213, , z, c, , y, , P(a, b, c), , P(a, b), , b, , y, b, x, a, , O, , Figure 4.4.1, , a, , x, , Coordinates of P in a rectangular, coordinate system in 2-space., , Coordinates of P in a rectangular, coordinate system in 3-space., , z, c, , y, , P(a, b, c), , P(a, b), , b, , y, b, x, a, , O, , Figure 4.4.2, , a, , x, , Coordinates of P in a nonrectangular, coordinate system in 2-space., , Coordinates of P in a nonrectangular, coordinate system in 3-space., , In linear algebra coordinate systems are commonly specified using vectors rather, than coordinate axes. For example, in Figure 4.4.3 we have re-created the coordinate, systems in Figure 4.4.2 by using unit vectors to identify the positive directions and then, attaching coordinates to a point P using the scalar coefficients in the equations, , −→, −→, OP = a u1 + bu2 and OP = a u1 + bu2 + cu3, , cu3, bu2, , u3, , P(a, b), , P(a, b, c), , u2, , O, u1, , Figure 4.4.3, , O, , u1, , au1, , u2, , bu2, , au1, , Units of measurement are essential ingredients of any coordinate system. In geometry problems one tries to use the same unit of measurement on all axes to avoid, distorting the shapes of figures. This is less important in applications where coordinates, represent physical quantities with diverse units (for example, time in seconds on one axis, and temperature in degrees Celsius on another axis). To allow for this level of generality,, we will relax the requirement that unit vectors be used to identify the positive directions, and require only that those vectors be linearly independent. We will refer to these as the, “basis vectors” for the coordinate system. In summary, it is the directions of the basis, vectors that establish the positive directions, and it is the lengths of the basis vectors that, establish the spacing between the integer points on the axes (Figure 4.4.4).

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214, , Chapter 4 General Vector Spaces, y, 4, 3, 2, 1, , –3 –2 –1, –1, , y, , y, , y, 2, , 4, , 2, 3, , x, 1, , 2 3, , 1, , x, –3 –2 –1, , –2, –3, –4, Equal spacing, Perpendicular axes, , 1, , 2, , 1, , 1, , 2 3, , –3, , –2, , –1, –1, , x, , x, 1, , 2, , –3 –2 –1, , 3, , –1, , –2, , –1, , 1 2 3, , –3, –4, , –2, Unequal spacing, Perpendicular axes, , Equal spacing, Skew axes, , –2, Unequal spacing, Skew axes, , Figure 4.4.4, , Basis for a Vector Space, , Our next goal is to extend the concepts of “basis vectors” and “coordinate systems” to, general vector spaces, and for that purpose we will need some definitions. Vector spaces, fall into two categories: A vector space V is said to be finite-dimensional if there is a, finite set of vectors in V that spans V and is said to be infinite-dimensional if no such set, exists., , = {v1 , v2 , . . . , vn } is a set of vectors in a finite-dimensional vector, space V , then S is called a basis for V if:, , DEFINITION 1 If S, , (a) S spans V., (b) S is linearly independent., If you think of a basis as describing a coordinate system for a finite-dimensional, vector space V , then part (a) of this definition guarantees that there are enough basis, vectors to provide coordinates for all vectors in V , and part (b) guarantees that there is, no interrelationship between the basis vectors. Here are some examples., E X A M P L E 1 The Standard Basis for R n, , Recall from Example 11 of Section 4.2 that the standard unit vectors, e1 = (1, 0, 0, . . . , 0), e2 = (0, 1, 0, . . . , 0), . . . , en = (0, 0, 0, . . . , 1), n, , span R and from Example 1 of Section 4.3 that they are linearly independent. Thus,, they form a basis for R n that we call the standard basis for R n . In particular,, i = (1, 0, 0), j = (0, 1, 0), k = (0, 0, 1), is the standard basis for R 3 ., E X A M P L E 2 The Standard Basis for Pn, , Show that S = {1, x, x 2 , . . . , x n } is a basis for the vector space Pn of polynomials of, degree n or less., Solution We must show that the polynomials in, , S are linearly independent and span, , Pn . Let us denote these polynomials by, p0 = 1, p1 = x, p2 = x 2 , . . . , pn = x n, We showed in Example 13 of Section 4.2 that these vectors span Pn and in Example 4, of Section 4.3 that they are linearly independent. Thus, they form a basis for Pn that we, call the standard basis for Pn .

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4.4 Coordinates and Basis, , 215, , E X A M P L E 3 Another Basis for R 3, , Show that the vectors v1 = (1, 2, 1), v2 = (2, 9, 0), and v3 = (3, 3, 4) form a basis for R 3 ., Solution We must show that these vectors are linearly independent and span, , R 3 . To, , prove linear independence we must show that the vector equation, , c1 v1 + c2 v2 + c3 v3 = 0, , (1), , has only the trivial solution; and to prove that the vectors span R 3 we must show that, every vector b = (b1 , b2 , b3 ) in R 3 can be expressed as, , c1 v1 + c2 v2 + c3 v3 = b, , (2), , By equating corresponding components on the two sides, these two equations can be, expressed as the linear systems, , c1 + 2c2 + 3c3 = 0, 2c1 + 9c2 + 3c3 = 0, , c1, , c1 + 2c2 + 3c3 = b1, 2c1 + 9c2 + 3c3 = b2, , and, , + 4 c3 = 0, , (3), , + 4 c3 = b 3, , c1, , (verify). Thus, we have reduced the problem to showing that in (3) the homogeneous, system has only the trivial solution and that the nonhomogeneous system is consistent, for all values of b1 , b2 , and b3 . But the two systems have the same coefficient matrix, , ⎡, , 1, ⎢, A = ⎣2, 1, From Examples 1 and 3 you, can see that a vector space can, have more than one basis., , ⎤, , 2, 9, 0, , 3, ⎥, 3⎦, 4, , so it follows from parts (b), (e), and (g) of Theorem 2.3.8 that we can prove both results, at the same time by showing that det(A)  = 0. We leave it for you to confirm that, det(A) = −1, which proves that the vectors v1 , v2 , and v3 form a basis for R 3 ., E X A M P L E 4 The Standard Basis for M mn, , Show that the matrices, 1 0, 0, M1 =, , M2 =, 0 0, 0, , 1, 0, , M3 =, 0, 1, , 0, 0, , M4 =, 0, 0, , 0, 1, , form a basis for the vector space M22 of 2 × 2 matrices., Solution We must show that the matrices are linearly independent and span M22 . To, prove linear independence we must show that the equation, , c1 M1 + c2 M2 + c3 M3 + c4 M4 = 0, , (4), , has only the trivial solution, where 0 is the 2 × 2 zero matrix; and to prove that the, matrices span M22 we must show that every 2 × 2 matrix, , B=, , a, c, , b, d, , can be expressed as, , c1 M1 + c2 M2 + c3 M3 + c4 M4 = B, , (5), , The matrix forms of Equations (4) and (5) are, , c1, , 1, 0, , 0, 0, + c2, 0, 0, , 1, 0, + c3, 0, 1, , 0, 0, + c4, 0, 0, , 0, 0, =, 1, 0, , 0, 0, , c1, , 1, 0, , 0, 0, + c2, 0, 0, , 1, 0, + c3, 0, 1, , 0, 0, + c4, 0, 0, , 0, a, =, 1, c, , b, d, , and

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216, , Chapter 4 General Vector Spaces, , which can be rewritten as, , c1, c3, , c2, 0 0, =, 0 0, c4, , and, , c1, c3, , c2, a, =, c4, c, , b, d, , Since the first equation has only the trivial solution, , c1 = c2 = c3 = c4 = 0, the matrices are linearly independent, and since the second equation has the solution, , c1 = a, c2 = b, c3 = c, c4 = d, the matrices span M22 . This proves that the matrices M1 , M2 , M3 , M4 form a basis for, M22 . More generally, the mn different matrices whose entries are zero except for a single, entry of 1 form a basis for Mmn called the standard basis for Mmn ., The simplest of all vector spaces is the zero vector space V = {0}. This space is, finite-dimensional because it is spanned by the vector 0. However, it has no basis in the, sense of Definition 1 because {0} is not a linearly independent set (why?). However, we, will find it useful to define the empty set Ø to be a basis for this vector space., E X A M P L E 5 An Infinite-Dimensional Vector Space, , Show that the vector space of P⬁ of all polynomials with real coefficients is infinitedimensional by showing that it has no finite spanning set., Solution If there were a finite spanning set, say S = {p1 , p2 , . . . , pr }, then the degrees, of the polynomials in S would have a maximum value, say n; and this in turn would, imply that any linear combination of the polynomials in S would have degree at most n., Thus, there would be no way to express the polynomial x n+1 as a linear combination of, the polynomials in S , contradicting the fact that the vectors in S span P⬁ ., , E X A M P L E 6 Some Finite- and Infinite-Dimensional Spaces, , In Examples 1, 2, and 4 we found bases for R n , Pn , and Mmn , so these vector spaces, are finite-dimensional. We showed in Example 5 that the vector space P⬁ is not spanned, by finitely many vectors and hence is infinite-dimensional. Some other examples of, infinite-dimensional vector spaces are R ⬁ , F (−⬁, ⬁), C(−⬁, ⬁), C m (−⬁, ⬁), and, C ⬁ (−⬁, ⬁)., Coordinates Relative to a, Basis, , Earlier in this section we drew an informal analogy between basis vectors and coordinate, systems. Our next goal is to make this informal idea precise by defining the notion of a, coordinate system in a general vector space. The following theorem will be our first step, in that direction., THEOREM 4.4.1 Uniqueness of Basis Representation, , If S = {v1 , v2 , . . . , vn } is a basis for a vector space V, then every vector v in V can be, expressed in the form v = c1 v1 + c2 v2 + · · · + cn vn in exactly one way., Proof Since S spans V, it follows from the definition of a spanning set that every vector, in V is expressible as a linear combination of the vectors in S . To see that there is only, one way to express a vector as a linear combination of the vectors in S , suppose that, some vector v can be written as, , v = c1 v1 + c2 v2 + · · · + cn vn

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4.4 Coordinates and Basis, , 217, , and also as, v = k1 v1 + k2 v2 + · · · + kn vn, Subtracting the second equation from the first gives, 0 = (c1 − k1 )v1 + (c2 − k2 )v2 + · · · + (cn − kn )vn, Since the right side of this equation is a linear combination of vectors in S , the linear, independence of S implies that, , c1 − k1 = 0, c2 − k2 = 0, . . . , cn − kn = 0, z, , that is,, , ck, k, , c1 = k1 , c2 = k2 , . . . , cn = kn, (0, 0, 1), , Thus, the two expressions for v are the same., (a, b, c), y, , j, i, x, , (1, 0, 0), , ai, , bj, , We now have all of the ingredients required to define the notion of “coordinates” in a, general vector space V . For motivation, observe that in R 3 , for example, the coordinates, (a, b, c) of a vector v are precisely the coefficients in the formula, v = a i + bj + ck, , (0, 1, 0), , Figure 4.4.5, , that expresses v as a linear combination of the standard basis vectors for R 3 (see Figure 4.4.5). The following definition generalizes this idea., DEFINITION 2 If S, , Sometimes it will be desirable, to write a coordinate vector as, a column matrix or row matrix, in which case we will denote it with square brackets as, [v]S . We will refer to this as the, matrix form of the coordinate, vector and (6) as the commadelimited form., , = {v1 , v2 , . . . , vn } is a basis for a vector space V , and, v = c1 v1 + c2 v2 + · · · + cn vn, , is the expression for a vector v in terms of the basis S , then the scalars c1 , c2 , . . . , cn, are called the coordinates of v relative to the basis S . The vector (c1 , c2 , . . . , cn ) in, R n constructed from these coordinates is called the coordinate vector of v relative to, S; it is denoted by, (v)S = (c1 , c2 , . . . , cn ), (6), , Remark It is standard to regard two sets to be the same if they have the same members, even if, those members are written in a different order. In particular, in a basis for a vector space V , which, is a set of linearly independent vectors that span V , the order in which those vectors are listed, does not generally matter. However, the order in which they are listed is critical for coordinate, vectors, since changing the order of the basis vectors changes the coordinate vectors [for example,, in R 2 the coordinate pair (1, 2) is not the same as the coordinate pair (2, 1)]. To deal with this, complication, many authors define an ordered basis to be one in which the listing order of the, basis vectors remains fixed. In all discussions involving coordinate vectors we will assume that the, underlying basis is ordered, even though we may not say so explicitly., , Observe that (v)S is a vector in R n , so that once an ordered basis S is given for a, vector space V , Theorem 4.4.1 establishes a one-to-one correspondence between vectors, in V and vectors in R n (Figure 4.4.6)., , A one-to-one correspondence, , (v)S, , v, , Figure 4.4.6, , V, , Rn

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218, , Chapter 4 General Vector Spaces, , E X A M P L E 7 Coordinates Relative to the Standard Basis for R n, , In the special case where V = R n and S is the standard basis, the coordinate vector (v)S, and the vector v are the same; that is,, v = (v)S, For example, in R 3 the representation of a vector v = (a, b, c) as a linear combination, of the vectors in the standard basis S = {i, j, k} is, v = a i + bj + ck, so the coordinate vector relative to this basis is (v)S = (a, b, c), which is the same as the, vector v., E X A M P L E 8 Coordinate Vectors Relative to Standard Bases, , (a) Find the coordinate vector for the polynomial, p(x) = c0 + c1 x + c2 x 2 + · · · + cn x n, relative to the standard basis for the vector space Pn ., (b) Find the coordinate vector of, , B=, , a, c, , b, d, , relative to the standard basis for M22 ., Solution (a) The given formula for p(x) expresses this polynomial as a linear combina-, , tion of the standard basis vectors S = {1, x, x 2 , . . . , x n }. Thus, the coordinate vector, for p relative to S is, (p)S = (c0 , c1 , c2 , . . . , cn ), Solution (b) We showed in Example 4 that the representation of a vector, , B=, , a, c, , b, d, , as a linear combination of the standard basis vectors is, , B=, , a, c, , 1 0, 0 1, 0 0, 0 0, b, =a, +b, +c, +d, 0 0, 0 0, 1 0, 0 1, d, , so the coordinate vector of B relative to S is, , (B)S = (a, b, c, d), E X A M P L E 9 Coordinates in R 3, , (a) We showed in Example 3 that the vectors, v1 = (1, 2, 1), v2 = (2, 9, 0), v3 = (3, 3, 4), form a basis for R 3 . Find the coordinate vector of v = (5, −1, 9) relative to the, basis S = {v1 , v2 , v3 }., (b) Find the vector v in R 3 whose coordinate vector relative to S is (v)S = (−1, 3, 2)., Solution (a) To find (v)S we must first express v as a linear combination of the vectors, in S ; that is, we must find values of c1 , c2 , and c3 such that, , v = c1 v1 + c2 v2 + c3 v3

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4.4 Coordinates and Basis, , 219, , or, in terms of components,, , (5, −1, 9) = c1 (1, 2, 1) + c2 (2, 9, 0) + c3 (3, 3, 4), Equating corresponding components gives, , c1 + 2c2 + 3c3 = 5, 2c1 + 9c2 + 3c3 = −1, c1, + 4 c3 = 9, Solving this system we obtain c1 = 1, c2 = −1, c3 = 2 (verify). Therefore,, , (v)S = (1, −1, 2), Solution (b) Using the definition of (v)S , we obtain, , v = (−1)v1 + 3v2 + 2v3, = (−1)(1, 2, 1) + 3(2, 9, 0) + 2(3, 3, 4) = (11, 31, 7), , Exercise Set 4.4, 1. Use the method of Example 3 to show that the following set, of vectors forms a basis for R 2 ., , ), *, (2, 1), (3, 0), 2. Use the method of Example 3 to show that the following set, of vectors forms a basis for R 3 ., , ), , (3, 1, −4), (2, 5, 6), (1, 4, 8), , *, , 4. Show that the following polynomials form a basis for P3 ., , 5. Show that the following matrices form a basis for M22 ., 3, , 6, , 3, , −6, , ,, , 0, , −1, , −1, , 0, , , , , , ,, , 0, , −12, , , , , , −8, ,, −4, , 1, , 0, , −1, , 2, , , , 6. Show that the following matrices form a basis for M22 ., , , , 1, , 1, , 1, , 1, , , , , , ,, , , , 1, , −1, , 0, , 0, , , , ,, , , , 0, , −1, , 1, , 0, , , , ,, , 1, , 0, , 0, , 0, , , , 7. In each part, show that the set of vectors is not a basis for R 3 ., , ), , *, , ), , *, , (a) (2, −3, 1), (4, 1, 1), (0, −7, 1), , (b) (1, 6, 4), (2, 4, −1), (−1, 2, 5), , 8. Show that the following vectors do not form a basis for P2 ., 1 − 3x + 2 x 2 ,, , 1 + x + 4x 2 ,, , 0, , 1, , 1, , , , ,, , 2, , −2, , 3, , 2, , , , , , ,, , 1, , , −1, , 1, , 0, , , , ,, , 0, , , −1, , 1, , 1, , 10. Let V be the space spanned by v1 = cos2 x , v2 = sin2 x ,, v3 = cos 2x ., (b) Find a basis for V ., , (a) u1 = (2, −4), u2 = (3, 8); w = (1, 1), (b) u1 = (1, 1), u2 = (0, 2); w = (a, b), , 1 + x, 1 − x, 1 − x 2 , 1 − x 3, , , , , , 1, , 11. Find the coordinate vector of w relative to the basis, S = {u1 , u2 } for R 2 ., , x 2 + 1, x 2 − 1, 2x − 1, , , , , , (a) Show that S = {v1 , v2 , v3 } is not a basis for V., , 3. Show that the following polynomials form a basis for P2 ., , , , 9. Show that the following matrices do not form a basis for M22 ., , 1 − 7x, , 12. Find the coordinate vector of w relative to the basis, S = {u1 , u2 } for R 2 ., (a) u1 = (1, −1), u2 = (1, 1); w = (1, 0), (b) u1 = (1, −1), u2 = (1, 1); w = (0, 1), 13. Find the coordinate vector of v relative to the basis, S = {v1 , v2 , v3 } for R 3 ., (a) v = (2, −1, 3); v1 = (1, 0, 0), v2 = (2, 2, 0),, v3 = (3, 3, 3), (b) v = (5, −12, 3); v1 = (1, 2, 3), v2 = (−4, 5, 6),, v3 = (7, −8, 9), 14. Find the coordinate vector of p relative to the basis, S = {p1 , p2 , p3 } for P2 ., (a) p = 4 − 3x + x 2 ; p1 = 1, p2 = x , p3 = x 2, (b) p = 2 − x + x 2 ; p1 = 1 + x , p2 = 1 + x 2 , p3 = x + x 2

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220, , Chapter 4 General Vector Spaces, , In Exercises 15–16, first show that the set S = {A1 , A2 , A3 , A4 }, is a basis for M22 , then express A as a linear combination of the, vectors in S , and then find the coordinate vector of A relative, to S ., 15. A1 =, , 1, 1, , 1, 0, , A2 =, 1, 1, , A4 =, , 0, 0, , 0, 1, ; A=, 1, 1, , 16. A1 =, , 1, 1, , 0, 1, , A2 =, 0, 0, , 0, 1, , 0, 6, ; A=, 0, 5, , A4 =, , 1, 0, , A3 =, 1, 1, , and u2 . Find the x y -coordinates of the points whose xy coordinates are given., , √, , (a) ( 3, 1), , (b) (1, 0), , (c) (0, 1), , (d) (a, b), , y and y', , 0, ,, 1, , x', , 0, 0, , j and u2, , 1, 1, , A3 =, 0, 0, , u1, , 0, ,, 1, , x, , 30°, i, , Figure Ex-23, , 2, 3, , In Exercises 17–18, first show that the set S = {p1 , p2 , p3 } is a, basis for P2 , then express p as a linear combination of the vectors, in S , and then find the coordinate vector of p relative to S ., 17. p1 = 1 + x + x 2 , p2 = x + x 2 , p3 = x 2 ;, p = 7 − x + 2x 2, , 24. The accompanying figure shows a rectangular xy -coordinate, system and an x y -coordinate system with skewed axes. Assuming that 1-unit scales are used on all the axes, find the x y coordinates of the points whose xy -coordinates are given., (a) (1, 1), , 18. p1 = 1 + 2x + x 2 , p2 = 2 + 9x, p3 = 3 + 3x + 4x 2 ;, p = 2 + 17x − 3x 2, , (b) (1, 0), , y, , (c) (0, 1), , (d) (a, b), , y´, , 19. In words, explain why the sets of vectors in parts (a) to (d) are, not bases for the indicated vector spaces., (a) u1 = (1, 2), u2 = (0, 3), u3 = (1, 5) for R 2, , 45°, , x and x´, , (b) u1 = (−1, 3, 2), u2 = (6, 1, 1) for R 3, , Figure Ex-24, , (c) p1 = 1 + x + x 2 , p2 = x for P2, , , , (d) A =, , , , 1, , 0, , 2, , 3, , 5, , 0, , 4, , 2, , , D=, , , , , B=, , , 6, , 0, , −1, , 4, , , , , , C=, , 3, , 0, , 1, , 7, , , 25. The first four Hermite polynomials [named for the French, mathematician Charles Hermite (1822–1901)] are, , ,, , 1, 2t, −2 + 4t 2 , −12t + 8t 3, , for M22, , 20. In any vector space a set that contains the zero vector must be, linearly dependent. Explain why this is so., 21. In each part, let TA : R 3 →R 3 be multiplication by A, and let, {e1 , e2 , e3 } be the standard basis for R 3 . Determine whether, the set {TA (e1 ), TA (e2 ), TA (e3 )} is linearly independent in R 2 ., , ⎡, , 1, , 1, , ⎢, (a) A = ⎣ 0, −1, , 1, , ⎤, , 1, , ⎥, −3⎦, , 2, , 0, , ⎡, , 1, , 1, , ⎢, (b) A = ⎣ 0, −1, , 2, , 2, , −1, , (a) A = ⎣1, , 1, , 0, , −1, , ⎢, , 0, , ⎤, ⎥, , 1⎦, 2, , ⎡, , 1, , ⎥, 1⎦, , 2, , 1, , ⎤, , 0, , 1, , 0, , (b) A = ⎣1, , 0, , 1⎦, , 0, , 0, , 1, , ⎢, , (a) Show that the first four Hermite polynomials form a basis, for P3 ., (b) Let B be the basis in part (a). Find the coordinate vector, of the polynomial, , ⎤, , 22. In each part, let TA : R 3 →R 3 be multiplication by A, and let, u = (1, −2, −1). Find the coordinate vector of TA (u) relative, to the basis S = {(1, 1, 0), (0, 1, 1), (1, 1, 1)} for R 3 ., , ⎡, , These polynomials have a wide variety of applications in, physics and engineering., , ⎥, , 23. The accompanying figure shows a rectangular xy -coordinate system determined by the unit basis vectors i and j and, an x y -coordinate system determined by unit basis vectors u1, , p(t) = −1 − 4t + 8t 2 + 8t 3, relative to B ., 26. The first four Laguerre polynomials [named for the French, mathematician Edmond Laguerre (1834–1886)] are, 1, 1 − t, 2 − 4t + t 2 , 6 − 18t + 9t 2 − t 3, (a) Show that the first four Laguerre polynomials form a basis, for P3 ., (b) Let B be the basis in part (a). Find the coordinate vector, of the polynomial, p(t) = −10t + 9t 2 − t 3, relative to B .

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4.5 Dimension, , 27. Consider the coordinate vectors, , ⎡ ⎤, −8, 6, 3, ⎢ 7⎥, ⎢ ⎥, ⎢ ⎥, ⎢ ⎥, [w]S = ⎣−1⎦ , [q]S = ⎣0⎦ , [B]S = ⎢ ⎥, ⎣ 6⎦, ⎡, , ⎤, , ⎡ ⎤, , 4, , 4, , 3, , 221, , (c) If {v1 , v2 , . . . , vn } is a basis for a vector space V, then every vector in V can be expressed as a linear combination of, v1 , v2 , . . . , vn ., (d) The coordinate vector of a vector x in R n relative to the standard basis for R n is x., , (a) Find w if S is the basis in Exercise 2., (b) Find q if S is the basis in Exercise 3., (c) Find B if S is the basis in Exercise 5., 28. The basis that we gave for M22 in Example 4 consisted of noninvertible matrices. Do you think that there is a basis for M22, consisting of invertible matrices? Justify your answer., , (e) Every basis of P4 contains at least one polynomial of degree 3, or less., , Working withTechnology, T1. Let V be the subspace of P3 spanned by the vectors, p1 = 1 + 5x − 3x 2 − 11x 3 , p2 = 7 + 4x − x 2 + 2x 3 ,, p3 = 5 + x + 9x 2 + 2x 3 ,, , Working with Proofs, , p4 = 3 − x + 7x 2 + 5x 3, , 29. Prove that R ⬁ is an infinite-dimensional vector space., , (a) Find a basis S for V ., , 30. Let TA : R →R be multiplication by an invertible matrix, A, and let {u1 , u2 , . . . , un } be a basis for R n . Prove that, {TA (u1 ), TA (u2 ), . . . , TA (un )} is also a basis for R n ., , (b) Find the coordinate vector of p = 19 + 18x − 13x 2 − 10x 3, relative to the basis S you obtained in part (a)., , n, , n, , 31. Prove that if V is a subspace of a vector space W and if V is, infinite-dimensional, then so is W ., , True-False Exercises, TF. In parts (a)–(e) determine whether the statement is true or, false, and justify your answer., (a) If V = span{v1 , . . . , vn }, then {v1 , . . . , vn } is a basis for V., (b) Every linearly independent subset of a vector space V is a, basis for V., , T2. Let V be the subspace of C ⬁ (−⬁, ⬁) spanned by the vectors, in the set, , B = {1, cos x, cos2 x, cos3 x, cos4 x, cos5 x}, and accept without proof that B is a basis for V . Confirm that, the following vectors are in V , and find their coordinate vectors, relative to B ., f0 = 1, f1 = cos x, f2 = cos 2x, f3 = cos 3x,, f4 = cos 4x, f5 = cos 5x, , 4.5 Dimension, We showed in the previous section that the standard basis for R n has n vectors and hence, that the standard basis for R 3 has three vectors, the standard basis for R 2 has two vectors, and, the standard basis for R 1 (= R) has one vector. Since we think of space as three-dimensional,, a plane as two-dimensional, and a line as one-dimensional, there seems to be a link between, the number of vectors in a basis and the dimension of a vector space. We will develop this, idea in this section., , Number of Vectors in a, Basis, , Our first goal in this section is to establish the following fundamental theorem., , THEOREM 4.5.1 All bases for a finite-dimensional vector space have the same number, , of vectors., , To prove this theorem we will need the following preliminary result, whose proof is, deferred to the end of the section.

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222, , Chapter 4 General Vector Spaces, THEOREM 4.5.2 Let, , V be an n-dimensional vector space, and let {v1 , v2 , . . . , vn } be, , any basis., (a) If a set in V has more than n vectors, then it is linearly dependent., (b) If a set in V has fewer than n vectors, then it does not span V., , We can now see rather easily why Theorem 4.5.1 is true; for if, , S = {v1 , v2 , . . . , vn }, is an arbitrary basis for V , then the linear independence of S implies that any set in V, with more than n vectors is linearly dependent and any set in V with fewer than n vectors, does not span V . Thus, unless a set in V has exactly n vectors it cannot be a basis., We noted in the introduction to this section that for certain familiar vector spaces, the intuitive notion of dimension coincides with the number of vectors in a basis. The, following definition makes this idea precise., DEFINITION 1 The dimension of a finite-dimensional vector space, , V is denoted by, dim(V ) and is defined to be the number of vectors in a basis for V . In addition, the, zero vector space is defined to have dimension zero., , Engineers often use the term, degrees of freedom as a synonym for dimension., , E X A M P L E 1 Dimensions of Some Familiar Vector Spaces, , dim(R n ) = n, , [ The standard basis has n vectors. ], , dim(Pn ) = n + 1 [ The standard basis has n + 1 vectors. ], dim(Mmn ) = mn [ The standard basis has mn vectors. ], E X A M P L E 2 Dimension of Span(S), , If S = {v1 , v2 , . . . , vr } then every vector in span(S) is expressible as a linear combination, of the vectors in S . Thus, if the vectors in S are linearly independent, they automatically, form a basis for span(S), from which we can conclude that, dim[span{v1 , v2 , . . . , vr }] = r, In words, the dimension of the space spanned by a linearly independent set of vectors is, equal to the number of vectors in that set., E X A M P L E 3 Dimension of a Solution Space, , Find a basis for and the dimension of the solution space of the homogeneous system, , x1 + 3x2 − 2x3, + 2 x5, =0, 2x1 + 6x2 − 5x3 − 2x4 + 4x5 − 3x6 = 0, 5x3 + 10x4, 2x1 + 6x2, , + 15x6 = 0, + 8x4 + 4x5 + 18x6 = 0, , Solution In Example 6 of Section 1.2 we found the solution of this system to be, , x1 = −3r − 4s − 2t, x2 = r, x3 = −2s, x4 = s, x5 = t, x6 = 0, which can be written in vector form as, , (x1 , x2 , x3 , x4 , x5 , x6 ) = (−3r − 4s − 2t, r, −2s, s, t, 0)

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4.5 Dimension, , 223, , or, alternatively, as, , (x1 , x2 , x3 , x4 , x5 , x6 ) = r(−3, 1, 0, 0, 0, 0) + s(−4, 0, −2, 1, 0, 0) + t(−2, 0, 0, 0, 1, 0), This shows that the vectors, v1 = (−3, 1, 0, 0, 0, 0), v2 = (−4, 0, −2, 1, 0, 0), v3 = (−2, 0, 0, 0, 1, 0), span the solution space. We leave it for you to check that these vectors are linearly, independent by showing that none of them is a linear combination of the other two (but, see the remark that follows). Thus, the solution space has dimension 3., Remark It can be shown that for any homogeneous linear system, the method of the last example, always produces a basis for the solution space of the system. We omit the formal proof., , Some Fundamental, Theorems, , We will devote the remainder of this section to a series of theorems that reveal the subtle, interrelationships among the concepts of linear independence, spanning sets, basis, and, dimension. These theorems are not simply exercises in mathematical theory—they are, essential to the understanding of vector spaces and the applications that build on them., We will start with a theorem (proved at the end of this section) that is concerned with, the effect on linear independence and spanning if a vector is added to or removed from, a nonempty set of vectors. Informally stated, if you start with a linearly independent set, S and adjoin to it a vector that is not a linear combination of those already in S , then, the enlarged set will still be linearly independent. Also, if you start with a set S of two, or more vectors in which one of the vectors is a linear combination of the others, then, that vector can be removed from S without affecting span(S ) (Figure 4.5.1)., , The vector outside the plane, can be adjoined to the other, two without affecting their, linear independence., , Any of the vectors can, be removed, and the, remaining two will still, span the plane., , Either of the collinear, vectors can be removed,, and the remaining two, will still span the plane., , Figure 4.5.1, , THEOREM 4.5.3 Plus/Minus Theorem, , Let S be a nonempty set of vectors in a vector space V., (a) If S is a linearly independent set, and if v is a vector in V that is outside of, span(S), then the set S ∪ {v} that results by inserting v into S is still linearly, independent., (b) If v is a vector in S that is expressible as a linear combination of other vectors, in S, and if S − {v} denotes the set obtained by removing v from S, then S and, S − {v} span the same space; that is,, span(S) = span(S − {v})

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224, , Chapter 4 General Vector Spaces, , E X A M P L E 4 Applying the Plus/Minus Theorem, , Show that p1 = 1 − x 2 , p2 = 2 − x 2 , and p3 = x 3 are linearly independent vectors., Solution The set S = {p1 , p2 } is linearly independent since neither vector in S is a scalar, multiple of the other. Since the vector p3 cannot be expressed as a linear combination, of the vectors in S (why?), it can be adjoined to S to produce a linearly independent set, S ∪ {p3 } = {p1 , p2 , p3 }., , In general, to show that a set of vectors {v1 , v2 , . . . , vn } is a basis for a vector space V,, one must show that the vectors are linearly independent and span V. However, if we, happen to know that V has dimension n (so that {v1 , v2 , . . . , vn } contains the right, number of vectors for a basis), then it suffices to check either linear independence or, spanning—the remaining condition will hold automatically. This is the content of the, following theorem., THEOREM 4.5.4 Let, , V be an n-dimensional vector space, and let S be a set in V, with exactly n vectors. Then S is a basis for V if and only if S spans V or S is linearly, independent., , S has exactly n vectors and spans V. To prove that S is a basis, we, must show that S is a linearly independent set. But if this is not so, then some vector v in, S is a linear combination of the remaining vectors. If we remove this vector from S , then, it follows from Theorem 4.5.3(b) that the remaining set of n − 1 vectors still spans V., But this is impossible since Theorem 4.5.2(b) states that no set with fewer than n vectors, can span an n-dimensional vector space. Thus S is linearly independent., Assume that S has exactly n vectors and is a linearly independent set. To prove, that S is a basis, we must show that S spans V. But if this is not so, then there is, some vector v in V that is not in span(S). If we insert this vector into S , then it follows from Theorem 4.5.3(a) that this set of n + 1 vectors is still linearly independent., But this is impossible, since Theorem 4.5.2(a) states that no set with more than n vectors in an n-dimensional vector space can be linearly independent. Thus S spans V., Proof Assume that, , E X A M P L E 5 Bases by Inspection, , (a) Explain why the vectors v1 = (−3, 7) and v2 = (5, 5) form a basis for R 2 ., (b) Explain why the vectors v1 = (2, 0, −1), v2 = (4, 0, 7), and v3 = (−1, 1, 4) form a, basis for R 3 ., Solution (a) Since neither vector is a scalar multiple of the other, the two vectors form, , a linearly independent set in the two-dimensional space R 2 , and hence they form a basis, by Theorem 4.5.4., Solution (b) The vectors v1 and v2 form a linearly independent set in the xz-plane (why?)., The vector v3 is outside of the xz-plane, so the set {v1 , v2 , v3 } is also linearly independent., Since R 3 is three-dimensional, Theorem 4.5.4 implies that {v1 , v2 , v3 } is a basis for the, vector space R 3 ., , The next theorem (whose proof is deferred to the end of this section) reveals two, important facts about the vectors in a finite-dimensional vector space V :

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4.5 Dimension, , 225, , 1. Every spanning set for a subspace is either a basis for that subspace or has a basis, as a subset., 2. Every linearly independent set in a subspace is either a basis for that subspace or, can be extended to a basis for it., , THEOREM 4.5.5 Let S be a finite set of vectors in a finite-dimensional vector space V., , (a) If S spans V but is not a basis for V, then S can be reduced to a basis for V by, removing appropriate vectors from S ., (b) If S is a linearly independent set that is not already a basis for V, then S can be, enlarged to a basis for V by inserting appropriate vectors into S ., , We conclude this section with a theorem that relates the dimension of a vector space, to the dimensions of its subspaces., THEOREM 4.5.6 If W is a subspace of a finite-dimensional vector space V, then:, , (a) W is finite-dimensional., (b) dim(W ) ≤ dim(V )., (c) W = V if and only if dim(W ) = dim(V )., Proof (a) We will leave the proof of this part as an exercise., Proof (b) Part (a) shows that W is finite-dimensional, so it has a basis, , S = {w1 , w2 , . . . , wm }, Either S is also a basis for V or it is not. If so, then dim(V ) = m, which means that, dim(V ) = dim(W ). If not, then because S is a linearly independent set it can be enlarged, to a basis for V by part (b) of Theorem 4.5.5. But this implies that dim(W ) < dim(V ),, so we have shown that dim(W ) ≤ dim(V ) in all cases., Proof (c) Assume that dim(W ), , = dim(V ) and that, S = {w1 , w2 , . . . , wm }, , is a basis for W . If S is not also a basis for V , then being linearly independent S can, be extended to a basis for V by part (b) of Theorem 4.5.5. But this would mean that, dim(V ) > dim(W ), which contradicts our hypothesis. Thus S must also be a basis for, V, which means that W = V . The converse is obvious., Figure 4.5.2 illustrates the geometric relationship between the subspaces of R 3 in, order of increasing dimension., Line through the origin, (1-dimensional), , The origin, (0-dimensional), , Figure 4.5.2, , Plane through, the origin, (2-dimensional), R3, (3-dimensional)

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4.5 Dimension, , 227, , The proof of Theorem 4.5.2(b) closely parallels that of Theorem 4.5.2(a) and will be, omitted., , S = {v1 , v2 , . . . , vr } is a linearly independent, set of vectors in V, and v is a vector in V that is outside of span(S). To show that, S = {v1 , v2 , . . . , vr , v} is a linearly independent set, we must show that the only scalars, that satisfy, (5), k1 v1 + k2 v2 + · · · + kr vr + kr+1 v = 0, Proof of Theorem 4.5.3 (a) Assume that, , are k1 = k2 = · · · = kr = kr+1 = 0. But it must be true that kr+1 = 0 for otherwise we, could solve (5) for v as a linear combination of v1 , v2 , . . . , vr , contradicting the assumption that v is outside of span(S). Thus, (5) simplifies to, , k1 v1 + k2 v2 + · · · + kr vr = 0, , (6), , which, by the linear independence of {v1 , v2 , . . . , vr }, implies that, , k1 = k2 = · · · = kr = 0, Proof of Theorem 4.5.3 (b) Assume that S = {v1 , v2 , . . . , vr } is a set of vectors in V, and, (to be specific) suppose that vr is a linear combination of v1 , v2 , . . . , vr−1 , say, , vr = c1 v1 + c2 v2 + · · · + cr−1 vr−1, , (7), , We want to show that if vr is removed from S , then the remaining set of vectors, {v1 , v2 , . . . , vr−1 } still spans S ; that is, we must show that every vector w in span(S), is expressible as a linear combination of {v1 , v2 , . . . , vr−1 }. But if w is in span(S), then, w is expressible in the form, w = k1 v1 + k2 v2 + · · · + kr−1 vr−1 + kr vr, or, on substituting (7),, w = k1 v1 + k2 v2 + · · · + kr−1 vr−1 + kr (c1 v1 + c2 v2 + · · · + cr−1 vr−1 ), which expresses w as a linear combination of v1 , v2 , . . . , vr−1 ., Proof of Theorem 4.5.5 (a) If S is a set of vectors that spans V but is not a basis for V,, then S is a linearly dependent set. Thus some vector v in S is expressible as a linear, combination of the other vectors in S . By the Plus/Minus Theorem (4.5.3b), we can, remove v from S , and the resulting set S will still span V. If S is linearly independent,, then S is a basis for V, and we are done. If S is linearly dependent, then we can remove, some appropriate vector from S to produce a set S that still spans V. We can continue, removing vectors in this way until we finally arrive at a set of vectors in S that is linearly, independent and spans V. This subset of S is a basis for V., Proof of Theorem 4.5.5 (b) Suppose that dim(V ) = n. If S is a linearly independent set, that is not already a basis for V, then S fails to span V, so there is some vector v in V, that is not in span(S). By the Plus/Minus Theorem (4.5.3a), we can insert v into S , and, the resulting set S will still be linearly independent. If S spans V, then S is a basis for, V, and we are finished. If S does not span V, then we can insert an appropriate vector, into S to produce a set S that is still linearly independent. We can continue inserting, vectors in this way until we reach a set with n linearly independent vectors in V. This set, will be a basis for V by Theorem 4.5.4.

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228, , Chapter 4 General Vector Spaces, , Exercise Set 4.5, In Exercises 1–6, find a basis for the solution space of the homogeneous linear system, and find the dimension of that space., 1., , x1 + x2 − x3 = 0, −2x1 − x2 + 2x3 = 0, −x1, + x3 = 0, , 2. 3x1 + x2 + x3 + x4 = 0, 5x1 − x2 + x3 − x4 = 0, , 3. 2x1 + x2 + 3x3 = 0, x1, + 5x3 = 0, x2 + x3 = 0, , 4. x1 − 4x2 + 3x3 − x4 = 0, 2x1 − 8x2 + 6x3 − 2x4 = 0, , 5. x1 − 3x2 + x3 = 0, 2x1 − 6x2 + 2x3 = 0, 3x1 − 9x2 + 3x3 = 0, , 6. x, 3x, 4x, 6x, , + y, + 2y, + 3y, + 5y, , + z=0, − 2z = 0, − z=0, + z=0, , 7. In each part, find a basis for the given subspace of R 3 , and, state its dimension., (a) The plane 3x − 2y + 5z = 0., (b) The plane x − y = 0., (c) The line x = 2t, y = −t, z = 4t ., (d) All vectors of the form (a, b, c), where b = a + c., 8. In each part, find a basis for the given subspace of R 4 , and, state its dimension., , 14. Let {v1 , v2 , v3 } be a basis for a vector space V. Show that, {u1 , u2 , u3 } is also a basis, where u1 = v1 , u2 = v1 + v2 , and, u3 = v1 + v2 + v3 ., 15. The vectors v1 = (1, −2, 3) and v2 = (0, 5, −3) are linearly, independent. Enlarge {v1 , v2 } to a basis for R 3 ., 16. The vectors v1 = (1, 0, 0, 0) and v2 = (1, 1, 0, 0) are linearly, independent. Enlarge {v1 , v2 } to a basis for R 4 ., 17. Find a basis for the subspace of R 3 that is spanned by the, vectors, v1 = (1, 0, 0), v2 = (1, 0, 1), v3 = (2, 0, 1), v4 = (0, 0, −1), 18. Find a basis for the subspace of R 4 that is spanned by the, vectors, v1 = (1, 1, 1, 1), v2 = (2, 2, 2, 0), v3 = (0, 0, 0, 3),, v4 = (3, 3, 3, 4), 19. In each part, let TA : R 3 →R 3 be multiplication by A and find, the dimension of the subspace of R 3 consisting of all vectors, x for which TA (x) = 0., , ⎡, , 1, , 0, , (a) A = ⎣1, , 0, , 1⎦, , 1, , 0, , 1, , ⎢, ⎡, , (a) All vectors of the form (a, b, c, 0)., (b) All vectors of the form (a, b, c, d), where d = a + b and, c = a − b., , (a) The vector space of all diagonal n × n matrices., (b) The vector space of all symmetric n × n matrices., (c) The vector space of all upper triangular n × n matrices., 10. Find the dimension of the subspace of P3 consisting of all, polynomials a0 + a1 x + a2 x 2 + a3 x 3 for which a0 = 0., 11. (a) Show that the set W of all polynomials in P2 such that, p(1) = 0 is a subspace of P2 ., (b) Make a conjecture about the dimension of W ., (c) Confirm your conjecture by finding a basis for W ., 12. Find a standard basis vector for R 3 that can be added to the, set {v1 , v2 } to produce a basis for R 3 ., (a) v1 = (−1, 2, 3), v2 = (1, −2, −2), (b) v1 = (1, −1, 0), v2 = (3, 1, −2), 13. Find standard basis vectors for R 4 that can be added to the, set {v1 , v2 } to produce a basis for R 4 ., v1 = (1, −4, 2, −3), v2 = (−3, 8, −4, 6), , ⎡, , ⎥, ⎤, , 1, , 0, , 0, , 1, , 0⎦, , 1, , 1, , 1, , ⎢, , ⎤, , 1, , 2, , 0, , (b) A = ⎣1, , 2, , 0⎦, , 1, , 2, , 0, , ⎢, , (c) A = ⎣−1, , (c) All vectors of the form (a, b, c, d), where a = b = c = d ., 9. Find the dimension of each of the following vector spaces., , ⎤, , 1, , ⎥, , ⎥, , 20. In each part, let TA be multiplication by A and find the dimension of the subspace R 4 consisting of all vectors x for which, TA (x) = 0., , , (a) A =, , 1, , 0, , 2, , −1, , −1, , 4, , 0, , 0, , , , ⎡, , 0, , ⎢, (b) A = ⎣−1, 1, , ⎤, , 0, , 1, , 1, , 1, , 0, , 0⎦, , 0, , 0, , 1, , ⎥, , Working with Proofs, 21. (a) Prove that for every positive integer n, one can find n + 1, linearly independent vectors in F (−⬁, ⬁). [Hint: Look, for polynomials.], (b) Use the result in part (a) to prove that F (−⬁, ⬁) is infinitedimensional., (c) Prove that C(−⬁, ⬁), C m (−⬁, ⬁), and C ⬁ (−⬁, ⬁) are, infinite-dimensional., 22. Let S be a basis for an n-dimensional vector space V. Prove, that if v1 , v2 , . . . , vr form a linearly independent set of vectors, in V, then the coordinate vectors (v1 )S , (v2 )S , . . . , (vr )S form, a linearly independent set in R n , and conversely.

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4.6 Change of Basis, , 23. Let S = {v1 , v2 , . . . , vr } be a nonempty set of vectors in an, n-dimensional vector space V . Prove that if the vectors in, S span V, then the coordinate vectors (v1 )S , (v2 )S , . . . , (vr )S, span R n , and conversely., , 25. Prove: A subspace of a finite-dimensional vector space is, finite-dimensional., 26. State the two parts of Theorem 4.5.2 in contrapositive form., 27. In each part, let S be the standard basis for P2 . Use the results, proved in Exercises 22 and 23 to find a basis for the subspace, of P2 spanned by the given vectors., (a) −1 + x − 2x , 3 + 3x + 6x , 9, 2, , (b) 1 + x , x , 2 + 2x + 3x, 2, , (g) Every linearly independent set of vectors in R n is contained in, some basis for R n ., (h) There is a basis for M22 consisting of invertible matrices., 2, , 24. Prove part (a) of Theorem 4.5.6., , 2, , 229, , (i) If A has size n × n and In , A, A2 , . . . , An are distinct matri2, ces, then {In , A, A2 , . . . , An } is a linearly dependent set., ( j) There are at least two distinct three-dimensional subspaces, of P2 ., (k) There are only three distinct two-dimensional subspaces of P2 ., , Working withTechnology, T1. Devise three different procedures for using your technology, utility to determine the dimension of the subspace spanned by a, set of vectors in R n , and then use each of those procedures to, determine the dimension of the subspace of R 5 spanned by the, vectors, , 2, , (c) 1 + x − 3x 2 , 2 + 2x − 6x 2 , 3 + 3x − 9x 2, , True-False Exercises, , v1 = (2, 2, −1, 0, 1),, , TF. In parts (a)–( k) determine whether the statement is true or, false, and justify your answer., , (b) There is a set of 17 linearly independent vectors in R 17 ., , ⎡, , (c) There is a set of 11 vectors that span R 17 ., , 3.4, , (d) Every linearly independent set of five vectors in R 5 is a basis, for R 5 ., 5, , (e) Every set of five vectors that spans R is a basis for R ., n, , v3 = (1, 1, −2, 0, −1), v4 = (0, 0, 1, 1, 1), T2. Find a basis for the row space of A by starting at the top and, successively removing each row that is a linear combination of its, predecessors., , (a) The zero vector space has dimension zero., , 5, , v2 = (−1, −1, 2, −3, 1),, , ⎢ 2 .1, ⎢, ⎢, A=⎢, ⎢8.9, ⎢, ⎣7.6, 1.0, , n, , (f ) Every set of vectors that spans R contains a basis for R ., , −1.8, , ⎤, , 2 .2, , 1.0, , 3 .6, , 4.0, , 8 .0, , 6.0, , 9.4, , 9.0, , −3.4⎥, ⎥, ⎥, 7.0⎥, ⎥, ⎥, −8.6⎦, , 2 .2, , 0.0, , 2 .2, , 4.6 Change of Basis, A basis that is suitable for one problem may not be suitable for another, so it is a common, process in the study of vector spaces to change from one basis to another. Because a basis is, the vector space generalization of a coordinate system, changing bases is akin to changing, coordinate axes in R2 and R3 . In this section we will study problems related to changing, bases., , Coordinate Maps, , If S = {v1 , v2 , . . . , vn } is a basis for a finite-dimensional vector space V , and if, , (v)S = (c1 , c2 , . . . , cn ), is the coordinate vector of v relative to S , then, as illustrated in Figure 4.4.6, the mapping, v → (v)S, , (1), , creates a connection (a one-to-one correspondence) between vectors in the general vector, space V and vectors in the Euclidean vector space R n . We call (1) the coordinate map, relative to S from V to R n . In this section we will find it convenient to express coordinate

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230, , Chapter 4 General Vector Spaces, , [ ]S, v, , c1, c2, ., ., ., cn, Rn, , V, , ⎡ ⎤, c1, ⎢c2 ⎥, ⎢ ⎥, [v]S = ⎢ . ⎥, ⎣ .. ⎦, cn, , vectors in the matrix form, , Coordinate map, , (2), , where the square brackets emphasize the matrix notation (Figure 4.6.1)., , Figure 4.6.1, , Change of Basis, , There are many applications in which it is necessary to work with more than one coordinate system. In such cases it becomes important to know how the coordinates of a, fixed vector relative to each coordinate system are related. This leads to the following, problem., , V,, and if we change the basis for V from a basis B to a basis B , how are the coordinate, vectors [v]B and [v]B related?, , The Change-of-Basis Problem If v is a vector in a finite-dimensional vector space, , Remark To solve this problem, it will be convenient to refer to B as the “old basis” and B as, the “new basis.” Thus, our objective is to find a relationship between the old and new coordinates, of a fixed vector v in V., For simplicity, we will solve this problem for two-dimensional spaces. The solution, for n-dimensional spaces is similar. Let, , B = {u1 , u2 } and B = {u1 , u2 }, be the old and new bases, respectively. We will need the coordinate vectors for the new, basis vectors relative to the old basis. Suppose they are, , a, b, , [u1 ]B =, That is,, , and [u2 ]B =, , c, d, , u1 = a u1 + bu2, u2 = cu1 + d u2, , (3), , (4), , Now let v be any vector in V , and let, , [v]B =, , k1, k2, , (5), , be the new coordinate vector, so that, v = k1 u1 + k2 u2, , (6), , In order to find the old coordinates of v, we must express v in terms of the old basis B ., To do this, we substitute (4) into (6). This yields, v = k1 (a u1 + bu2 ) + k2 (cu1 + d u2 ), or, v = (k1 a + k2 c)u1 + (k1 b + k2 d)u2, Thus, the old coordinate vector for v is, , [v]B =, , k1 a + k2 c, k1 b + k2 d

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4.6 Change of Basis, , 231, , which, by using (5), can be written as, , [v]B =, , a, b, , c, d, , k1, a, =, k2, b, , c, [v]B, d, , This equation states that the old coordinate vector [v]B results when we multiply the new, coordinate vector [v]B on the left by the matrix, , P =, , a, b, , c, d, , Since the columns of this matrix are the coordinates of the new basis vectors relative to, the old basis [see (3)], we have the following solution of the change-of-basis problem., Solution of the Change-of-Basis Problem If we change the basis for a vector space V, , from an old basis B = {u1 , u2 , . . . , un } to a new basis B = {u1 , u2 , . . . , un }, then for, each vector v in V , the old coordinate vector [v]B is related to the new coordinate, vector [v]B by the equation, [v]B = P [v]B, (7), where the columns of P are the coordinate vectors of the new basis vectors relative, to the old basis; that is, the column vectors of P are, , [u1 ]B , [u2 ]B , . . . , [un ]B, , Transition Matrices, , (8), , The matrix P in Equation (7) is called the transition matrix from B to B . For emphasis,, we will often denote it by PB →B . It follows from (8) that this matrix can be expressed, in terms of its column vectors as, , , , PB →B = [u1 ]B | [u2 ]B | · · · | [un ]B, , (9), , Similarly, the transition matrix from B to B can be expressed in terms of its column, vectors as, , , , PB→B = [u1 ]B | [u2 ]B | · · · | [un ]B, , (10), , Remark There is a simple way to remember both of these formulas using the terms “old basis”, and “new basis” defined earlier in this section: In Formula (9) the old basis is B and the new basis, is B , whereas in Formula (10) the old basis is B and the new basis is B . Thus, both formulas can, be restated as follows:, , The columns of the transition matrix from an old basis to a new basis are the coordinate, vectors of the old basis relative to the new basis., , E X A M P L E 1 Finding Transition Matrices, , Consider the bases B = {u1 , u2 } and B = {u1 , u2 } for R 2 , where, u1 = (1, 0), u2 = (0, 1), u1 = (1, 1), u2 = (2, 1), (a) Find the transition matrix PB →B from B to B ., (b) Find the transition matrix PB→B from B to B .

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232, , Chapter 4 General Vector Spaces, Solution (a) Here the old basis vectors are u1 and u2 and the new basis vectors are u1, and u2 . We want to find the coordinate matrices of the old basis vectors u1 and u2 relative, to the new basis vectors u1 and u2 . To do this, observe that, , u1 = u1 + u2, u2 = 2u1 + u2, from which it follows that, , [u1 ]B =, , 1, 1, , and [u2 ]B =, , 2, 1, , and hence that, , PB →B =, , 1, 1, , 2, 1, , Solution (b) Here the old basis vectors are u1 and u2 and the new basis vectors are u1, and u2 . As in part (a), we want to find the coordinate matrices of the old basis vectors, u1 and u2 relative to the new basis vectors u1 and u2 . To do this, observe that, , u1 = −u1 + u2, u2 = 2u1 − u2, from which it follows that, , [u1 ]B =, , −1, 1, , and [u2 ]B =, , and hence that, , PB→B =, , −1, , 2, , 1, , −1, , 2, , −1, , Suppose now that B and B are bases for a finite-dimensional vector space V . Since, multiplication by PB →B maps coordinate vectors relative to the basis B into coordinate, vectors relative to a basis B , and PB→B maps coordinate vectors relative to B into, coordinate vectors relative to B , it follows that for every vector v in V we have, , [v]B = PB →B [v]B, , (11), , [v]B = PB→B [v]B, , (12), , E X A M P L E 2 Computing Coordinate Vectors, , Let B and B be the bases in Example 1. Use an appropriate formula to find [v]B given, that, −3, [v]B =, 5, Solution To find [v]B we need to make the transition from B to B . It follows from, Formula (11) and part (a) of Example 1 that, , [v]B = PB →B [v]B =, , Invertibility of Transition, Matrices, , 1, 1, , 2, 1, , −3, 5, , =, , 7, 2, , If B and B are bases for a finite-dimensional vector space V , then, , (PB →B )(PB→B ) = PB→B

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4.6 Change of Basis, , 233, , because multiplication by the product (PB →B )(PB→B ) first maps the B -coordinates of a, vector into its B -coordinates, and then maps those B -coordinates back into the original, B -coordinates. Since the net effect of the two operations is to leave each coordinate vector, unchanged, we are led to conclude that PB→B must be the identity matrix, that is,, , (PB →B )(PB→B ) = I, , (13), , (we omit the formal proof). For example, for the transition matrices obtained in Example, 1 we have, 1 2 −1, 2, 1 0, =, =I, (PB →B )(PB→B ) =, 1 1, 1 −1, 0 1, It follows from (13) that PB →B is invertible and that its inverse is PB→B . Thus, we, have the following theorem., THEOREM 4.6.1 If P is the transition matrix from a basis B to a basis B for a finite-, , dimensional vector space V, then P is invertible and P −1 is the transition matrix from, B to B ., , An Efficient Method for, ComputingTransition, Matrices for Rn, , Our next objective is to develop an efficient procedure for computing transition matrices, between bases for R n . As illustrated in Example 1, the first step in computing a transition, matrix is to express each new basis vector as a linear combination of the old basis vectors., For R n this involves solving n linear systems of n equations in n unknowns, each of which, has the same coefficient matrix (why?). An efficient way to do this is by the method, illustrated in Example 2 of Section 1.6, which is as follows:, A Procedure for Computing PB→B , , Step 1. Form the matrix [B | B]., Step 2. Use elementary row operations to reduce the matrix in Step 1 to reduced row, echelon form., Step 3. The resulting matrix will be [I | PB→B ]., Step 4. Extract the matrix PB→B from the right side of the matrix in Step 3., This procedure is captured in the following diagram., , [new basis | old basis], , row operations, , −→, , [I | transition from old to new], , (14), , E X A M P L E 3 Example 1 Revisited, , In Example 1 we considered the bases B = {u1 , u2 } and B = {u1 , u2 } for R 2 , where, u1 = (1, 0), u2 = (0, 1), u1 = (1, 1), u2 = (2, 1), (a) Use Formula (14) to find the transition matrix from B to B ., (b) Use Formula (14) to find the transition matrix from B to B ., Solution (a) Here B is the old basis and B is the new basis, so, , [new basis | old basis] =, , 1, 0, , 0, 1, , 1, 1, , 2, 1

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234, , Chapter 4 General Vector Spaces, , Since the left side is already the identity matrix, no reduction is needed. We see by, inspection that the transition matrix is, , PB →B =, , 1, 1, , 2, 1, , which agrees with the result in Example 1., Solution (b) Here B is the old basis and B is the new basis, so, , [new basis | old basis] =, , 1, 1, , 2, 1, , 1, 0, , 0, 1, , By reducing this matrix, so the left side becomes the identity, we obtain (verify), 1, 0, , [I | transition from old to new] =, so the transition matrix is, , PB→B =, , −1, , 2, , 1, , −1, , 0, 1, , −1, 1, , 2, −1, , which also agrees with the result in Example 1., , Transition to the Standard, Basis for R n, , Note that in part (a) of the last example the column vectors of the matrix that made, the transition from the basis B to the standard basis turned out to be the vectors in B, written in column form. This illustrates the following general result., , B = {u1 , u2 , . . . , un } be any basis for the vector space R n and, let S = {e1 , e2 , . . . , en } be the standard basis for R n . If the vectors in these bases are, written in column form, then, , THEOREM 4.6.2 Let, , PB →S = [u1 | u2 | · · · | un ], , (15), , It follows from this theorem that if, , A = [u1 | u2 | · · · | un ], is any invertible n × n matrix, then A can be viewed as the transition matrix from the, basis {u1 , u2 , . . . , un } for R n to the standard basis for R n . Thus, for example, the matrix, , ⎡, , 1, ⎢, A = ⎣2, 1, , 2, 5, 0, , ⎤, , 3, ⎥, 3⎦, 8, , which was shown to be invertible in Example 4 of Section 1.5, is the transition matrix, from the basis, u1 = (1, 2, 1), u2 = (2, 5, 0), u3 = (3, 3, 8), to the basis, e1 = (1, 0, 0), e2 = (0, 1, 0), e3 = (0, 0, 1)

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4.6 Change of Basis, , 235, , Exercise Set 4.6, 1. Consider the bases B = {u1 , u2 } and B = {u1 , u2 } for R 2 ,, where, ,  , , u1 =, , 2, 2, , , , , u2 =, , 4, ,  , , , , 1, , , u1 =, , −1, , 3, , , , −1, −1, , , u2 =, , , , (a) Find the transition matrix from B to B ., , , , 3, , p1 = 6 + 3x, p2 = 10 + 2x, q1 = 2, q2 = 3 + 2x, , −5, , (a) Find the transition matrix from B to B ., , and use (12) to compute [w]B ., , (b) Find the transition matrix from B to B ., , (d) Check your work by computing [w]B directly., 2. Repeat the directions of Exercise 1 with the same vector w but, with, ,  , , u1 =, , 1, 0, ,  , , , u2 =, , 0, 1, ,  , , , u1 =, , 2, , 1, , , u2 =, ,  , −3, 4, , 3. Consider the bases B = {u1 , u2 , u3 } and B = {u1 , u2 , u3 } for, R 3 , where, , ⎡ ⎤, , ⎡, , ⎢ ⎥, , ⎢, , 2, , 2, , 1, , ⎤, , ⎢ ⎥, , (c) Confirm that PB2 →B1 and PB1 →B2 are inverses of one, another., , 1, , (d) Let w = (0, 1). Find [w]B1 and then use the matrix PB1 →B2, to compute [w]B2 from [w]B1 ., , 1, , ⎤, , 1, , (a) Find the transition matrix PB→S by inspection., , ⎤, −5, ⎢ ⎥, w = ⎣ 8⎦, −5, , (b) Use Formula (14) to find the transition matrix PS→B ., (c) Confirm that PB→S and PS→B are inverses of one another., (d) Let w = (5, −3). Find [w]B and then use Formula (11) to, compute [w]S ., , and use (12) to compute [w]B ., (c) Check your work by computing [w]B directly., 4. Repeat the directions of Exercise 3 with the same vector w, but, with, ⎡ ⎤, ⎡ ⎤, ⎡ ⎤, 1, −3, −3, , ⎥, , ⎢, , ⎢, , ⎥, , u1 = ⎣ 0⎦,, , u2 = ⎣ 2⎦,, , −3, ⎡ ⎤, −6, ⎢ ⎥, u1 = ⎣−6⎦,, , −1, −1, ⎡ ⎤, ⎡ ⎤, −2, −2, ⎢ ⎥, ⎢ ⎥, u2 = ⎣−6⎦, u3 = ⎣−3⎦, 4, , (e) Let w = (2, 5). Find [w]B2 and then use the matrix PB2 →B1, to compute [w]B1 from [w]B2 ., 8. Let S be the standard basis for R 2 , and let B = {v1 , v2 } be the, basis in which v1 = (2, 1) and v2 = (−3, 4)., , ⎡, , 0, , (a) Use Formula (14) to find the transition matrix PB2 →B1 ., , ⎥, , (b) Compute the coordinate vector [w]B , where, , ⎥, , 7. Let B1 = {u1 , u2 } and B2 = {v1 , v2 } be the bases for R 2 in, which u1 = (1, 2), u2 = (2, 3), v1 = (1, 3), and v2 = (1, 4)., (b) Use Formula (14) to find the transition matrix PB1 →B2 ., , 1, , (a) Find the transition matrix B to B ., , ⎢, , (d) Check your work by computing [p]B directly., , ⎡ ⎤, , ⎡ ⎤, −1, ⎢ ⎥, ⎢ ⎥, ⎢ ⎥, u1 = ⎣ 1⎦, u2 = ⎣ 1⎦, u3 = ⎣ 0⎦, 2, −5, −3, 3, , ⎡, , (c) Compute the coordinate vector [p]B , where p = −4 + x ,, and use (12) to compute [p]B ., , ⎤, , u1 = ⎣1⎦, u2 = ⎣−1⎦, u3 = ⎣2⎦, , ⎡, , (d) Compute the coordinate vector [h]B , where, h = 2 sin x − 5 cos x , and use (12) to obtain [h]B ., , 6. Consider the bases B = {p1 , p2 } and B = {q1 , q2 } for P1 ,, where, , (c) Compute the coordinate vector [w]B , where, w=, , (c) Find the transition matrix from B to B ., , (e) Check your work by computing [h]B directly., , (b) Find the transition matrix from B to B ., , , , (b) Find the transition matrix from B = {g1 , g2 } to, B = {f1 , f2 }., , u 3 = ⎣ 6⎦, , 7, , 5. Let V be the space spanned by f1 = sin x and f2 = cos x ., (a) Show that g1 = 2 sin x + cos x and g2 = 3 cos x form a, basis for V., , (e) Let w = (3, −5). Find [w]S and then use Formula (12) to, compute [w]B ., 9. Let S be the standard basis for R 3 , and let B = {v1 , v2 , v3 }, be the basis in which v1 = (1, 2, 1), v2 = (2, 5, 0), and, v3 = (3, 3, 8)., (a) Find the transition matrix PB→S by inspection., (b) Use Formula (14) to find the transition matrix PS→B ., (c) Confirm that PB→S and PS→B are inverses of one another., (d) Let w = (5, −3, 1). Find [w]B and then use Formula (11), to compute [w]S ., (e) Let w = (3, −5, 0). Find [w]S and then use Formula (12), to compute [w]B .

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236, , Chapter 4 General Vector Spaces, , 10. Let S = {e1 , e2 } be the standard basis for R 2 , and let, B = {v1 , v2 } be the basis that results when the vectors in S are, reflected about the line y = x ., (a) Find the transition matrix PB→S ., (b) Let P = PB→S and show that P T = PS→B ., 11. Let S = {e1 , e2 } be the standard basis for R 2 , and let, B = {v1 , v2 } be the basis that results when the vectors in S are, reflected about the line that makes an angle θ with the positive, x -axis., (a) Find the transition matrix PB→S ., , 12. If B1 , B2 , and B3 are bases for R 2 , and if, 3, 5, , then PB3 →B1 =, , 1, 2, , and PB2 →B3 =, , 7, 4, , 2, , −1, , ., , 13. If P is the transition matrix from a basis B to a basis B , and, Q is the transition matrix from B to a basis C , what is the, transition matrix from B to C ? What is the transition matrix, from C to B ?, 14. To write the coordinate vector for a vector, it is necessary to, specify an order for the vectors in the basis. If P is the transition matrix from a basis B to a basis B , what is the effect, on P if we reverse the order of vectors in B from v1 , . . . , vn to, vn , . . . , v1 ? What is the effect on P if we reverse the order of, vectors in both B and B ?, 15. Consider the matrix, , ⎡, , 1, ⎢, P = ⎣1, 0, , 1, 0, 2, , is applied to each vector in S . Find the transition matrix PB→S ., 19. If [w]B = w holds for all vectors w in R n , what can you say, about the basis B ?, , Working with Proofs, 20. Let B be a basis for R n . Prove that the vectors v1 , v2 , . . . , vk, span R n if and only if the vectors [v1 ]B , [v2 ]B , . . . , [vk ]B, span R n ., 21. Let B be a basis for R n . Prove that the vectors v1 , v2 , . . . , vk, form a linearly independent set in R n if and only if the vectors, [v1 ]B , [v2 ]B , . . . , [vk ]B form a linearly independent set in R n ., , (b) Let P = PB→S and show that P T = PS→B ., , PB1 →B2 =, , T (x1 , x2 , x3 ) = (x1 + x2 , 2x1 − x2 + 4x3 , x2 + 3x3 ), , ⎤, , 0, ⎥, 2⎦, 1, , True-False Exercises, TF. In parts (a)–(f ) determine whether the statement is true or, false, and justify your answer., (a) If B1 and B2 are bases for a vector space V, then there exists a, transition matrix from B1 to B2 ., (b) Transition matrices are invertible., (c) If B is a basis for a vector space R n , then PB→B is the identity, matrix., (d) If PB1 →B2 is a diagonal matrix, then each vector in B2 is a, scalar multiple of some vector in B1 ., (e) If each vector in B2 is a scalar multiple of some vector in B1 ,, then PB1 →B2 is a diagonal matrix., (f ) If A is a square matrix, then A = PB1 →B2 for some bases B1, and B2 for R n ., , Working withTechnology, T1. Let, , (a) P is the transition matrix from what basis B to the standard basis S = {e1 , e2 , e3 } for R 3 ?, (b) P is the transition matrix from the standard basis, S = {e1 , e2 , e3 } to what basis B for R 3 ?, 16. The matrix, , ⎡, , 1, P = ⎣0, 0, , 0, 3, 1, , ⎤, , 0, 2⎦, 1, , what basis B to the basis, *, )is the transition matrix from, (1, 1, 1), (1, 1, 0), (1, 0, 0) for R 3 ?, 17. Let S = {e1 , e2 } be the standard basis for R 2 , and let, B = {v1 , v2 } be the basis that results when the linear transformation defined by, , T (x1 , x2 ) = (2x1 + 3x2 , 5x1 − x2 ), is applied to each vector in S . Find the transition matrix PB→S ., 18. Let S = {e1 , e2 , e3 } be the standard basis for R 3 , and let, B = {v1 , v2 , v3 } be the basis that results when the linear transformation defined by, , and, , ⎡, , 5, , 8, , 6, , ⎢3, ⎢, P =⎢, ⎣0, , −1, , 0, , 1, , −1, , 2, , 4, , 3, , v1 = (2, 4, 3, −5),, , −13, , ⎤, , −9⎥, ⎥, ⎥, 0⎦, −5, , v2 = (0, 1, −1, 0),, , v3 = (3, −1, 0, −9), v4 = (5, 8, 6, −13), Find a basis B = {u1 , u2 , u3 , u4 } for R 4 for which P is the transition matrix from B to B = {v1 , v2 , v3 , v4 }., T2. Given that the matrix for a linear transformation T : R 4 →R 4, relative to the standard basis B = {e1 , e2 , e3 , e4 } for R 4 is, , ⎡, , 1, , ⎢3, ⎢, ⎢, ⎣2, 1, , 1, , ⎤, , 2, , 0, , 0, , −1, , 5, , 3, , 1⎦, , 2, , 1, , 3, , 2⎥, ⎥, , ⎥, , find the matrix for T relative to the basis, , B = {e1 , e1 + e2 , e1 + e2 + e3 , e1 + e2 + e3 + e4 }

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4.7 Row Space, Column Space, and Null Space, , 237, , 4.7 Row Space, Column Space, and Null Space, In this section we will study some important vector spaces that are associated with matrices., Our work here will provide us with a deeper understanding of the relationships between the, solutions of a linear system and properties of its coefficient matrix., , Row Space, Column Space,, and Null Space, , Recall that vectors can be written in comma-delimited form or in matrix form as either, row vectors or column vectors. In this section we will use the latter two., DEFINITION 1 For an m × n matrix, , ⎡, , the vectors, , ⎤, a1n, a2 n ⎥, ⎥, .. ⎥, . ⎦, amn, , a11, ⎢a, ⎢ 21, A=⎢ ., ⎣ .., am1, , a12, a22, .., ., am2, , ···, ···, , r1 = [a11, , a12, , · · · a 1n ], , r2 = [a21, , a22, , .., ., , rm = [am1, , ···, , .., ., , · · · a2 n ], , am2 · · · amn ], in R that are formed from the rows of A are called the row vectors of A, and the, vectors, ⎤, ⎤, ⎤, ⎡, ⎡, ⎡, a11, a12, a 1n, ⎢a ⎥, ⎢a ⎥, ⎢a ⎥, ⎢ 21 ⎥, ⎢ 22 ⎥, ⎢ 2n ⎥, c1 = ⎢ . ⎥, c2 = ⎢ . ⎥, . . . , cn = ⎢ . ⎥, ⎣ .. ⎦, ⎣ .. ⎦, ⎣ .. ⎦, am1, am2, amn, in R m formed from the columns of A are called the column vectors of A., n, , E X A M P L E 1 Row and Column Vectors of a 2 × 3 Matrix, , Let, , A=, , 2, 3, , 1, , −1, , 0, 4, , The row vectors of A are, r1 = [2 1 0] and r2 = [3, , −1, , 4], , and the column vectors of A are, c1 =, , 2, 1, 0, , c2 =, , and c3 =, 3, 4, −1, , The following definition defines three important vector spaces associated with a, matrix., We will sometimes denote the, row space of A, the column, space of A, and the null space, of A by row(A), col(A), and, null(A), respectively., , A is an m × n matrix, then the subspace of R n spanned by the, row vectors of A is called the row space of A, and the subspace of R m spanned by, the column vectors of A is called the column space of A. The solution space of the, homogeneous system of equations Ax = 0, which is a subspace of R n , is called the, null space of A., DEFINITION 2 If

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238, , Chapter 4 General Vector Spaces, , In this section and the next we will be concerned with two general questions:, Question 1. What relationships exist among the solutions of a linear system Ax = b, and the row space, column space, and null space of the coefficient matrix A?, Question 2. What relationships exist among the row space, column space, and null, space of a matrix?, Starting with the first question, suppose that, , ⎡, , a12, a22, .., ., am 2, , a11, ⎢a, ⎢ 21, A=⎢ ., ⎣ .., a m1, , ···, ···, ···, , ⎤, ⎡ ⎤, a1n, x1, ⎥, ⎥, ⎢, a2 n ⎥, ⎢ x2 ⎥, and, x, =, ⎥, ⎢, .., .⎥, ⎣ .. ⎦, . ⎦, amn, xn, , It follows from Formula (10) of Section 1.3 that if c1 , c2 , . . . , cn denote the column, vectors of A, then the product Ax can be expressed as a linear combination of these, vectors with coefficients from x; that is,, , Ax = x1 c1 + x2 c2 + · · · + xn cn, , (1), , Thus, a linear system, Ax = b, of m equations in n unknowns can be written as, , x1 c1 + x2 c2 + · · · + xn cn = b, , (2), , from which we conclude that Ax = b is consistent if and only if b is expressible as a linear, combination of the column vectors of A. This yields the following theorem., THEOREM 4.7.1 A system of linear equations Ax, , = b is consistent if and only if b is in, , the column space of A., , E X A M P L E 2 A Vector b in the Column Space of A, , Let Ax = b be the linear system, , ⎡, −1, ⎢, ⎣ 1, 2, , ⎤⎡ ⎤ ⎡ ⎤, 1, x1, ⎥⎢ ⎥ ⎢ ⎥, 2 −3⎦ ⎣x2 ⎦ = ⎣−9⎦, x3, −3, 1 −2, 3, , 2, , Show that b is in the column space of A by expressing it as a linear combination of the, column vectors of A., Solution Solving the system by Gaussian elimination yields (verify), , x1 = 2, x2 = −1, x3 = 3, It follows from this and Formula (2) that, , ⎡, , ⎤, , ⎡ ⎤, , ⎡, , ⎢, , ⎥, , ⎢ ⎥, , ⎢, , −1, , 3, , 2, , ⎤, , ⎡, , ⎥, , ⎢, , ⎤, , 1, , ⎥, , 2 ⎣ 1⎦ − ⎣2⎦ + 3 ⎣−3⎦ = ⎣−9⎦, 2, , 1, , −2, , −3, , Recall from Theorem 3.4.4 that the general solution of a consistent linear system, , Ax = b can be obtained by adding any specific solution of the system to the general, solution of the corresponding homogeneous system Ax = 0. Keeping in mind that the, null space of A is the same as the solution space of Ax = 0, we can rephrase that theorem, in the following vector form.

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4.7 Row Space, Column Space, and Null Space, , 239, , Ax = b, and if, S = {v1 , v2 , . . . , vk } is a basis for the null space of A, then every solution of Ax = b can, , THEOREM 4.7.2 If x0 is any solution of a consistent linear system, , be expressed in the form, x = x0 + c1 v1 + c2 v2 + · · · + ck vk, , (3), , Conversely, for all choices of scalars c1 , c2 , . . . , ck , the vector x in this formula is a, solution of Ax = b., The vector x0 in Formula (3) is called a particular solution of Ax = b, and the remaining part of the formula is called the general solution of Ax = 0. With this terminology, Theorem 4.7.2 can be rephrased as:, , The general solution of a consistent linear system can be expressed as the sum of a particular solution of that system and the general solution of the corresponding homogeneous, system., Geometrically, the solution set of Ax = b can be viewed as the translation by x0 of the, solution space of Ax = 0 (Figure 4.7.1)., , y, , x0 + x, , x, , x, , x0, , Solution set, of Ax = b, , Figure 4.7.1, , Solution space, of Ax = 0, , E X A M P L E 3 General Solution of a Linear System Ax = b, , In the concluding subsection of Section 3.4 we compared solutions of the linear systems, , ⎡, , 1, , ⎢2, ⎢, ⎢, ⎣0, 2, , 3, 6, 0, 6, , −2, −5, 5, 0, , 0, −2, 10, 8, , ⎡ ⎤, x1, ⎡ ⎤, ⎡, ⎥, 2, 0 ⎢, 1, 0, x, ⎢ 2⎥, ⎥, ⎢, ⎥, ⎢, ⎥, ⎢, 4 −3⎥ ⎢x3 ⎥ ⎢0⎥, ⎢2, ⎥ ⎢ ⎥ = ⎢ ⎥ and ⎢, 0 15⎦ ⎢x4 ⎥ ⎣0⎦, ⎣0, ⎢ ⎥, ⎣, ⎦, 4 18, 2, 0, x5, x6, ⎤, , 3, 6, 0, 6, , −2, 0, −5 − 2, 5, 0, , 10, 8, , ⎡ ⎤, x1, ⎡ ⎤, ⎥, 2, 0 ⎢, 0, x, ⎢ 2⎥, ⎥, ⎢, ⎥, ⎢, 4 −3⎥ ⎢x3 ⎥ ⎢−1⎥, ⎥, ⎥⎢ ⎥ = ⎢ ⎥, 0 15⎦ ⎢x4 ⎥ ⎣ 5⎦, ⎢ ⎥, 4 18 ⎣x5 ⎦, 6, x6, ⎤, , and deduced that the general solution x of the nonhomogeneous system and the general, solution xh of the corresponding homogeneous system (when written in column-vector, form) are related by

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240, , Chapter 4 General Vector Spaces, , ⎡ ⎤ ⎡, ⎤ ⎡ ⎤, ⎡ ⎤, ⎡ ⎤, ⎡ ⎤, 0, −3r − 4s − 2t, x1, −3, −4, −2, ⎢x2 ⎥ ⎢, ⎥ ⎢0 ⎥, ⎢ 1⎥, ⎢ 0⎥, ⎢ 0⎥, r, ⎢ ⎥ ⎢, ⎥ ⎢ ⎥, ⎢ ⎥, ⎢ ⎥, ⎢ ⎥, ⎢x ⎥ ⎢, ⎥ ⎢0 ⎥, ⎢ 0⎥, ⎢ −2 ⎥, ⎢ 0⎥, −2s, ⎢ 3⎥ ⎢, ⎥ ⎢ ⎥, ⎢ ⎥, ⎢ ⎥, ⎢ ⎥, ⎢ ⎥=⎢, ⎥ = ⎢ ⎥+ r ⎢ ⎥ + s ⎢ ⎥ + t ⎢ ⎥, ⎢x4 ⎥ ⎢, ⎥ ⎢0 ⎥, ⎢ 0⎥, ⎢ 1⎥, ⎢ 0⎥, s, ⎢ ⎥ ⎢, ⎥ ⎢ ⎥, ⎢ ⎥, ⎢ ⎥, ⎢ ⎥, ⎣x5 ⎦ ⎣, ⎦, ⎣, ⎦, ⎣, ⎦, ⎣, ⎦, ⎣ 1⎦, 0, t, 0, 0, 1, 1, 0, 0, 0, x6, 3, 3,   , , ,  ,  , x, , xh, , x0, , Recall from the Remark following Example 3 of Section 4.5 that the vectors in xh, form a basis for the solution space of Ax = 0., Bases for Row Spaces,, Column Spaces, and Null, Spaces, , We know that performing elementary row operations on the augmented matrix [A | b], of a linear system does not change the solution set of that system. This is true, in, particular, if the system is homogeneous, in which case the augmented matrix is [A | 0]., But elementary row operations have no effect on the column of zeros, so it follows that, the solution set of Ax = 0 is unaffected by performing elementary row operations on A, itself. Thus, we have the following theorem., THEOREM 4.7.3 Elementary row operations do not change the null space of a matrix., , The following theorem, whose proof is left as an exercise, is a companion to Theorem 4.7.3., THEOREM 4.7.4 Elementary row operations do not change the row space of a matrix., , Theorems 4.7.3 and 4.7.4 might tempt you into incorrectly believing that elementary, row operations do not change the column space of a matrix. To see why this is not true,, compare the matrices, 1 3, 1 3, and B =, A=, 2 6, 0 0, The matrix B can be obtained from A by adding −2 times the first row to the second., However, this operation has changed the column space of A, since that column space, consists of all scalar multiples of, 1, 2, whereas the column space of B consists of all scalar multiples of, 1, 0, and the two are different spaces., E X A M P L E 4 Finding a Basis for the Null Space of a Matrix, , Find a basis for the null space of the matrix, ⎡, 1, 3 −2, ⎢2, 6 −5, ⎢, A=⎢, 0, 5, ⎣0, 2, 6, 0, , 0, −2, 10, 8, , 2, 4, 0, 4, , ⎤, , 0, −3⎥, ⎥, ⎥, 15⎦, 18

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4.7 Row Space, Column Space, and Null Space, , 241, , A is the solution space of the homogeneous linear system, Ax = 0, which, as shown in Example 3, has the basis, ⎡ ⎤, ⎡ ⎤, ⎡ ⎤, −3, −4, −2, ⎢ 1⎥, ⎢ 0⎥, ⎢ 0⎥, ⎢ ⎥, ⎢ ⎥, ⎢ ⎥, ⎢ 0⎥, ⎢ −2 ⎥, ⎢ 0⎥, ⎢ ⎥, ⎢ ⎥, ⎢ ⎥, v1 = ⎢ ⎥, v2 = ⎢ ⎥, v3 = ⎢ ⎥, ⎢ 0⎥, ⎢ 1⎥, ⎢ 0⎥, ⎢ ⎥, ⎢ ⎥, ⎢ ⎥, ⎣ 0⎦, ⎣ 0⎦, ⎣ 1⎦, , Solution The null space of, , 0, , 0, , 0, , Remark Observe that the basis vectors v1 , v2 , and v3 in the last example are the vectors that result, by successively setting one of the parameters in the general solution equal to 1 and the others equal, to 0., The following theorem makes it possible to find bases for the row and column spaces, of a matrix in row echelon form by inspection., , THEOREM 4.7.5 If a matrix, , R is in row echelon form, then the row vectors with the, leading 1’s (the nonzero row vectors) form a basis for the row space of R, and the column, vectors with the leading 1’s of the row vectors form a basis for the column space of R ., , The proof essentially involves an analysis of the positions of the 0’s and 1’s of R . We, omit the details., , E X A M P L E 5 Bases for the Row and Column Spaces of a Matrix in Row, Echelon Form, , Find bases for the row and column spaces of the matrix, , ⎡, , 1, ⎢0, ⎢, R=⎢, ⎣0, 0, , −2, , 5, 3, 0, 0, , 1, 0, 0, , 0, 0, 1, 0, , ⎤, , 3, 0⎥, ⎥, ⎥, 0⎦, 0, , Solution Since the matrix R is in row echelon form, it follows from Theorem 4.7.5 that, the vectors, , r1 = [1, , −2, , 5, , 0, , 3], , r2 = [0, , 1, , 3, , 0, , 0], , r3 = [0, , 0, , 0, , 1, , 0], , form a basis for the row space of R , and the vectors, , ⎡ ⎤, , ⎡, , ⎤, , ⎡ ⎤, , 1, 0, −2, ⎢0⎥, ⎢ 1⎥, ⎢0⎥, ⎢ ⎥, ⎢ ⎥, ⎢ ⎥, c1 = ⎢ ⎥, c2 = ⎢ ⎥, c4 = ⎢ ⎥, ⎣0⎦, ⎣ 0⎦, ⎣1⎦, 0, 0, 0, form a basis for the column space of R .

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242, , Chapter 4 General Vector Spaces, , E X A M P L E 6 Basis for a Row Space by Row Reduction, , Find a basis for the row space of the matrix, , ⎡, , 1, ⎢ 2, ⎢, A=⎢, ⎣ 2, −1, , −3, −6, −6, 3, , 4, 9, 9, −4, , −2, −1, −1, 2, , 5, 8, 9, −5, , ⎤, , 4, 2⎥, ⎥, ⎥, 7⎦, −4, , Solution Since elementary row operations do not change the row space of a matrix, we, , can find a basis for the row space of A by finding a basis for the row space of any row, echelon form of A. Reducing A to row echelon form, we obtain (verify), , ⎡, , 1, ⎢0, ⎢, R=⎢, ⎣0, 0, , −3, 0, 0, 0, , 4, 1, 0, 0, , −2, 3, 0, 0, , 5, −2, 1, 0, , ⎤, , 4, −6⎥, ⎥, ⎥, 5⎦, 0, , By Theorem 4.7.5, the nonzero row vectors of R form a basis for the row space of R and, hence form a basis for the row space of A. These basis vectors are, , Basis for the Column, Space of a Matrix, , r1 = [1, , −3, , 4, , −2, , 5, , 4], , r2 = [0, , 0, , 1, , 3, , −2, , −6], , r3 = [0, , 0, , 0, , 0, , 1, , 5], , The problem of finding a basis for the column space of a matrix A in Example 6 is, complicated by the fact that an elementary row operation can alter its column space., However, the good news is that elementary row operations do not alter dependence relationships among the column vectors. To make this more precise, suppose that w1 , w2 , . . . , wk, are linearly dependent column vectors of A, so there are scalars c1 , c2 , . . . , ck that are, not all zero and such that, , c1 w1 + c2 w2 + · · · + ck wk = 0, , (4), , If we perform an elementary row operation on A, then these vectors will be changed, into new column vectors w1 , w2 , . . . , wk . At first glance it would seem possible that the, transformed vectors might be linearly independent. However, this is not so, since it can, be proved that these new column vectors are linearly dependent and, in fact, related by, an equation, c1 w1 + c2 w2 + · · · + ck wk = 0, that has exactly the same coefficients as (4). It can also be proved that elementary row, operations do not alter the linear independence of a set of column vectors. All of these, results are summarized in the following theorem., , Although elementary row operations can change the column space of a matrix, it, follows from Theorem 4.7.6(b), that they do not change the, dimension of its column space., , THEOREM 4.7.6 If A and B are row equivalent matrices, then:, , (a) A given set of column vectors of A is linearly independent if and only if the corresponding column vectors of B are linearly independent., (b) A given set of column vectors of A forms a basis for the column space of A if and, only if the corresponding column vectors of B form a basis for the column space, of B .

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4.7 Row Space, Column Space, and Null Space, , 243, , E X A M P L E 7 Basis for a Column Space by Row Reduction, , Find a basis for the column space of the matrix, , ⎡, , 1, ⎢ 2, ⎢, A=⎢, ⎣ 2, −1, , −3, −6, −6, 3, , 4, 9, 9, −4, , −2, −1, −1, , 5, 8, 9, −5, , 2, , ⎤, , 4, 2⎥, ⎥, ⎥, 7⎦, −4, , that consists of column vectors of A., Solution We observed in Example 6 that the matrix, , ⎡, , 1, ⎢0, ⎢, R=⎢, ⎣0, 0, , −3, 0, 0, 0, , 4, 1, 0, 0, , −2, 3, 0, 0, , 5, −2, 1, 0, , ⎤, , 4, −6⎥, ⎥, ⎥, 5⎦, 0, , is a row echelon form of A. Keeping in mind that A and R can have different column, spaces, we cannot find a basis for the column space of A directly from the, column vectors of R . However, it follows from Theorem 4.7.6(b) that if we can find, a set of column vectors of R that forms a basis for the column space of R , then the, corresponding column vectors of A will form a basis for the column space of A., Since the first, third, and fifth columns of R contain the leading 1’s of the row vectors,, the vectors, ⎡ ⎤, ⎡ ⎤, ⎡ ⎤, 1, 4, 5, ⎢0⎥, ⎢ 1⎥, ⎢−2⎥, ⎢ ⎥, ⎢ ⎥, ⎢ ⎥, c1 = ⎢ ⎥, c3 = ⎢ ⎥, c5 = ⎢ ⎥, ⎣0⎦, ⎣ 0⎦, ⎣ 1⎦, 0, 0, 0, form a basis for the column space of R . Thus, the corresponding column vectors of A,, which are, ⎡ ⎤, ⎡ ⎤, ⎡ ⎤, 1, 4, 5, ⎢ 2⎥, ⎢ 9⎥, ⎢ 8⎥, ⎢ ⎥, ⎢ ⎥, ⎢ ⎥, c1 = ⎢ ⎥, c3 = ⎢ ⎥, c5 = ⎢ ⎥, ⎣ 2⎦, ⎣ 9⎦, ⎣ 9⎦, −1, −4, −5, form a basis for the column space of A., Up to now we have focused on methods for finding bases associated with matrices., Those methods can readily be adapted to the more general problem of finding a basis, for the subspace spanned by a set of vectors in R n ., E X A M P L E 8 Basis for the Space Spanned by a Set of Vectors, , The following vectors span a subspace of R 4 . Find a subset of these vectors that forms, a basis of this subspace., v1 = (1, 2, 2, −1),, , v2 = (−3, −6, −6, 3),, , v3 = (4, 9, 9, −4),, , v4 = (−2, −1, −1, 2),, , v5 = (5, 8, 9, −5),, , v6 = (4, 2, 7, −4), , Solution If we rewrite these vectors in column form and construct the matrix that has, , those vectors as its successive columns, then we obtain the matrix A in Example 7 (verify)., Thus,, span{v1 , v2 , v3 , v4 , v5 , v6 } = col(A)

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244, , Chapter 4 General Vector Spaces, , Proceeding as in that example (and adjusting the notation appropriately), we see that, the vectors v1 , v3 , and v5 form a basis for, span{v1 , v2 , v3 , v4 , v5 , v6 }, Bases Formed from Row, and Column Vectors of a, Matrix, , In Example 6, we found a basis for the row space of a matrix by reducing that matrix, to row echelon form. However, the basis vectors produced by that method were not all, row vectors of the original matrix. The following adaptation of the technique used in, Example 7 shows how to find a basis for the row space of a matrix that consists entirely, of row vectors of that matrix., E X A M P L E 9 Basis for the Row Space of a Matrix, , Find a basis for the row space of, , ⎡, , 1, ⎢2, ⎢, A=⎢, ⎣0, 2, , ⎤, , 0, 0, −2, − 5 −3 − 2, 5, 6, , 15, 18, , 3, 6⎥, ⎥, ⎥, 0⎦, 6, , 10, 8, , consisting entirely of row vectors from A., Solution We will transpose A, thereby converting the row space of A into the column, space of AT ; then we will use the method of Example 7 to find a basis for the column, space of AT ; and then we will transpose again to convert column vectors back to row, vectors., Transposing A yields, , ⎡, , 1, ⎢−2, ⎢, ⎢, AT = ⎢ 0, ⎢, ⎣ 0, 3, , 2, −5, −3, −2, 6, , ⎤, , 0, 5, 15, 10, 0, , 2, 6⎥, ⎥, ⎥, 18⎥, ⎥, 8⎦, 6, , and then reducing this matrix to row echelon form we obtain, , ⎡, , 1, ⎢0, ⎢, ⎢, ⎢0, ⎢, ⎣0, 0, , 2, 1, 0, 0, 0, , 0, −5, 0, 0, 0, , ⎤, , 2, −10⎥, ⎥, ⎥, 1⎥, ⎥, 0⎦, 0, , The first, second, and fourth columns contain the leading 1’s, so the corresponding, column vectors in AT form a basis for the column space of AT ; these are, , ⎡, , ⎤, , ⎡, , ⎤, , ⎡ ⎤, , 1, 2, 2, ⎢−2⎥, ⎢−5⎥, ⎢ 6⎥, ⎢ ⎥, ⎢ ⎥, ⎢ ⎥, ⎢ ⎥, ⎢ ⎥, ⎢ ⎥, c1 = ⎢ 0⎥, c2 = ⎢−3⎥, and c4 = ⎢18⎥, ⎢ ⎥, ⎢ ⎥, ⎢ ⎥, ⎣ 0⎦, ⎣−2⎦, ⎣ 8⎦, 6, 3, 6, Transposing again and adjusting the notation appropriately yields the basis vectors, r1 = [1, , −2, , 0, , 0, , 3],, , r4 = [2, for the row space of A., , 6, , r2 = [2, 18, , 8, , −5, 6], , −3, , −2, , 6],

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4.7 Row Space, Column Space, and Null Space, , 245, , Next we will give an example that adapts the method of Example 7 to solve the, following general problem in R n :, , S = {v1 , v2 , . . . , vk } in R n , find a subset of these, vectors that forms a basis for span(S), and express each vector that is not in that basis, Problem Given a set of vectors, , as a linear combination of the basis vectors., , E X A M P L E 10 Basis and Linear Combinations, , (a) Find a subset of the vectors, v1 = (1, −2, 0, 3), v2 = (2, −5, −3, 6),, v3 = (0, 1, 3, 0), v4 = (2, −1, 4, −7), v5 = (5, −8, 1, 2), that forms a basis for the subspace of R 4 spanned by these vectors., (b) Express each vector not in the basis as a linear combination of the basis vectors., Solution (a) We begin by constructing a matrix that has v1 , v2 , . . . , v5 as its column, , vectors:, , Had we only been interested, in part (a) of this example, it, would have sufficed to reduce, the matrix to row echelon, form. It is for part (b) that, the reduced row echelon form, is most useful., , ⎡, , 1, ⎢− 2, ⎢, ⎢, ⎣ 0, 3, , 2, −5, −3, 6, , 0, 1, 3, 0, , 2, −1, 4, −7, , ⎤, , 5, −8⎥, ⎥, ⎥, 1⎦, 2, , ↑, , ↑, , ↑, , ↑, , ↑, , v1, , v2, , v3, , v4, , v5, , (5), , The first part of our problem can be solved by finding a basis for the column space of, this matrix. Reducing the matrix to reduced row echelon form and denoting the column, vectors of the resulting matrix by w1 , w2 , w3 , w4 , and w5 yields, , ⎡, , 1, ⎢0, ⎢, ⎢, ⎣0, 0, , 0, 1, 0, 0, , 2, −1, 0, 0, , ⎤, , 0, 0, 1, 0, , 1, 1⎥, ⎥, ⎥, 1⎦, 0, , ↑, , ↑, , ↑, , ↑, , ↑, , w1, , w2, , w3, , w4, , w5, , (6), , The leading 1’s occur in columns 1, 2, and 4, so by Theorem 4.7.5,, , {w1 , w2 , w4 }, is a basis for the column space of (6), and consequently,, , {v1 , v2 , v4 }, is a basis for the column space of (5)., Solution (b) We will start by expressing w3 and w5 as linear combinations of the basis, , vectors w1 , w2 , w4 . The simplest way of doing this is to express w3 and w5 in terms, of basis vectors with smaller subscripts. Accordingly, we will express w3 as a linear, combination of w1 and w2 , and we will express w5 as a linear combination of w1 , w2 ,, and w4 . By inspection of (6), these linear combinations are, w3 = 2w1 − w2, w5 = w1 + w2 + w4

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246, , Chapter 4 General Vector Spaces, , We call these the dependency equations. The corresponding relationships in (5) are, v3 = 2v1 − v2, v5 = v1 + v2 + v4, The following is a summary of the steps that we followed in our last example to solve, the problem posed above., Basis for the Space Spanned by a Set of Vectors, Step 1. Form the matrix A whose columns are the vectors in the set S = {v1 , v2 , . . . , vk }., Step 2. Reduce the matrix A to reduced row echelon form R ., Step 3. Denote the column vectors of R by w1 , w2 , . . . , wk ., Step 4. Identify the columns of R that contain the leading 1’s. The corresponding, column vectors of A form a basis for span(S)., This completes the first part of the problem., Step 5. Obtain a set of dependency equations for the column vectors w1 , w2 , . . . , wk, of R by successively expressing each wi that does not contain a leading 1 of, R as a linear combination of predecessors that do., Step 6. In each dependency equation obtained in Step 5, replace the vector wi by the, vector vi for i = 1, 2, . . . , k ., This completes the second part of the problem., , Exercise Set 4.7, , ⎡, , In Exercises 1–2, express the product Ax as a linear combination of the column vectors of A., , ⎡, , 1. (a), , 2, , 3, 4, , −1, ⎡, , −3, , 1, 2, , 1, , 0, 6, −1, , 2, (b), 6, , 1, 3, , ⎤, , 2 ⎡ ⎤, −1, 0⎥, ⎥⎢ ⎥, ⎥ ⎣ 2⎦, − 1⎦, 5, 3, , 6, −4, 3, 8, , ⎢ 5, ⎢, 2. (a) ⎢, ⎣ 2, , 4, ⎢, (b) ⎣3, 0, , −1, , ⎤⎡, , ⎤, , ⎥⎢, , ⎥, , −2, , 2 ⎦ ⎣ 3⎦, 5, 4, , ⎡, , ⎤, , 3, 5 ⎢ ⎥, ⎣ 0⎦, −8, −5, , 1, ⎢, 3. (a) A = ⎣1, 2, , ⎡, , 1, ⎢, (b) A = ⎣9, 1, , ⎡, , 1, ⎢, 4. (a) A = ⎣−1, −1, , ⎤, , ⎡, , ⎤, , 2, −1, ⎥, ⎢ ⎥, 1⎦; b = ⎣ 0⎦, 3, 2, , 1, 0, 1, , −1, 3, 1, , −1, 1, −1, , ⎡, , ⎤, , ⎤, , 1, 5, ⎢ ⎥, ⎥, 1⎦; b = ⎣ 1⎦, 1, −1, , ⎤, , 1, , ⎡ ⎤, 2, , ⎢ ⎥, ⎥, −1⎦; b = ⎣0⎦, 1, , 2, 1, 2, 1, , 0, 2, 1, 2, , ⎤, , ⎡ ⎤, , 1, 4, ⎢ 3⎥, 1⎥, ⎥, ⎢ ⎥, ⎥; b = ⎢ ⎥, 3⎦, ⎣ 5⎦, 2, 7, , 5. Suppose that x1 = 3, x2 = 0, x3 = −1, x4 = 5 is a solution of, a nonhomogeneous linear system Ax = b and that the solution set of the homogeneous system Ax = 0 is given by the, formulas, , x1 = 5r − 2s, x2 = s, x3 = s + t, x4 = t, (a) Find a vector form of the general solution of Ax = 0., , In Exercises 3–4, determine whether b is in the column space, of A, and if so, express b as a linear combination of the column, vectors of A, , ⎡, , 1, ⎢0, ⎢, (b) A = ⎢, ⎣1, 0, , 0, , (b) Find a vector form of the general solution of Ax = b., 6. Suppose that x1 = −1, x2 = 2, x3 = 4, x4 = −3 is a solution, of a nonhomogeneous linear system Ax = b and that the solution set of the homogeneous system Ax = 0 is given by the, formulas, , x1 = −3r + 4s, x2 = r − s, x3 = r, x4 = s, (a) Find a vector form of the general solution of Ax = 0., (b) Find a vector form of the general solution of Ax = b., In Exercises 7–8, find the vector form of the general solution, of the linear system Ax = b, and then use that result to find the, vector form of the general solution of Ax = 0.

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4.7 Row Space, Column Space, and Null Space, , 7. (a) x1 − 3x2 = 1, 2x1 − 6x2 = 2, 8. (a), , (b), , x1 − 2x2, 2x1 − 4x2, −x1 + 2x2, 3x1 − 6x2, x1, −2x1, −x1, 4 x1, , (b) x1 + x2 + 2x3 =, , 5, , x1, + x3 = −2, 2x1 + x2 + 3x3 = 3, , + x3, + 2x3, − x3, + 3x3, , + 2x4, + 4x4, − 2x4, + 6x4, , = −1, = −2, = 1, = −3, , + 2x2 − 3x3 + x4, + x2 + 2x3 + x4, + 3x2 − x3 + 2x4, − 7x2, − 5x4, , 17. v1 = (1, −1, 5, 2), v2 = (−2, 3, 1, 0),, v3 = (4, −5, 9, 4), v4 = (0, 4, 2, −3),, v5 = (−7, 18, 2, −8), , = 4, = −1, = 3, = −5, , −1, −4, −6, , 1, ⎢, 9. (a) A = ⎣5, 7, , ⎡, , 1, ⎢, 10. (a) A = ⎣ 2, −1, , ⎡, , 3, , 1, , 2, ⎢, (b) A = ⎣4, 0, , 2, 5, 3, 2, , 4, −2, 0, 3, , 2, , ⎡, , ⎥, −4⎦, , 4, 1, 3, , ⎢ 3, ⎢, (b) A = ⎢, ⎣−1, , ⎤, , 0, 0, 0, , ⎤, , −1, ⎥, −2⎦, 0, , ⎤, , 2, ⎥, 0⎦, 2, 5, 1, −1, 5, , 6, 4, −2, 7, , 1, , ⎢, 11. (a) ⎣0, 0, , ⎡, , 1, , ⎢0, ⎢, ⎢, 12. (a) ⎢0, ⎢, ⎣0, 0, , 0, 0, 0, , 1, ⎢0, ⎢, (b) ⎢, ⎣0, 0, , 2, ⎥, 1⎦, 0, 2, 1, 0, 0, 0, , 4, , −3, 1, 0, 0, , ⎤, , 20. Construct a matrix whose null space consists of all linear, combinations of the⎡vectors, ⎤, ⎡ ⎤, 1, 2, ⎢−1⎥, ⎢ 0⎥, ⎢ ⎥, ⎢ ⎥, v1 = ⎢ ⎥ and v2 = ⎢ ⎥, ⎣ 3⎦, ⎣−2⎦, 4, 2, , (a) b = (0, 0), , 1, ⎢−2, A=⎢, ⎣−1, −3, , −3, 1, 0, 0, , 1, ⎢0, ⎢, (b) ⎢, ⎣0, 0, , −2, 5, 3, 8, , 5, −7, −2, −9, , 2, , 0, , 1, , −1, , 4, , , . For the given vector b,, , 2, 1, 0, 0, , ⎤, , 0, 0, 0, 0, , 0, 0⎥, ⎥, ⎥, 0⎦, 0, , −1, , 5, 3⎥, ⎥, ⎥, −7⎦, 1, , 4, 1, 0, , ⎤, , 0, 0, 1, 1, , (c) b = (−1, 1), , ⎤, , 0, , 1⎥, ⎥, , ⎥. For the given vector b, find, , 1⎦, , 2 0, the general form of all vectors x in R 2 for which TA (x) = b if, such vectors exist., (a) b = (0, 0, 0, 0), , (b) b = (1, 1, −1, −1), , (c) b = (2, 0, 0, 2), 23. (a) Let, , ⎡, , 0, , ⎢, A = ⎣1, , ⎤, , 1, , 0, , 0, , 0⎦, , ⎥, , 0 0 0, Show that relative to an xyz-coordinate system in 3-space, the null space of A consists of all points on the z-axis and, that the column space consists of all points in the xy -plane, (see the accompanying figure)., (b) Find a 3 × 3 matrix whose null space is the x -axis and, whose column space is the yz-plane., z, , (b) Use the method of Example 9 to find a basis for the row, space of A that consists entirely of row vectors of A., , Null space of A, , In Exercises 14–15, find a basis for the subspace of R 4 that is, spanned by the given vectors., , y, , 14. (1, 1, −4, −3), (2, 0, 2, −2), (2, −1, 3, 2), 15. (1, 1, 0, 0), (0, 0, 1, 1), (−2, 0, 2, 2), (0, −3, 0, 3), , ⎡, , 2, , ⎤, , 3, −6⎥, ⎥, −3⎦, −9, , (b) b = (1, 3), , ⎢0, ⎢, 22. In each part, let A = ⎢, ⎣1, , 13. (a) Use the methods of Examples 6 and 7 to find bases for the, row space and column space of the matrix, , ⎡, , 1, , find the general form of all vectors x in R 3 for which TA (x) = b, if such vectors exist., , ⎡, , 5, 0⎥, ⎥, ⎥, −3⎥, ⎥, 1⎦, 0, , 19. The matrix in Exercise 10(b)., , 21. In each part, let A =, , 9, −1⎥, ⎥, ⎥, −1⎦, 8, , ⎡, , ⎤, , In Exercises 18–19, find a basis for the row space of A that, consists entirely of row vectors of A., 18. The matrix in Exercise 10(a)., , , , ⎤, , In Exercises 11–12, a matrix in row echelon form is given. By, inspection, find a basis for the row space and for the column space, of that matrix., , ⎡, , In Exericses 16–17, find a subset of the given vectors that forms, a basis for the space spanned by those vectors, and then express, each vector that is not in the basis as a linear combination of the, basis vectors., 16. v1 = (1, 0, 1, 1), v2 = (−3, 3, 7, 1),, v3 = (−1, 3, 9, 3), v4 = (−5, 3, 5, −1), , In Exercises 9–10, find bases for the null space and row space, of A., , ⎡, , 247, , x, , Column space, of A, , Figure Ex-23

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248, , Chapter 4 General Vector Spaces, , 24. Find a 3 × 3 matrix whose null space is, (a) a point., , (b) a line., , (d) The set of nonzero row vectors of a matrix A is a basis for the, row space of A., , (c) a plane., , (e) If A and B are n × n matrices that have the same row space,, then A and B have the same column space., , 25. (a) Find all 2 × 2 matrices whose null space is the line, 3x − 5y = 0., (b) Describe the null spaces of the following matrices:, , , , A=, , 1, , 4, , 0, , 5, , , , , , , B=, , 1, , 0, , 0, , 5, , , , , , , C=, , 6, , 2, , 3, , 1, , , , , D=, , , , 0, , 0, , 0, , 0, , , , (f ) If E is an m × m elementary matrix and A is an m × n matrix,, then the null space of EA is the same as the null space of A., (g) If E is an m × m elementary matrix and A is an m × n matrix,, then the row space of EA is the same as the row space of A., (h) If E is an m × m elementary matrix and A is an m × n matrix,, then the column space of EA is the same as the column space, of A., , Working with Proofs, 26. Prove Theorem 4.7.4., 27. Prove that the row vectors of an n × n invertible matrix A, form a basis for R n ., , (i) The system Ax = b is inconsistent if and only if b is not in the, column space of A., , 28. Suppose that A and B are n × n matrices and A is invertible., Invent and prove a theorem that describes how the row spaces, of AB and B are related., , ( j) There is an invertible matrix A and a singular matrix B such, that the row spaces of A and B are the same., , Working withTechnology, True-False Exercises, , T1. Find a basis for the column space of, , TF. In parts (a)–( j) determine whether the statement is true or, false, and justify your answer., (a) The span of v1 , . . . , vn is the column space of the matrix, whose column vectors are v1 , . . . , vn ., (b) The column space of a matrix A is the set of solutions of, Ax = b., (c) If R is the reduced row echelon form of A, then those column, vectors of R that contain the leading 1’s form a basis for the, column space of A., , ⎡, , 2, , ⎢3, ⎢, ⎢, A=⎢, ⎢3, ⎢, ⎣2, 1, , 4, , ⎤, , 6, , 0, , 8, , 4, , 12, , 9, , −2, , 8, , 6, , 18, , 9, , −7, , −2, , 6, , −3, , 6, , 5, , 18, , 4, , 33, , ⎥, ⎥, −1⎥, ⎥, ⎥, 11⎦, , 3, , −2, , 0, , 2, , 6, , 2, , 6⎥, , that consists of column vectors of A., T2. Find a basis for the row space of the matrix A in Exercise T1, that consists of row vectors of A., , 4.8 Rank, Nullity, and the Fundamental Matrix Spaces, In the last section we investigated relationships between a system of linear equations and, the row space, column space, and null space of its coefficient matrix. In this section we will, be concerned with the dimensions of those spaces. The results we obtain will provide a, deeper insight into the relationship between a linear system and its coefficient matrix., , Row and Column Spaces, Have Equal Dimensions, , In Examples 6 and 7 of Section 4.7 we found that the row and column spaces of the, matrix, ⎤, ⎡, 1 −3, 4 −2, 5, 4, ⎢ 2 −6, 9 −1, 8, 2⎥, ⎥, ⎢, A=⎢, ⎥, 9 −1, 9, 7⎦, ⎣ 2 −6, −1, 3 −4, 2 − 5 −4, both have three basis vectors and hence are both three-dimensional. The fact that these, spaces have the same dimension is not accidental, but rather a consequence of the following theorem.

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4.8 Rank, Nullity, and the Fundamental Matrix Spaces, THEOREM 4.8.1 The row space and the column space of a matrix, , 249, , A have the same, , dimension., Proof It follows from Theorems 4.7.4 and 4.7.6 (b) that elementary row operations do, not change the dimension of the row space or of the column space of a matrix. Thus, if, R is any row echelon form of A, it must be true that, , The proof of Theorem 4.8.1, shows that the rank of A can, be interpreted as the number, of leading 1’s in any row echelon form of A., , Rank and Nullity, , dim(row space of A) = dim(row space of R), dim(column space of A) = dim(column space of R), so it suffices to show that the row and column spaces of R have the same dimension. But, the dimension of the row space of R is the number of nonzero rows, and by Theorem, 4.7.5 the dimension of the column space of R is the number of leading 1’s. Since these, two numbers are the same, the row and column space have the same dimension., , The dimensions of the row space, column space, and null space of a matrix are such, important numbers that there is some notation and terminology associated with them., DEFINITION 1 The common dimension of the row space and column space of a, , matrix A is called the rank of A and is denoted by rank(A); the dimension of the null, space of A is called the nullity of A and is denoted by nullity(A)., , E X A M P L E 1 Rank and Nullity of a 4 × 6 Matrix, , Find the rank and nullity of the matrix, , ⎡, , −1, , ⎢ 3, ⎢, A=⎢, ⎣ 2, 4, , 2, −7, −5, −9, , 0, 2, 2, 2, , 4, 0, 4, −4, , 5, 1, 6, −4, , −3, , 4⎥, ⎥, ⎥, 1⎦, 7, , Solution The reduced row echelon form of A is, , ⎡, , 1, ⎢0, ⎢, ⎢, ⎣0, 0, , 0, 1, 0, 0, , −4, −2, 0, 0, , −28 −37, −12 −16, 0, 0, , 0, 0, , ⎤, , ⎤, , 13, 5⎥, ⎥, ⎥, 0⎦, 0, , (1), , (verify). Since this matrix has two leading 1’s, its row and column spaces are twodimensional and rank(A) = 2. To find the nullity of A, we must find the dimension of, the solution space of the linear system Ax = 0. This system can be solved by reducing, its augmented matrix to reduced row echelon form. The resulting matrix will be identical to (1), except that it will have an additional last column of zeros, and hence the, corresponding system of equations will be, , x1 − 4x3 − 28x4 − 37x5 + 13x6 = 0, x2 − 2x3 − 12x4 − 16x5 + 5x6 = 0, Solving these equations for the leading variables yields, , x1 = 4x3 + 28x4 + 37x5 − 13x6, x2 = 2x3 + 12x4 + 16x5 − 5x6, , (2)

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250, , Chapter 4 General Vector Spaces, , from which we obtain the general solution, , x1 = 4r + 28s + 37t − 13u, x2 = 2r + 12s + 16t − 5u, x3 = r, x4 = s, x5 = t, x6 = u, or in column vector form, , ⎡ ⎤, ⎡ ⎤, ⎡ ⎤, ⎡ ⎤, ⎡, ⎤, 4, 28, 37, x1, −13, ⎢ ⎥, ⎢ ⎥, ⎢ ⎥, ⎢ ⎥, ⎢, ⎥, ⎢ x2 ⎥, ⎢2 ⎥, ⎢12⎥, ⎢16⎥, ⎢ −5⎥, ⎢ ⎥, ⎢ ⎥, ⎢ ⎥, ⎢ ⎥, ⎢, ⎥, ⎢x3 ⎥, ⎢ ⎥, ⎢ ⎥, ⎢ ⎥, ⎢, ⎥, ⎢ ⎥ = r ⎢1⎥ + s ⎢ 0⎥ + t ⎢ 0⎥ + u ⎢ 0⎥, ⎢x4 ⎥, ⎢ 0⎥, ⎢ 1⎥, ⎢ 0⎥, ⎢ 0⎥, ⎢ ⎥, ⎢ ⎥, ⎢ ⎥, ⎢ ⎥, ⎢, ⎥, ⎢ ⎥, ⎢ ⎥, ⎢ ⎥, ⎢ ⎥, ⎢, ⎥, ⎣x5 ⎦, ⎣ 0⎦, ⎣ 0⎦, ⎣ 1⎦, ⎣ 0⎦, 0, 0, 0, 1, x6, , (3), , Because the four vectors on the right side of (3) form a basis for the solution space,, nullity(A) = 4., E X A M P L E 2 Maximum Value for Rank, , What is the maximum possible rank of an m × n matrix A that is not square?, , A lie in R n and the column vectors in R m , the row, space of A is at most n-dimensional and the column space is at most m-dimensional., Since the rank of A is the common dimension of its row and column space, it follows, that the rank is at most the smaller of m and n. We denote this by writing, , Solution Since the row vectors of, , rank(A) ≤ min(m, n), in which min(m, n) is the minimum of m and n., The following theorem establishes a fundamental relationship between the rank and, nullity of a matrix., THEOREM 4.8.2 Dimension Theorem for Matrices, , If A is a matrix with n columns, then, rank(A) + nullity(A) = n, , (4), , = 0 has n unknowns, (variables). These fall into two distinct categories: the leading variables and the free, variables. Thus,, number of leading, number of free, +, =n, variables, variables, But the number of leading variables is the same as the number of leading 1’s in any row, echelon form of A, which is the same as the dimension of the row space of A, which is, the same as the rank of A. Also, the number of free variables in the general solution of, Ax = 0 is the same as the number of parameters in that solution, which is the same as, the dimension of the solution space of Ax = 0, which is the same as the nullity of A., This yields Formula (4)., Proof Since A has n columns, the homogeneous linear system Ax

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4.8 Rank, Nullity, and the Fundamental Matrix Spaces, , 251, , E X A M P L E 3 The Sum of Rank and Nullity, , ⎡, , The matrix, , −1, , ⎢ 3, ⎢, A=⎢, ⎣ 2, 4, , 2, −7, −5, −9, , 0, 2, 2, 2, , 4, 0, 4, −4, , 5, 1, 6, −4, , −3, , ⎤, , 4⎥, ⎥, ⎥, 1⎦, 7, , has 6 columns, so, rank(A) + nullity(A) = 6, This is consistent with Example 1, where we showed that, rank(A) = 2 and nullity(A) = 4, , The following theorem, which summarizes results already obtained, interprets rank, and nullity in the context of a homogeneous linear system., THEOREM 4.8.3 If A is an m × n matrix, then, , (a) rank(A) = the number of leading variables in the general solution of Ax = 0., (b) nullity(A) = the number of parameters in the general solution of Ax = 0., , E X A M P L E 4 Rank, Nullity, and Linear Systems, , (a) Find the number of parameters in the general solution of Ax = 0 if A is a 5 × 7, matrix of rank 3., (b) Find the rank of a 5 × 7 matrix A for which Ax = 0 has a two-dimensional solution, space., Solution (a) From (4),, , nullity(A) = n − rank(A) = 7 − 3 = 4, Thus, there are four parameters., Solution (b) The matrix A has nullity 2, so, , rank(A) = n − nullity(A) = 7 − 2 = 5, Recall from Section 4.7 that if Ax = b is a consistent linear system, then its general, solution can be expressed as the sum of a particular solution of this system and the general, solution of Ax = 0. We leave it as an exercise for you to use this fact and Theorem 4.8.3, to prove the following result., , = b is a consistent linear system of m equations in n unknowns,, and if A has rank r , then the general solution of the system contains n − r parameters., , THEOREM 4.8.4 If Ax, , The Fundamental Spaces of, a Matrix, , There are six important vector spaces associated with a matrix A and its transpose AT :, row space of A, , row space of AT, , column space of A, , column space of AT, , null space of A, , null space of AT

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252, , Chapter 4 General Vector Spaces, , If A is an m × n matrix, then, the row space and null space, of A are subspaces of R n , and, the column space of A and the, null space of AT are subspaces, of R m ., , However, transposing a matrix converts row vectors into column vectors and conversely,, so except for a difference in notation, the row space of AT is the same as the column, space of A, and the column space of AT is the same as the row space of A. Thus, of the, six spaces listed above, only the following four are distinct:, row space of A, , column space of A, , null space of A, , null space of AT, , These are called the fundamental spaces of a matrix A. We will now consider how these, four subspaces are related., Let us focus for a moment on the matrix AT . Since the row space and column space, of a matrix have the same dimension, and since transposing a matrix converts its columns, to rows and its rows to columns, the following result should not be surprising., THEOREM 4.8.5 If A is any matrix, then rank(A), , = rank(AT )., , Proof, , rank(A) = dim(row space of A) = dim(column space of AT ) = rank(AT )., This result has some important implications. For example, if A is an m × n matrix,, then applying Formula (4) to the matrix AT and using the fact that this matrix has m, columns yields, rank(AT ) + nullity(AT ) = m, which, by virtue of Theorem 4.8.5, can be rewritten as, rank(A) + nullity(AT ) = m, , (5), , This alternative form of Formula (4) makes it possible to express the dimensions of all, four fundamental spaces in terms of the size and rank of A. Specifically, if rank(A) = r ,, then, , A Geometric Link Between, the Fundamental Spaces, , dim[row(A)] = r, , dim[col(A)] = r, , dim[null(A)] = n − r, , dim[null(AT )] = m − r, , (6), , The four formulas in (6) provide an algebraic relationship between the size of a matrix, and the dimensions of its fundamental spaces. Our next objective is to find a geometric, relationship between the fundamental spaces themselves. For this purpose recall from, Theorem 3.4.3 that if A is an m × n matrix, then the null space of A consists of those, vectors that are orthogonal to each of the row vectors of A. To develop that idea in more, detail, we make the following definition., , W is a subspace of R n , then the set of all vectors in R n that are, orthogonal to every vector in W is called the orthogonal complement of W and is, denoted by the symbol W ⊥ ., DEFINITION 2 If, , The following theorem lists three basic properties of orthogonal complements. We, will omit the formal proof because a more general version of this theorem will be proved, later in the text.

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4.8 Rank, Nullity, and the Fundamental Matrix Spaces, , Part (b) of Theorem 4.8.6 can, be expressed as, , THEOREM 4.8.6 If W is a subspace of R n , then:, , W ∩ W ⊥ = {0}, , (b) The only vector common to W and W ⊥ is 0., , and part (c) as, , 253, , (a) W ⊥ is a subspace of R n ., (c) The orthogonal complement of W ⊥ is W ., , (W ⊥ )⊥ = W, , E X A M P L E 5 Orthogonal Complements, , In R 2 the orthogonal complement of a line W through the origin is the line through the, origin that is perpendicular to W (Figure 4.8.1a); and in R 3 the orthogonal complement, of a plane W through the origin is the line through the origin that is perpendicular to, that plane (Figure 4.8.1b)., y, , y, , W⊥, W, W, x, x, , Explain why {0} and R n are, orthogonal complements., , W⊥, , z, , (a), , Figure 4.8.1, , (b), , The next theorem will provide a geometric link between the fundamental spaces of, a matrix. In the exercises we will ask you to prove that if a vector in R n is orthogonal, to each vector in a basis for a subspace of R n , then it is orthogonal to every vector in, that subspace. Thus, part (a) of the following theorem is essentially a restatement of, Theorem 3.4.3 in the language of orthogonal complements; it is illustrated in Example 6, of Section 3.4. The proof of part (b), which is left as an exercise, follows from part (a)., The essential idea of the theorem is illustrated in Figure 4.8.2., z, , z, , y, T, , 0, , Figure 4.8.2, , Row, , A, , ll A, Nu, , ll A, , Nu, , x, , y, , x, , 0, Col, , A, , THEOREM 4.8.7 If A is an m × n matrix, then:, , (a) The null space of A and the row space of A are orthogonal complements in R n ., (b) The null space of AT and the column space of A are orthogonal complements in R m .

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254, , Chapter 4 General Vector Spaces, , More on the Equivalence, Theorem, , In Theorem 2.3.8 we listed six results that are equivalent to the invertibility of a square, matrix A. We are now in a position to add ten more statements to that list to produce a, single theorem that summarizes and links together all of the topics that we have covered, thus far. We will prove some of the equivalences and leave others as exercises., THEOREM 4.8.8 Equivalent Statements, , If A is an n × n matrix, then the following statements are equivalent., (a), , A is invertible., , (b), (c), , Ax = 0 has only the trivial solution., The reduced row echelon form of A is In ., , (d ), , A is expressible as a product of elementary matrices., , Ax = b is consistent for every n × 1 matrix b., ( f ) Ax = b has exactly one solution for every n × 1 matrix b., ( g) det(A)  = 0., (h) The column vectors of A are linearly independent., (i ) The row vectors of A are linearly independent., ( j) The column vectors of A span R n ., , (e), , (k), , The row vectors of A span R n ., , (l ), , The column vectors of A form a basis for R n ., , (m) The row vectors of A form a basis for R n ., (n), (o), , A has rank n., A has nullity 0., , ( p) The orthogonal complement of the null space of A is R n ., (q), , The orthogonal complement of the row space of A is {0}., , (h) through (m) follows from Theorem 4.5.4 (we omit the, details). To complete the proof we will show that (b), (n), and (o) are equivalent by, proving the chain of implications (b) ⇒ (o) ⇒ (n) ⇒ (b)., , Proof The equivalence of, , (b) ⇒ (o) If Ax = 0 has only the trivial solution, then there are no parameters in that, solution, so nullity(A) = 0 by Theorem 4.8.3(b)., (o) ⇒ (n) Theorem 4.8.2., (n) ⇒ (b) If A has rank n, then Theorem 4.8.3(a) implies that there are n leading variables, , (hence no free variables) in the general solution of Ax = 0. This leaves the trivial solution, as the only possibility., , Applications of Rank, , The advent of the Internet has stimulated research on finding efficient methods for transmitting large amounts of digital data over communications lines with limited bandwidths., Digital data are commonly stored in matrix form, and many techniques for improving, transmission speed use the rank of a matrix in some way. Rank plays a role because it, measures the “redundancy” in a matrix in the sense that if A is an m × n matrix of rank, k , then n − k of the column vectors and m − k of the row vectors can be expressed in, terms of k linearly independent column or row vectors. The essential idea in many data, compression schemes is to approximate the original data set by a data set with smaller, rank that conveys nearly the same information, then eliminate redundant vectors in the, approximating set to speed up the transmission time.

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4.8 Rank, Nullity, and the Fundamental Matrix Spaces, O PT I O N A L, , Overdetermined and, Underdetermined Systems, , In engineering and physics,, the occurrence of an overdetermined or underdetermined, linear system often signals that, one or more variables were, omitted in formulating the, problem or that extraneous, variables were included. This, often leads to some kind of, complication., , 255, , In many applications the equations in a linear system correspond to physical constraints, or conditions that must be satisfied. In general, the most desirable systems are those that, have the same number of constraints as unknowns since such systems often have a unique, solution. Unfortunately, it is not always possible to match the number of constraints and, unknowns, so researchers are often faced with linear systems that have more constraints, than unknowns, called overdetermined systems, or with fewer constraints than unknowns,, called underdetermined systems. The following theorem will help us to analyze both, overdetermined and underdetermined systems., THEOREM 4.8.9 Let A be an m × n matrix., , (a) (Overdetermined Case). If m > n, then the linear system Ax = b is inconsistent, for at least one vector b in R n ., (b) (Underdetermined Case). If m < n, then for each vector b in R m the linear system, Ax = b is either inconsistent or has infinitely many solutions., Proof (a) Assume that m > n, in which case the column vectors of A cannot span R m, (fewer vectors than the dimension of R m ). Thus, there is at least one vector b in R m that, is not in the column space of A, and for any such b the system Ax = b is inconsistent by, Theorem 4.7.1., Proof (b) Assume that m, , < n. For each vector b in R n there are two possibilities: either, , the system Ax = b is consistent or it is inconsistent. If it is inconsistent, then the proof, is complete. If it is consistent, then Theorem 4.8.4 implies that the general solution has, n − r parameters, where r = rank(A). But we know from Example 2 that rank(A) is at, most the smaller of m and n (which is m), so, , n−r ≥n−m>0, This means that the general solution has at least one parameter and hence there are, infinitely many solutions., E X A M P L E 6 Overdetermined and Underdetermined Systems, , (a) What can you say about the solutions of an overdetermined system Ax = b of 7, equations in 5 unknowns in which A has rank r = 4?, (b) What can you say about the solutions of an underdetermined system Ax = b of 5, equations in 7 unknowns in which A has rank r = 4?, , R 7 , and for any such b the, number of parameters in the general solution is n − r = 5 − 4 = 1., Solution (a) The system is consistent for some vector b in, , Solution (b) The system may be consistent or inconsistent, but if it is consistent for the, , vector b in R 5 , then the general solution has n − r = 7 − 4 = 3 parameters., E X A M P L E 7 An Overdetermined System, , The linear system, , x1, x1, x1, x1, x1, , − 2 x2, − x2, + x2, + 2 x2, + 3x 2, , = b1, = b2, = b3, = b4, = b5, , is overdetermined, so it cannot be consistent for all possible values of b1 , b2 , b3 , b4 , and, b5 . Conditions under which the system is consistent can be obtained by solving the linear

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256, , Chapter 4 General Vector Spaces, , system by Gauss–Jordan elimination. We leave it for you to show that the augmented, matrix is row equivalent to, , ⎡, , 1, ⎢0, ⎢, ⎢, ⎢0, ⎢, ⎣0, 0, , 2b2 − b1, , 0, 1, 0, 0, 0, , b2, b3 − 3b2, b4 − 4b2, b5 − 5b2, , ⎤, , − b1 ⎥, ⎥, ⎥, + 2 b1 ⎥, ⎥, + 3b1 ⎦, + 4 b1, , (7), , Thus, the system is consistent if and only if b1 , b2 , b3 , b4 , and b5 satisfy the conditions, , =0, 2b1 − 3b2 + b3, 3b1 − 4b2, + b4, =0, 4b1 − 5b2, + b5 = 0, Solving this homogeneous linear system yields, , b1 = 5r − 4s, b2 = 4r − 3s, b3 = 2r − s, b4 = r, b5 = s, where r and s are arbitrary., Remark The coefficient matrix for the given linear system in the last example has n = 2 columns,, and it has rank r = 2 because there are two nonzero rows in its reduced row echelon form. This, implies that when the system is consistent its general solution will contain n − r = 0 parameters;, that is, the solution will be unique. With a moment’s thought, you should be able to see that this, is so from (7)., , Exercise Set 4.8, In Exercises 1–2, find the rank and nullity of the matrix A by, reducing it to row echelon form., , ⎡, , 1, , ⎢2, ⎢, 1. (a) A = ⎢, ⎣3, , 4, , 4, , 8, , ⎡, , −1, −2, −3, −4, , 2, 6, , 1, , ⎤, ⎥, , 4, 2, , 3, , (b) A = ⎣−3, , 6, , −1, , 1, , 2, , −4, , 5, , 8, , 1, , 0, , −2, −3, , 1, , 1, , −1, , ⎡, , (c) Find the number of leading variables and the number, of parameters in the general solution of Ax = 0 without, solving the system., , 3⎦, , −2, , ⎢, , (b) Confirm that the rank and nullity satisfy Formula (4)., , 2⎥, ⎥, , 1, , ⎤, −1, ⎥, − 7⎦, −4, ⎤, 0, , ⎢ 0, ⎢, 2. (a) A = ⎢, ⎣−2, , −1, −1, , 0, , 1, , 3, , 0, , 1, , 3, , 1, , 3, , 1, , 1, , ⎡, , ⎢ 0, ⎢, ⎢, (b) A = ⎢−3, ⎢, ⎣ 3, , 0, , 6, , 4, , 2, , 0, , −2, −4, , (a) By inspection of the matrix R , find the rank and nullity, of A., , 1, , ⎤, , 3⎥, ⎥, ⎥, 3⎦, , −4, , 0⎥, ⎥, , ⎥, −1⎥, ⎥, 1⎦, −2, , In Exercises 3–6, the matrix R is the reduced row echelon form, of the matrix A., , ⎡, , ⎤, ⎡, −3, 1, ⎥, ⎢, −3⎦; R = ⎣0, , 2, , −1, , 3. A = ⎣−1, , 2, , 1, , 1, , 2, , −1, , 4. A = ⎣−1, , 2, , 1, , 1, , 2, , −1, , 5. A = ⎣−2, , 1, , −4, , 2, , 6, , 2, , 2, , 0, , −1, , 3, , 1, , 1, , 3, , ⎢, ⎡, ⎢, ⎡, ⎢, ⎡, , 0, , ⎢ 1, ⎢, ⎣ 2, −2, , 6. A = ⎢, , 4, , 0, , ⎡, ⎤, 1, −3, ⎢, ⎥, −3⎦; R = ⎣0, 0, −6, , ⎤, , 0, , 0, , 1, , 0⎦, , 0, , 1, , ⎥, , 1, , ⎤, −3, ⎥, −3⎦, , 0, , 0, , ⎡, ⎤, 1, −3, ⎢, ⎥, 3⎦; R = ⎣0, , − 21, , − 23, , 0, , 0⎦, , 0, , 0, , 0, , 1, , 0, , −1, , 1, , 1, , 0, , 0, , 1⎦, , 0, , 0, , 0, , 4, , ⎤, , 0, , ⎡, , ⎢0, −3⎥, ⎢, ⎥, ⎥; R = ⎢, ⎣0, 1⎦, 0, −2, , ⎤, ⎥, ⎤, , 0, , 0⎥, ⎥, , ⎥

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4.8 Rank, Nullity, and the Fundamental Matrix Spaces, , 7. In each part, find the largest possible value for the rank of A, and the smallest possible value for the nullity of A., (a) A is 4 × 4, , (b) A is 3 × 5, , 15. Are there values of r and s for which, , ⎡, , 1, ⎢0, ⎢, ⎢, ⎣0, 0, , (c) A is 5 × 3, , 8. If A is an m × n matrix, what is the largest possible value for, its rank and the smallest possible value for its nullity?, , 0, , r −2, s−1, 0, , 257, , ⎤, , 0, 2 ⎥, ⎥, ⎥, r + 2⎦, 3, , has rank 1? Has rank 2? If so, find those values., 9. In each part, use the information in the table to:, (i) find the dimensions of the row space of A, column space, of A, null space of A, and null space of AT ;, (ii) determine whether or not the linear system Ax = b is, consistent;, (iii) find the number of parameters in the general solution of, each system in (ii) that is consistent., (a), , (b), , (c), , (d), , (e), , (f ), , (g), , Size of A, 3×3 3×3 3×3 5×9 5×9 4×4 6×2, 3, 2, 1, 2, 2, 0, 2, Rank(A), 3, 1, 2, 3, 0, 2, Rank[A | b] 3, , 1, ⎢, A = ⎣−3, −2, , 2, 1, 3, , 4, 5, 9, , ⎤, , (c) What kind of geometric object is the row space of your, matrix?, 17. Suppose that A is a 3 × 3 matrix whose null space is a line, through the origin in 3-space. Can the row or column space, of A also be a line through the origin? Explain., 18. (a) If A is a 3 × 5 matrix, then the rank of A is at most, . Why?, , (c) If A is a 3 × 5 matrix, then the rank of AT is at most, . Why?, , 0, ⎥, 2⎦, 2, , (d) If A is a 3 × 5 matrix, then the nullity of AT is at most, . Why?, , 11. (a) Find an equation relating nullity(A) and nullity(AT ) for, the matrix in Exercise 10., (b) Find an equation relating nullity(A) and nullity(AT ) for, a general m × n matrix., 12. Let T : R 2 →R 3 be the linear transformation defined by the, formula, , T (x1 , x2 ) = (x1 + 3x2 , x1 − x2 , x1 ), , 19. (a) If A is a 3 × 5 matrix, then the number of leading 1’s in, ., the reduced row echelon form of A is at most, Why?, (b) If A is a 3 × 5 matrix, then the number of parameters in, . Why?, the general solution of Ax = 0 is at most, (c) If A is a 5 × 3 matrix, then the number of leading 1’s in, ., the reduced row echelon form of A is at most, Why?, (d) If A is a 5 × 3 matrix, then the number of parameters in, . Why?, the general solution of Ax = 0 is at most, , (a) Find the rank of the standard matrix for T ., (b) Find the nullity of the standard matrix for T ., 13. Let T : R →R be the linear transformation defined by the, formula, 5, , (b) What kind of geometric object is the null space of your, matrix?, , (b) If A is a 3 × 5 matrix, then the nullity of A is at most, . Why?, , 10. Verify that rank(A) = rank(AT )., , ⎡, , 16. (a) Give an example of a 3 × 3 matrix whose column space is, a plane through the origin in 3-space., , 3, , 20. Let A be a 7 × 6 matrix such that Ax = 0 has only the trivial, solution. Find the rank and nullity of A., 21. Let A be a 5 × 7 matrix with rank 4., (a) What is the dimension of the solution space of Ax = 0 ?, , T (x1 , x2 , x3 , x4 , x5 ) = (x1 + x2 , x2 + x3 + x4 , x4 + x5 ), , (b) Is Ax = b consistent for all vectors b in R 5 ? Explain., , (a) Find the rank of the standard matrix for T ., , 22. Let, , (b) Find the nullity of the standard matrix for T ., 14. Discuss how the rank of A varies with t ., , ⎡, , ⎤, , 1, ⎢, (a) A = ⎣1, , 1, , t, , t, , t, , 1, , 1⎦, 1, , ⎥, , ⎡, ⎢, , t, , (b) A = ⎣ 3, −1, , A=, , 3, 6, −3, , ⎤, , −1, ⎥, −2⎦, t, , a11, a21, , a12, a22, , a13, a23, , Show that A has rank 2 if and only if one or more of the following determinants is nonzero., , , a11, , a, 21, , , a12 , ,, a22 , , , a11, , a, 21, , , a13 , ,, a23 , , , a12, , a, 22, , , a13 , a23 

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258, , Chapter 4 General Vector Spaces, , 23. Use the result in Exercise 22 to show that the set of points, (x, y, z) in R 3 for which the matrix, , x, 1, , y, x, , z, y, , has rank 1 is the curve with parametric equations x = t ,, , y = t 2, z = t 3., , Working with Proofs, 29. Prove: If k  = 0, then A and kA have the same rank., 30. Prove: If a matrix A is not square, then either the row vectors, or the column vectors of A are linearly dependent., 31. Use Theorem 4.8.3 to prove Theorem 4.8.4., , 24. Find matrices A and B for which rank(A) = rank(B), but, rank(A2 )  = rank(B 2 )., , 32. Prove Theorem 4.8.7(b)., , 25. In Example 6 of Section 3.4 we showed that the row space and, the null space of the matrix, , 33. Prove: If a vector v in R n is orthogonal to each vector in a, basis for a subspace W of R n , then v is orthogonal to every, vector in W ., , ⎡, , 1, , ⎢2, ⎢, A=⎢, ⎣0, 2, , ⎤, , 0, , 2, , 6, , −2, −5, , −2, , 4, , 0, , 5, , 10, , 0, , −3⎥, ⎥, ⎥, 15⎦, , 6, , 0, , 8, , 4, , 18, , 3, , 0, , are orthogonal complements in R 6 , as guaranteed by part (a), of Theorem 4.8.7. Show that null space of AT and the column, space of A are orthogonal complements in R 4 , as guaranteed, by part (b) of Theorem 4.8.7. [Suggestion: Show that each, column vector of A is orthogonal to each vector in a basis for, the null space of AT .], , True-False Exercises, TF. In parts (a)–( j) determine whether the statement is true or, false, and justify your answer., (a) Either the row vectors or the column vectors of a square matrix, are linearly independent., (b) A matrix with linearly independent row vectors and linearly, independent column vectors is square., (c) The nullity of a nonzero m × n matrix is at most m., , 26. Confirm the results stated in Theorem 4.8.7 for the matrix., , ⎡, −2, ⎢ 1, ⎢, A=⎢, ⎣ 3, 1, , −5, , 8, , 0, , 3, , −5, −19, −13, , 1, , 11, 7, , 7, 5, , ⎤, −17, 5⎥, ⎥, ⎥, 1⎦, −3, , 27. In each part, state whether the system is overdetermined or, underdetermined. If overdetermined, find all values of the b’s, for which it is inconsistent, and if underdetermined, find all, values of the b’s for which it is inconsistent and all values for, which it has infinitely many solutions., , ⎡, ⎢, , 1, , (a) ⎣−3, 0, , , (b), , −2, , , (c), , 1, , 1, , −1, , ⎡ ⎤, ⎤, b1, −1  , ⎢ ⎥, ⎥ x, 1⎦, = ⎣b2 ⎦, y, 1, b3, ⎡ ⎤,  ,  x, b1, 4 ⎢ ⎥, −3, ⎣y ⎦ =, 8, −6, b2, z, ⎡ ⎤,  ,  x, b1, 0 ⎢ ⎥, −3, ⎣y ⎦ =, 1, 1, b2, z, , 28. What conditions must be satisfied by b1 , b2 , b3 , b4 , and b5 for, the overdetermined linear system, , x1 − 3x2 = b1, x1 − 2x2 = b2, x1 + x2 = b3, x1 − 4x2 = b4, x1 + 5x2 = b5, to be consistent?, , (d) Adding one additional column to a matrix increases its rank, by one., (e) The nullity of a square matrix with linearly dependent rows is, at least one., (f ) If A is square and Ax = b is inconsistent for some vector b,, then the nullity of A is zero., (g) If a matrix A has more rows than columns, then the dimension, of the row space is greater than the dimension of the column, space., (h) If rank(AT ) = rank(A), then A is square., (i) There is no 3 × 3 matrix whose row space and null space are, both lines in 3-space., ( j) If V is a subspace of R n and W is a subspace of V, then W ⊥, is a subspace of V ⊥ ., , Working withTechnology, T1. It can be proved that a nonzero matrix A has rank k if and, only if some k × k submatrix has a nonzero determinant and all, square submatrices of larger size have determinant zero. Use this, fact to find the rank of, , ⎡, , 3, , ⎢5, ⎢, A=⎢, ⎣1, 7, , ⎤, , −1, , 3, , 2, , 5, , −3, , 2, , 3, , 4⎥, ⎥, , −3, , −5, , 0, , ⎥, −7⎦, , −5, , 1, , 4, , 1, , Check your result by computing the rank of A in a different way.

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4.9 Basic Matrix Transformations in R 2 and R 3, , T2. Sylvester’s inequality states that if A and B are n × n matrices, with rank rA and rB , respectively, then the rank rAB of AB satisfies, the inequality, , 259, , where min(rA , rB ) denotes the smaller of rA and rB or their common value if the two ranks are the same. Use your technology, utility to confirm this result for some matrices of your choice., , rA + rB − n ≤ rAB ≤ min(rA , rB ), , 4.9 Basic Matrix Transformations in R 2 and R 3, In this section we will continue our study of linear transformations by considering some, basic types of matrix transformations in R 2 and R 3 that have simple geometric, interpretations. The transformations we will study here are important in such fields as, computer graphics, engineering, and physics., , There are many ways to transform the vector spaces R 2 and R 3 , some of the most, important of which can be accomplished by matrix transformations using the methods, introduced in Section 1.8. For example, rotations about the origin, reflections about, lines and planes through the origin, and projections onto lines and planes through the, origin can all be accomplished using a linear operator TA in which A is an appropriate, 2 × 2 or 3 × 3 matrix., , Reflection Operators, , Some of the most basic matrix operators on R 2 and R 3 are those that map each point into, its symmetric image about a fixed line or a fixed plane that contains the origin; these are, called reflection operators. Table 1 shows the standard matrices for the reflections about, the coordinate axes in R 2 , and Table 2 shows the standard matrices for the reflections, about the coordinate planes in R 3 . In each case the standard matrix was obtained using, the following procedure introduced in Section 1.8: Find the images of the standard basis, vectors, convert those images to column vectors, and then use those column vectors as, successive columns of the standard matrix., , Table 1, Operator, , Illustration, y, , Standard Matrix, , (x, y), , x, , Reflection about, the x -axis, , T (x, y) = (x, −y), , Images of e1 and e2, , x, , T (e1 ) = T (1, 0) = (1, 0), T (e2 ) = T (0, 1) = (0, −1), , 1, 0, , T (e1 ) = T (1, 0) = (−1, 0), T (e2 ) = T (0, 1) = (0, 1), , −1, , T (e1 ) = T (1, 0) = (0, 1), T (e2 ) = T (0, 1) = (1, 0), , 0, 1, , 0, −1, , T(x), (x, –y), y, , Reflection about, the y -axis, , T (x, y) = (−x, y), , (–x, y), , (x, y), T(x), , x, , y, , Reflection about, the line y = x, , T (x, y) = (y, x), , (y, x), , 0, 1, , x, y=x, , T(x), x, , 0, , (x, y) x, , 1, 0

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260, , Chapter 4 General Vector Spaces, Table 2, Operator, , Images of e1 , e2 , e3, , Illustration, , Standard Matrix, , z, (x, y, z), , Reflection about, the xy -plane, , T (e1 ) = T (1, 0, 0) = (1, 0, 0), T (e2 ) = T (0, 1, 0) = (0, 1, 0), T (e3 ) = T (0, 0, 1) = (0, 0, −1), , x, y, , T (x, y, z) = (x, y, −z), x, , T(x), , ⎡, , 1, ⎢, ⎣0, 0, , ⎤, , 0, 1, 0, , 0, ⎥, 0⎦, −1, , 0, −1, 0, , 0, ⎥, 0⎦, 1, , (x, y, –z), z, , (x, –y, z), , Reflection about, the xz-plane, , (x, y, z), x, , T(x), , y, , T (x, y, z) = (x, −y, z), , T (e1 ) = T (1, 0, 0) = (1, 0, 0), T (e2 ) = T (0, 1, 0) = (0, −1, 0), T (e3 ) = T (0, 0, 1) = (0, 0, 1), , ⎡, , 1, ⎢, ⎣0, 0, , ⎤, , x, , z, , Reflection about, the yz-plane, , (–x, y, z), , T(x), , (x, y, z), , T (x, y, z) = (−x, y, z), , y, , x, , T (e1 ) = T (1, 0, 0) = (−1, 0, 0), T (e2 ) = T (0, 1, 0) = (0, 1, 0), T (e3 ) = T (0, 0, 1) = (0, 0, 1), , ⎡, −1, ⎢, ⎣ 0, 0, , 0, 1, 0, , ⎤, , 0, ⎥, 0⎦, 1, , x, , Projection Operators, , Matrix operators on R 2 and R 3 that map each point into its orthogonal projection onto, a fixed line or plane through the origin are called projection operators (or more precisely,, orthogonal projection operators). Table 3 shows the standard matrices for the orthogonal, projections onto the coordinate axes in R 2 , and Table 4 shows the standard matrices for, the orthogonal projections onto the coordinate planes in R 3 ., , Table 3, Operator, , Illustration, , Images of e1 and e2, , Standard Matrix, , y, (x, y), , Orthogonal projection, onto the x -axis, , x, , T (x, y) = (x, 0), , (x, 0) x, , T (e1 ) = T (1, 0) = (1, 0), T (e2 ) = T (0, 1) = (0, 0), , 1, 0, , 0, 0, , T (e1 ) = T (1, 0) = (0, 0), T (e2 ) = T (0, 1) = (0, 1), , 0, 0, , 0, 1, , T(x), y, , Orthogonal projection, onto the y -axis, , T (x, y) = (0, y), , (0, y), T(x), , (x, y), x, , x

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4.9 Basic Matrix Transformations in R 2 and R 3, , 261, , Table 4, Operator, , Images of e1 , e2 , e3, , Illustration, z, , Orthogonal projection, onto the xy -plane, , (x, y, z), , x, , T (x, y, z) = (x, y, 0), , y, , T(x), , x, , T (e1 ) = T (1, 0, 0) = (1, 0, 0), T (e2 ) = T (0, 1, 0) = (0, 1, 0), T (e3 ) = T (0, 0, 1) = (0, 0, 0), , Standard Matrix, , ⎡, , 1, ⎢, ⎣0, 0, , ⎤, , 0, 1, 0, , 0, ⎥, 0⎦, 0, , 0, 0, 0, , 0, ⎥, 0⎦, 1, , 0, 1, 0, , 0, ⎥, 0⎦, 1, , (x, y, 0), , z, , Orthogonal projection, onto the xz-plane, , (x, 0, z), , (x, y, z), , x, , y, , T(x), , T (x, y, z) = (x, 0, z), , T (e1 ) = T (1, 0, 0) = (1, 0, 0), T (e2 ) = T (0, 1, 0) = (0, 0, 0), T (e3 ) = T (0, 0, 1) = (0, 0, 1), , ⎡, , 1, ⎢, ⎣0, 0, , ⎤, , x, z, , (0, y, z), T(x), , Orthogonal projection, onto the yz-plane, , (x, y, z), y, , x, , T (x, y, z) = (0, y, z), , T (e1 ) = T (1, 0, 0) = (0, 0, 0), T (e2 ) = T (0, 1, 0) = (0, 1, 0), T (e3 ) = T (0, 0, 1) = (0, 0, 1), , ⎡, , 0, ⎢, ⎣0, 0, , ⎤, , x, , Rotation Operators, , Matrix operators on R 2 and R 3 that move points along arcs of circles centered at the, origin are called rotation operators. Let us consider how to find the standard matrix for, the rotation operator T : R 2 →R 2 that moves points counterclockwise about the origin, through a positive angle θ . As illustrated in Figure 4.9.1, the images of the standard, basis vectors are, , T (e1 ) = T (1, 0) = (cos θ, sin θ) and T (e2 ) = T (0, 1) = (− sin θ, cos θ ), so it follows from Formula (14) of Section 1.8 that the standard matrix for T is, , A = [T (e1 ) | T (e2 )] =, , (–sin θ, cos θ), , cos θ, sin θ, , − sin θ, cos θ, , y, e2, , T, , (cos θ, sin θ), 1, , u, , 1, u, , T, , x, , e1, , Figure 4.9.1, , In keeping with common usage we will denote this operator by Rθ and call, , Rθ =, , cos θ, sin θ, , − sin θ, cos θ, , (1)

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262, , Chapter 4 General Vector Spaces, , In the plane, counterclockwise, angles are positive and clockwise angles are negative. The, rotation matrix for a clockwise, rotation of −θ radians can be, obtained by replacing θ by −θ, in (1). After simplification this, yields, , , , R−θ =, , cos θ, , sin θ, , − sin θ, , cos θ, , the rotation matrix for R 2 . If x = (x, y) is a vector in R 2 , and if w = (w1 , w2 ) is its, image under the rotation, then the relationship w = Rθ x can be written in component, form as, w1 = x cos θ − y sin θ, (2), w2 = x sin θ + y cos θ, These are called the rotation equations for R 2 . These ideas are summarized in Table 5., Table 5, , , , Operator, , Illustration, y, , Counterclockwise, rotation about the, origin through an, angle θ, , Rotation Equations, , Standard Matrix, , (w1, w2), , w1 = x cos θ − y sin θ, w2 = x sin θ + y cos θ, , w, (x, y), , θ, , − sin θ, cos θ, , cos θ, sin θ, , x, , x, , E X A M P L E 1 A Rotation Operator, , Find the image of x = (1, 1) under a rotation of π/6 radians (= 30◦ ) about the origin., Solution It follows from (1) with θ, ,  √3, , Rπ/6 x =, , 2, 1, 2, , = π/6 that, ,  √3−1 , − 21, 1, 0.37, 2, =, ≈, √, √, 1+ 3, 3, 1, 1.37, 2, , 2, , or in comma-delimited notation, Rπ/6 (1, 1) ≈ (0.37, 1.37)., Rotations in R 3, , A rotation of vectors in R 3 is commonly described in relation to a line through the origin, called the axis of rotation and a unit vector u along that line (Figure 4.9.2a). The unit, vector and what is called the right-hand rule can be used to establish a sign for the angle of, rotation by cupping the fingers of your right hand so they curl in the direction of rotation, and observing the direction of your thumb. If your thumb points in the direction of u,, then the angle of rotation is regarded to be positive relative to u, and if it points in the, direction opposite to u, then it is regarded to be negative relative to u (Figure 4.9.2b)., z, , z, x, , Figure 4.9.2, , (a) Angle of rotation, , u, , y, , y, x, , x, , Negative, rotation, , u, , θ, w, , z, , Positive, rotation, , Axis of rotation, l, , y, , x, , (b) Right-hand rule, , For rotations about the coordinate axes in R 3 , we will take the unit vectors to be i, j,, and k, in which case an angle of rotation will be positive if it is counterclockwise looking, toward the origin along the positive coordinate axis and will be negative if it is clockwise., Table 6 shows the standard matrices for the rotation operators on R 3 that rotate each, vector about one of the coordinate axes through an angle θ . You will find it instructive, to compare these matrices to that in Table 5.

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4.9 Basic Matrix Transformations in R 2 and R 3, , 263, , Table 6, Operator, , Rotation Equations, , Illustration, , Standard Matrix, , z, , Counterclockwise, rotation about the, positive x -axis through, an angle θ, , y, w, , w1 = x, w2 = y cos θ − z sin θ, w3 = y sin θ + z cos θ, , ⎡, , 1, ⎢, ⎣0, 0, , ⎤, , 0, cos θ, sin θ, , 0, ⎥, − sin θ ⎦, cos θ, , x, , θ, , x, z, , Counterclockwise, rotation about the, positive y -axis through, an angle θ, , x, θ, , x, , w1 = x cos θ + z sin θ, w2 = y, w3 = −x sin θ + z cos θ, , y, , ⎡, , cos θ, ⎢, ⎣ 0, − sin θ, , 0, 1, 0, , ⎤, , sin θ, ⎥, 0 ⎦, cos θ, , w, z, , Counterclockwise, rotation about the, positive z-axis through, an angle θ, , θ, , x, , w1 = x cos θ − y sin θ, w2 = x sin θ + y cos θ, w3 = z, , w, y, , ⎡, , cos θ, ⎢, ⎣ sin θ, 0, , − sin θ, cos θ, 0, , ⎤, , 0, ⎥, 0⎦, 1, , x, , Yaw, Pitch, and Roll, In aeronautics and astronautics, the orientation of an aircraft or, space shuttle relative to an xyz-coordinate system is often described, in terms of angles called yaw, pitch, and roll. If, for example, an, aircraft is flying along the y -axis and the xy -plane defines the horizontal, then the aircraft’s angle of rotation about the z-axis is called, the yaw, its angle of rotation about the x -axis is called the pitch, and, its angle of rotation about the y -axis is called the roll. A combination of yaw, pitch, and roll can be achieved by a single rotation, about some axis through the origin. This is, in fact, how a space, shuttle makes attitude adjustments—it doesn’t perform each rotation separately; it calculates one axis, and rotates about that axis, to get the correct orientation. Such rotation maneuvers are used to, , align an antenna, point the nose toward a celestial object, or position, a payload bay for docking., z, Yaw, , y, , x, Pitch, , Roll, , For completeness, we note that the standard matrix for a counterclockwise rotation, through an angle θ about an axis in R 3 , which is determined by an arbitrary unit vector, u = (a, b, c) that has its initial point at the origin, is, , ⎡, , ⎤, a 2 (1 − cos θ) + cos θ ab(1 − cos θ) − c sin θ ac(1 − cos θ) + b sin θ, ⎢, ⎥, bc(1 − cos θ) − a sin θ ⎦, ⎣ab(1 − cos θ) + c sin θ b2 (1 − cos θ) + cos θ, ac(1 − cos θ) − b sin θ bc(1 − cos θ) + a sin θ c2 (1 − cos θ) + cos θ, , (3), , The derivation can be found in the book Principles of Interactive Computer Graphics, by, W. M. Newman and R. F. Sproull (New York: McGraw-Hill, 1979). You may find it, instructive to derive the results in Table 6 as special cases of this more general result.

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264, , Chapter 4 General Vector Spaces, , Dilations and Contractions, , If k is a nonnegative scalar, then the operator T(x) = k x on R 2 or R 3 has the effect of, increasing or decreasing the length of each vector by a factor of k . If 0 ≤ k < 1 the, operator is called a contraction with factor k , and if k > 1 it is called a dilation with, factor k (Figure 4.9.3). Tables 7 and 8 illustrate these operators. If k = 1, then T is the, identity operator., x, , T(x) = kx, x, , T(x) = kx, , (a) 0 ≤ k < 1, , Figure 4.9.3, , (b) k > 1, , Table 7, Illustration, , Effect on the, Unit Square, , T (x, y) = (kx, ky), , Operator, , y, , Contraction with, factor k in R 2, , x, T(x), , (0 ≤ k < 1), , (0, 1), , (x, y), , (kx, ky), , (0, k), , x, , (1, 0), y, , Dilation with, factor k in R 2, , T(x), x, , Standard, Matrix, , (kx, ky), , (k, 0), (0, k), , (0, 1), , (x, y), , (k > 1), , x, , (1, 0), , (k, 0), , Table 8, Illustration, , Standard, Matrix, , T (x, y, z) = (kx, ky, kz), , Operator, , z, , Contraction with, factor k in R 3, , x, T(x), , (x, y, z), , (kx, ky, kz), , (0 ≤ k < 1), , y, , x, z, , (kx, ky, kz), T(x), , Dilation with, factor k in R 3, , x, , (x, y, z), y, , (k > 1), x, , ⎡, , k, , 0, , ⎤, , ⎢, ⎣0, , k, , 0, ⎥, 0⎦, , 0, , 0, , k, , k, , 0, , 0, , k

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4.9 Basic Matrix Transformations in R 2 and R 3, , Expansions and, Compressions, , 265, , In a dilation or contraction of R 2 or R 3 , all coordinates are multiplied by a nonnegative, factor k . If only one coordinate is multiplied by k , then, depending on the value of k ,, the resulting operator is called a compression or expansion with factor k in the direction, of a coordinate axis. This is illustrated in Table 9 for R 2 . The extension to R 3 is left as, an exercise., , Table 9, Illustration, , Effect on the, Unit Square, , T (x, y) = (kx, y), , Operator, , Compression in the, x -direction, with factor k in R 2, , y, (kx, y), , x, (1, 0), y, , (x, y), , (kx, y), , (0, 1), , T(x), , (k, 0), , Effect on the, Unit Square, , Standard, Matrix, , y, (0, 1), , (x, y), x, , (0, k), , (x, ky), x, (1, 0), , T(x), y, , Shears, , (0, 1), , (1, 0), , (0 ≤ k < 1), , (k > 1), , 0, 1, , x, , Illustration, , Expansion in the, y -direction, with factor k in R 2, , k, , x, , T (x, y) = (x, ky), , Compression in the, y -direction, with factor k in R 2, , (k, 0), , 0, , (k > 1), , Operator, , (0, 1), , x, , (0 ≤ k < 1), , Expansion in the, x -direction, with factor k in R 2, , (0, 1), , (x, y), , T(x), , Standard, Matrix, , (x, ky), T(x), , (x, y), x, , (1, 0), , 1, 0, , 0, , k, , (0, k), , (0, 1), , x, (1, 0), , (1, 0), , A matrix operator of the form T (x, y) = (x + ky, y) translates a point (x, y) in the, xy -plane parallel to the x -axis by an amount ky that is proportional to the y -coordinate, of the point. This operator leaves the points on the x -axis fixed (since y = 0), but, as we progress away from the x -axis, the translation distance increases. We call this, operator the shear in the x-direction by a factor k. Similarly, a matrix operator of the, form T (x, y) = (x, y + kx) is called the shear in the y-direction by a factor k. Table 10,, which illustrates the basic information about shears in R 2 , shows that a shear is in the, positive direction if k > 0 and the negative direction if k < 0.

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266, , Chapter 4 General Vector Spaces, Table 10, Operator, , Effect on the Unit Square, , (k, 1), , (k, 1), , (0, 1), , Shear in the, x -direction by a, factor k in R 2, , (k > 0), , (0, 1), , Shear in the, y -direction by a, factor k in R 2, , k, , 1, , 0, 1, , 1, , (k < 0), , (0, 1), , (0, 1), (1, k), , k, , (1, 0), , T (x, y) = (x, y + kx), , 1, 0, (1, 0), , (1, 0), , (1, 0), , T (x, y) = (x + ky, y), , Standard Matrix, , (1, k), (k > 0), , (k < 0), , E X A M P L E 2 Effect of Matrix Operators on the Unit Square, , In each part, describe the matrix operator whose standard matrix is shown, and show, its effect on the unit square., (a), , 1, 0, , A1 =, , 2, 1, , (b), , A2 =, , −2, , 1, 0, , 2, 0, , A3 =, , (c), , 1, , 0, 2, , 2, 0, , A4 =, , (d), , 0, 1, , Solution By comparing the forms of these matrices to those in Tables 7, 9, and 10, we, , see that the matrix A1 corresponds to a shear in the x -direction by a factor 2, the matrix, A2 corresponds to a shear in the x -direction by a factor −2, the matrix A3 corresponds, to a dilation with factor 2, and the matrix A4 corresponds to an expansion in the x direction with factor 2. The effects of these operators on the unit square are shown in, Figure 4.9.4., y, , y, , y, 3, , 3, , 3, , 2, , 2, , 2, , 2, , 1, , 1, , 0, , Orthogonal Projections, onto LinesThrough the, Origin, y, L, , x, T(x), θ, , Figure 4.9.5, , x, , 1, , 2, , 3, , 1, , 1, , x, , Figure 4.9.4, , y, , 3, , x, –2, , A1, , –1, , 0, , 1, , x, , x, 0, , 1, , A2, , 2, , 0, , 3, , 1, , A3, , 2, , 3, , A4, , In Table 3 we listed the standard matrices for the orthogonal projections onto the coordinate axes in R 2 . These are special cases of the more general matrix operator TA : R 2 →R 2, that maps each point into its orthogonal projection onto a line L through the origin that, makes an angle θ with the positive x -axis (Figure 4.9.5). In Example 4 of Section 3.3, we used Formula (10) of that section to find the orthogonal projections of the standard, basis vectors for R 2 onto that line. Expressed in matrix form, we found those projections, to be, sin θ cos θ, cos2 θ, T (e1 ) =, and T (e2 ) =, sin θ cos θ, sin2 θ, Thus, the standard matrix for TA is, , A = [T (e1 ) | T (e2 )] =, , cos2 θ, sin θ cos θ, , sin θ cos θ, 2, , sin θ, , =, , cos2 θ, 1, 2, , sin 2θ, , 1, 2, , sin 2θ, , sin2 θ

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4.9 Basic Matrix Transformations in R 2 and R 3, , 267, , In keeping with common usage, we will denote this operator by, We have included two versions, of Formula (4) because both, are commonly used. Whereas, the first version involves only, the angle θ , the second involves both θ and 2θ ., , Pθ =, , cos2 θ, , sin θ cos θ, , sin θ cos θ, , sin θ, , 1, 2, , cos2 θ, , =, , 2, , 1, 2, , sin 2θ, , (4), , sin2 θ, , sin 2θ, , E X A M P L E 3 Orthogonal Projection onto a Line Through the Origin, , Use Formula (4) to find the orthogonal projection of the vector x = (1, 5) onto the line, through the origin that makes an angle of π/6 (= 30◦ ) with the positive x -axis., Solution Since sin(π/6), , = 1/2 and cos(π/6) =, , √, , 3/2, it follows from (4) that the stan-, , dard matrix for this projection is, , Pπ/6 =, , cos2 (π/6), , sin(π/6) cos(π/6), , sin(π/6) cos(π/6), , sin2 (π/6), , Thus,, , , , √, , 3, 4, , 3, 4, , Pπ/6 x =, , √, , , , 1, 4, , 3, 4, , 1, =, 5, , , =, ,  3+5√3 , √, , √, , 3, 4, , √, , 3, 4, , , , 1, 4, , 2.91, 1.68, , ≈, , 4, 3+5, 4, , 3, 4, , or in comma-delimited notation, Pπ/6 (1, 5) ≈ (2.91, 1.68)., Reflections About Lines, Through the Origin, y, , Hθx, L, θ, , x, , x, , Figure 4.9.6, , y, , In Table 1 we listed the reflections about the coordinate axes in R 2 . These are special cases, of the more general operator Hθ : R 2 →R 2 that maps each point into its reflection about, a line L through the origin that makes an angle θ with the positive x -axis (Figure 4.9.6)., We could find the standard matrix for Hθ by finding the images of the standard basis, vectors, but instead we will take advantage of our work on orthogonal projections by, using Formula (4) for Pθ to find a formula for Hθ ., You should be able to see from Figure 4.9.7 that for every vector x in R n, , Pθ x − x = 21 (Hθ x − x) or equivalently Hθ x = (2Pθ − I )x, Thus, it follows from Theorem 1.8.4 that, , Hθ = 2 Pθ − I, Hθx, , and hence from (4) that, L, , θ, , cos 2θ, sin 2θ, , Hθ =, , Pθx, x, , Figure 4.9.7, , (5), , x, , sin 2θ, − cos 2θ, , (6), , E X A M P L E 4 Reflection About a Line Through the Origin, , Find the reflection of the vector x = (1, 5) about the line through the origin that makes, an angle of π/6 (= 30◦ ) with the x -axis., , Hπ/6, , =, , √, , 3/2 and cos(π/3) = 1/2, it follows from (6) that the standard matrix for this reflection is, Solution Since sin(π/3), , cos(π/3), =, sin(π/3), , Thus,, , , Hπ/6 x =, , 1, 2, , √, , 3, 2, , √, , 3, 2, , − 21, , sin(π/3), =, − cos(π/3), , , , 1, =, 5, , , , 2, 3−5, 2, , 3, 2, , √, , ≈, , , , − 21, , 3, 2, ,  1+5√3 , √, , √, , 1, 2, , 4.83, −1.63, , or in comma-delimited notation, Hπ/6 (1, 5) ≈ (4.83, −1.63).

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268, , Chapter 4 General Vector Spaces, , Exercise Set 4.9, 1. Use matrix multiplication to find the reflection of (−1, 2), about the, (a) x -axis., , (b) y -axis., , (c) line y = x ., , 2. Use matrix multiplication to find the reflection of (a, b) about, the, (a) x -axis., , (b) y -axis., , (c) line y = x ., , 3. Use matrix multiplication to find the reflection of (2, −5, 3), about the, (a) xy -plane., , (b) xz-plane., , (c) yz-plane., , 4. Use matrix multiplication to find the reflection of (a, b, c), about the, (a) xy -plane., , (b) xz-plane., , (c) yz-plane., , 5. Use matrix multiplication to find the orthogonal projection of, (2, −5) onto the, (a) x -axis., , (b) y -axis., , 6. Use matrix multiplication to find the orthogonal projection of, (a, b) onto the, (a) x -axis., , (b) y -axis., , 7. Use matrix multiplication to find the orthogonal projection of, (−2, 1, 3) onto the, (a) xy -plane., , (b) xz-plane., , (c) yz-plane., , 8. Use matrix multiplication to find the orthogonal projection of, (a, b, c) onto the, (a) xy -plane., , (b) xz-plane., , (c) yz-plane., , 9. Use matrix multiplication to find the image of the vector, (3, −4) when it is rotated about the origin through an angle, of, (a) θ = 30◦ ., , (b) θ = −60◦ ., , (c) θ = 45◦ ., , (d) θ = 90◦ ., , 10. Use matrix multiplication to find the image of the nonzero, vector v = (v1 , v2 ) when it is rotated about the origin through, (a) a positive angle α ., , (b) a negative angle −α ., , 11. Use matrix multiplication to find the image of the vector, (2, −1, 2) if it is rotated, (a) 30◦ clockwise about the positive x -axis., (b) 30◦ counterclockwise about the positive y -axis., (c) 45◦ clockwise about the positive y -axis., , (c) 45◦ counterclockwise about the positive y -axis., (d) 90◦ clockwise about the positive z-axis., 13. (a) Use matrix multiplication to find the contraction of, (−1, 2) with factor k = 21 ., (b) Use matrix multiplication to find the dilation of (−1, 2), with factor k = 3., 14. (a) Use matrix multiplication to find the contraction of (a, b), with factor k = 1/α , where α > 1., (b) Use matrix multiplication to find the dilation of (a, b) with, factor k = α , where α > 1., 15. (a) Use matrix multiplication to find the contraction of, (2, −1, 3) with factor k = 41 ., (b) Use matrix multiplication to find the dilation of (2, −1, 3), with factor k = 2., 16. (a) Use matrix multiplication to find the contraction of, (a, b, c) with factor k = 1/α , where α > 1., (b) Use matrix multiplication to find the dilation of (a, b, c), with factor k = α , where α > 1., 17. (a) Use matrix multiplication to find the compression of, (−1, 2) in the x -direction with factor k = 21 ., (b) Use matrix multiplication to find the compression of, (−1, 2) in the y -direction with factor k = 21 ., 18. (a) Use matrix multiplication to find the expansion of (−1, 2), in the x -direction with factor k = 3., (b) Use matrix multiplication to find the expansion of (−1, 2), in the y -direction with factor k = 3., 19. (a) Use matrix multiplication to find the compression of (a, b), in the x -direction with factor k = 1/α , where α > 1., (b) Use matrix multiplication to find the expansion of (a, b), in the y -direction with factor k = α , where α > 1., 20. Based on Table 9, make a conjecture about the standard matrices for the compressions with factor k in the directions of, the coordinate axes in R 3 ., Exercises 21–22 Using Example 2 as a model, describe the matrix operator whose standard matrix is given, and then show in a, coordinate system its effect on the unit square., , (d) 90◦ counterclockwise about the positive z-axis., 12. Use matrix multiplication to find the image of the vector, (2, −1, 2) if it is rotated, (a) 30◦ counterclockwise about the positive x -axis., (b) 30◦ clockwise about the positive y -axis., , , 21. (a) A1 =, , , (c) A3 =, , , , 1, 2, , 0, , 0, , 1, 2, , 1, , 0, , 1, 2, , 1, , , , , (b) A2 =, , , , 1, , 0, , 0, , 1, 2, , , (d) A4 =, , 1, , 0, , − 21, , 1, , 

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4.9 Basic Matrix Transformations in R 2 and R 3, , , 22. (a) A1 =, , , , 3, , 0, , 0, , 3, , , (c) A3 =, , , (b) A2 =, , , , 1, , 0, , 3, , 1, , 1, , 0, , 0, , 3, , , (d) A4 =, , , , 1, , 0, , −3, , 1, , 269, , 32. In each part of the accompanying figure, find the standard, matrix for the pictured operator., , , , z, , z, , z, , (x, y, z), , (z, y, x), (y, x, z), , In each part of Exercises 23–24, the effect of some matrix operator on the unit square is shown. Find the standard matrix for, an operator with that effect., 23. (a), , y, , (b), , 3, 2, , x, , y, 3, , 2, , 0, , (b), , 3, , 1, , 2, , 3, , x, , x, –2, , –1, , 0, , 1, , (b), , (c), , 33. Use Formula (3) to find the standard matrix for a rotation, of 180◦ about the axis determined by the vector v = (2, 2, 1)., [Note: Formula (3) requires that the vector defining the axis, of rotation have length 1.], , y, , 1, , 1, , x, , 3, , 2, , 2, , y, , x, , 3, y, , 24. (a), , (x, y, z), , (x, y, z), , Figure Ex-32, , 1, 1, , x, , (a), , x, 0, , y, , (x, z, y), , 2, , 1, , y, , 0, , 1, , 2, , 3, , In Exercises 25–26, find the standard matrix for the orthogonal, projection of R 2 onto the stated line, and then use that matrix to, find the orthogonal projection of the given point onto that line., 25. The orthogonal projection of (3, 4) onto the line that makes, an angle of π/3 (= 60◦ ) with the positive x -axis., 26. The orthogonal projection of (1, 2) onto the line that makes, an angle of π/4 (= 45◦ ) with the positive x -axis., In Exercises 27–28, find the standard matrix for the reflection, of R 2 about the stated line, and then use that matrix to find the, reflection of the given point about that line., 27. The reflection of (3, 4) about the line that makes an angle of, π/3 (= 60◦ ) with the positive x -axis., 28. The reflection of (1, 2) about the line that makes an angle of, π/4 (= 45◦ ) with the positive x -axis., 29. For each reflection operator in Table 2 use the standard matrix, to compute T (1, 2, 3), and convince yourself that your result, makes sense geometrically., 30. For each orthogonal projection operator in Table 4 use the, standard matrix to compute T (1, 2, 3), and convince yourself, that your result makes sense geometrically., 31. Find the standard matrix for the operator T : R →R that, 3, , 3, , (a) rotates each vector 30◦ counterclockwise about the z-axis, (looking along the positive z-axis toward the origin)., (b) rotates each vector 45◦ counterclockwise about the x -axis, (looking along the positive x -axis toward the origin)., (c) rotates each vector 90◦ counterclockwise about the y -axis, (looking along the positive y -axis toward the origin)., , 34. Use Formula (3) to find the standard matrix for a rotation, of π/2 radians about the axis determined by v = (1, 1, 1)., [Note: Formula (3) requires that the vector defining the axis, of rotation have length 1.], 35. Use Formula (3) to derive the standard matrices for the rotations about the x -axis, the y -axis, and the z-axis through an, angle of 90◦ in R 3 ., 36. Show that the standard matrices listed in Tables 1 and 3 are, special cases of Formulas (4) and (6)., 37. In a sentence, describe the geometric effect of multiplying a, vector x by the matrix, , , , A=, , cos2 θ − sin2 θ, 2 sin θ cos θ, , −2 sin θ cos θ, cos2 θ − sin2 θ, , , , 38. If multiplication by A rotates a vector x in the xy -plane, through an angle θ , what is the effect of multiplying x by AT ?, Explain your reasoning., 39. Let x0 be a nonzero column vector in R 2 , and suppose that, T : R 2 →R 2 is the transformation defined by the formula, T (x) = x0 + Rθ x, where Rθ is the standard matrix of the rotation of R 2 about the origin through the angle θ . Give a, geometric description of this transformation. Is it a matrix, transformation? Explain., 40. In R 3 the orthogonal projections onto the x -axis, y -axis, and, z-axis are, , T1 (x, y, z) = (x, 0, 0), T2 (x, y, z) = (0, y, 0),, T3 (x, y, z) = (0, 0, z), respectively., (a) Show that the orthogonal projections onto the coordinate, axes are matrix operators, and then find their standard, matrices.

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270, , Chapter 4 General Vector Spaces, , (b) Show that if T : R 3 →R 3 is an orthogonal projection onto, one of the coordinate axes, then for every vector x in R 3 ,, the vectors T(x) and x − T(x) are orthogonal., , (c) Make a sketch showing x and x − T(x) in the case where, T is the orthogonal projection onto the x -axis., , 4.10 Properties of MatrixTransformations, In this section we will discuss properties of matrix transformations. We will show, for, example, that if several matrix transformations are performed in succession, then the same, result can be obtained by a single matrix transformation that is chosen appropriately. We, will also explore the relationship between the invertibility of a matrix and properties of the, corresponding transformation., , Compositions of Matrix, Transformations, , Suppose that TA is a matrix transformation from R n to R k and TB is a matrix transformation from R k to R m . If x is a vector in R n , then TA maps this vector into a vector TA (x), in R k , and TB , in turn, maps that vector into the vector TB (TA (x)) in R m . This process, creates a transformation from R n to R m that we call the composition of TB with TA and, denote by the symbol, , TB ◦ TA, which is read “TB circle TA .” As illustrated in Figure 4.10.1, the transformation TA in, the formula is performed first; that is,, , (TB ◦ TA )(x) = TB (TA (x)), , (1), , This composition is itself a matrix transformation since, , (TB ◦ TA )(x) = TB (TA (x)) = B(TA (x)) = B(Ax) = (BA)x, which shows that it is multiplication by BA. This is expressed by the formula, , TB ◦ TA = TBA, TA, Rn, , Figure 4.10.1, , x, , (2), TB, , Rk, , TA(x), , Rm, , TB (TA (x)), , TB ° TA, , Compositions can be defined for any finite succession of matrix transformations, whose domains and ranges have the appropriate dimensions. For example, to extend, Formula (2) to three factors, consider the matrix transformations, , TA : R n → R k , TB : R k → R l , TC : R l → R m, We define the composition (TC ◦ TB ◦ TA ): R n →R m by, , (TC ◦ TB ◦ TA )(x) = TC (TB (TA (x))), As above, it can be shown that this is a matrix transformation whose standard matrix is, CBA and that, TC ◦ TB ◦ TA = TCBA, (3), Sometimes we will want to refer to the standard matrix for a matrix transformation, , T : R n →R m without giving a name to the matrix itself. In such cases we will denote the, standard matrix for T by the symbol [T ]. Thus, the equation, T (x) = [T ]x

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4.10 Properties of Matrix Transformations, , ⎡, , ⎤, , − sin θ, cos θ, , cos θ, ⎢, [T1 ] = ⎣ sin θ, 0, , ⎡, , 0, −1, ⎥, ⎢, 0⎦, [T2 ] = ⎣ 0, 0, 1, , 0, , 0, 1, 0, , ⎤, , ⎡, , 0, 1, ⎥, ⎢, 0⎦, [T3 ] = ⎣0, 1, 0, , Thus, it follows from (5) that the standard matrix for T is, , ⎡, , 1, ⎢, [T ] = ⎣0, 0, , ⎤⎡, , −1, 0, ⎥⎢, 0⎦ ⎣ 0, 0, 0, , 0, 1, 0, , ⎡, − cos θ, ⎢, = ⎣ sin θ, 0, , One-to-One Matrix, Transformations, , ⎤⎡, , 0, cos θ, ⎥⎢, 0⎦ ⎣ sin θ, 1, 0, , 0, 1, 0, , − sin θ, cos θ, 0, , 0, 1, 0, , 273, , ⎤, , 0, ⎥, 0⎦, 0, , ⎤, , 0, ⎥, 0⎦, 1, , ⎤, , sin θ, cos θ, 0, , 0, ⎥, 0⎦, 0, , Our next objective is to establish a link between the invertibility of a matrix A and, properties of the corresponding matrix transformation TA ., DEFINITION 1 A matrix transformation TA : R n →R m is said to be one-to-one if TA, , maps distinct vectors (points) in R n into distinct vectors (points) in R m ., , (See Figure 4.10.5.) This idea can be expressed in various ways. For example, you should, be able to see that the following are just restatements of Definition 1:, 1. TA is one-to-one if for each vector b in the range of A there is exactly one vector x in, R n such that TA x = b., 2. TA is one-to-one if the equality TA (u) = TA (v) implies that u = v., , Rn, , Rm, , Figure 4.10.5, , Rn, , Rm, Not one-to-one, , One-to-one, , Rotation operators on R 2 are one-to-one since distinct vectors that are rotated, through the same angle have distinct images (Figure 4.10.6). In contrast, the orthogonal, projection of R 2 onto the x-axis is not one-to-one because it maps distinct points on the, same vertical line into the same point (Figure 4.10.7)., y, , T(v), , y, , T(u), θ, θ, , P, v, u, , Q, x, , M, , Figure 4.10.6 Distinct, vectors u and v are rotated, into distinct vectors T (u), and T (v)., , x, , Figure 4.10.7 The, distinct points P and, Q are mapped into the, same point M .

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274, , Chapter 4 General Vector Spaces, , Kernel and Range, , In the discussion leading up to Theorem 4.2.5 we introduced the notion of the “kernel”, of a matrix transformation. The following definition formalizes this idea and defines the, companion notion of “range.”, DEFINITION 2 If TA : R n →R m is a matrix transformation, then the set of all vectors, , in R n that TA maps into 0 is called the kernel of TA and is denoted by ker(TA ). The set, of all vectors in R m that are images under this transformation of at least one vector, in R n is called the range of TA and is denoted by R(TA )., In brief:, ker(TA ) = null space of A, , (6), , R(TA ) = column space of A, , (7), , The key to solving a mathematical problem is often adopting the right point of view;, and this is why, in linear algebra, we develop different ways of thinking about the same, vector space. For example, if A is an m × n matrix, here are three ways of viewing the, same subspace of R n :, • Matrix view: the null space of A, • System view: the solution space of Ax = 0, • Transformation view: the kernel of TA, and here are three ways of viewing the same subspace of R m :, • Matrix view: the column space of A, • System view: all b in R m for which Ax = b is consistent, • Transformation view: the range of TA, In the special case of a linear operator TA : R n →R n , the following theorem establishes, fundamental relationships between the invertibility of A and properties of TA ., , THEOREM 4.10.1 If, , A is an n × n matrix and TA : R n →R n is the corresponding, , matrix operator, then the following statements are equivalent., (a) A is invertible., (b) The kernel of TA is {0}., (c) The range of TA is R n ., (d ) TA is one-to-one., Proof We can prove this theorem by establishing the chain of implications (a ) ⇒ (b) ⇒, (c) ⇒ (d ) ⇒ (a ). We will prove the first two implications and leave the rest as exercises., (a) ⇒ (b) Assume that A is invertible. It follows from parts (a) and (b) of Theorem 4.8.8, that the system Ax = 0 has only the trivial solution and hence that the null space of A, is {0}. Formula (6) now implies that the kernel of TA is {0}., (b) ⇒ (c) Assume that the kernel of TA is {0}. It follows from Formula (6) that the null, , space of A is {0} and hence that A has nullity 0. This in turn implies that the rank of A, is n and hence that the column space of A is all of R n . Formula (7) now implies that the, range of TA is R n .

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4.10 Properties of Matrix Transformations, , 275, , E X A M P L E 5 The Rotation Operator on R 2 Is One-to-One, , As was illustrated in Figure 4.10.6, the operator T : R 2 →R 2 that rotates vectors through, an angle θ is one-to-one. In accordance with parts (a) and (d) of Theorem 4.10.1, show, that the standard matrix for T is invertible., Solution We will show that the standard matrix for T is invertible by showing that its, determinant is nonzero. From Table 5 of Section 4.9 the standard matrix for T is, , [T ] =, , − sin θ, cos θ, , cos θ, sin θ, , This matrix is invertible because, , ,  cos θ, det[T ] = , sin θ, , , − sin θ , = cos2 θ + sin2 θ = 1  = 0, cos θ , , E X A M P L E 6 Projection Operators Are Not One-to-One, , As illustrated in Figure 4.10.7, the operator T : R 2 →R 2 that projects onto the x -axis in, the xy -plane is not one-to-one. In accordance with parts (a) and (d) of Theorem 4.10.1,, show that the standard matrix for T is not invertible., Solution We will show that the standard matrix for T is not invertible by showing that, its determinant is zero. From Table 3 of Section 4.9 the standard matrix for T is, , , [T ] =, , , , 1, , 0, , 0, , 0, , Since det[T ] = 0, the operator T is not one-to-one., , Inverse of a One-to-One, Matrix Operator, , If TA : R n →R n is a one-to-one matrix operator, then it follows from Theorem 4.10.1 that, A is invertible. The matrix operator, , TA−1 : R n →R n, that corresponds to A−1 is called the inverse operator or (more simply) the inverse of TA ., This terminology is appropriate because TA and TA−1 cancel the effect of each other in, the sense that if x is any vector in R n , then, , TA (TA−1 (x)) = AA−1 x = I x = x, TA−1 (TA (x)) = A−1 Ax = I x = x, y, , or, equivalently,, TA, , s x to w, m ap, , x, TA–1 maps w, , w, , to x, x, , Figure 4.10.8, , TA ◦ TA−1 = TAA−1 = TI, TA−1 ◦ TA = TA−1 A = TI, From a more geometric viewpoint, if w is the image of x under TA , then TA−1 maps w, backinto x, since, , TA−1 (w) = TA−1 (TA (x)) = x, This is illustrated in Figure 4.10.8 for R 2 .

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276, , Chapter 4 General Vector Spaces, , Before considering examples, it will be helpful to touch on some notational matters., If TA : R n →R n is a one-to-one matrix operator, and if TA−1 : R n →R n is its inverse, then, the standard matrices for these operators are related by the equation, , TA−1 = TA−1, , (8), , In cases where it is preferable not to assign a name to the matrix, we can express this, equation as, , [T −1 ] = [T ]−1, , (9), , E X A M P L E 7 Standard Matrix for T −1, , Let T : R 2 →R 2 be the operator that rotates each vector in R 2 through the angle θ , so, from Table 5 of Section 4.9,, , − sin θ, cos θ, , cos θ, sin θ, , [T ] =, , (10), , It is evident geometrically that to undo the effect of T , one must rotate each vector in R 2, through the angle −θ . But this is exactly what the operator T −1 does, since the standard, matrix for T −1 is, , [T −1 ] = [T ]−1 =, , cos θ, − sin θ, , sin θ, cos(−θ), =, cos θ, sin(−θ), , − sin(−θ), cos(−θ), , (verify), which is the standard matrix for a rotation through the angle −θ ., E X A M P L E 8 Finding T −1, , Show that the operator T : R 2 →R 2 defined by the equations, , w1 = 2x1 + x2, w2 = 3x1 + 4x2, is one-to-one, and find T −1 (w1 , w2 )., Solution The matrix form of these equations is, , w1, 2, =, 3, w2, , x1, x2, , 1, 4, , so the standard matrix for T is, 2, 3, , [T ] =, , 1, 4, , This matrix is invertible (so T is one-to-one) and the standard matrix for T −1 is, , ⎡, , 4, 5, , [T −1 ] = [T ]−1 = ⎣, − 35, Thus, , ⎡, 4, w, 1, 5, =⎣, [T −1 ], w2, −3, T −1 (w1 , w2 ) =, , 2, 5, , ⎤, ⎦, , ⎡, ⎤, 4, 1, w, −, w, 1, 2, w1, 5, 5, ⎦, ⎦, =⎣, 2, 3, 2, w2, −, w, +, w, 5, 5 1, 5 2, , − 15, , 5, , from which we conclude that, , − 15, , 4, 5, , ⎤, , w1 − 15 w2 , − 35 w1 + 25 w2, , 

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4.10 Properties of Matrix Transformations, , More on the Equivalence, Theorem, , 277, , As our final result in this section, we will add parts (b), (c), and (d) of Theorem 4.10.1, to Theorem 4.8.8., THEOREM 4.10.2 Equivalent Statements, , If A is an n × n matrix, then the following statements are equivalent., , (c), , A is invertible., Ax = 0 has only the trivial solution., The reduced row echelon form of A is In ., , (d ), , A is expressible as a product of elementary matrices., , (e), , Ax = b is consistent for every n × 1 matrix b., , (a), (b), , Ax = b has exactly one solution for every n × 1 matrix b., ( g) det(A)  = 0., (h) The column vectors of A are linearly independent., (i ) The row vectors of A are linearly independent., ( j) The column vectors of A span R n ., ( f), , (k), , The row vectors of A span R n ., , (l ), , The column vectors of A form a basis for R n ., , (m) The row vectors of A form a basis for R n ., (n), (o), , A has rank n., A has nullity 0., , ( p) The orthogonal complement of the null space of A is R n ., (q), , The orthogonal complement of the row space of A is {0}., , (r), , The kernel of TA is {0}., , (s), , The range of TA is R n ., , (t), , TA is one-to-one., , Exercise Set 4.10, In Exercises 1–4, determine whether the operators T1 and T2, commute; that is, whether T1 ◦ T2 = T2 ◦ T1 ., 1. (a) T1 : R 2 →R 2 is the reflection about the line y = x , and, T2 : R 2 →R 2 is the orthogonal projection onto the x -axis., (b) T1 : R 2 →R 2 is the reflection about the x -axis, and, T2 : R 2 →R 2 is the reflection about the line y = x ., 2. (a) T1 : R →R is the orthogonal projection onto the x -axis,, and T2 : R 2 →R 2 is the orthogonal projection onto the, y -axis., 2, , 2, , (b) T1 : R →R is the rotation about the origin through an, angle of π/4, and T2 : R 2 →R 2 is the reflection about the, y -axis., 2, , 2, , 3. T1 : R 3 →R 3 is a dilation with factor k , and T2 : R 3 →R 3 is a, contraction with factor 1/k ., 4. T1 : R 3 →R 3 is the rotation about the x -axis through an angle, θ1 , and T2 : R 3 →R 3 is the rotation about the z-axis through, an angle θ2 ., , In Exercises 5–6, let TA and TB bet the operators whose standard matrices are given. Find the standard matrices for TB ◦ TA, and TA ◦ TB ., 5. A =, , 1, 4, , ⎡, , 6, ⎢, 6. A = ⎣2, 4, , −2, 1, 3, 0, −3, , , B=, ⎤, , 2, 5, , −3, 0, , ⎡, , 4, −1, ⎥, ⎢−1, 1⎦ , B = ⎣, 6, , 2, , 0, 5, −3, , ⎤, , 4, 2⎥, ⎦, 8, , 7. Find the standard matrix for the stated composition in R 2 ., (a) A rotation of 90◦ , followed by a reflection about the line, y = x., (b) An orthogonal projection onto the y -axis, followed by a, contraction with factor k = 21 ., (c) A reflection about the x -axis, followed by a dilation with, factor k = 3, followed by a rotation about the origin, of 60◦ .

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4.10 Properties of Matrix Transformations, , In Exercises 23–24, let T be multiplication by the matrix A., Find, (a) a basis for the range of T ., (c) the rank and nullity of T ., , Working with Proofs, , (d) the rank and nullity of A., 1, , −1, , 7, , 6, 4, , ⎢, 23. A = ⎣5, , ⎡, , ⎤, , 3, ⎥, −4 ⎦, 2, , 2, , ⎢, 24. A = ⎣ 4, , 20, , 0, 0, 0, , ⎤, , −1, ⎥, −2 ⎦, 0, , In Exercises 25–26, let TA : R →R be multiplication by A., Find a basis for the kernel of TA , and then find a basis for the, range of TA that consists of column vectors of A., 4, , ⎡, , 1, ⎢, 25. A = ⎣−3, −3, , ⎡, , 1, ⎢, 26. A = ⎣−2, −1, , −1, , 2, 1, 8, 1, 4, 8, , −2, , 0, 2, 3, , ⎤, , 2, , 27. Let A be an n × n matrix such that det(A) = 0, and let, T : R n →R n be multiplication by A., (a) What can you say about the range of the matrix operator, T ? Give an example that illustrates your conclusion., (b) What can you say about the number of vectors that T maps, into 0?, 28. Answer the questions in Exercise 27 in the case where, det(A)  = 0., 29. (a) Is a composition of one-to-one matrix transformations, one-to-one? Justify your conclusion., (b) Can the composition of a one-to-one matrix transformation and a matrix transformation that is not one-to-one, be one-to-one? Account for both possible orders of composition and justify your conclusion., 30. Let TA : R 2 →R 2 be multiplication by, , A=, , 2 sin θ cos θ, , 35. Prove the implication (d) ⇒ (a) in Theorem 4.10.1., , True-False Exercises, , (a) If TA and TB are matrix operators on R n , then, TA (TB (x)) = TB (TA (x)) for every vector x in R n ., , 1, ⎥, 2⎦, 5, , cos2 θ − sin2 θ, , 34. Prove the implication (c) ⇒ (d) in Theorem 4.10.1., , TF. In parts (a)–(g) determine whether the statement is true or, false, and justify your answer., , ⎤, , , , 33. Prove that the matrix transformations TA and TB commute if, and only if the matrices A and B commute., , 3, , ⎥, 4⎦, , 3, 4, , 32. (a) The inverse transformation for a reflections about a coordinate axis is a reflection about that axis., (b) The inverse transformation for a shear along a coordinate, axis is a shear along that axis., , (b) a basis for the kernel of T ., , ⎡, , 279, , −2 sin θ cos θ, cos2 θ − sin2 θ, , , , (a) What is the geometric effect of applying this transformation to a vector x in R 2 ?, (b) Express the operator TA as a composition of two linear, operators on R 2 ., In Exercises 31–32, use matrix inversion to confirm the stated, result in R 2 ., 31. (a) The inverse transformation for a reflection about y = x is, a reflection about y = x ., (b) The inverse transformation for a compression along an, axis is an expansion along that axis., , (b) If T1 and T2 are matrix operators on R n , then, [T2 ◦ T1 ] = [T2 ][T1 ]., (c) A composition of two rotation operators about the origin of, R 2 is another rotation about the origin., (d) A composition of two reflection operators in R 2 is another, reflection operator., (e) The kernel of a matrix transformation TA : R n →R m is the, same as the null space of A., (f ) If there is a nonzero vector in the kernel of the matrix operator, TA : R n →R n , then this operator is not one-to-one., (g) If A is an n × n matrix and if the linear system Ax = 0 has, a nontrivial solution, then the range of the matrix operator is, not R n ., , Working withTechnology, T1. (a) Find the standard matrix for the linear operator on R 3, that performs a counterclockwise rotation of 47◦ about, the x -axis, followed by a counterclockwise rotation of 68◦, about the y -axis, followed by a counterclockwise rotation, of 33◦ about the z-axis., (b) Find the image of the point (1, 1, 1) under the operator, in part (a)., T2. Find the standard matrix for the linear operator on R 2 that, first reflects each point in the plane about the line through the origin that makes an angle of 27◦ with the positive x -axis and then, projects the resulting point orthogonally onto the line through the, origin that makes an angle of 51◦ with the positive x -axis.

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280, , Chapter 4 General Vector Spaces, , 4.11 Geometry of Matrix Operators on R 2, In applications such as computer graphics it is important to understand not only how linear, operators on R 2 and R 3 affect individual vectors but also how they affect two-dimensional, or three-dimensional regions. That is the focus of this section., , Transformations of Regions, , Figure 4.11.1 shows a famous picture of Albert Einstein that has been transformed in, various ways using matrix operators on R 2 . The original image was scanned and then, digitized to decompose it into a rectangular array of pixels. Those pixels were then, transformed as follows:, • The program MATLAB was used to assign coordinates and a gray level to each pixel., • The coordinates of the pixels were transformed by matrix multiplication., • The pixels were then assigned their original gray levels to produce the transformed, picture., In computer games a perception of motion is created by using matrices to rapidly, and repeatedly transform the arrays of pixels that form the visual images., , Digitized scan, , Rotated, , Sheared horizontally, , Compressed horizontally, , Figure 4.11.1 [Image: ARTHUR SASSE/AFP/Getty Images], , The effect of a matrix operator on R 2 can often be deduced by studying how it transforms, the points that form the unit square. The following theorem, which we state without, proof, shows that if the operator is invertible, then it maps each line segment in the unit, square into the line segment connecting the images of its endpoints. In particular, the, edges of the unit square get mapped into edges of the image (see Figure 4.11.2 in which, the edges of a unit square and the corresponding edges of its image have been numbered)., , Images of Lines Under, Matrix Operators, , y, , y, , e2, , 3, , 4, , 4, x, , 1, , 2, 1, , Figure 4.11.2, , 2, , 4, , x, , 1, , Unit square rotated, , 3, x, , x, 1, , e1, , Unit square, , 3, , 2, , (1, 1), 2, , y, , y, , 3, , Unit square reflected, about the y-axis, , 4, Unit square reflected, about the line y = x

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4.11 Geometry of Matrix Operators on R 2, , 281, , T : R 2 →R 2 is multiplication by an invertible matrix, then:, , THEOREM 4.11.1 If, , (a) The image of a straight line is a straight line., (b) The image of a line through the origin is a line through the origin., (c), , The images of parallel lines are parallel lines., , (d ) The image of the line segment joining points P and Q is the line segment joining, the images of P and Q., (e), , The images of three points lie on a line if and only if the points themselves lie on a, line., , E X A M P L E 1 Image of a Line, , According to Theorem 4.11.1, the invertible matrix, 3, 2, , A=, , 1, 1, , maps the line y = 2x + 1 into another line. Find its equation., , = 2x + 1, and let (x , y ) be its image under, , Solution Let (x, y) be a point on the line y, , multiplication by A. Then, , x, y, , 3, =, 2, , x, y, , 1, 1, , and, , x, 3, =, 2, y, , 1, 1, , −1, , x, y, , =, , 1, , −1, , −2, , 3, , x, y, , so, , x=, x − y, y = −2x + 3y, Substituting these expressions in y = 2x + 1 yields, −2x + 3y = 2(x − y ) + 1, or, equivalently,, , y = 45 x +, , 1, 5, , E X A M P L E 2 Transformation of the Unit Square, , Sketch the image of the unit square under multiplication by the invertible matrix, , , , y, 3, , (0, 1), , 2, 1, , (0, 0), , A=, , (1, 1), , 4, , x, , Solution Since, , , , 0, , 1, , 2, , 1, , y, (1, 3), 2, , , 3, , 1, , (1, 1), 4, , 1, , 2, , 1, , , , Label the vertices of the image with their coordinates, and number the edges of the unit, square and their corresponding images (as in Figure 4.11.2)., , (1, 0), , (0, 2), , 0, , x, , 0, , 1, ,  , 0, 0, ,  , =, ,  , 0, , 0, 0, , , ,, ,  , =, , 1, , 0, , 1, , 2, , 1, , , ,, , 0, , 1, ,  , 1, 0, ,  , =, ,  , 1, , 0, , 2, , ,, ,  , =, , 1, , 2 1 1, 1, 2 1 1, 3, the image of the unit square is a parallelogram with vertices (0, 0), (0, 2), (1, 1), and, (1, 3) (Figure 4.11.3)., , (0, 0), , Figure 4.11.3, , The next example illustrates a transformation of the unit square under a composition, of matrix operators.

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282, , Chapter 4 General Vector Spaces, , E X A M P L E 3 Transformation of the Unit Square, , (a) Find the standard matrix for the operator on R 2 that first shears by a factor of 2 in, the x -direction and then reflects the result about the line y = x . Sketch the image, of the unit square under this operator., (b) Find the standard matrix for the operator on R 2 that first reflects about y = x and, then shears by a factor of 2 in the x -direction. Sketch the image of the unit square, under this operator., (c) Confirm that the shear and the reflection in parts (a) and (b) do not commute., Solution (a) The standard matrix for the shear is, , A1 =, , 1, 0, , 2, 1, , A2 =, , 0, 1, , 1, 0, , and for the reflection is, , Thus, the standard matrix for the shear followed by the reflection is, , A2 A1 =, , 0, 1, , 1, 0, , 1, 0, , 2, 0, =, 1, 1, , 1, 2, , Solution (b) The standard matrix for the reflection followed by the shear is, , A1 A2 =, , 1, 0, , 2, 1, , 0, 1, , 1, 2, =, 0, 1, , 1, 0, , Solution (c) The computations in Solutions (a) and (b) show that A1 A2  = A2 A1 , so, the standard matrices, and hence the operators, do not commute. The same conclusion, follows from Figures 4.11.4 and 4.11.5 since the two operators produce different images, of the unit square., , y, , y, , y, , y=x, , (3, 1), (1, 1), , (1, 1), x, , x, , x, , Shear in the, x-direction by a, factor k = 2, , Reflection, about y = x, , Figure 4.11.4, y, , y, , y, , y=x, , (1, 3), , y=x, , (3, 1), (1, 1), x, , Figure 4.11.5, , x, , Shear in the, x-direction by a, factor k = 2, , x, , Reflection, about y = x

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4.11 Geometry of Matrix Operators on R 2, , 283, , In Example 3 we illustrated the effect on the unit square in R 2 of a composition of shears, and reflections. Our next objective is to show how to decompose any 2 × 2 invertible, matrix into a product of matrices in Table 1, thereby allowing us to analyze the geometric, effect of a matrix operator in R 2 as a composition of simpler matrix operators. The next, theorem is our first step in this direction., , Geometry of Invertible, Matrix Operators, , Table 1, , Operator, , Standard, Matrix, , Effect on the Unit Square, y, , Reflection about, the y -axis, , −1, 0, , y, (1, 1), , 0, 1, , (––1, 1), , x, , x, , y, , y, (1, 1), , Reflection about, the x -axis, , 1, 0, , 0, , x, , x, , −1, , (1, –1), , y, , 1, 0, , Reflection about, the line y = x, , Rotation about the, origin through a, positive angle θ, , Compression in the, x -direction with, factor k, , 0, 1, , y, (1, 1), , − sin θ, cos θ, , cos θ, sin θ, , (1, 1), , x, , y, , k, 0, , 0, 1, , y, , (k > 1), , (k, 1), , y, (1, 1), , 0, , (1, k), , k, , 0, , 0, 1, , x, , x, , y, , k, , x, , x, , y, , 1, 0, , x, , θ, , (1, 1), , (0 < k < 1), , Expansion in the, x -direction with, factor k, , x, , (cos θ – sin θ, sin θ + cos θ), y, , y, , (0 < k < 1), , Compression in the, y -direction with, factor k, , (1, 1), , x, , y, (k, 1), , (1, 1), , x, , x, , (Continued on the following page.)

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284, , Chapter 4 General Vector Spaces, , Standard, Matrix, , Operator, , Effect on the Unit Square, y, , y, (1, k), , Expansion in the, y -direction with, factor k, , 1, 0, , 0, , (1, 1), , k, , (k > 1), , x, , x, , Shear in the, positive x -direction, by a factor k, , 1, 0, , (1, 1), , k, 1, , (k, 1), , (1, 1), , k, 1, , k, , x, , y, , y, , 1, , (1, 1 + k), , (1, 1), , 0, 1, , (1, k), , x, , (k > 0), , 1, , k, , 0, 1, , x, , y, , y, , Shear in the, negative y -direction, by a factor k, , y, (k + 1, 1), , x, , (k < 0), , Shear in the, positive y -direction, by a factor k, , x, , y, , 1, 0, , (1 + k, 1), , x, , (k > 0), , Shear in the, negative x -direction, by a factor k, , y, (k, 1), , y, , (1, 1), , (1, 1 + k), x, x, , (k < 0), , (1, k), , THEOREM 4.11.2 If E is an elementary matrtix, then TE : R 2 →R 2 is one of the, , following:, (a), , A shear along a coordinate axis., , (b), , A reflection about y = x ., , (c), , A compression along a coordinate axis., , (d ), , An expansion along a coordinate axis., , (e), , A reflection about a coordinate axis., , ( f), , A compression or expansion along a coordinate axis followed by a reflection about, a coordinate axis.

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4.11 Geometry of Matrix Operators on R 2, , 285, , Proof Because a 2 × 2 elementary matrix results from performing a single elementary, row operation on the 2 × 2 identity matrix, such a matrix must have one of the following, forms (verify):, , 1, , k, , 0, ,, 1, , 1, 0, , k, 1, , 0, 1, , ,, , k, , 1, ,, 0, , 0, ,, 1, , 0, , 1, 0, , 0, , k, , The first two matrices represent shears along coordinate axes, and the third represents, a reflection about y = x . If k > 0, the last two matrices represent compressions or, expansions along coordinate axes, depending on whether 0 ≤ k < 1 or k > 1. If k < 0,, and if we express k in the form k = −k1 , where k1 > 0, then the last two matrices can be, written as, k 0, −k1 0, −1 0 k1 0, (1), =, =, 0 1, 0 1, 0 1 0 1, 1, 0, , 0, , k, , =, , 1, 0, , 0, , −k1, , =, , 1, 0, , 0, , −1, , 1, 0, , 0, , (2), , k1, , Since k1 > 0, the product in (1) represents a compression or expansion along the, x -axis followed by a reflection about the y -axis, and (2) represents a compression or, expansion along the y -axis followed by a reflection about the x -axis. In the case where, k = −1, transformations (1) and (2) are simply reflections about the y -axis and x -axis,, respectively., We know from Theorem 4.10.2(d) that an invertible matrix can be expressed as a, product of elementary matrices, so Theorem 4.11.2 implies the following result., , TA : R 2 →R 2 is multiplication by an invertible matrix A, then the, geometric effect of TA is the same as an appropriate succession of shears, compressions,, expansions, and reflections., , THEOREM 4.11.3 If, , The next example will illustrate how Theorems 4.11.2 and 4.11.3 together with, Table 1 can be used to analyze the geometric effect of multiplication by a 2 × 2 invertible, matrix., E X A M P L E 4 Decomposing a Matrix Operator, , In Example 2 we illustrated the effect on the unit square of multiplication by, , A=, , 0, 2, , 1, 1, , (see Figure 4.11.3). Express this matrix as a product of elementary matrices, and then, describe the effect of multiplication by A in terms of shears, compressions, expansions,, and reflections., Solution The matrix A can be reduced to the identity matrix as follows:, , 0, 2, , 1, 2, −→, 0, 1, , Multiply the, first row by 21 ., , 1, 2, , 1, , −→, , 1, 0, , , , , , , Interchange the, first and second, rows., , 1, 1, −→, 1, 0, , Add − 21 times, the second row, to the first., , 0, 1

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286, , Chapter 4 General Vector Spaces, , These three successive row operations can be performed by multiplying A on the left, successively by, , , , E1 =, , 0, , 1, , 1, , 0, , , , , , , E2 =, , 1, 2, , 0, , 0, , 1, , , , , , , E3 =, , , , 1, , − 21, , 0, , 1, , Inverting these matrices and using Formula (4) of Section 1.5 yields, , , , A=, , , , 0, , 1, , 2, , 1, , , , −1, , −1, , −1, , = E1 E2 E3 =, , 0, , 1, , 1, , 0, , , , 2, , 0, , 0, , 1, , , , , , 1, , 1, 2, , 0, , 1, , Reading from right to left we can now see that the geometric effect of multiplying by A, is equivalent to successively, 1. shearing by a factor of, , 1, 2, , in the x -direction;, , 2. expanding by a factor of 2 in the x -direction;, 3. reflecting about the line y = x ., This is illustrated in Figure 4.11.6, whose end result agrees with that in Example 2., , y, , y, , y, , y, , y=x, , (1, 3), , y=x, , (0, 2), (3, 1), , ( 32 , 1), , (1, 1), x, , (1, 1), x, , x, , x, (0, 0), , Figure 4.11.6, , E X A M P L E 5 Transformations with Diagonal Matrices, , Discuss the geometric effect on the unit square of multiplication by a diagonal matrix, , A=, , k1, , 0, , 0, , k2, , in which the entries k1 and k2 are positive real numbers ( = 1)., Solution The matrix A is invertible and can be expressed as, , A=, , k1, , 0, , 0, , k2, , 1, 0, , =, , 0, , k1, , k2, , 0, , 0, 1, , which show that multiplication by A causes a compression or expansion of the unit, square by a factor of k1 in the x -direction followed by an expansion or compression of, the unit square by a factor of k2 in the y -direction., , y, (1, 1), x, , E X A M P L E 6 Reflection About the Origin, , As illustrated in Figure 4.11.7, multiplication by the matrix, (–1, –1), , Figure 4.11.7, , A=, , −1, , 0, , 0, , −1

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4.11 Geometry of Matrix Operators on R 2, , 287, , has the geometric effect of reflecting the unit square about the origin. Note, however,, that the matrix equation, , A=, , −1, 0, , −1, 0, =, 0, −1, , 0, 1, , 1, 0, , 0, −1, , together with Table 1 shows that the same result can be obtained by first reflecting the, unit square about the x -axis and then reflecting that result about the y -axis. You should, be able to see this as well from Figure 4.11.7., , y, (1, 1), , E X A M P L E 7 Reflection About the Line y = –x, x, , We leave it for you to verify that multiplication by the matrix, , A=, (–1, –1), , 0, , −1, , −1, , 0, , reflects the unit square about the line y = −x (Figure 4.11.8)., , Figure 4.11.8, , Exercise Set 4.11, 1. Use the method of Example 1 to find an equation for the image, of the line y = 4x under multiplication by the matrix, , , , A=, , 5, , 2, , 2, , 1, , , , (b) Rotates through 30◦ about the origin, then shears by a factor of −2 in the y -direction, and then expands by a factor, of 3 in the y -direction., , 2. Use the method of Example 1 to find an equation for the image, of the line y = −4x + 3 under multiplication by the matrix, 4, 3, , A=, , 3. A shear by a factor 3 in the x -direction., 1, 2, , in the y -direction., , In Exercises 5–6, sketch the image of the unit square under, multiplication by the given invertible matrix. As in Example 2,, number the edges of the unit square and its image so it is clear, how those edges correspond., 5., , 3, 1, , −1, −2, , 6., , 2, , −1, , In each part of Exercises 9–10, determine whether the stated, operators commute., 9. (a) A reflection about the x -axis and a compression in the, x -direction with factor 13 ., , −3, −2, , In Exercises 3–4, find an equation for the image of the line, y = 2x that results from the stated transformation., , 4. A compression with factor, , 8. (a) Reflects about the y -axis, then expands by a factor of 5 in, the x -direction, and then reflects about y = x ., , 1, 2, , In each part of Exercises 7–8, find the standard matrix for a, single operator that performs the stated succession of operations., 7. (a) Compresses by a factor of 21 in the x -direction, then expands by a factor of 5 in the y -direction., , (b) A reflection about the line y = x and an expansion in the, x -direction with factor 2., 10. (a) A shear in the y -direction by a factor, y -direction by a factor 35 ., , 1, 4, , and a shear in the, , (b) A shear in the y -direction by a factor, x -direction by a factor 35 ., , 1, 4, , and a shear in the, , In Exercises 11–14, express the matrix as a product of elementary matrices, and then describe the effect of multiplication by A, in terms of shears, compressions, expansions, and reflections., 11. A =, , 4, 0, , 4, −2, , 13. A =, , 0, 4, , −2, 0, , 12. A =, , 1, 2, , 4, 9, , 14. A =, , 1, 4, , −3, 6, , In each part of Exercises 15–16, describe, in words, the effect on, the unit square of multiplication by the given diagonal matrix., , (b) Expands by a factor of 5 in the y -direction, then shears by, a factor of 2 in the y -direction., , 15. (a) A =, , (c) Reflects about y = x , then rotates through an angle of, 180◦ about the origin., , 16. (a) A =, , 3, 0, , 0, 1, , −2, 0, , (b) A =, 0, 1, , (b) A =, , 1, 0, , 0, , −5, , −3, , 0, , 0, , −1

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288, , Chapter 4 General Vector Spaces, , 17. (a) Show that multiplication by, 3, 6, , A=, , 1, 2, , maps each point in the plane onto the line y = 2x ., (b) It follows from part (a) that the noncollinear points (1, 0),, (0, 1), (−1, 0) are mapped onto a line. Does this violate, part (e) of Theorem 4.11.1?, 18. Find the matrix for a shear in the x -direction that transforms, the triangle with vertices (0, 0), (2, 1), and (3, 0) into a right, triangle with the right angle at the origin., , 25. Find the image of the triangle with vertices (0, 0), (1, 1), (2, 0), under multiplication by, 2 −1, A=, 0, 0, Does your answer violate part (e) of Theorem 4.11.1? Explain., 26. In R 3 the shear in the xy-direction by a factor k is the matrix, transformation that moves each point (x, y, z) parallel to the, xy -plane to the new position (x + kz, y + kz, z). (See the, accompanying figure.), (a) Find the standard matrix for the shear in the xy -direction, by a factor k ., (b) How would you define the shear in the xz-direction by a, factor k and the shear in the yz-direction by a factor k ?, What are the standard matrices for these matrix transformations?, , 19. In accordance with part (c) of Theorem 4.11.1, show that, multiplication by the invertible matrix, 3, 1, , A=, , 2, 1, , z, , maps the parallel lines y = 3x + 1 and y = 3x − 2 into parallel lines., , z, , (x, y, z), , 20. Draw a figure that shows the image of the triangle with vertices, (0, 0), (1, 0), and (0.5, 1) under a shear by a factor of 2 in the, x -direction., 21. (a) Draw a figure that shows the image of the triangle with, vertices (0, 0), (1, 0), and (0.5, 1) under multiplication by, , A=, , 1, 1, , 1, , 22. Find the endpoints of the line segment that results when the, line segment from P (1, 2) to Q(3, 4) is transformed by, 1, 2, , y, , x, , y, , x, , Figure Ex-26, , −1, , (b) Find a succession of shears, compressions, expansions,, and reflections that produces the same image., , (a) a compression with factor, , (x + kz, y + kz, z), , in the y -direction., , (b) a rotation of 30◦ about the origin., 23. Draw a figure showing the italicized letter “T ” that results, when the letter in the accompanying figure is sheared by a, factor 41 in the x -direction., y, , Working with Proofs, 27. Prove part (a) of Theorem 4.11.1. [Hint: A line in the plane, has an equation of the form Ax + By + C = 0, where A and, B are not both zero. Use the method of Example 1 to show, that the image of this line under multiplication by the invertible, matrix, , a b, c d, has the equation A x + B y + C = 0, where, A = (dA − cB)/(ad − bc), and, , B = (−bA + aB)/(ad − bc), Then show that A and B are not both zero to conclude that, the image is a line.], 28. Use the hint in Exercise 27 to prove parts (b) and (c) of Theorem 4.11.1., , 1, (0, .90), , True-False Exercises, TF. In parts (a)–(g) determine whether the statement is true or, false, and justify your answer., , x, 1, (.45, 0) (.55, 0), , Figure Ex-23, , 24. Can an invertible matrix operator on R 2 map a square region, into a triangular region? Justify your answer., , (a) The image of the unit square under a one-to-one matrix operator is a square., (b) A 2 × 2 invertible matrix operator has the geometric effect of, a succession of shears, compressions, expansions, and reflections.

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Chapter 4 Supplementary Exercises, , (c) The image of a line under an invertible matrix operator is a, line., , (f ) The matrix, , 1, 2, , −2, , (g) The matrix, , 1, 0, , 0, represents an expansion., 3, , 1, , 289, , represents a shear., , (d) Every reflection operator on R 2 is its own inverse., (e) The matrix, , 1, 1, , 1, , represents reflection about a line., , −1, , Chapter 4 Supplementary Exercises, 1. Let V be the set of all ordered triples of real numbers, and, consider the following addition and scalar multiplication operations on u = (u1 , u2 , u3 ) and v = (v1 , v2 , v3 ):, u + v = (u1 + v1 , u2 + v2 , u3 + v3 ), k u = (ku1 , 0, 0), (a) Compute u + v and k u for u = (3, −2, 4),, v = (1, 5, −2), and k = −1., (b) In words, explain why V is closed under addition and, scalar multiplication., (c) Since the addition operation on V is the standard addition, operation on R 3 , certain vector space axioms hold for V, because they are known to hold for R 3 . Which axioms in, Definition 1 of Section 4.1 are they?, (d) Show that Axioms 7, 8, and 9 hold., , 5. Let W be the space spanned by f = sin x and g = cos x ., (a) Show that for any value of θ , f1 = sin(x + θ) and, g1 = cos(x + θ) are vectors in W ., (b) Show that f1 and g1 form a basis for W ., 6. (a) Express v = (1, 1) as a linear combination of, v1 = (1, −1), v2 = (3, 0), and v3 = (2, 1) in two different, ways., (b) Explain why this does not violate Theorem 4.4.1., 7. Let A be an n × n matrix, and let v1 , v2 , . . . , vn be linearly, independent vectors in R n expressed as n × 1 matrices. What, must be true about A for Av1 , Av2 , . . . , Avn to be linearly independent?, , (e) Show that Axiom 10 fails for the given operations., 2. In each part, the solution space of the system is a subspace of, R 3 and so must be a line through the origin, a plane through, the origin, all of R 3 , or the origin only. For each system, determine which is the case. If the subspace is a plane, find an, equation for it, and if it is a line, find parametric equations., (a) 0x + 0y + 0z = 0, , (c), , x − 2y + 7z = 0, −4x + 8y + 5z = 0, 2x − 4y + 3z = 0, , (b), , 8. Must a basis for Pn contain a polynomial of degree k for each, k = 0, 1, 2, . . . , n? Justify your answer., 9. For the purpose of this exercise, let us define a “checkerboard, matrix” to be a square matrix A = [aij ] such that, , , , 2 x − 3y + z = 0, 6 x − 9y + 3 z = 0, −4 x + 6y − 2 z = 0, , (d) x + 4y + 8z = 0, 2x + 5y + 6z = 0, 3x + y − 4 z = 0, , 3. For what values of s is the solution space of, , x1 + x2 + sx3 = 0, x1 + sx2 + x3 = 0, sx1 + x2 + x3 = 0, the origin only, a line through the origin, a plane through the, origin, or all of R 3 ?, 4. (a) Express (4a, a − b, a + 2b) as a linear combination of, (4, 1, 1) and (0, −1, 2)., , aij =, , 1, , if i + j is even, , 0, , if i + j is odd, , Find the rank and nullity of the following checkerboard, matrices., (a) The 3 × 3 checkerboard matrix., (b) The 4 × 4 checkerboard matrix., (c) The n × n checkerboard matrix., 10. For the purpose of this exercise, let us define an “X -matrix” to, be a square matrix with an odd number of rows and columns, that has 0’s everywhere except on the two diagonals where it, has 1’s. Find the rank and nullity of the following X -matrices., , ⎡, , ⎡, , 0, 1, 0, , ⎤, , 1, ⎥, 0⎦, 1, , 1, ⎢, ⎢0, ⎢, (b) ⎢, ⎢0, ⎢0, ⎣, 1, , 0, 1, 0, 1, 0, , (b) Express (3a + b + 3c, −a + 4b − c, 2a + b + 2c) as a, linear combination of (3, −1, 2) and (1, 4, 1)., , 1, ⎢, (a) ⎣0, 1, , (c) Express (2a − b + 4c, 3a − c, 4b + c) as a linear combination of three nonzero vectors., , (c) the X -matrix of size (2n + 1) × (2n + 1), , 0, 0, 1, 0, 0, , 0, 1, 0, 1, 0, , ⎤, , 1, ⎥, 0⎥, ⎥, 0⎥, ⎥, 0⎥, ⎦, 1

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290, , Chapter 4 General Vector Spaces, , 11. In each part, show that the stated set of polynomials is a subspace of Pn and find a basis for it., (a) All polynomials in Pn such that p(−x) = p(x)., (b) All polynomials in Pn such that p(0) = p(1)., 12. (Calculus required ) Show that the set of all polynomials in Pn, that have a horizontal tangent at x = 0 is a subspace of Pn ., Find a basis for this subspace., 13. (a) Find a basis for the vector space of all 3 × 3 symmetric, matrices., (b) Find a basis for the vector space of all 3 × 3 skewsymmetric matrices., 14. Various advanced texts in linear algebra prove the following, determinant criterion for rank: The rank of a matrix A is r if, and only if A has some r × r submatrix with a nonzero determinant, and all square submatrices of larger size have determinant, zero. [Note: A submatrix of A is any matrix obtained by, deleting rows or columns of A. The matrix A itself is also, considered to be a submatrix of A.] In each part, use this, criterion to find the rank of the matrix., (a), , 1, 2, , ⎡, , 1, ⎢, (c) ⎣2, 3, , 2, 4, 0, −1, −1, , 0, , (b), , −1, ⎤, , 1, ⎥, 3⎦, 4, , 1, 2, , ⎡, , 1, ⎢, (d) ⎣ 3, −1, , 2, 4, , 3, 6, , −1, 1, 2, , 2, 0, 4, , ⎤, , 0, ⎥, 0⎦, 0, , 15. Use the result in Exercise 14 above to find the possible ranks, for matrices of the form, , ⎡, , 0, 0, 0, 0, , ⎢, ⎢, ⎢, ⎢, ⎢, ⎢, ⎣, a51, , 0, 0, 0, 0, , 0, 0, 0, 0, , 0, 0, 0, 0, , 0, 0, 0, 0, , a52, , a53, , a54, , a55, , ⎤, a16, ⎥, a26 ⎥, ⎥, a36 ⎥, ⎥, a46 ⎥, ⎦, a56, , 16. Prove: If S is a basis for a vector space V, then for any vectors, u and v in V and any scalar k , the following relationships hold., (a) (u + v)S = (u)S + (v)S, , (b) (k u)S = k(u)S, , 17. Let Dk , Rθ , and Sk be a dilation of R 2 with factor k , a counterclockwise rotation about the origin of R 2 through an angle, θ , and a shear of R 2 by a factor k , respectively., (a) Do Dk and Rθ commute?, (b) Do Rθ and Sk commute?, (c) Do Dk and Sk commute?, 18. A vector space V is said to be the direct sum of its subspaces, U and W , written V = U ⊕W , if every vector in V can be, expressed in exactly one way as v = u + w, where u is a vector, in U and w is a vector in W ., (a) Prove that V = U ⊕W if and only if every vector in V is, the sum of some vector in U and some vector in W and, U ∩ W = {0}., (b) Let U be the xy -plane and W the z-axis in R 3 . Is it true, that R 3 = U ⊕W ? Explain., (c) Let U be the xy -plane and W the yz-plane in R 3 . Can every vector in R 3 be expressed as the sum of a vector in U, and a vector in W ? Is it true that R 3 = U ⊕W ? Explain.

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CHAPTER, , 5, , Eigenvalues and, Eigenvectors, CHAPTER CONTENTS, , 5.1 Eigenvalues and Eigenvectors, 5.2 Diagonalization, , 291, , 302, , 5.3 Complex Vector Spaces, 5.4 Differential Equations, , 313, 326, , 5.5 Dynamical Systems and Markov Chains, INTRODUCTION, , 332, , In this chapter we will focus on classes of scalars and vectors known as “eigenvalues”, and “eigenvectors,” terms derived from the German word eigen, meaning “own,”, “peculiar to,” “characteristic,” or “individual.” The underlying idea first appeared in, the study of rotational motion but was later used to classify various kinds of surfaces, and to describe solutions of certain differential equations. In the early 1900s it was, applied to matrices and matrix transformations, and today it has applications in such, diverse fields as computer graphics, mechanical vibrations, heat flow, population, dynamics, quantum mechanics, and economics, to name just a few., , 5.1 Eigenvalues and Eigenvectors, In this section we will define the notions of “eigenvalue” and “eigenvector” and discuss, some of their basic properties., , Definition of Eigenvalue, and Eigenvector, , We begin with the main definition in this section., , A is an n × n matrix, then a nonzero vector x in R n is called an, eigenvector of A (or of the matrix operator TA ) if Ax is a scalar multiple of x; that is,, DEFINITION 1 If, , Ax = λx, for some scalar λ. The scalar λ is called an eigenvalue of A (or of TA ), and x is said, to be an eigenvector corresponding to λ., The requirement that an eigenvector be nonzero is imposed, to avoid the unimportant case, A0 = λ0, which holds for every A and λ., , In general, the image of a vector x under multiplication by a square matrix A differs from x in both magnitude and direction. However, in the special case where x is, an eigenvector of A, multiplication by A leaves the direction unchanged. For example,, in R 2 or R 3 multiplication by A maps each eigenvector x of A (if any) along the same, line through the origin as x. Depending on the sign and magnitude of the eigenvalue λ, 291

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292, , Chapter 5 Eigenvalues and Eigenvectors, , corresponding to x, the operation Ax = λx compresses or stretches x by a factor of λ,, with a reversal of direction in the case where λ is negative (Figure 5.1.1)., λx, x, , x, , x, , x, , λx, 0, , 0, , 0, , 0, , λx, λx, , Figure 5.1.1, , (a) 0 ≤ λ ≤ 1, , (b) λ ≥ 1, , (c) –1 ≤ λ ≤ 0, , (d ) λ ≤ –1, , E X A M P L E 1 Eigenvector of a 2 × 2 Matrix, , The vector x =, , 1, is an eigenvector of, 2, , y, 6, , A=, , 3x, , 3, 8, , 0, , −1, , corresponding to the eigenvalue λ = 3, since, 2, , Ax =, , x, x, 1, , 3, , 3, 8, , 0, , 1, 3, =, = 3x, 2, 6, , −1, , Geometrically, multiplication by A has stretched the vector x by a factor of 3 (Figure, 5.1.2)., , Figure 5.1.2, , Computing Eigenvalues, and Eigenvectors, , Our next objective is to obtain a general procedure for finding eigenvalues and eigenvectors of an n × n matrix A. We will begin with the problem of finding the eigenvalues of A., Note first that the equation Ax = λx can be rewritten as Ax = λI x, or equivalently, as, , (λI − A)x = 0, , Note that if (A)ij = aij , then, formula (1) can be written in, expanded form as, , , , λ − a11 a12 · · · −a1n , , ,  −a21 λ − a22 · · · −a2n , , ,  ., .. , ..,  .., , ., ., , ,  −a, −an2 · · · λ − ann , n1, =0, , For λ to be an eigenvalue of A this equation must have a nonzero solution for x. But, it follows from parts (b) and (g) of Theorem 4.10.2 that this is so if and only if the, coefficient matrix λI − A has a zero determinant. Thus, we have the following result., THEOREM 5.1.1 If, , A is an n × n matrix, then λ is an eigenvalue of A if and only if it, , satisfies the equation, det(λI − A) = 0, , (1), , This is called the characteristic equation of A., , E X A M P L E 2 Finding Eigenvalues, , In Example 1 we observed that λ = 3 is an eigenvalue of the matrix, , A=, , 3, 8, , 0, −1, , but we did not explain how we found it. Use the characteristic equation to find all, eigenvalues of this matrix.

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5.1 Eigenvalues and Eigenvectors, , 293, , Solution It follows from Formula (1) that the eigenvalues of A are the solutions of the, equation det(λI − A) = 0, which we can write as, , , , λ − 3, 0 , ,  −8 λ + 1 = 0, from which we obtain, , (λ − 3)(λ + 1) = 0, , (2), , This shows that the eigenvalues of A are λ = 3 and λ = −1. Thus, in addition to, the eigenvalue λ = 3 noted in Example 1, we have discovered a second eigenvalue, λ = −1., , When the determinant det(λI − A) in (1) is expanded, the characteristic equation, of A takes the form, , λn + c1 λn−1 + · · · + cn = 0, , (3), , where the left side of this equation is a polynomial of degree n in which the coefficient, of λn is 1 (Exercise 37). The polynomial, , p(λ) = λn + c1 λn−1 + · · · + cn, , (4), , is called the characteristic polynomial of A. For example, it follows from (2) that the, characteristic polynomial of the 2 × 2 matrix in Example 2 is, , p(λ) = (λ − 3)(λ + 1) = λ2 − 2λ − 3, which is a polynomial of degree 2., Since a polynomial of degree n has at most n distinct roots, it follows from (3) that, the characteristic equation of an n × n matrix A has at most n distinct solutions and, consequently the matrix has at most n distinct eigenvalues. Since some of these solutions, may be complex numbers, it is possible for a matrix to have complex eigenvalues, even if, that matrix itself has real entries. We will discuss this issue in more detail later, but for, now we will focus on examples in which the eigenvalues are real numbers., , E X A M P L E 3 Eigenvalues of a 3 × 3 Matrix, , Find the eigenvalues of, , ⎡, , 0, ⎢, A = ⎣0, 4, , 1, 0, −17, , ⎤, , 0, ⎥, 1⎦, 8, , Solution The characteristic polynomial of A is, , ⎡, ⎢, , λ, , det(λI − A) = det ⎣ 0, −4, , −1, λ, 17, , ⎤, , 0, ⎥, −1 ⎦ = λ3 − 8λ2 + 17λ − 4, λ−8, , The eigenvalues of A must therefore satisfy the cubic equation, , λ3 − 8λ2 + 17λ − 4 = 0, , (5)

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294, , Chapter 5 Eigenvalues and Eigenvectors, , To solve this equation, we will begin by searching for integer solutions. This task can be, simplified by exploiting the fact that all integer solutions (if there are any) of a polynomial, equation with integer coefficients, , λn + c1 λn−1 + · · · + cn = 0, must be divisors of the constant term, cn . Thus, the only possible integer solutions of (5), are the divisors of −4, that is, ±1, ±2, ±4. Successively substituting these values in (5), shows that λ = 4 is an integer solution and hence that λ − 4 is a factor of the left side, of (5). Dividing λ − 4 into λ3 − 8λ2 + 17λ − 4 shows that (5) can be rewritten as, In applications involving large, matrices it is often not feasible to compute the characteristic equation directly, so other, methods must be used to find, eigenvalues. We will consider, such methods in Chapter 9., , (λ − 4)(λ2 − 4λ + 1) = 0, Thus, the remaining solutions of (5) satisfy the quadratic equation, , λ2 − 4 λ + 1 = 0, which can be solved by the quadratic formula. Thus, the eigenvalues of A are, , λ = 4, λ = 2 +, , √, , 3, and λ = 2 −, , √, , 3, , E X A M P L E 4 Eigenvalues of an Upper Triangular Matrix, , Find the eigenvalues of the upper triangular matrix, , ⎡, , a11, , ⎢0, ⎢, A=⎢, ⎣0, , a12, a22, , 0, , 0, 0, , a13, a23, a33, 0, , ⎤, a14, a24 ⎥, ⎥, ⎥, a34 ⎦, a44, , Solution Recalling that the determinant of a triangular matrix is the product of the, , entries on the main diagonal (Theorem 2.1.2), we obtain, , ⎡, , λ − a11, , ⎢, ⎢, ⎣, , det(λI − A) = det ⎢, , 0, 0, 0, , −a12, λ − a22, 0, 0, , −a13, −a23, λ − a33, 0, , ⎤, −a14, −a24 ⎥, ⎥, ⎥, −a34 ⎦, λ − a44, , = (λ − a11 )(λ − a22 )(λ − a33 )(λ − a44 ), Thus, the characteristic equation is, , (λ − a11 )(λ − a22 )(λ − a33 )(λ − a44 ) = 0, and the eigenvalues are, , λ = a11 , λ = a22 , λ = a33 , λ = a44, which are precisely the diagonal entries of A., , The following general theorem should be evident from the computations in the preceding example., THEOREM 5.1.2 If A is an n × n triangular matrix (upper triangular, lower triangular,, , or diagonal ), then the eigenvalues of A are the entries on the main diagonal of A.

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5.1 Eigenvalues and Eigenvectors, , 295, , E X A M P L E 5 Eigenvalues of a Lower Triangular Matrix, , By inspection, the eigenvalues of the lower triangular matrix, , ⎡, , Had Theorem 5.1.2 been available earlier, we could have anticipated the result obtained in, Example 2., , 1, 2, , ⎤, , 0, , ⎢, A = ⎣−1, , 0, ⎥, 0⎦, , 2, 3, , −8, , 5, , − 41, , are λ = 21 , λ = 23 , and λ = − 41 ., The following theorem gives some alternative ways of describing eigenvalues., THEOREM 5.1.3 If A is an n × n matrix, the following statements are equivalent., , (a) λ is an eigenvalue of A., (b) λ is a solution of the characteristic equation det(λI − A) = 0., (c), , The system of equations (λI − A)x = 0 has nontrivial solutions., , (d ) There is a nonzero vector x such that Ax = λx., , Finding Eigenvectors and, Bases for Eigenspaces, , Now that we know how to find the eigenvalues of a matrix, we will consider the, problem of finding the corresponding eigenvectors. By definition, the eigenvectors of A, corresponding to an eigenvalue λ are the nonzero vectors that satisfy, , (λI − A)x = 0, Notice that x = 0 is in every, eigenspace but is not an eigenvector (see Definition 1). In, the exercises we will ask you to, show that this is the only vector, that distinct eigenspaces have, in common., , Thus, we can find the eigenvectors of A corresponding to λ by finding the nonzero, vectors in the solution space of this linear system. This solution space, which is called, the eigenspace of A corresponding to λ, can also be viewed as:, 1. the null space of the matrix λI − A, 2. the kernel of the matrix operator TλI −A : R n →R n, 3. the set of vectors for which Ax = λx, E X A M P L E 6 Bases for Eigenspaces, , Find bases for the eigenspaces of the matrix, , , , A=, , −1 3, 2, , , , 0, , Historical Note Methods of linear algebra are used in the emerging field of computerized face recognition. Researchers are working, with the idea that every human face in a racial group is a combination of a few dozen primary shapes. For example, by analyzing threedimensional scans of many faces, researchers at Rockefeller University, have produced both an average head shape in the Caucasian group—, dubbed the meanhead (top row left in the figure to the left)—and a set, of standardized variations from that shape, called eigenheads (15 of, which are shown in the picture). These are so named because they are, eigenvectors of a certain matrix that stores digitized facial information., Face shapes are represented mathematically as linear combinations of, the eigenheads., [Image: © Dr. Joseph J. Atick, adapted from Scientific American]

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296, , Chapter 5 Eigenvalues and Eigenvectors, Solution The characteristic equation of A is, , , λ + 1, , ,  −2, , , −3,  = λ(λ + 1) − 6 = (λ − 2)(λ + 3) = 0, λ, , so the eigenvalues of A are λ = 2 and λ = −3. Thus, there are two eigenspaces of A,, one for each eigenvalue., By definition,,  , , x1, x2, , x=, , is an eigenvector of A corresponding to an eigenvalue λ if and only if (λI − A)x = 0,, that is,, ,    , 0, λ + 1 −3 x1, , −2, , λ, , x2, , =, , 0, , In the case where λ = 2 this equation becomes, , , , 3, ,    , x1, 0, =, 2 x2, 0, , −3, , −2, whose general solution is, , x1 = t, x2 = t, (verify). Since this can be written in matrix form as, ,   ,  , x1, t, 1, =, =t, x2, t, 1,  , , it follows that, , 1, 1, , is a basis for the eigenspace corresponding to λ = 2. We leave it for you to follow the, pattern of these computations and show that, , , , − 23, , , , 1, is a basis for the eigenspace corresponding to λ = −3., Figure 5.1.3 illustrates the geometric effect of multiplication by the matrix A in, Example 6. The eigenspace corresponding to λ = 2 is the line L1 through the origin and, the point (1, 1), and the eigenspace corresponding to λ = 3 is the line L2 through the, origin and the point (− 23 , 1). As indicated in the figure, multiplication by A maps each, vector in L1 back into L1 , scaling it by a factor of 2, and it maps each vector in L2 back, into L2 , scaling it by a factor of −3., E X A M P L E 7 Eigenvectors and Bases for Eigenspaces, , Find bases for the eigenspaces of, , ⎡, , 0, A = ⎣1, 1, , 0, 2, 0, , −2, , ⎤, , 1⎦, 3

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5.1 Eigenvalues and Eigenvectors, , 297, , y, , L1, , L2, , (– 32 , 1), , (2, 2), Multiplication, (1, 1), by λ = 2, x, , ( 92 , –3), Multiplication, by λ = –3, , Figure 5.1.3, , A is λ3 − 5λ2 + 8λ − 4 = 0, or in factored, form, (λ − 1)(λ − 2) = 0 (verify). Thus, the distinct eigenvalues of A are λ = 1 and, λ = 2, so there are two eigenspaces of A., Solution The characteristic equation of, 2, , By definition,, , ⎡ ⎤, x1, x = ⎣x2 ⎦, x3, , is an eigenvector of A corresponding to λ if and only if x is a nontrivial solution of, (λI − A)x = 0, or in matrix form,, , ⎡, , 0, λ, ⎣−1 λ − 2, −1, 0, , ⎤⎡ ⎤, , ⎡ ⎤, , 2, 0, x1, ⎦, ⎣, ⎦, ⎣, −1, x2 = 0 ⎦, λ−3, x3, 0, , (6), , In the case where λ = 2, Formula (6) becomes, , ⎡, , 2, ⎣−1, −1, , 0, 0, 0, , ⎤⎡ ⎤, , ⎡ ⎤, , x1, 2, 0, ⎦, ⎣, ⎦, ⎣, −1, x2 = 0⎦, −1, x3, 0, , Solving this system using Gaussian elimination yields (verify), , x1 = −s, x2 = t, x3 = s, Thus, the eigenvectors of A corresponding to λ = 2 are the nonzero vectors of the form, , ⎡, , ⎤ ⎡ ⎤ ⎡ ⎤, ⎡ ⎤, ⎡ ⎤, 0, 0, −s, −s, −1, x = ⎣ t ⎦ = ⎣ 0⎦ + ⎣ t ⎦ = s ⎣ 0⎦ + t ⎣1⎦, 0, s, s, 1, 0, Since, , ⎡, , ⎤, ⎡ ⎤, 0, −1, ⎣ 0⎦ and ⎣1⎦, 1, , 0, , are linearly independent (why?), these vectors form a basis for the eigenspace corresponding to λ = 2.

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298, , Chapter 5 Eigenvalues and Eigenvectors, , If λ = 1, then (6) becomes, , ⎡, , 1, ⎣ −1, −1, , 0, −1, 0, , ⎤⎡ ⎤, , ⎡ ⎤, , x1, 2, 0, −1⎦ ⎣x2 ⎦ = ⎣0⎦, x3, −2, 0, , Solving this system yields (verify), , x1 = −2s, x2 = s, x3 = s, Thus, the eigenvectors corresponding to λ = 1 are the nonzero vectors of the form, , ⎡, , ⎤, ⎡ ⎤, ⎡ ⎤, −2s, −2, −2, ⎣ s ⎦ = s ⎣ 1⎦ so that ⎣ 1⎦, 1, 1, s, , is a basis for the eigenspace corresponding to λ = 1., , Eigenvalues and, Invertibility, , The next theorem establishes a relationship between the eigenvalues and the invertibility, of a matrix., , THEOREM 5.1.4, , A square matrix A is invertible if and only if λ = 0 is not an eigen-, , value of A., , Proof Assume that A is an n × n matrix and observe first that λ, , = 0 is a solution of the, , characteristic equation, , λn + c1 λn−1 + · · · + cn = 0, if and only if the constant term cn is zero. Thus, it suffices to prove that A is invertible, if and only if cn  = 0. But, det(λI − A) = λn + c1 λn−1 + · · · + cn, or, on setting λ = 0,, det(−A) = cn or (−1)n det(A) = cn, It follows from the last equation that det(A) = 0 if and only if cn = 0, and this in turn, implies that A is invertible if and only if cn  = 0., , E X A M P L E 8 Eigenvalues and Invertibility, , The matrix A in Example 7 is invertible since it has eigenvalues λ = 1 and λ = 2, neither of which is zero. We leave it for you to check this conclusion by showing that, det(A)  = 0., , More on the Equivalence, Theorem, , As our final result in this section, we will use Theorem 5.1.4 to add one additional part, to Theorem 4.10.2.

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5.1 Eigenvalues and Eigenvectors, , 299, , THEOREM 5.1.5 Equivalent Statements, , If A is an n × n matrix, then the following statements are equivalent., (a) A is invertible., (b), (c), , Ax = 0 has only the trivial solution., The reduced row echelon form of A is In ., , A is expressible as a product of elementary matrices., Ax = b is consistent for every n × 1 matrix b., ( f ) Ax = b has exactly one solution for every n × 1 matrix b., ( g) det(A)  = 0., (h) The column vectors of A are linearly independent., (d ), (e), , (i ), , The row vectors of A are linearly independent., , ( j), , The column vectors of A span R n ., , (k), , The row vectors of A span R n ., , (l ), , The column vectors of A form a basis for R n ., , (m) The row vectors of A form a basis for R n ., (n), (o), , A has rank n., A has nullity 0., , ( p) The orthogonal complement of the null space of A is R n ., , Eigenvalues of General, LinearTransformations, , (q), , The orthogonal complement of the row space of A is {0}., , (r), , The kernel of TA is {0}., , (s), , The range of TA is R n ., , (t), , TA is one-to-one., , (u), , λ = 0 is not an eigenvalue of A., , Thus far, we have only defined eigenvalues and eigenvectors for matrices and linear, operators on R n . The following definition, which parallels Definition 1, extends this, concept to general vector spaces., , →V is a linear operator on a vector space V , then a nonzero, vector x in V is called an eigenvector of T if T(x) is a scalar multiple of x; that is,, T(x) = λx, for some scalar λ. The scalar λ is called an eigenvalue of T , and x is said to be an, eigenvector corresponding to λ., DEFINITION 2 If T : V, , As with matrix operators, we call the kernel of the operator λI − A the eigenspace of, T corresponding to λ. Stated another way, this is the subspace of all vectors in V for, which T(x) = λx., CA L C U L U S R E Q U I R E D, , In vector spaces of functions, eigenvectors are commonly referred to as eigenfunctions., , E X A M P L E 9 Eigenvalue of a Differentiation Operator, , If D : C ⬁ →C ⬁ is the differentiation operator on the vector space of functions with, continuous derivatives of all orders on the interval (−⬁, ⬁), and if λ is a constant, then, , D(eλx ) = λeλx, so that λ is an eigenvalue of D and eλx is a corresponding eigenvector.

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300, , Chapter 5 Eigenvalues and Eigenvectors, , Exercise Set 5.1, In Exercises 1–4, confirm by multiplication that x is an eigenvector of A, and find the corresponding eigenvalue., 1. A =, , 1, 3, , 2, 1, ; x=, 2, −1, , 2. A =, , 5, 1, , −1, , ; x=, , 3, , ⎡, , 4, ⎢, 3. A = ⎣2, 1, , ⎤, , 0, 3, 0, , ⎡, , 2, ⎢, 4. A = ⎣−1, −1, , 15. T (x, y) = (x + 4y, 2x + 3y), 16. T (x, y, z) = (2x − y − z, x − z, −x + y + 2z), , 1, 1, , 17. (Calculus required ) Let D 2 : C ⬁ (−⬁, ⬁) →C ⬁ (−⬁, ⬁) be the, operator that maps a function into its second derivative., , ⎡ ⎤, , 1, 1, ⎥, ⎢ ⎥, 2⎦ ; x = ⎣2⎦, 4, 1, , −1, 2, , −1, , (a) Show that D 2 is linear., , 1, , In each part of Exercises 5–6, find the characteristic equation,, the eigenvalues, and bases for the eigenspaces of the matrix., , −2, , −7, , 1, , 2, , 5. (a), , 1, 2, , 4, 3, , (b), , (c), , 1, 0, , 0, 1, , (d), , 1, 0, , −2, , 6. (a), , 2, 1, , 1, 2, , (b), , 2, 0, , −3, , (c), , 2, 0, , 0, 2, , (d), , 1, −2, , 4, ⎢, 7. ⎣−2, −2, , ⎡, , 6, ⎢, 9. ⎣0, 1, , ⎡, , 4, , ⎢, 11. ⎣0, 1, , 0, 1, 0, 3, , −2, 0, 0, 3, 0, , ⎤, , 19. (a) Reflection about the line y = x ., , 1, , (b) Orthogonal projection onto the x -axis., (c) Rotation about the origin through a positive angle of 90◦ ., , 2, , ⎡, , 1, ⎥, 0⎦, 1, , −8, , 1, ⎢, 8. ⎣ 0, −2, , (d) Contraction with factor k (0 ≤ k < 1)., , 2, −1, , (e) Shear in the x -direction by a factor k (k  = 0)., , ⎤, ⎥, , 0⎦, −3, , ⎤, , −1, ⎥, 0⎦, 2, , ⎡, , 0, ⎢, 10. ⎣1, 1, , ⎡, , 1, , ⎢, 12. ⎣3, 6, , −2, , 0, 0, 0, , (c) Dilation with factor k (k > 1)., , ⎥, , 0⎦, 4, , (e) Shear in the y -direction by a factor k (k  = 0)., In each part of Exercises 21–22, find the eigenvalues and the, corresponding eigenspaces of the stated matrix operator on R 3 ., Refer to the tables in Section 4.9 and use geometric reasoning to, find the answers. No computations are needed., 21. (a) Reflection about the xy -plane., , ⎤, , (b) Orthogonal projection onto the xz-plane., , 3, ⎥, 3⎦, 4, , (c) Counterclockwise rotation about the positive x -axis, through an angle of 90◦ ., , In Exercises 13–14, find the characteristic equation of the, matrix by inspection., , ⎡, , 3, ⎢, 13. ⎣−2, 4, , 0, 7, 8, , ⎤, , 0, ⎥, 0⎦, 1, , ⎡, , 9, , ⎢0, ⎢, 14. ⎢, ⎣0, 0, , −8, −1, 0, 0, , (b) Rotation about the origin through a positive angle of 180◦ ., (d) Expansion in the y -direction with factor k (k > 1)., , 1, ⎥, 1⎦, 0, , −3, −5, −6, , 20. (a) Reflection about the y -axis., , ⎤, , ⎤, , 1, 0, 1, , 18. (Calculus required ) Let D 2 : C ⬁ →C ⬁ be the linear operator, in Exercise 17. Show that if ω is a positive constant, then, √, √, sinh ωx and cosh ωx are eigenvectors of D 2 , and find their, corresponding eigenvalues., In each part of Exercises 19–20, find the eigenvalues and the, corresponding eigenspaces of the stated matrix operator on R 2 ., Refer to the tables in Section 4.9 and use geometric reasoning to, find the answers. No computations are needed., , In Exercises 7–12, find the characteristic equation, the eigenvalues, and bases for the eigenspaces of the matrix., , ⎡, , √, , (b) Show that if ω is a positive constant, then sin ωx and, √, cos ωx are eigenvectors of D 2 , and find their corresponding eigenvalues., , ⎡ ⎤, ⎤, 1, −1, ⎢ ⎥, ⎥, −1⎦ ; x = ⎣1⎦, 2, , In Exercises 15–16, find the eigenvalues and a basis for each, eigenspace of the linear operator defined by the stated formula., [Suggestion: Work with the standard matrix for the operator.], , 6, 0, 3, 0, , ⎤, , 3, 0⎥, ⎥, ⎥, 0⎦, 7, , (d) Contraction with factor k (0 ≤ k < 1)., 22. (a) Reflection about the xz-plane., (b) Orthogonal projection onto the yz-plane., (c) Counterclockwise rotation about the positive y -axis, through an angle of 180◦ ., (d) Dilation with factor k (k > 1).

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5.1 Eigenvalues and Eigenvectors, , 23. Let A be a 2 × 2 matrix, and call a line through the origin of, R 2 invariant under A if Ax lies on the line when x does. Find, equations for all lines in R 2 , if any, that are invariant under, the given matrix., (a) A =, , 4, 2, , −1, , (b) A =, , 1, , 0, −1, , 1, 0, , 24. Find det(A) given that A has p(λ) as its characteristic polynomial., , 30. Let A be the matrix in Exercise 29. Show that if b  = 0, then, , λ1 =, , 1, 2, , λ2 =, , 1, 2, , and, , %, , (a + d) +, , %, , (a + d) −, , ", , (a − d)2 + 4bc, , ", , (a − d)2 + 4bc, , &, , &, , p(λ) = λ2 + c1 λ + c2, , (a) What is the size of A?, , is the characteristic polynomial of a 2 × 2 matrix, then, , p(A) = A2 + c1 A + c2 I = 0, (Stated informally, A satisfies its characteristic equation. This, result is true as well for n × n matrices.), , (b) Is A invertible?, , 32. Prove: If a , b, c, and d are integers such that a + b = c + d ,, then, , (c) How many eigenspaces does A have?, 26. The eigenvectors that we have been studying are sometimes, called right eigenvectors to distinguish them from left eigenvectors, which are n × 1 column matrices x that satisfy the, equation xTA = μxT for some scalar μ. For a given matrix A,, how are the right eigenvectors and their corresponding eigenvalues related to the left eigenvectors and their corresponding, eigenvalues?, 27. Find a 3 × 3 matrix A that has eigenvalues 1, −1, and 0, and, for which, ⎡ ⎤ ⎡ ⎤ ⎡ ⎤, 1, 1, 1, ⎢ ⎥ ⎢ ⎥ ⎢ ⎥, ⎣−1⎦ , ⎣1⎦ , ⎣−1⎦, 0, 1, 0, are their corresponding eigenvectors., , A=, , a, c, , b, d, , has integer eigenvalues., 33. Prove: If λ is an eigenvalue of an invertible matrix A and x is, a corresponding eigenvector, then 1/λ is an eigenvalue of A−1, and x is a corresponding eigenvector., 34. Prove: If λ is an eigenvalue of A, x is a corresponding eigenvector, and s is a scalar, then λ − s is an eigenvalue of A − sI, and x is a corresponding eigenvector., 35. Prove: If λ is an eigenvalue of A and x is a corresponding, eigenvector, then sλ is an eigenvalue of sA for every scalar s, and x is a corresponding eigenvector., 36. Find the eigenvalues and bases for the eigenspaces of, , ⎡, , Working with Proofs, , −2, ⎢, A = ⎣−2, −4, , 28. Prove that the characteristic equation of a 2 × 2 matrix A can, be expressed as λ2 − tr(A)λ + det(A) = 0, where tr(A) is the, trace of A., , A=, , a, c, , (a) A−1, , b, d, , then the solutions of the characteristic equation of A are, , (a + d) ±, , ", , (a − d)2 + 4bc, , 2, 3, 2, , ⎤, , 3, ⎥, 2⎦, 5, , and then use Exercises 33 and 34 to find the eigenvalues and, bases for the eigenspaces of, , 29. Use the result in Exercise 28 to show that if, , %, , −b, a − λ2, , 31. Use the result of Exercise 28 to prove that if, , 25. Suppose that the characteristic polynomial of some matrix A, is found to be p(λ) = (λ − 1)(λ − 3)2 (λ − 4)3 . In each part,, answer the question and explain your reasoning., , 1, 2, , and x2 =, , are eigenvectors of A that correspond, respectively, to the, eigenvalues, , [Hint: See the proof of Theorem 5.1.4.], , λ=, , −b, a − λ1, , x1 =, , (a) p(λ) = λ3 − 2λ2 + λ + 5, (b) p(λ) = λ4 − λ3 + 7, , 301, , &, , Use this result to show that A has, (a) two distinct real eigenvalues if (a − d)2 + 4bc > 0., (b) two repeated real eigenvalues if (a − d)2 + 4bc = 0., (c) complex conjugate eigenvalues if (a − d)2 + 4bc < 0., , (b) A − 3I, , (c) A + 2I, , 37. Prove that the characteristic polynomial of an n × n matrix A, has degree n and that the coefficient of λn in that polynomial, is 1., 38. (a) Prove that if A is a square matrix, then A and AT have, the same eigenvalues. [Hint: Look at the characteristic, equation det(λI − A) = 0.], (b) Show that A and AT need not have the same eigenspaces., [Hint: Use the result in Exercise 30 to find a 2 × 2 matrix, for which A and AT have different eigenspaces.]

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302, , Chapter 5 Eigenvalues and Eigenvectors, , Working withTechnology, , 39. Prove that the intersection of any two distinct eigenspaces of, a matrix A is {0}., , T1. For the given matrix A, find the characteristic polynomial, and the eigenvalues, and then use the method of Example 7 to find, bases for the eigenspaces., , True-False Exercises, , ⎡, , TF. In parts (a)–(f) determine whether the statement is true or, false, and justify your answer., , −8, , ⎢ 0, ⎢, ⎢, A=⎢, ⎢ 0, ⎢, ⎣ 0, , (a) If A is a square matrix and Ax = λx for some nonzero scalar, λ, then x is an eigenvector of A., (b) If λ is an eigenvalue of a matrix A, then the linear system, (λI − A)x = 0 has only the trivial solution., , 4, , 33, , 38, , 173, , 0, , −1, , −4, , 0, , −5, , −25, , 0, , 1, , 5, , −16, , −19, , −86, , ⎤, −30, ⎥, 0⎥, ⎥, 1⎥, ⎥, ⎥, 0⎦, 15, , T2. The Cayley–Hamilton Theorem states that every square matrix satisfies its characteristic equation; that is, if A is an n × n, matrix whose characteristic equation is, , (c) If the characteristic polynomial of a matrix A is, p(λ) = λ2 + 1, then A is invertible., , λ + c1 λn−1 + · · · + cn = 0, , (d) If λ is an eigenvalue of a matrix A, then the eigenspace of A, corresponding to λ is the set of eigenvectors of A corresponding to λ., , then An + c1 An−1 + · · · + cn = 0., (a) Verify the Cayley–Hamilton Theorem for the matrix, , ⎡, , (e) The eigenvalues of a matrix A are the same as the eigenvalues, of the reduced row echelon form of A., , 0, , ⎢, A = ⎣0, 2, , (f ) If 0 is an eigenvalue of a matrix A, then the set of columns of, A is linearly independent., , ⎤, , 1, , 0, , 0, , 1⎦, , −5, , ⎥, , 4, , (b) Use the result in Exercise 28 to prove the Cayley–Hamilton, Theorem for 2 × 2 matrices., , 5.2 Diagonalization, In this section we will be concerned with the problem of finding a basis for Rn that consists, of eigenvectors of an n × n matrix A. Such bases can be used to study geometric properties, of A and to simplify various numerical computations. These bases are also of physical, significance in a wide variety of applications, some of which will be considered later in this, text., , The Matrix Diagonalization, Problem, , Products of the form P −1AP in which A and P are n × n matrices and P is invertible, will be our main topic of study in this section. There are various ways to think about, such products, one of which is to view them as transformations, , A →P −1AP, in which the matrix A is mapped into the matrix P −1AP . These are called similarity, transformations. Such transformations are important because they preserve many properties of the matrix A. For example, if we let B = P −1AP , then A and B have the same, determinant since, det(B) = det(P −1AP ) = det(P −1 ) det(A) det(P ), , =, , 1, det(A) det(P ) = det(A), det(P )

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5.2 Diagonalization, , 303, , In general, any property that is preserved by a similarity transformation is called a, similarity invariant and is said to be invariant under similarity. Table 1 lists the most, important similarity invariants. The proofs of some of these are given as exercises., , Table 1 Similarity Invariants, Property, , Description, , Determinant, , A and P −1AP have the same determinant., , Invertibility, , A is invertible if and only if P −1AP is invertible., , Rank, , A and P −1AP have the same rank., , Nullity, , A and P −1AP have the same nullity., , Trace, , A and P −1AP have the same trace., , Characteristic polynomial, , A and P −1AP have the same characteristic polynomial., , Eigenvalues, , A and P −1AP have the same eigenvalues., , Eigenspace dimension, , If λ is an eigenvalue of A (and hence of P −1AP ) then the eigenspace, of A corresponding to λ and the eigenspace of P −1AP, corresponding to λ have the same dimension., , We will find the following terminology useful in our study of similarity transformations., DEFINITION 1 If, , A and B are square matrices, then we say that B is similar to A if, there is an invertible matrix P such that B = P −1AP ., , Note that if B is similar to A, then it is also true that A is similar to B since we can, express A as A = Q−1 BQ by taking Q = P −1 . This being the case, we will usually say, that A and B are similar matrices if either is similar to the other., Because diagonal matrices have such a simple form, it is natural to inquire whether, a given n × n matrix A is similar to a matrix of this type. Should this turn out to be, the case, and should we be able to actually find a diagonal matrix D that is similar to, A, then we would be able to ascertain many of the similarity invariant properties of A, directly from the diagonal entries of D . For example, the diagonal entries of D will, be the eigenvalues of A (Theorem 5.1.2), and the product of the diagonal entries of D, will be the determinant of A (Theorem 2.1.2). This leads us to introduce the following, terminology., DEFINITION 2 A square matrix A is said to be diagonalizable if it is similar to some, diagonal matrix; that is, if there exists an invertible matrix P such that P −1AP is, diagonal. In this case the matrix P is said to diagonalize A., , The following theorem and the ideas used in its proof will provide us with a roadmap, for devising a technique for determining whether a matrix is diagonalizable and, if so,, for finding a matrix P that will perform the diagonalization.

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304, , Chapter 5 Eigenvalues and Eigenvectors, , Part (b) of Theorem 5.2.1 is, equivalent to saying that there, is a basis for R n consisting of, eigenvectors of A. Why?, , THEOREM 5.2.1 If A is an n × n matrix, the following statements are equivalent., , (a) A is diagonalizable., (b) A has n linearly independent eigenvectors., Proof (a) ⇒ (b) Since A is assumed to be diagonalizable, it follows that there exist an, invertible matrix P and a diagonal matrix D such that P −1AP = D or, equivalently,, , AP = PD, , (1), , If we denote the column vectors of P by p1 , p2 , . . . , pn , and if we assume that the, diagonal entries of D are λ1 , λ2 , . . . , λn , then by Formula (6) of Section 1.3 the left side, of (1) can be expressed as, , AP = A[p1 p2, , · · · pn ] = [Ap1 Ap2, , · · · Apn ], , and, as noted in the comment following Example 1 of Section 1.7, the right side of (1), can be expressed as, PD = [λ1 p1 λ2 p2 · · · λn pn ], Thus, it follows from (1) that, , Ap1 = λ1 p1 , Ap2 = λ2 p2 , . . . , Apn = λn pn, , (2), , Since P is invertible, we know from Theorem 5.1.5 that its column vectors p1 , p2 , . . . , pn, are linearly independent (and hence nonzero). Thus, it follows from (2) that these n, column vectors are eigenvectors of A., Proof (b) ⇒ (a) Assume that A has n linearly independent eigenvectors, p1 , p2 , . . . , pn ,, and that λ1 , λ2 , . . . , λn are the corresponding eigenvalues. If we let, , P = [p1 p2, , · · · pn ], , and if we let D be the diagonal matrix that has λ1 , λ2 , . . . , λn as its successive diagonal, entries, then, , AP = A[p1 p2 · · · pn ] = [Ap1 Ap2, = [λ1 p1 λ2 p2 · · · λn pn ] = PD, , · · · Apn ], , Since the column vectors of P are linearly independent, it follows from Theorem 5.1.5, that P is invertible, so that this last equation can be rewritten as P −1AP = D , which, shows that A is diagonalizable., Whereas Theorem 5.2.1 tells us that we need to find n linearly independent eigenvectors to diagonalize a matrix, the following theorem tells us where such vectors might, be found. Part (a) is proved at the end of this section, and part (b) is an immediate, consequence of part (a) and Theorem 5.2.1 (why?)., THEOREM 5.2.2, , (a) If λ1 , λ2 , . . . , λk are distinct eigenvalues of a matrix A, and if v1 , v2 , . . . , vk are, corresponding eigenvectors, then {v1 , v2 , . . . , vk } is a linearly independent set., (b) An n × n matrix with n distinct eigenvalues is diagonalizable., Remark Part (a) of Theorem 5.2.2 is a special case of a more general result: Specifically, if, λ1 , λ2 , . . . , λk are distinct eigenvalues, and if S1 , S2 , . . . , Sk are corresponding sets of linearly, independent eigenvectors, then the union of these sets is linearly independent.

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5.2 Diagonalization, , Procedure for, Diagonalizing a Matrix, , 305, , Theorem 5.2.1 guarantees that an n × n matrix A with n linearly independent eigenvectors is diagonalizable, and the proof of that theorem together with Theorem 5.2.2, suggests the following procedure for diagonalizing A., A Procedure for Diagonalizing an n × n Matrix, Step 1. Determine first whether the matrix is actually diagonalizable by searching for, n linearly independent eigenvectors. One way to do this is to find a basis for, each eigenspace and count the total number of vectors obtained. If there is, a total of n vectors, then the matrix is diagonalizable, and if the total is less, than n, then it is not., Step 2. If you ascertained that the matrix is diagonalizable, then form the matrix, P = [p1 p2 · · · pn ] whose column vectors are the n basis vectors you obtained in Step 1., Step 3. P −1AP will be a diagonal matrix whose successive diagonal entries are the, eigenvalues λ1 , λ2 , . . . , λn that correspond to the successive columns of P ., , E X A M P L E 1 Finding a Matrix P That Diagonalizes a Matrix A, , Find a matrix P that diagonalizes, , ⎡, , 0, ⎣, A= 1, 1, , −2, , 0, 2, 0, , ⎤, , 1⎦, 3, , Solution In Example 7 of the preceding section we found the characteristic equation of, , A to be, , (λ − 1)(λ − 2)2 = 0, , and we found the following bases for the eigenspaces:, , ⎡, , −1, , ⎤, , ⎡ ⎤, , ⎡, , −2, , 0, λ = 2: p1 = ⎣ 0⎦, p2 = ⎣1⎦ ;, 1, 0, , λ = 1: p3 = ⎣ 1⎦, 1, , There are three basis vectors in total, so the matrix, , ⎡, , −1, , P =⎣ 0, 1, , 0, 1, 0, , −2, , ⎤, , 1⎦, 1, , diagonalizes A. As a check, you should verify that, , ⎡, , 1, P −1AP = ⎣ 1, −1, , 0, 1, 0, , ⎤⎡, , 2, 0, 1⎦ ⎣1, 1, −1, , 0, 2, 0, , −2, , ⎤, , ⎤⎡, , −1, , 1⎦ ⎣ 0, 3, 1, , 0, 1, 0, , −2, , ⎤, , ⎡, , 2, 1⎦ = ⎣0, 1, 0, , 0, 2, 0, , ⎤, , 0, 0⎦, 1, , In general, there is no preferred order for the columns of P . Since the i th diagonal, entry of P −1AP is an eigenvalue for the i th column vector of P , changing the order of, the columns of P just changes the order of the eigenvalues on the diagonal of P −1AP ., Thus, had we written, ⎡, ⎤, −1 −2, 0, 1, 1⎦, P =⎣ 0, 1, 1, 0

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306, , Chapter 5 Eigenvalues and Eigenvectors, , in the preceding example, we would have obtained, , ⎡, , 2, ⎢, P −1AP = ⎣0, 0, , ⎤, , 0, 1, 0, , 0, ⎥, 0⎦, 2, , E X A M P L E 2 A Matrix That Is Not Diagonalizable, , Show that the following matrix is not diagonalizable:, , ⎡, , 1, ⎢, A=⎣ 1, −3, , 0, 2, 5, , ⎤, , 0, ⎥, 0⎦, 2, , Solution The characteristic polynomial of A is, , , λ−1, , , det(λI − A) =  −1, ,  3, , 0, , λ−2, −5, , , , , ,  = (λ − 1)(λ − 2)2, , λ−2, 0, 0, , so the characteristic equation is, , (λ − 1)(λ − 2)2 = 0, and the distinct eigenvalues of A are λ = 1 and λ = 2. We leave it for you to show that, bases for the eigenspaces are, , ⎡, , λ = 1: p1 =, , ⎤, , 1, ⎢ 81 ⎥, ⎢− ⎥ ;, ⎣ 8⎦, , ⎡ ⎤, 0, , ⎢ ⎥, λ = 2: p2 = ⎣0⎦, 1, , 1, , Since A is a 3 × 3 matrix and there are only two basis vectors in total, A is not diagonalizable., Alternative Solution If you are concerned only in determining whether a matrix is di-, , agonalizable and not with actually finding a diagonalizing matrix P , then it is not necessary to compute bases for the eigenspaces—it suffices to find the dimensions of the, eigenspaces. For this example, the eigenspace corresponding to λ = 1 is the solution, space of the system, ⎡, ⎤⎡ ⎤ ⎡ ⎤, 0, 0, 0, 0, x1, ⎢, ⎥⎢ ⎥ ⎢ ⎥, 0⎦ ⎣x2 ⎦ = ⎣0⎦, ⎣−1 −1, 0, 3 −5 −1, x3, Since the coefficient matrix has rank 2 (verify), the nullity of this matrix is 1 by Theorem 4.8.2, and hence the eigenspace corresponding to λ = 1 is one-dimensional., The eigenspace corresponding to λ = 2 is the solution space of the system, , ⎡, , 1, ⎢, ⎣−1, 3, , 0, 0, −5, , ⎤⎡ ⎤, , ⎡ ⎤, , 0, 0, x1, ⎥⎢ ⎥ ⎢ ⎥, 0⎦ ⎣x2 ⎦ = ⎣0⎦, 0, x3, 0, , This coefficient matrix also has rank 2 and nullity 1 (verify), so the eigenspace corresponding to λ = 2 is also one-dimensional. Since the eigenspaces produce a total of two, basis vectors, and since three are needed, the matrix A is not diagonalizable.

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5.2 Diagonalization, , 307, , E X A M P L E 3 Recognizing Diagonalizability, , We saw in Example 3 of the preceding section that, , ⎡, , 0, ⎢, A = ⎣0, 4, , 1, 0, −17, , has three distinct eigenvalues: λ = 4, λ = 2 +, diagonalizable and, ⎡, 4, 0, , √, , √, , ⎢, P −1AP = ⎣0, , 2+, , 0, , 0, , ⎤, , 0, ⎥, 1⎦, 8, 3, and λ = 2 −, 0, 3, , 0, 2−, , √, , √, , 3. Therefore, A is, , ⎤, ⎥, ⎦, 3, , for some invertible matrix P . If needed, the matrix P can be found using the method, shown in Example 1 of this section., E X A M P L E 4 Diagonalizability of Triangular Matrices, , From Theorem 5.1.2, the eigenvalues of a triangular matrix are the entries on its main, diagonal. Thus, a triangular matrix with distinct entries on the main diagonal is diagonalizable. For example,, ⎤, ⎡, 2, 4, 0, −1, ⎢ 0, 3, 1, 7⎥, ⎥, ⎢, A=⎢, ⎥, 0, 5, 8⎦, ⎣ 0, 0, 0, 0 −2, is a diagonalizable matrix with eigenvalues λ1 = −1, λ2 = 3, λ3 = 5, λ4 = −2., Eigenvalues of Powers of a, Matrix, , Since there are many applications in which it is necessary to compute high powers of a, square matrix A, we will now turn our attention to that important problem. As we will, see, the most efficient way to compute Ak , particularly for large values of k , is to first, diagonalize A. But because diagonalizing a matrix A involves finding its eigenvalues and, eigenvectors, we will need to know how these quantities are related to those of Ak . As an, illustration, suppose that λ is an eigenvalue of A and x is a corresponding eigenvector., Then, A2 x = A(Ax) = A(λx) = λ(Ax) = λ(λx) = λ2 x, which shows not only that λ2 is a eigenvalue of A2 but that x is a corresponding eigenvector. In general, we have the following result., , Note that diagonalizability is, not a requirement in Theorem 5.2.3., , THEOREM 5.2.3, , If k is a positive integer, λ is an eigenvalue of a matrix A, and x is, a corresponding eigenvector, then λk is an eigenvalue of Ak and x is a corresponding, eigenvector., , E X A M P L E 5 Eigenvalues and Eigenvectors of Matrix Powers, , In Example 2 we found the eigenvalues and corresponding eigenvectors of the matrix, , ⎡, , 1, ⎢, A=⎣ 1, −3, Do the same for A7 ., , 0, 2, 5, , ⎤, , 0, ⎥, 0⎦, 2

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308, , Chapter 5 Eigenvalues and Eigenvectors, , A are λ = 1 and λ = 2, so, the eigenvalues of A7 are λ = 17 = 1 and λ = 27 = 128. The eigenvectors p1 and p2, obtained in Example 1 corresponding to the eigenvalues λ = 1 and λ = 2 of A are also, the eigenvectors corresponding to the eigenvalues λ = 1 and λ = 128 of A7 ., Solution We know from Example 2 that the eigenvalues of, , Computing Powers of a, Matrix, , The problem of computing powers of a matrix is greatly simplified when the matrix is, diagonalizable. To see why this is so, suppose that A is a diagonalizable n × n matrix,, that P diagonalizes A, and that, , ⎡, , λ1, , ⎢0, ⎢, P −1AP = ⎢ .., ⎣., , λ2, .., ., , 0, , 0, , ⎤, , ···, ···, , 0, , 0, 0⎥, ⎥, , .. ⎥ = D, .⎦, · · · λn, , Squaring both sides of this equation yields, , ⎡, , λ21, , ⎢0, ⎢, (P −1AP )2 = ⎢ .., ⎣., , λ22, .., ., , 0, , 0, , ⎤, , ···, ···, , 0, , 0, 0⎥, ⎥, , .. ⎥ = D 2, .⎦, · · · λ2n, , We can rewrite the left side of this equation as, , (P −1AP )2 = P −1APP −1AP = P −1 AIAP = P −1A2 P, from which we obtain the relationship P −1A2P = D 2 . More generally, if k is a positive, integer, then a similar computation will show that, , ⎡, , λk1, , ⎢0, ⎢, P −1AkP = D k = ⎢ .., ⎣., , λ2k, .., ., , 0, , 0, , which we can rewrite as, , ⎡, , λ1k, , Formula (3) reveals that raising a diagonalizable matrix A, to a positive integer power has, the effect of raising its eigenvalues to that power., , ···, ···, , 0, , .. ⎥, .⎦, · · · λkn, , λk2, .., ., , 0, , 0, , ⎤, , ···, ···, , 0, , ⎢0, ⎢, Ak = PD kP −1 = P ⎢ .., ⎣., , ⎤, , 0, 0⎥, ⎥, , 0, 0⎥, ⎥, , .. ⎥ P −1, .⎦, · · · λkn, , E X A M P L E 6 Powers of a Matrix, , Use (3) to find A13 , where, , ⎡, , 0, ⎢, A = ⎣1, 1, , 0, 2, 0, , −2, , ⎤, ⎥, , 1⎦, 3, , Solution We showed in Example 1 that the matrix A is diagonalized by, , ⎡, −1, ⎢, P =⎣ 0, 1, , and that, , 0, 1, 0, , ⎡, , 2, ⎢, −1, D = P AP = ⎣0, 0, , −2, , ⎤, ⎥, , 1⎦, 1, 0, 2, 0, , ⎤, , 0, ⎥, 0⎦, 1, , (3)

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5.2 Diagonalization, , Thus, it follows from (3) that, , ⎡, , −1, , 0, 1, 0, , ⎢, A13 = PD 13 P −1 = ⎣ 0, 1, , ⎡, , −8190, , ⎢, = ⎣ 8191, 8191, , −2, , ⎤⎡, , 213, ⎥⎢, 1⎦ ⎣0, 1, 0, , 0, 8192, 0, , 0, 213, 0, , −16382, , ⎤, , ⎤⎡, , 0, 1, ⎥⎢, 0 ⎦⎣ 1, 113, −1, , 0, 1, 0, , 309, , ⎤, , 2, ⎥, 1⎦, −1, , (4), , ⎥, , 8191 ⎦, 16383, , Remark With the method in the preceding example, most of the work is in diagonalizing A., Once that work is done, it can be used to compute any power of A. Thus, to compute A1000 we, need only change the exponents from 13 to 1000 in (4)., Geometric and Algebraic, Multiplicity, , Theorem 5.2.2(b) does not completely settle the diagonalizability question since it only, guarantees that a square matrix with n distinct eigenvalues is diagonalizable; it does not, preclude the possibility that there may exist diagonalizable matrices with fewer than n, distinct eigenvalues. The following example shows that this is indeed the case., E X A M P L E 7 The Converse of Theorem 5.2.2(b) Is False, , Consider the matrices, , ⎡, , 1, ⎢, I = ⎣0, 0, , 0, 1, 0, , ⎤, , ⎡, , 0, 1, ⎥, ⎢, 0⎦ and J = ⎣0, 1, 0, , 1, 1, 0, , ⎤, , 0, ⎥, 1⎦, 1, , It follows from Theorem 5.1.2 that both of these matrices have only one distinct eigenvalue, namely λ = 1, and hence only one eigenspace. We leave it as an exercise for you, to solve the characteristic equations, , (λI − I )x = 0 and (λI − J )x = 0, with λ = 1 and show that for I the eigenspace is three-dimensional (all of R 3 ) and for J, it is one-dimensional, consisting of all scalar multiples of, , ⎡ ⎤, , 1, ⎢ ⎥, x = ⎣0⎦, 0, This shows that the converse of Theorem 5.2.2(b) is false, since we have produced two, 3 × 3 matrices with fewer than three distinct eigenvalues, one of which is diagonalizable, and the other of which is not., A full excursion into the study of diagonalizability is left for more advanced courses,, but we will touch on one theorem that is important for a fuller understanding of diagonalizability. It can be proved that if λ0 is an eigenvalue of A, then the dimension of the, eigenspace corresponding to λ0 cannot exceed the number of times that λ − λ0 appears, as a factor of the characteristic polynomial of A. For example, in Examples 1 and 2 the, characteristic polynomial is, (λ − 1)(λ − 2)2, Thus, the eigenspace corresponding to λ = 1 is at most (hence exactly) one-dimensional,, and the eigenspace corresponding to λ = 2 is at most two-dimensional. In Example 1

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5.2 Diagonalization, , 311, , Since the eigenvector vr+1 is nonzero, it follows that, , cr+1 = 0, , (8), , But equations (7) and (8) contradict the fact that c1 , c2 , . . . , cr+1 are not all zero so the, proof is complete., , Exercise Set 5.2, In Exercises 1–4, show that A and B are not similar matrices., 1, 1. A =, 3, 2. A =, , 4, 2, , ⎡, , 1, ⎢, 3. A = ⎣0, 0, , ⎡, , 1, ⎢, 4. A = ⎣2, 3, , 1, 1, ,B =, 2, 3, , −1, , 4, 2, , ,B =, , 4, , 0, −2, , 1, , 2, 1, 0, , 3, ⎢1, ⎥, 2 ⎦, B = ⎢, ⎣2, 1, 0, , 0, 0, 0, , 1, 1, ⎥, ⎢, 2⎦, B = ⎣2, 3, 0, , ⎤, , ⎡, , 0, , 1, , 0⎥, ⎦, , 0, , 1, , 1, 2, 1, , 0, ⎥, 0⎦, 1, , 5. A =, , 1, , 0, , 6, , −1, , ⎡, , −2, , 2, , 0, , 7. A = ⎣0, , 3, , 0⎦, , 0, , 0, , 3, , ⎢, , −14, −20, , 12, , , , 17, , ⎤, , 1, , 0, , 0, , 8. A = ⎣0, , 1, , 1⎦, , 0, , 1, , 1, , ⎡, , 4, , 0, , ⎢, A = ⎣2, 1, , 1, , ⎥, , 3, , ⎥, 2⎦, , 0, , 4, , (b) For each eigenvalue λ, find the rank of the matrix λI − A., (c) Is A diagonalizable? Justify your conclusion., 10. Follow the directions in Exercise 9 for the matrix, 3, , ⎢, ⎣0, 0, , ⎤, , 0, , 0, , 2, , 0⎦, , 1, , 2, , 3, , 17, , ⎤, , 0, , 0, , 0, , 0, , 0⎦, , 3, , 0, , 1, , ⎡, , ⎥, , 5, , 0, , 14. A = ⎣1, , 5, , 0⎦, , 0, , 1, , 5, , ⎢, , 0, , ⎥, , In Exercises 17–18, use the method of Example 6 to compute, the matrix A10 ., , , 17. A =, , ⎥, , 0, , 3, , 2, , −1, , 19. Let, , ⎡, , , , , 18. A =, , −1, , 1, , 0, , 15, , 0, , −1, , 2, , ⎤, , 1, , 1, , 0, , 1⎦, , 0, , 5, , ⎥, , Confirm that P diagonalizes A, and then compute A11 ., 20. Let, , ⎡, , ⎤, , ⎡, , ⎥, , ⎢, , ⎢, A = ⎣0, , −2, −1, , 8, , 0, , 0, , −1, , 1, , ⎤, , 1, , −4, , 0⎦ and P = ⎣1, , 0, , 0⎦, , 1, , 0, , 0, , 1, , ⎥, , Confirm that P diagonalizes A, and then compute each of the, following powers of A., (a) A1000, , (b) A−1000, , (d) A−2301, , (c) A2301, , 21. Find An if n is a positive integer and, , ⎡, , In Exercises 11–14, find the geometric and algebraic multiplicity of each eigenvalue of the matrix A, and determine whether A, is diagonalizable. If A is diagonalizable, then find a matrix P that, diagonalizes A, and find P −1AP ., , , , 1, , ⎡, ⎤, 1, −1, ⎢, ⎥, 0⎦ and P = ⎣0, 1, −2, , 7, , ⎢, A=⎣ 0, , ⎤, , (a) Find the eigenvalues of A., , ⎡, , 1, , (b) λ3 − 3λ2 + 3λ − 1 = 0, , ⎢, , 9. Let, , 12. A = ⎣25, , 16. (a) λ3 (λ2 − 5λ − 6) = 0, , ⎡, , ⎥, , ⎢, , ⎥, , 0⎦, , 4, , ⎤, −9 −6, ⎥, −11 −9⎦, −9 −4, ⎤, , (b) λ2 (λ − 1)(λ − 2)3 = 0, , , , ⎤, , 19, , 15. (a) (λ − 1)(λ + 3)(λ − 5) = 0, , ⎤, , 6. A =, , ⎡, , ⎤, , In each part of Exercises 15–16, the characteristic equation of, a matrix A is given. Find the size of the matrix and the possible, dimensions of its eigenspaces., , ⎥, , , , −2, , 4, , 13. A = ⎣0, , ⎤, , 2, , In Exercises 5–8, find a matrix P that diagonalizes A, and, check your work by computing P −1AP ., , , , −1, ⎢, 11. A = ⎣−3, −3, ⎡, ⎢, , 1, 4, , ⎡, , ⎤, , ⎡, , 3, , ⎢, A = ⎣−1, 0, , −1, 2, , −1, , 0, , ⎤, , ⎥, −1⎦, 3

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312, , Chapter 5 Eigenvalues and Eigenvectors, , 22. Show that the matrices, , ⎡, , 1, , ⎢, A = ⎣1, 1, , ⎤, , ⎡, , ⎥, , ⎢, , ⎤, , 1, , 1, , 3, , 0, , 0, , 1, , 1⎦ and B = ⎣0, 1, 0, , 0, , 0⎦, , 0, , 0, , 1, , ⎥, , are similar., 23. We know from Table 1 that similar matrices have the same, rank. Show that the converse is false by showing that the, matrices, , , , , 1 0, 0 1, and B =, A=, 0 0, 0 0, have the same rank but are not similar. [Suggestion: If they, were similar, then there would be an invertible 2 × 2 matrix P, for which AP = PB . Show that there is no such matrix.], 24. We know from Table 1 that similar matrices have the same, eigenvalues. Use the method of Exercise 23 to show that the, converse is false by showing that the matrices, , , A=, , 1, , 1, , 0, , 1, , , , , and B =, , 1, , 0, , 0, , 1, , , , have the same eigenvalues but are not similar., 25. If A, B , and C are n × n matrices such that A is similar to B, and B is similar to C , do you think that A must be similar to, C ? Justify your answer., 26. (a) Is it possible for an n × n matrix to be similar to itself ?, Justify your answer., (b) What can you say about an n × n matrix that is similar to, 0n×n ? Justify your answer., (c) Is is possible for a nonsingular matrix to be similar to a, singular matrix? Justify your answer., 27. Suppose that the characteristic polynomial of some matrix A, is found to be p(λ) = (λ − 1)(λ − 3)2 (λ − 4)3 . In each part,, answer the question and explain your reasoning., (a) What can you say about the dimensions of the eigenspaces, of A?, (b) What can you say about the dimensions of the eigenspaces, if you know that A is diagonalizable?, (c) If {v1 , v2 , v3 } is a linearly independent set of eigenvectors, of A, all of which correspond to the same eigenvalue of A,, what can you say about that eigenvalue?, 28. Let, , a, A=, c, , b, d, , Show that, (a) A is diagonalizable if (a − d)2 + 4bc > 0., (b) A is not diagonalizable if (a − d)2 + 4bc < 0., [Hint: See Exercise 29 of Section 5.1.], , 29. In the case where the matrix A in Exercise 28 is diagonalizable,, find a matrix P that diagonalizes A. [Hint: See Exercise 30 of, Section 5.1.], In Exercises 30–33, find the standard matrix A for the given linear operator, and determine whether that matrix is diagonalizable., If diagonalizable, find a matrix P that diagonalizes A., 30. T (x1 , x2 ) = (2x1 − x2 , x1 + x2 ), 31. T (x1 , x2 ) = (−x2 , −x1 ), 32. T (x1 , x2 , x3 ) = (8x1 + 3x2 − 4x3 , −3x1 + x2 + 3x3 ,, 4x1 + 3x2 ), 33. T (x1 , x2 , x3 ) = (3x1 , x2 , x1 − x2 ), 34. If P is a fixed n × n matrix, then the similarity transformation, , A →P −1AP, can be viewed as an operator SP (A) = P −1AP on the vector, space Mnn of n × n matrices., (a) Show that SP is a linear operator., (b) Find the kernel of SP ., (c) Find the rank of SP ., , Working with Proofs, 35. Prove that similar matrices have the same rank and nullity., 36. Prove that similar matrices have the same trace., 37. Prove that if A is diagonalizable, then so is Ak for every positive, integer k., 38. We know from Table 1 that similar matrices, A and B , have, the same eigenvalues. However, it is not true that those eigenvalues have the same corresponding eigenvectors for the two, matrices. Prove that if B = P −1AP , and v is an eigenvector of, B corresponding to the eigenvalue λ, then P v is the eigenvector of A corresponding to λ., 39. Let A be an n × n matrix, and let q(A) be the matrix, , q(A) = an An + an−1 An−1 + · · · + a1 A + a0 In, (a) Prove that if B = P −1AP , then q(B) = P −1 q(A)P ., (b) Prove that if A is diagonalizable, then so is q(A)., 40. Prove that if A is a diagonalizable matrix, then the rank of A, is the number of nonzero eigenvalues of A., 41. This problem will lead you through a proof of the fact that, the algebraic multiplicity of an eigenvalue of an n × n matrix, A is greater than or equal to the geometric multiplicity. For, this purpose, assume that λ0 is an eigenvalue with geometric, multiplicity k., (a) Prove that there is a basis B = {u1 , u2 , . . . , un } for R n, in which the first k vectors of B form a basis for the, eigenspace corresponding to λ0 .

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5.3 Complex Vector Spaces, , (b) Let P be the matrix having the vectors in B as columns. Prove that the product AP can be expressed as, , AP = P, , λ0 Ik, , X, Y, , 0, , (c) Use the result in part (b) to prove that A is similar to, , C=, , 0, , (g) If there is a basis for R n consisting of eigenvectors of an n × n, matrix A, then A is diagonalizable., (h) If every eigenvalue of a matrix A has algebraic multiplicity 1,, then A is diagonalizable., , [Hint: Compare the first k column vectors on both sides.], , λ0 Ik, , 313, , X, Y, , and hence that A and C have the same characteristic polynomial., (d) By considering det(λI − C), prove that the characteristic polynomial of C (and hence A) contains the factor, (λ − λ0 ) at least k times, thereby proving that the algebraic, multiplicity of λ0 is greater than or equal to the geometric, multiplicity k., , (i) If 0 is an eigenvalue of a matrix A, then A2 is singular., , Working withTechnology, T1. Generate a random 4 × 4 matrix A and an invertible 4 × 4, matrix P and then confirm, as stated in Table 1, that P −1AP and, A have the same, (a) determinant., (b) rank., (c) nullity., (d) trace., (e) characteristic polynomial., (f ) eigenvalues., , True-False Exercises, TF. In parts (a)–(i) determine whether the statement is true or, false, and justify your answer., , T2. (a) Use Theorem 5.2.1 to show that the following matrix is, diagonalizable., , ⎡, , (a) An n × n matrix with fewer than n distinct eigenvalues is not, diagonalizable., , −13, , ⎢, A = ⎣ 10, −5, , −60, , −60, , ⎤, ⎥, , 40⎦, , 42, , −20, , −18, , (b) An n × n matrix with fewer than n linearly independent eigenvectors is not diagonalizable., , (b) Find a matrix P that diagonalizes A., , (c) If A and B are similar n × n matrices, then there exists an, invertible n × n matrix P such that PA = BP ., , (c) Use the method of Example 6 to compute A10 , and check your, result by computing A10 directly., , (d) If A is diagonalizable, then there is a unique matrix P such, that P −1AP is diagonal., , T3. Use Theorem 5.2.1 to show that the following matrix is not, diagonalizable., , (e) If A is diagonalizable and invertible, then A−1 is diagonalizable., , ⎤, , ⎡, −10, ⎢, A = ⎣−15, , 11, 16, , ⎥, −10⎦, , −3, , 3, , −2, , (f ) If A is diagonalizable, then AT is diagonalizable., , −6, , 5.3 Complex Vector Spaces, Because the characteristic equation of any square matrix can have complex solutions, the, notions of complex eigenvalues and eigenvectors arise naturally, even within the context of, matrices with real entries. In this section we will discuss this idea and apply our results to, study symmetric matrices in more detail. A review of the essentials of complex numbers, appears in the back of this text., , Review of Complex, Numbers, , Recall that if z = a + bi is a complex number, then:, • Re(z) = a and Im(z) = b are called the real part of z and the imaginary part of z,, respectively,, • |z| =, , √, a 2 + b2 is called the modulus (or absolute value) of z,, , • z = a − bi is called the complex conjugate of z,

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CHAPTER, , 6, , Inner Product Spaces, CHAPTER CONTENTS, , 6.1 Inner Products, , 345, , 6.2 Angle and Orthogonality in Inner Product Spaces, 6.3 Gram–Schmidt Process; QR-Decomposition, 6.4 Best Approximation; Least Squares, , 378, , 6.5 Mathematical Modeling Using Least Squares, 6.6 Function Approximation; Fourier Series, INTRODUCTION, , 355, , 364, , 387, , 394, , In Chapter 3 we defined the dot product of vectors in R n , and we used that concept to, define notions of length, angle, distance, and orthogonality. In this chapter we will, generalize those ideas so they are applicable in any vector space, not just R n . We will, also discuss various applications of these ideas., , 6.1 Inner Products, In this section we will use the most important properties of the dot product on R n as, axioms, which, if satisfied by the vectors in a vector space V, will enable us to extend the, notions of length, distance, angle, and perpendicularity to general vector spaces., , General Inner Products, , Note that Definition 1 applies, only to real vector spaces. A, definition of inner products on, complex vector spaces is given, in the exercises. Since we will, have little need for complex, vector spaces from this point, on, you can assume that all, vector spaces under discussion, are real, even though some of, the theorems are also valid in, complex vector spaces., , In Definition 4 of Section 3.2 we defined the dot product of two vectors in R n , and in, Theorem 3.2.2 we listed four fundamental properties of such products. Our first goal, in this section is to extend the notion of a dot product to general real vector spaces by, using those four properties as axioms. We make the following definition., DEFINITION 1 An inner product on a real vector space V is a function that associates, a real number u, v with each pair of vectors in V in such a way that the following, axioms are satisfied for all vectors u, v, and w in V and all scalars k ., , 1. u, v = v, u, , [ Symmetry axiom ], , 2. u + v, w = u, w + v, w, , [ Additivity axiom ], , 3. k u, v = ku, v, , [ Homogeneity axiom ], , 4. v, v ≥ 0 and v, v = 0 if and only if v = 0 [ Positivity axiom ], A real vector space with an inner product is called a real inner product space., Because the axioms for a real inner product space are based on properties of the dot, product, these inner product space axioms will be satisfied automatically if we define the, inner product of two vectors u and v in R n to be, , u, v = u · v = u1 v1 + u2 v2 + · · · + un vn, , (1), 345

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346, , Chapter 6 Inner Product Spaces, , This inner product is commonly called the Euclidean inner product (or the standard inner, product) on R n to distinguish it from other possible inner products that might be defined, on R n . We call R n with the Euclidean inner product Euclidean n-space., Inner products can be used to define notions of norm and distance in a general inner, product space just as we did with dot products in R n . Recall from Formulas (11) and (19), of Section 3.2 that if u and v are vectors in Euclidean n-space, then norm and distance, can be expressed in terms of the dot product as, v =, , √, , v · v and d(u, v) = u − v =, , ", (u − v) · (u − v), , Motivated by these formulas, we make the following definition., DEFINITION 2 If V is a real inner product space, then the norm (or length) of a vector, v in V is denoted by v and is defined by, , v =, , ", v, v, , and the distance between two vectors is denoted by d(u, v) and is defined by, , d(u, v) = u − v =, , ", u − v, u − v, , A vector of norm 1 is called a unit vector., The following theorem, whose proof is left for the exercises, shows that norms and, distances in real inner product spaces have many of the properties that you might expect., THEOREM 6.1.1 If u and v are vectors in a real inner product space, , V, and if k is a, , scalar, then:, (a), , v ≥ 0 with equality if and only if v = 0., , (b), , k v = |k| v ., , (c), , d(u, v) = d(v, u)., , (d ) d(u, v) ≥ 0 with equality if and only if u = v., Although the Euclidean inner product is the most important inner product on R n ,, there are various applications in which it is desirable to modify it by weighting each term, differently. More precisely, if, , w1 , w2 , . . . , wn, are positive real numbers, which we will call weights, and if u = (u1 , u2 , . . . , un ) and, v = (v1 , v2 , . . . , vn ) are vectors in R n , then it can be shown that the formula, , u, v = w1 u1 v1 + w2 u2 v2 + · · · + wn un vn, , (2), , defines an inner product on R n that we call the weighted Euclidean inner product with, weights w1 , w2 , . . . , wn ., Note that the standard Euclidean inner product in Formula (1) is the special case, of the weighted Euclidean inner product in which all the, weights are 1., , E X A M P L E 1 Weighted Euclidean Inner Product, , Let u = (u1 , u2 ) and v = (v1 , v2 ) be vectors in R 2 . Verify that the weighted Euclidean, inner product, (3), u, v = 3u1 v1 + 2u2 v2, satisfies the four inner product axioms.

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348, , Chapter 6 Inner Product Spaces, , and, , Unit Circles and Spheres in, Inner Product Spaces, , d(u, v) = u − v = (1, −1), (1, −1)1/2, √, = [3(1)(1) + 2(−1)(−1)]1/2 = 5, , DEFINITION 3 If V is an inner product space, then the set of points in V that satisfy, , u =1, is called the unit sphere or sometimes the unit circle in V ., , y, , E X A M P L E 3 Unusual Unit Circles in R 2, , ||u|| = 1, x, 1, , (a) Sketch the unit circle in an xy-coordinate system in R 2 using the Euclidean inner, product u, v = u1 v1 + u2 v2 ., (b) Sketch the unit circle in an xy-coordinate system in R 2 using the weighted Euclidean, inner product u, v = 19 u1 v1 + 41 u2 v2 ., , (a) The unit circle using, the standard Euclidean, inner product., , = (x, y), then u = u, u1/2 =, ", circle is x 2 + y 2 = 1, or on squaring both sides,, , Solution (a) If u, , x2 + y2 = 1, , y, 2, , ", x 2 + y 2 , so the equation of the unit, , ||u|| = 1, , As expected, the graph of this equation is a circle of radius 1 centered at the origin, (Figure 6.1.1a)., x, 3, , Solution (b) If u, , #, , unit circle is, (b) The unit circle using, a weighted Euclidean, inner product., , Figure 6.1.1, , Inner Products Generated, by Matrices, , = (x, y), then u = u, u1/2 =, , 1 2, x, 9, , #, , 1 2, x, 9, , + 41 y 2 , so the equation of the, , + 41 y 2 = 1, or on squaring both sides,, x2, 9, , +, , y2, 4, , =1, , The graph of this equation is the ellipse shown in Figure 6.1.1b. Though this may seem, odd when viewed geometrically, it makes sense algebraically since all points on the ellipse, are 1 unit away from the origin relative to the given weighted Euclidean inner product. In, short, weighting has the effect of distorting the space that we are used to seeing through, “unweighted Euclidean eyes.”, , The Euclidean inner product and the weighted Euclidean inner products are special cases, of a general class of inner products on R n called matrix inner products. To define this, class of inner products, let u and v be vectors in R n that are expressed in column form,, and let A be an invertible n × n matrix. It can be shown (Exercise 47) that if u · v is the, Euclidean inner product on R n , then the formula, , u, v = Au · Av, , (5), , also defines an inner product; it is called the inner product on R n generated by A., Recall from Table 1 of Section 3.2 that if u and v are in column form, then u · v can, be written as vTu from which it follows that (5) can be expressed as, , u, v = (Av)TAu

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6.1 Inner Products, , 349, , or equivalently as, , u, v = vTATAu, , (6), , E X A M P L E 4 Matrices Generating Weighted Euclidean Inner Products, , The standard Euclidean and weighted Euclidean inner products are special cases of, matrix inner products. The standard Euclidean inner product on R n is generated by the, n × n identity matrix, since setting A = I in Formula (5) yields, , u, v = I u · I v = u · v, and the weighted Euclidean inner product, , u, v = w1 u1 v1 + w2 u2 v2 + · · · + wn un vn, is generated by the matrix, , ⎡√, w1, ⎢, ⎢ 0, A=⎢, ⎢ .., ⎣ ., , 0, , √, w2, .., ., , 0, , 0, , 0, 0, , (7), , ⎤, , ···, ···, , 0, ⎥, 0 ⎥, , .. ⎥, ⎥, . ⎦, √, wn, 0 ···, .., ., , This can be seen by observing that ATA is the n × n diagonal matrix whose diagonal, entries are the weights w1 , w2 , . . . , wn ., E X A M P L E 5 Example 1 Revisited, Every diagonal matrix with, positive diagonal entries generates a weighted inner product. Why?, , Other Examples of Inner, Products, , The weighted Euclidean inner product u, v = 3u1 v1 + 2u2 v2 discussed in Example 1, is the inner product on R 2 generated by, , √, , A=, , 3, 0, , , , 0, √, 2, , So far, we have only considered examples of inner products on R n . We will now consider, examples of inner products on some of the other kinds of vector spaces that we discussed, earlier., E X A M P L E 6 The Standard Inner Product on M nn, , If u = U and v = V are matrices in the vector space Mnn , then the formula, , u, v = tr(U TV ), , (8), , defines an inner product on Mnn called the standard inner product on that space (see, Definition 8 of Section 1.3 for a definition of trace). This can be proved by confirming, that the four inner product space axioms are satisfied, but we can see why this is so by, computing (8) for the 2 × 2 matrices, , U=, , u1, u3, , u2, u4, , and V =, , v1, v3, , v2, v4, , This yields, , u, v = tr(U TV ) = u1 v1 + u2 v2 + u3 v3 + u4 v4

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350, , Chapter 6 Inner Product Spaces, , which is just the dot product of the corresponding entries in the two matrices. And it, follows from this that, u =, , #, ", ", u, u = trU T U  = u21 + u22 + u23 + u24, , For example, if, 1, 3, , u=U =, , 2, 4, , and v = V =, , −1, 3, , 0, 2, , then, , u, v = tr(U T V ) = 1(−1) + 2(0) + 3(3) + 4(2) = 16, and, , u =, v =, , √, √, , u, u =, v, v =, , ", ", , tr(U T U ) =, , √, , 1 2 + 2 2 + 32 + 4 2 =, , √, , 30, , ", √, tr(V T V ) = (−1)2 + 02 + 32 + 22 = 14, , E X A M P L E 7 The Standard Inner Product on Pn, , If, p = a0 + a1 x + · · · + an x n and q = b0 + b1 x + · · · + bn x n, are polynomials in Pn , then the following formula defines an inner product on Pn (verify), that we will call the standard inner product on this space:, , p, q = a0 b0 + a1 b1 + · · · + an bn, , (9), , The norm of a polynomial p relative to this inner product is, p =, , #, ", p, p = a02 + a12 + · · · + an2, , E X A M P L E 8 The Evaluation Inner Product on Pn, , If, p = p(x) = a0 + a1 x + · · · + an x n and q = q(x) = b0 + b1 x + · · · + bn x n, are polynomials in Pn , and if x0 , x1 , . . . , xn are distinct real numbers (called sample, points), then the formula, , p, q = p(x0 )q(x0 ) + p(x1 )q(x1 ) + · · · + p(xn )q(xn ), , (10), , defines an inner product on Pn called the evaluation inner product at x0 , x1 , . . . , xn ., Algebraically, this can be viewed as the dot product in R n of the n-tuples, , , , , , p(x0 ), p(x1 ), . . . , p(xn ), , and, , , , , , q(x0 ), q(x1 ), . . . , q(xn ), , and hence the first three inner product axioms follow from properties of the dot product., The fourth inner product axiom follows from the fact that, , p, p = [p(x0 )]2 + [p(x1 )]2 + · · · + [p(xn )]2 ≥ 0, with equality holding if and only if, , p(x0 ) = p(x1 ) = · · · = p(xn ) = 0, But a nonzero polynomial of degree n or less can have at most n distinct roots, so it must, be that p = 0, which proves that the fourth inner product axiom holds.

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6.1 Inner Products, , 351, , The norm of a polynomial p relative to the evaluation inner product is, p =, , ", ", p, p = [p(x0 )]2 + [p(x1 )]2 + · · · + [p(xn )]2, , (11), , E X A M P L E 9 Working with the Evaluation Inner Product, , Let P2 have the evaluation inner product at the points, , x0 = −2, x1 = 0, and x2 = 2, Compute p, q and p for the polynomials p = p(x) = x 2 and q = q(x) = 1 + x ., Solution It follows from (10) and (11) that, , p, q = p(−2)q(−2) + p(0)q(0) + p(2)q(2) = (4)(−1) + (0)(1) + (4)(3) = 8, ", ", p = [p(x0 )]2 + [p(x1 )]2 + [p(x2 )]2 = [p(−2)]2 + [p(0)]2 + [p(2)]2, √, √, √, = 42 + 02 + 42 = 32 = 4 2, , CA L C U L U S R E Q U I R E D, , E X A M P L E 10 An Integral Inner Product on C [a, b], , Let f = f(x) and g = g(x) be two functions in C[a, b] and define, , , , b, , f, g =, , f(x)g(x) dx, , (12), , a, , We will show that this formula defines an inner product on C[a, b] by verifying the four, inner product axioms for functions f = f(x), g = g(x), and h = h(x) in C[a, b]:, , , , Axiom 1: f, g =, , , , b, , b, , f(x)g(x) dx =, g(x)f(x) dx = g, f , a , a, b, (f(x) + g(x))h(x) dx, Axiom 2: f + g, h =, , , a, b, , =, , , , b, , f(x)h(x) dx +, , g(x)h(x) dx, , a, , a, , = f, h + g, h,  b, , Axiom 3: k f, g =, kf(x)g(x) dx = k, a, , b, , f(x)g(x) dx = kf, g, , a, , Axiom 4: If f = f(x) is any function in C[a, b], then, , , , f, f  =, , b, , f 2 (x) dx ≥ 0, , (13), , a, , since f 2 (x) ≥ 0 for all x in the interval [a, b]. Moreover, because f is continuous on, [a, b], the equality in Formula (13) holds if and only if the function f is identically zero, on [a, b], that is, if and only if f = 0; and this proves that Axiom 4 holds., CA L C U L U S R E Q U I R E D, , E X A M P L E 11 Norm of a Vector in C [a, b], , If C[a, b] has the inner product that was defined in Example 10, then the norm of a, function f = f(x) relative to this inner product is, , +, , , f = f, f, , 1/ 2, , b, , =, , f 2 (x) dx, a, , (14)

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352, , Chapter 6 Inner Product Spaces, , and the unit sphere in this space consists of all functions f in C[a, b] that satisfy the, equation, , , , b, , f 2 (x) dx = 1, , a, , Remark Note that the vector space Pn is a subspace of C[a, b] because polynomials are continuous functions. Thus, Formula (12) defines an inner product on Pn that is different from both the, standard inner product and the evaluation inner product., , WARNING Recall from calculus that the arc length of a curve y, , is given by the formula, , , L=, , b, , = f(x) over an interval [a, b], , ", , 1 + [f (x)]2 dx, , (15), , a, , Do not confuse this concept of arc length with f , which is the length (norm) of f when f is, viewed as a vector in C[a, b]. Formulas (14) and (15) have different meanings., , Algebraic Properties of, Inner Products, , The following theorem lists some of the algebraic properties of inner products that follow, from the inner product axioms. This result is a generalization of Theorem 3.2.3, which, applied only to the dot product on R n ., THEOREM 6.1.2 If u, v, and w are vectors in a real inner product space V, and if k is a, , scalar, then:, (a) 0, v = v, 0 = 0, (b) u, v + w = u, v + u, w, (c), , u, v − w = u, v − u, w, , (d ) u − v, w = u, w − v, w, (e), , ku, v = u, k v, , Proof We will prove part (b) and leave the proofs of the remaining parts as exercises., , u, v + w = v + w, u, = v, u + w, u, = u, v + u, w, , [ By symmetry ], [ By additivity ], [ By symmetry ], , The following example illustrates how Theorem 6.1.2 and the defining properties of, inner products can be used to perform algebraic computations with inner products. As, you read through the example, you will find it instructive to justify the steps., E X A M P L E 1 2 Calculating with Inner Products, , u − 2v, 3u + 4v = u, 3u + 4v − 2v, 3u + 4v, = u, 3u + u, 4v − 2v, 3u − 2v, 4v, = 3u, u + 4u, v − 6v, u − 8v, v, = 3 u 2 + 4u, v − 6u, v − 8 v 2, = 3 u 2 − 2u, v − 8 v 2

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6.1 Inner Products, , 353, , Exercise Set 6.1, 1. Let R 2 have the weighted Euclidean inner product, , u, v = 2u1 v1 + 3u2 v2, , 16. x0 = −1, x1 = 0, x2 = 1, x3 = 2, , and let u = (1, 1), v = (3, 2), w = (0, −1), and k = 3. Compute the stated quantities., (a) u, v, , (b) k v, w, , (c) u + v, w, , (d) v, , (e) d(u, v), , (f ), , u − kv, , 2. Follow the directions of Exercise 1 using the weighted Euclidean inner product, , u, v = 21 u1 v1 + 5u2 v2, , 3. A =, , 2, , 1, , 1, , 1, , , , , 4. A =, , , , 1, , 0, , 2, , −1, , In Exercises 5–6, find a matrix that generates the stated, weighted inner product on R 2 ., 5. u, v = 2u1 v1 + 3u2 v2, , 7. A =, , 4, , 1, , 2, , −3, , , , −2, , 10. U =, , 8, , 8. A =, , 1, , −3, , , V =, , −1, , 2, 4, , V =, 0, 5, , 20. p = −5 + 2x + x 2 , q = 3 + 2x − 4x 2, In Exercises 21–22, find U and d(U, V ) relative to the standard inner product on M22 ., , 1, , −2, , 3, 4, , 21. U =, , 8, 1, , 22. U =, , , V =, , −1, , 2, 4, , V =, 0, 5, , −3, , 3, 1, , 1, 6, 8, , In Exercises 23–24, let, p = x + x 3 and q = 1 + x 2, Find p and d(p, q) relative to the evaluation inner product on, P3 at the stated sample points., , −1, , 3, , 23. x0 = −2, x1 = −1, x2 = 0, x3 = 1, , , , 3, 1, , 24. x0 = −1, x1 = 0, x2 = 1, x3 = 2, In Exercises 25–26, find u and d(u, v) for the vectors, u = (−1, 2) and v = (2, 5) relative to the inner product on R 2, generated by the matrix A., , , , 6, 8, , 11. p = −2 + x + 3x , q = 4 − 7x, , 25. A =, , 2, , 12. p = −5 + 2x + x 2 , q = 3 + 2x − 4x 2, In Exercises 13–14, a weighted Euclidean inner product on, R 2 is given for the vectors u = (u1 , u2 ) and v = (v1 , v2 ). Find a, matrix that generates it., 13. u, v = 3u1 v1 + 5u2 v2, , In Exercises 19–20, find p and d(p, q) relative to the standard, inner product on P2 ., , 1, , In Exercises 11–12, find the standard inner product on P2 of, the given polynomials., 2, , 18. u = (−1, 2) and v = (2, 5), , 2, , , , In Exercises 9–10, compute the standard inner product on M22, of the given matrices., 3, 9. U =, 4, , 17. u = (−3, 2) and v = (1, 7), , 6. u, v = 21 u1 v1 + 5u2 v2, , In Exercises 7–8, use the inner product on R 2 generated by the, matrix A to find u, v for the vectors u = (0, −3) and v = (6, 2)., , , , In Exercises 17–18, find u and d(u, v) relative to the weighted, Euclidean inner product u, v = 2u1 v1 + 3u2 v2 on R 2 ., , 19. p = −2 + x + 3x 2 , q = 4 − 7x 2, , In Exercises 3–4, compute the quantities in parts (a)–(f) of, Exercise 1 using the inner product on R 2 generated by A., , , , 15. x0 = −2, x1 = −1, x2 = 0, x3 = 1, , 14. u, v = 4u1 v1 + 6u2 v2, , In Exercises 15–16, a sequence of sample points is given. Use, the evaluation inner product on P3 at those sample points to find, p, q for the polynomials, p = x + x 3 and q = 1 + x 2, , 4, , 0, , 3, , 5, , , , , 26. A =, , , , 1, , 2, , −1, , 3, , In Exercises 27–28, suppose that u, v, and w are vectors in an, inner product space such that, , u, v = 2, v, w = −6, u, w = −3, u = 1,, , v = 2,, , w =7, , Evaluate the given expression., 27. (a) 2v − w, 3u + 2w, , (b) u + v, , 28. (a) u − v − 2w, 4u + v, , (b) 2w − v, , In Exercises 29–30, sketch the unit circle in R 2 using the given, inner product., 29. u, v = 41 u1 v1 +, , 1, uv, 16 2 2, , 30. u, v = 2u1 v1 + u2 v2

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354, , Chapter 6 Inner Product Spaces, , In Exercises 31–32, find a weighted Euclidean inner product, on R 2 for which the “unit circle” is the ellipse shown in the accompanying figure., y, , 31., , (b) What conditions must k1 and k2 satisfy for, u, v = k1 u1 v1 + k2 u2 v2 to define an inner product on, R 2 ? Justify your answer., , y, , 32., , 1, , 1, , x, , x, 3, 4, , 3, , 43. (a) Let u = (u1 , u2 ) and v = (v1 , v2 ). Prove that, u, v = 3u1 v1 + 5u2 v2 defines an inner product on R 2 by, showing that the inner product axioms hold., , 44. Prove that the following identity holds for vectors in any inner, product space., , u, v =, Figure Ex-31, , Figure Ex-31, , In Exercises 33–34, let u = (u1 , u2 , u3 ) and v = (v1 , v2 , v3 )., Show that the expression does not define an inner product on R 3 ,, and list all inner product axioms that fail to hold., 33. u, v =, , u21 v12, , +, , u22 v22, , +, , u23 v32, , 34. u, v = u1 v1 − u2 v2 + u3 v3, In Exercises 35–36, suppose that u and v are vectors in an inner product space. Rewrite the given expression in terms of u, v,, u 2 , and v 2 ., 35. 2v − 4u, u − 3v, , 36. 5u + 6v, 4v − 3u, , 37. (Calculus required ) Let the vector space P2 have the inner, product, , , , p, q =, , 1, , p(x)q(x) dx, −1, , Find the following for p = 1 and q = x 2 ., (a) p, q, , (b) d(p, q), , (c), , (d) q, , p, , 38. (Calculus required ) Let the vector space P3 have the inner, product, , , , p, q =, , 1, , 1, 4, , u+v, , 2, , −, , 1, 4, , u−v, , 2, , 45. Prove that the following identity holds for vectors in any inner, product space., u+v, , 2, , + u−v, , 2, , =2 u, , 2, , +2 v, , 2, , 46. The definition of a complex vector space was given in the first, margin note in Section 4.1. The definition of a complex inner, product on a complex vector space V is identical to that in, Definition 1 except that scalars are allowed to be complex, numbers, and Axiom 1 is replaced by u, v = v, u. The, remaining axioms are unchanged. A complex vector space, with a complex inner product is called a complex inner product, space. Prove that if V is a complex inner product space, then, u, k v = ku, v., 47. Prove that Formula (5) defines an inner product on R n ., 48. (a) Prove that if v is a fixed vector in a real inner product space, V , then the mapping T : V →R defined by T (x) = x, v, is a linear transformation., (b) Let V = R 3 have the Euclidean inner product, and let, v = (1, 0, 2). Compute T (1, 1, 1)., (c) Let V = P2 have the standard inner product, and let, v = 1 + x . Compute T (x + x 2 )., (d) Let V = P2 have the evaluation inner product at the points, x0 = 1, x1 = 0, x2 = −1, and let v = 1 + x . Compute, T (x + x 2 )., , p(x)q(x) dx, −1, , Find the following for p = 2x 3 and q = 1 − x 3 ., (a) p, q, , (b) d(p, q), , (c), , (d) q, , p, , True-False Exercises, TF. In parts (a)–(g) determine whether the statement is true or, false, and justify your answer., (a) The dot product on R 2 is an example of a weighted inner, product., , (Calculus required ) In Exericses 39–40, use the inner product, , , , 1, , f, g =, , f (x)g(x)dx, 0, , on C[0, 1] to compute f, g., 39. f = cos 2π x, g = sin 2πx 40. f = x, g = ex, , Working with Proofs, 41. Prove parts (a) and (b) of Theorem 6.1.1., 42. Prove parts (c) and (d) of Theorem 6.1.1., , (b) The inner product of two vectors cannot be a negative real, number., (c) u, v + w = v, u + w, u., (d) k u, k v = k 2 u, v., (e) If u, v = 0, then u = 0 or v = 0., (f ) If v, , 2, , = 0, then v = 0., , (g) If A is an n × n matrix, then u, v = Au · Av defines an inner, product on R n .

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6.2 Angle and Orthogonality in Inner Product Spaces, , Working withTechnology, , 355, , and let, , T1. (a) Confirm that the following matrix generates an inner, product., ⎡, ⎤, 5, 8, 6 −13, , ⎢3, ⎢, A=⎢, ⎣0, , −1, , 0, , 1, , −1, , 2, , 4, , 3, , −9⎥, ⎥, ⎥, 0⎦, , (a) Compute p, q, p , and q ., (b) Verify that the identities in Exercises 44 and 45 hold for the, vectors p and q., , −5, , (b) For the following vectors, use the inner product in part (a) to, compute u, v, first by Formula (5) and then by Formula (6)., ⎡ ⎤, ⎡ ⎤, 1, 0, , ⎢ 1⎥, ⎢−2⎥, ⎢ ⎥, ⎢ ⎥, u = ⎢ ⎥ and v = ⎢ ⎥, ⎣−1⎦, ⎣ 0⎦, 3, , p = p(x) = x + x 3 and q = q(x) = 1 + x 2 + x 4, , T3. Let the vector space M33 have the standard inner product and, let, , ⎡, , 1, , −2, , u = U = ⎣−2, 3, , ⎢, , ⎤, , ⎡, , ⎥, , ⎢, , ⎤, , 2, , −1, , 4, , 1⎦ and v = V = ⎣1, , 4, , 3⎦, , 1, , 0, , 1, , 0, , 2, , 3, , 0, , ⎥, , (a) Use Formula (8) to compute u, v, u , and v ., , 2, , T2. Let the vector space P4 have the evaluation inner product at, the points, −2, −1, 0, 1, 2, , (b) Verify that the identities in Exercises 44 and 45 hold for the, vectors u and v., , 6.2 Angle and Orthogonality in Inner Product Spaces, In Section 3.2 we defined the notion of “angle” between vectors in R n . In this section we, will extend this idea to general vector spaces. This will enable us to extend the notion of, orthogonality as well, thereby setting the groundwork for a variety of new applications., , Cauchy–Schwarz Inequality, , Recall from Formula (20) of Section 3.2 that the angle θ between two vectors u and v in, R n is, !, u·v, −1, (1), θ = cos, u v, We were assured that this formula was valid because it followed from the Cauchy–, Schwarz inequality (Theorem 3.2.4) that, , −1 ≤, , u·v, ≤1, u v, , (2), , as required for the inverse cosine to be defined. The following generalization of the, Cauchy–Schwarz inequality will enable us to define the angle between two vectors in any, real inner product space., , THEOREM 6.2.1 Cauchy–Schwarz Inequality, , If u and v are vectors in a real inner product space V, then, , |u, v| ≤ u v, , (3), , Proof We warn you in advance that the proof presented here depends on a clever trick, , that is not easy to motivate., In the case where u = 0 the two sides of (3) are equal since u, v and u are both, zero. Thus, we need only consider the case where u  = 0. Making this assumption, let, , a = u, u, b = 2u, v, c = v, v

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356, , Chapter 6 Inner Product Spaces, , and let t be any real number. Since the positivity axiom states that the inner product of, any vector with itself is nonnegative, it follows that, 0 ≤ t u + v, t u + v = u, ut 2 + 2u, vt + v, v, , = at 2 + bt + c, This inequality implies that the quadratic polynomial at 2 + bt + c has either no real, roots or a repeated real root. Therefore, its discriminant must satisfy the inequality, b2 − 4ac ≤ 0. Expressing the coefficients a, b, and c in terms of the vectors u and v, gives 4u, v2 − 4u, uv, v ≤ 0 or, equivalently,, , u, v2 ≤ u, uv, v, Taking square roots of both sides and using the fact that u, u and v, v are nonnegative, yields, |u, v| ≤ u, u1/2 v, v1/2 or equivalently |u, v| ≤ u v, which completes the proof., The following two alternative forms of the Cauchy–Schwarz inequality are useful to, know:, , u, v2 ≤ u, uv, v, , (4), , u, v2 ≤ u, , (5), , 2, , v, , 2, , The first of these formulas was obtained in the proof of Theorem 6.2.1, and the second, is a variation of the first., , Angle Between Vectors, , Our next goal is to define what is meant by the “angle” between vectors in a real inner, product space. As a first step, we leave it as an exercise for you to use the Cauchy–Schwarz, inequality to show that, , u, v, , −1 ≤, , ≤1, u v, This being the case, there is a unique angle θ in radian measure for which, u, v, , cos θ =, , u, , (6), , and 0 ≤ θ ≤ π, , v, , (7), , (Figure 6.2.1). This enables us to define the angle θ between u and v to be, , θ = cos, , −1, , u, v, u, , !, (8), , v, , y, 1, θ, –π, , Figure 6.2.1, , –π, 2, , π, 2, –1, , π, , 3π, 2, , 2π, , 5π, 2, , 3π

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6.2 Angle and Orthogonality in Inner Product Spaces, , 357, , E X A M P L E 1 Cosine of the Angle Between Vectors in M 22, , Let M22 have the standard inner product. Find the cosine of the angle between the, vectors, , , , , 1 2, −1 0, u=U =, and v = V =, 3 4, 3 2, Solution We showed in Example 6 of the previous section that, , u, v = 16,, , u =, , √, , √, , 30,, , v =, , 16, , ≈ 0.78, , 14, , from which it follows that, cos θ =, , Properties of Length and, Distance in General Inner, Product Spaces, , u, v, u, , v, , =√ √, , 30 14, , In Section 3.2 we used the dot product to extend the notions of length and distance to R n ,, and we showed that various basic geometry theorems remained valid (see Theorems 3.2.5,, 3.2.6, and 3.2.7). By making only minor adjustments to the proofs of those theorems,, one can show that they remain valid in any real inner product space. For example, here, is the generalization of Theorem 3.2.5 (the triangle inequalities)., THEOREM 6.2.2 If u, v, and w are vectors in a real inner product space, , V, and if k is, , any scalar, then:, (a), , u+v ≤ u + v, , (b) d(u, v) ≤ d(u, w) + d(w, v), , [ Triangle inequality for vectors ], [ Triangle inequality for distances ], , Proof (a), , u+v, , 2, , = u + v, u + v, = u, u + 2u, v + v, v, ≤ u, u + 2|u, v| + v, v, ≤ u, u + 2 u v + v, v, = u 2+2 u v + v 2, = ( u + v )2, , [ Property of absolute value ], [ By (3) ], , Taking square roots gives u + v ≤ u + v ., Proof (b) Identical to the proof of part (b) of Theorem 3.2.5., , Orthogonality, , Although Example 1 is a useful mathematical exercise, there is only an occasional need, to compute angles in vector spaces other than R 2 and R 3 . A problem of more interest, in general vector spaces is ascertaining whether the angle between vectors is π/2. You, should be able to see from Formula (8) that if u and v are nonzero vectors, then the angle, between them is θ = π/2 if and only if u, v = 0. Accordingly, we make the following, definition, which is a generalization of Definition 1 in Section 3.3 and is applicable even, if one or both of the vectors is zero., DEFINITION 1 Two vectors u and v in an inner product space V called orthogonal if, , u, v = 0.

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358, , Chapter 6 Inner Product Spaces, , As the following example shows, orthogonality depends on the inner product in the, sense that for different inner products two vectors can be orthogonal with respect to one, but not the other., E X A M P L E 2 Orthogonality Depends on the Inner Product, , The vectors u = (1, 1) and v = (1, −1) are orthogonal with respect to the Euclidean, inner product on R 2 since, u · v = (1)(1) + (1)(−1) = 0, However, they are not orthogonal with respect to the weighted Euclidean inner product, , u, v = 3u1 v1 + 2u2 v2 since, u, v = 3(1)(1) + 2(1)(−1) = 1  = 0, E X A M P L E 3 Orthogonal Vectors in M 22, , If M22 has the inner product of Example 6 in the preceding section, then the matrices, 1, 1, , U=, , 0, 1, , 0, 0, , and V =, , 2, 0, , are orthogonal since, , U, V  = 1(0) + 0(2) + 1(0) + 1(0) = 0, CA L C U L U S R E Q U I R E D, , E X A M P L E 4 Orthogonal Vectors in P 2, , Let P2 have the inner product, , , p, q =, , 1, , −1, , p(x)q(x) dx, , and let p = x and q = x 2 . Then, , , , p = p, p, , 1/2, , =, , −1, , , q = q, q, , 1/2, , ,  p , q =, , 1, , −1, , 1/2, , 1, , =, , , =, , xx dx, , 1/2, , 1, , −1, , , xx 2 dx =, , x x dx, , 1, , −1, , , =, , 2 2, , =, , 2, , −1, , ,, , 1/2, , 1, , x dx, , 4, , −1, , ,, , 1/2, , 1, , x dx, , 2, 3, , =, , 2, 5, , x 3 dx = 0, , Because p, q = 0, the vectors p = x and q = x 2 are orthogonal relative to the given, inner product., In Theorem 3.3.3 we proved the Theorem of Pythagoras for vectors in Euclidean, , n-space. The following theorem extends this result to vectors in any real inner product, space., THEOREM 6.2.3 Generalized Theorem of Pythagoras, , If u and v are orthogonal vectors in a real inner product space, then, u+v, , 2, , = u, , 2, , + v, , 2

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6.2 Angle and Orthogonality in Inner Product Spaces, Proof The orthogonality of u and v implies that u, v, , u+v, , = u + v, u + v = u, , 2, , = u, CA L C U L U S R E Q U I R E D, , 2, , + v, , = 0, so, , + 2u, v + v, , 2, , 359, , 2, , 2, , E X A M P L E 5 Theorem of Pythagoras in P 2, , In Example 4 we showed that p = x and q = x 2 are orthogonal with respect to the inner, product, , , , p, q =, , 1, , −1, , p(x)q(x) dx, , on P2 . It follows from Theorem 6.2.3 that, p+q, , 2, , = p, , + q, , 2, , 2, , Thus, from the computations in Example 4, we have, , , 2, , p+q, , 2, , =, , 2, 3, , ,  2, , +, , 2, 5, , =, , 2, 2, 16, + =, 3, 5, 15, , We can check this result by direct integration:, , , , p+q, , 2, , = p + q, p + q =, , =, , Orthogonal Complements, , 1, , , x dx + 2, , −1, 1, , 2, , −1, , 1, , (x + x 2 )(x + x 2 ) dx, , x dx +, , , , 1, , 3, , −1, , −1, , x 4 dx =, , 2, 2, 16, +0+ =, 3, 5, 15, , In Section 4.8 we defined the notion of an orthogonal complement for subspaces of R n ,, and we used that definition to establish a geometric link between the fundamental spaces, of a matrix. The following definition extends that idea to general inner product spaces., DEFINITION 2 If W is a subspace of a real inner product space V, then the set of, all vectors in V that are orthogonal to every vector in W is called the orthogonal, complement of W and is denoted by the symbol W ⊥ ., , In Theorem 4.8.6 we stated three properties of orthogonal complements in R n . The, following theorem generalizes parts (a) and (b) of that theorem to general real inner, product spaces., THEOREM 6.2.4 If W is a subspace of a real inner product space V, then:, , (a) W ⊥ is a subspace of V ., (b) W ∩ W ⊥ = {0}., Proof (a) The set W ⊥ contains at least the zero vector, since 0, w, , = 0 for every vector, w in W . Thus, it remains to show that W ⊥ is closed under addition and scalar multiplication. To do this, suppose that u and v are vectors in W ⊥ , so that for every vector w in, W we have u, w = 0 and v, w = 0. It follows from the additivity and homogeneity, axioms of inner products that, u + v, w = u, w + v, w = 0 + 0 = 0, k u, w = ku, w = k(0) = 0, which proves that u + v and k u are in W ⊥ .

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360, , Chapter 6 Inner Product Spaces, , W and W ⊥ , then v is orthogonal to itself; that is,, v, v = 0. It follows from the positivity axiom for inner products that v = 0., Proof (b) If v is any vector in both, , The next theorem, which we state without proof, generalizes part (c) of Theorem 4.8.6. Note, however, that this theorem applies only to finite-dimensional inner, product spaces, whereas Theorem 4.8.6 does not have this restriction., Theorem 6.2.5 implies that, in a finite-dimensional inner product space orthogonal, complements occur in pairs,, each being orthogonal to the, other (Figure 6.2.2)., , W⊥, , W, , THEOREM 6.2.5 If W is a subspace of a real finite-dimensional inner product space V,, , then the orthogonal complement of W ⊥ is W ; that is,, , (W ⊥ )⊥ = W, In our study of the fundamental spaces of a matrix in Section 4.8 we showed that the, row space and null space of a matrix are orthogonal complements with respect to the, Euclidean inner product on R n (Theorem 4.8.7). The following example takes advantage, of that fact., E X A M P L E 6 Basis for an Orthogonal Complement, , Let W be the subspace of R 6 spanned by the vectors, , Figure 6.2.2 Each vector in, W is orthogonal to each vector, in W ⊥ and conversely., , w1 = (1, 3, −2, 0, 2, 0),, , w2 = (2, 6, −5, −2, 4, −3),, , w3 = (0, 0, 5, 10, 0, 15),, , w4 = (2, 6, 0, 8, 4, 18), , Find a basis for the orthogonal complement of W ., Solution The subspace W is the same as the row space of the matrix, , ⎡, , 1, ⎢2, ⎢, A=⎢, ⎣0, 2, , 3, 6, 0, 6, , −2, −5, 5, 0, , 0, −2, 10, 8, , 2, 4, 0, 4, , ⎤, , 0, −3⎥, ⎥, ⎥, 15⎦, 18, , Since the row space and null space of A are orthogonal complements, our problem, reduces to finding a basis for the null space of this matrix. In Example 4 of Section 4.7, we showed that, ⎡ ⎤, ⎡ ⎤, ⎡ ⎤, −3, −4, −2, ⎢ 1⎥, ⎢ 0⎥, ⎢ 0⎥, ⎢ ⎥, ⎢ ⎥, ⎢ ⎥, ⎢ 0⎥, ⎢ −2 ⎥, ⎢ 0⎥, ⎢ ⎥, ⎢ ⎥, ⎢ ⎥, v1 = ⎢ ⎥, v2 = ⎢ ⎥, v3 = ⎢ ⎥, ⎢ 0⎥, ⎢ 1⎥, ⎢ 0⎥, ⎢ ⎥, ⎢ ⎥, ⎢ ⎥, ⎣ 0⎦, ⎣ 0⎦, ⎣ 1⎦, 0, 0, 0, form a basis for this null space. Expressing these vectors in comma-delimited form (to, match that of w1 , w2 , w3 , and w4 ), we obtain the basis vectors, v1 = (−3, 1, 0, 0, 0, 0), v2 = (−4, 0, −2, 1, 0, 0), v3 = (−2, 0, 0, 0, 1, 0), You may want to check that these vectors are orthogonal to w1 , w2 , w3 , and w4 by, computing the necessary dot products.

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6.2 Angle and Orthogonality in Inner Product Spaces, , 361, , Exercise Set 6.2, In Exercises 1–2, find the cosine of the angle between the vectors with respect to the Euclidean inner product., 1. (a) u = (1, −3), v = (2, 4), (b) u = (−1, 5, 2), v = (2, 4, −9), (c) u = (1, 0, 1, 0), v = (−3, −3, −3, −3), 2. (a) u = (−1, 0), v = (3, 8), (b) u = (4, 1, 8), v = (1, 0, −3), , p1 = 2 + kx + 6x 2 , p2 = l + 5x + 3x 2 , p3 = 1 + 2x + 3x 2, , In Exercises 3–4, find the cosine of the angle between the vectors with respect to the standard inner product on P2 ., 3. p = −1 + 5x + 2x , q = 2 + 4x − 9x, , 16. Let R 4 have the Euclidean inner product. Find two unit vectors that are orthogonal to all three of the vectors, u = (2, 1, −4, 0), v = (−1, −1, 2, 2), and w = (3, 2, 5, 4)., 17. Do there exist scalars k and l such that the vectors, , (c) u = (2, 1, 7, −1), v = (4, 0, 0, 0), , 2, , 15. If the vectors u = (1, 2) and v = (2, −4) are orthogonal, with respect to the weighted Euclidean inner product, u, v = w1 u1 v1 + w2 u2 v2 , what must be true of the weights, w1 and w2 ?, , 2, , are mutually orthogonal with respect to the standard inner, product on P2 ?, 18. Show that the vectors, , 4. p = x − x 2 , q = 7 + 3x + 3x 2, ,  , , u=, , In Exercises 5–6, find the cosine of the angle between A and B, with respect to the standard inner product on M22 ., 5. A =, , 2, 1, , 6, 3, , B=, 1, −3, , 2, 6. A =, −1, , 3, , , and v =, , 5, , , , −8, , are orthogonal with respect to the inner product on R 2 that is, generated by the matrix, , 2, 0, , 4, −3, , B=, 4, 3, , 3, , , , 1, 2, , A=, , In Exercises 7–8, determine whether the vectors are orthogonal, with respect to the Euclidean inner product., 7. (a) u = (−1, 3, 2), v = (4, 2, −1), (b) u = (−2, −2, −2), v = (1, 1, 1), (c) u = (a, b), v = (−b, a), 8. (a) u = (u1 , u2 , u3 ), v = (0, 0, 0), (b) u = (−4, 6, −10, 1), v = (2, 1, −2, 9), (c) u = (a, b, c), v = (−c, 0, a), In Exercises 9–10, show that the vectors are orthogonal with, respect to the standard inner product on P2 ., 9. p = −1 − x + 2x , q = 2x + x, 2, , 2, , 1, , 1, , 1, , , , [See Formulas (5) and (6) of Section 6.1.], 19. Let P2 have the evaluation inner product at the points, , x0 = −2, x1 = 0, x2 = 2, Show that the vectors p = x and q = x 2 are orthogonal with, respect to this inner product., 20. Let M22 have the standard inner product. Determine whether, the matrix A is in the subspace spanned by the matrices U, and V ., , A=, , , −1, , 1, , 0, , 2, , , , , , , U=, , 1, , , −1, , 3, , 0, , , , , V =, , 4, , 0, , 9, , 2, , , , 2, , 10. p = 2 − 3x + x 2 , q = 4 + 2x − 2x 2, , In Exercises 21–24, confirm that the Cauchy–Schwarz inequality holds for the given vectors using the stated inner product., , In Exercises 11–12, show that the matrices are orthogonal with, respect to the standard inner product on M22 ., , 21. u = (1, 0, 3), v = (2, 1, −1) using the weighted Euclidean inner product u, v = 2u1 v1 + 3u2 v2 + u3 v3 in R 3 ., , 11. U =, 12. U =, , 2, , −1, 5, 2, , 1, −3, , V =, 0, 3, , 0, 2, , 1, −1, , V =, −2, −1, , 3, 0, , In Exercises 13–14, show that the vectors are not orthogonal, with respect to the Euclidean inner product on R 2 , and then find, a value of k for which the vectors are orthogonal with respect to, the weighted Euclidean inner product u, v = 2u1 v1 + ku2 v2 ., 13. u = (1, 3), v = (2, −1), , 14. u = (2, −4), v = (0, 3), , 22. U =, , −1, 6, , 2, 1, , and V =, , 1, 3, , 0, 3, , using the standard inner product on M22 ., 23. p = −1 + 2x + x 2 and q = 2 − 4x 2 using the standard inner, product on P2 ., 24. The vectors, u=, , 1, 1, , and v =, , 1, −1, , with respect to the inner product in Exercise 18.

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362, , Chapter 6 Inner Product Spaces, , 25. Let R 4 have the Euclidean inner product, and let, u = (−1, 1, 0, 2). Determine whether the vector u is orthogonal to the subspace spanned by the vectors w1 = (1, −1, 3, 0), and w2 = (4, 0, 9, 2)., 26. Let P3 have the standard inner product, and let, p = −1 − x + 2x 2 + 4x 3, Determine whether p is orthogonal to the subspace spanned by, the polynomials w1 = 2 − x 2 + x 3 and w2 = 4x − 2x 2 + 2x 3 ., In Exercises 27–28, find a basis for the orthogonal complement, of the subspace of R n spanned by the vectors., , 35. (Calculus required ) Let C[0, 1] have the inner product in Exercise 31., (a) Show that the vectors, p = p(x) = 1 and q = q(x) =, , In Exercises 29–30, assume that R n has the Euclidean inner, product., 29. (a) Let W be the line in R 2 with equation y = 2x . Find an, equation for W ⊥ ., (b) Let W be the plane in R 3 with equation x − 2y − 3z = 0., Find parametric equations for W ⊥ ., 30. (a) Let W be the y -axis in an xyz-coordinate system in R 3 ., Describe the subspace W ⊥ ., , are orthogonal., , 36. (Calculus required ) Let C[−1, 1] have the inner product in, Exercise 33., (a) Show that the vectors, p = p(x) = x and q = q(x) = x 2 − 1, are orthogonal., (b) Show that the vectors in part (a) satisfy the Theorem of, Pythagoras., 37. Let V be an inner product space. Show that√if u and v are, orthogonal unit vectors in V, then u − v = 2., 38. Let V be an inner product space. Show that if w is orthogonal, to both u1 and u2 , then it is orthogonal to k1 u1 + k2 u2 for all, scalars k1 and k2 . Interpret this result geometrically in the case, where V is R 3 with the Euclidean inner product., 39. (Calculus required ) Let C[0, π ] have the inner product, , , , (b) Let W be the yz-plane of an xyz-coordinate system in R 3 ., Describe the subspace W ⊥ ., 31. (Calculus required ) Let C[0, 1] have the integral inner product, , , , 1, , p, q =, , p(x)q(x) dx, 0, , and let p = p(x) = x and q = q(x) = x 2 ., (a) Find p, q., (b) Find p and q ., , f(x)g(x) dx, 0, , and let fn = cos nx (n = 0, 1, 2, . . .). Show that if k  = l , then, fk and fl are orthogonal vectors., 40. As illustrated, in the accompanying, figure, the vectors, √, √, u = (1, 3 ) and v = (−1, 3 ) have norm 2 and an angle, of 60◦ between them relative to the Euclidean inner product., Find a weighted Euclidean inner product with respect to which, u and v are orthogonal unit vectors., (–1, √3), , (b) Find the distance between the vectors p and q in Exercise 31., , v, , y, , (1, √3), , 60°, u, , x, 2, , 33. (Calculus required ) Let C[−1, 1] have the integral inner, product, , , , π, , f, g =, , 32. (a) Find the cosine of the angle between the vectors p and q, in Exercise 31., , p, q =, , −x, , (b) Show that the vectors in part (a) satisfy the Theorem of, Pythagoras., , 27. v1 = (1, 4, 5, 2), v2 = (2, 1, 3, 0), v3 = (−1, 3, 2, 2), 28. v1 = (1, 4, 5, 6, 9), v2 = (3, −2, 1, 4, −1),, v3 = (−1, 0, −1, −2, −1), v4 = (2, 3, 5, 7, 8), , 1, 2, , 1, , Figure Ex-40, , p(x)q(x) dx, −1, , and let p = p(x) = x 2 − x and q = q(x) = x + 1., (a) Find p, q., (b) Find p and q ., 34. (a) Find the cosine of the angle between the vectors p and q, in Exercise 33., (b) Find the distance between the vectors p and q in Exercise 33., , Working with Proofs, 41. Let V be an inner product space. Prove that if w is orthogonal, to each of the vectors u1 , u2 , . . . , ur , then it is orthogonal to, every vector in span{u1 , u2 , . . . , ur }., 42. Let {v1 , v2 , . . . , vr } be a basis for an inner product space V ., Prove that the zero vector is the only vector in V that is orthogonal to all of the basis vectors.

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6.2 Angle and Orthogonality in Inner Product Spaces, , 43. Let {w1 , w2 , . . . , wk } be a basis for a subspace W of V . Prove, that W ⊥ consists of all vectors in V that are orthogonal to, every basis vector., 44. Prove the following generalization of Theorem 6.2.3: If, v1 , v2 , . . . , vr are pairwise orthogonal vectors in an inner, product space V, then, v1 + v2 + · · · + v r, , 2, , = v1, , 2, , + v2, , + · · · + vr, , 2, , 363, , (a) Assuming that P2 has the standard inner product, find all, vectors q in P2 such that p, q = T (p), T (q)., (b) Assuming that P2 has the evaluation inner product at the, points x0 = −1, x1 = 0, x2 = 1, find all vectors q in P2, such that p, q = T (p), T (q)., , True-False Exercises, , 2, , 45. Prove: If u and v are n × 1 matrices and A is an n × n matrix,, then, (vTATAu)2 ≤ (uTATAu)(vTATAv), , TF. In parts (a)–(f ) determine whether the statement is true or, false, and justify your answer., , 46. Use the Cauchy–Schwarz inequality to prove that for all real, values of a , b, and θ ,, , (b) If u is a vector in both W and W ⊥ , then u = 0., , (a cos θ + b sin θ )2 ≤ a 2 + b2, , (c) If u and v are vectors in W ⊥ , then u + v is in W ⊥ ., , 47. Prove: If w1 , w2 , . . . , wn are positive real numbers, and, if u = (u1 , u2 , . . . , un ) and v = (v1 , v2 , . . . , vn ) are any two, vectors in R n , then, , |w1 u1 v1 + w2 u2 v2 + · · · + wn un vn |, ≤, , (w1 u21, , + w2 u22 + · · · + wn un2 )1/2 (w1 v12 + w2 v22 + · · · + wn vn2 )1/2, , 48. Prove that equality holds in the Cauchy–Schwarz inequality if, and only if u and v are linearly dependent., 49. (Calculus required ) Let f(x) and g(x) be continuous functions, on [0, 1]. Prove:, , , , 2, , 1, , f(x)g(x) dx, , (a), , , , (b), , g 2 (x) dx, , 0, 1, , (d) If u is a vector in W ⊥ and k is a real number, then k u is in W ⊥ ., (e) If u and v are orthogonal, then |u, v| = u, , , , Working withTechnology, T1. (a) We know that the row space and null space of a matrix, are orthogonal complements relative to the Euclidean inner, product. Confirm this fact for the matrix, , ⎡, , ≤, , 1/2, , 1, , f 2 (x) dx, , 0, , 0, , , +, , 1/2, , 1, , v ., , (f ) If u and v are orthogonal, then u + v = u + v ., , 0, , 1/2, [f(x) + g(x)]2 dx, , , , 1, , f 2 (x) dx, , 0, , , , , , 1, , ≤, , (a) If u is orthogonal to every vector of a subspace W , then u = 0., , g 2 (x) dx, , 2, , −1, , 3, , 5, , ⎤, , ⎢4, ⎢, ⎢, A=⎢, ⎢3, ⎢, ⎣4, , −3, , 1, , −2, , 3, , −1, , 15, , ⎥, ⎥, 4⎥, ⎥, ⎥, 17⎦, , 7, , −6, , −7, , 0, , 0, , 3⎥, , [Hint: Use the Cauchy–Schwarz inequality.], 50. Prove that Formula (4) holds for all nonzero vectors u and v, in a real inner product space V ., 51. Let TA : R 2 →R 2 be multiplication by, , , , A=, , , , 1, , 1, , −1, , 1, , and let x = (1, 1)., 2, , (a) Assuming that R has the Euclidean inner product, find, all vectors v in R 2 such that x, v = TA (x), TA (v)., (b) Assuming that R 2 has the weighted Euclidean inner product u, v = 2u1 v1 + 3u2 v2 , find all vectors v in R 2 such, that x, v = TA (x), TA (v)., 52. Let T : P2 →P2 be the linear transformation defined by, , T (a + bx + cx 2 ) = 3a − cx 2, and let p = 1 + x ., , (b) Find a basis for the orthogonal complement of the column, space of A., T2. In each part, confirm that the vectors u and v satisfy the, Cauchy–Schwarz inequality relative to the stated inner product., (a) M44 with the standard inner product., , ⎡, , ⎤, , 1, , 0, , 2, , 0, , ⎢0, ⎢, u=⎢, ⎣3, , −1, , 0, , 1⎥, ⎥, , 0, , 0, , 0, , 4, , −3, , ⎡, , ⎤, , 2, , 2, , 1, , 3, , ⎢ 3, ⎢, ⎥ and v = ⎢, ⎣ 1, 2⎦, , −1, , 0, , 1⎥, ⎥, , 0, , 0, , ⎥, −2 ⎦, , −3, , 1, , 2, , 0, , 0, , (b) R 4 with the weighted Euclidean inner product with weights, w1 = 21 , w2 = 41 , w3 = 18 , w4 = 18 ., u = (1, −2, 2, 1) and v = (0, −3, 3, −2)

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364, , Chapter 6 Inner Product Spaces, , 6.3 Gram–Schmidt Process; QR-Decomposition, In many problems involving vector spaces, the problem solver is free to choose any basis for, the vector space that seems appropriate. In inner product spaces, the solution of a problem, can often be simplified by choosing a basis in which the vectors are orthogonal to one, another. In this section we will show how such bases can be obtained., , Orthogonal and, Orthonormal Sets, , Recall from Section 6.2 that two vectors in an inner product space are said to be orthogonal, if their inner product is zero. The following definition extends the notion of orthogonality, to sets of vectors in an inner product space., DEFINITION 1 A set of two or more vectors in a real inner product space is said to be, orthogonal if all pairs of distinct vectors in the set are orthogonal. An orthogonal set, in which each vector has norm 1 is said to be orthonormal., , E X A M P L E 1 An Orthogonal Set in R 3, , Let, v1 = (0, 1, 0), v2 = (1, 0, 1), v3 = (1, 0, −1), 3, , and assume that R has the Euclidean inner product. It follows that the set of vectors, S = {v1 , v2 , v3 } is orthogonal since v1 , v2  = v1 , v3  = v2 , v3  = 0., , Note that Formula (1) is identical to Formula (4) of Section 3.2, but whereas Formula (4) was valid only for vectors in R n with the Euclidean, inner product, Formula (1) is, valid in general inner product, spaces., , It frequently happens that one has found a set of orthogonal vectors in an inner, product space but what is actually needed is a set of orthonormal vectors. A simple way, to convert an orthogonal set of nonzero vectors into an orthonormal set is to multiply, each vector v in the orthogonal set by the reciprocal of its length to create a vector of, norm 1 (called a unit vector). To see why this works, suppose that v is a nonzero vector, in an inner product space, and let, 1, u=, v, (1), v, Then it follows from Theorem 6.1.1(b) with k = v that, , $, $ , , $ 1 $  1 , $, $, ,  v = 1, u =$, v$ = , v, v , v, , v =1, , This process of multiplying a vector v by the reciprocal of its length is called normalizing v., We leave it as an exercise to show that normalizing the vectors in an orthogonal set of, nonzero vectors preserves the orthogonality of the vectors and produces an orthonormal, set., E X A M P L E 2 Constructing an Orthonormal Set, , The Euclidean norms of the vectors in Example 1 are, v1 = 1,, , v2 =, , √, , 2,, , v3 =, , √, , 2, , Consequently, normalizing u1 , u2 , and u3 yields, u1 =, , 1, , v1, v2, = (0, 1, 0), u2 =, =, v1, v2, v3, u3 =, =, v3, , 1, , 1, , √ , 0, − √, 2, , 1, , √ , 0, √, , 2, , !, , 2, , 2, , !, ,

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6.3 Gram–Schmidt Process; QR -Decomposition, , 365, , We leave it for you to verify that the set S = {u1 , u2 , u3 } is orthonormal by showing that, , u1 , u2  = u1 , u3  = u2 , u3  = 0 and, , u1 = u2 = u3 = 1, , In R 2 any two nonzero perpendicular vectors are linearly independent because neither, is a scalar multiple of the other; and in R 3 any three nonzero mutually perpendicular, vectors are linearly independent because no one lies in the plane of the other two (and, hence is not expressible as a linear combination of the other two). The following theorem, generalizes these observations., , S = {v1 , v2 , . . . , vn } is an orthogonal set of nonzero vectors in an, inner product space, then S is linearly independent., , THEOREM 6.3.1 If, , Proof Assume that, , k1 v1 + k2 v2 + · · · + kn vn = 0, , (2), , To demonstrate that S = {v1 , v2 , . . . , vn } is linearly independent, we must prove that, k1 = k2 = · · · = kn = 0., For each vi in S , it follows from (2) that, , k1 v1 + k2 v2 + · · · + kn vn , vi  = 0, vi  = 0, or, equivalently,, , k1 v1 , vi  + k2 v2 , vi  + · · · + kn vn , vi  = 0, From the orthogonality of S it follows that vj , vi  = 0 when j  = i , so this equation, reduces to, k i v i , v i  = 0, Since the vectors in S are assumed to be nonzero, it follows from the positivity axiom, for inner products that vi , vi   = 0. Thus, the preceding equation implies that each ki in, Equation (2) is zero, which is what we wanted to prove., Since an orthonormal set is orthogonal, and since its vectors, are nonzero (norm 1), it follows from Theorem 6.3.1 that, every orthonormal set is linearly independent., , In an inner product space, a basis consisting of orthonormal vectors is called an, orthonormal basis, and a basis consisting of orthogonal vectors is called an orthogonal, basis. A familiar example of an orthonormal basis is the standard basis for R n with the, Euclidean inner product:, e1 = (1, 0, 0, . . . , 0), e2 = (0, 1, 0, . . . , 0), . . . , en = (0, 0, 0, . . . , 1), E X A M P L E 3 An Orthonormal Basis for Pn, , Recall from Example 7 of Section 6.1 that the standard inner product of the polynomials, p = a0 + a1 x + · · · + an x n and q = b0 + b1 x + · · · + bn x n, is, , p, q = a0 b0 + a1 b1 + · · · + an bn, and the norm of p relative to this inner product is, p =, , #, ", p, p = a02 + a12 + · · · + an2, , You should be able to see from these formulas that the standard basis, , *, ), S = 1, x, x 2 , . . . , x n, , is orthonormal with respect to this inner product.

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6.3 Gram–Schmidt Process; QR -Decomposition, , 367, , Using the terminology and notation from Definition 2 of Section 4.4, it follows from, Theorem 6.3.2 that the coordinate vector of a vector u in V relative to an orthogonal, basis S = {v1 , v2 , . . . , vn } is, , (u)S =, , u, v1  u, v2 , u, vn , ,, ,...,, 2, 2, 2, v1, , v2, , !, (6), , vn, , and relative to an orthonormal basis S = {v1 , v2 , . . . , vn } is, , (u)S = (u, v1 , u, v2 , . . . , u, vn ), , (7), , E X A M P L E 5 A Coordinate Vector Relative to an Orthonormal Basis, , Let, , , , , , 0, 45, It is easy to check that S = {v1 , v2 , v3 } is an orthonormal basis for R 3 with the Euclidean, inner product. Express the vector u = (1, 1, 1) as a linear combination of the vectors in, S , and find the coordinate vector (u)S ., v1 = (0, 1, 0), v2 = − 45 , 0,, , 3, 5, , , , 3, , , v3 =, , 5, , Solution We leave it for you to verify that, , u, v1  = 1, u, v2  = − 15 , and u, v3  =, , 7, 5, , Therefore, by Theorem 6.3.2 we have, u = v1 − 15 v2 + 75 v3, that is,, ,  4, , , , − 5 , 0, 35 + 75 35 , 0, 45, Thus, the coordinate vector of u relative to S is, , , (u)S = (u, v1 , u, v2 , u, v3 ) = 1, − 15 , 75, (1, 1, 1) = (0, 1, 0) −, , 1, 5, , E X A M P L E 6 An Orthonormal Basis from an Orthogonal Basis, , (a) Show that the vectors, w1 = (0, 2, 0), w2 = (3, 0, 3), w3 = (−4, 0, 4), form an orthogonal basis for R 3 with the Euclidean inner product, and use that, basis to find an orthonormal basis by normalizing each vector., (b) Express the vector u = (1, 2, 4) as a linear combination of the orthonormal basis, vectors obtained in part (a)., Solution (a) The given vectors form an orthogonal set since, , w1 , w2  = 0, w1 , w3  = 0, w2 , w3  = 0, It follows from Theorem 6.3.1 that these vectors are linearly independent and hence form, a basis for R 3 by Theorem 4.5.4. We leave it for you to calculate the norms of w1 , w2 ,, and w3 and then obtain the orthonormal basis, v1 =, , 1, , w1, w2, = (0, 1, 0), v2 =, =, w1, w2, v3 =, , w3, =, w3, , 1, , 1, , − √ , 0, √, 2, , 1, , √ , 0, √, , 2, , !, , 2, , 2, , !, ,

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368, , Chapter 6 Inner Product Spaces, Solution (b) It follows from Formula (4) that, , u = u, v1 v1 + u, v2 v2 + u, v3 v3, We leave it for you to confirm that, , u, v1  = (1, 2, 4) · (0, 1, 0) = 2, 1, , u, v2  = (1, 2, 4) ·, , 1, , !, , 5, , =√, 2, 2, 2, !, 1, 1, 3, u, v3  = (1, 2, 4) · − √ , 0, √, =√, √ , 0, √, , 2, , 2, , and hence that, 5, , 1, , (1, 2, 4) = 2(0, 1, 0) + √, , √ , 0, √, , 2, , Orthogonal Projections, , 1, , 2, , 2, , !, , 2, , 3, , +√, , 2, , 1, , 1, , !, , − √ , 0, √, 2, , 2, , Many applied problems are best solved by working with orthogonal or orthonormal, basis vectors. Such bases are typically found by starting with some simple basis (say a, standard basis) and then converting that basis into an orthogonal or orthonormal basis., To explain exactly how that is done will require some preliminary ideas about orthogonal, projections., In Section 3.3 we proved a result called the Projection Theorem (see Theorem 3.3.2), that dealt with the problem of decomposing a vector u in R n into a sum of two terms,, w1 and w2 , in which w1 is the orthogonal projection of u on some nonzero vector a and, w2 is orthogonal to w1 (Figure 3.3.2). That result is a special case of the following more, general theorem, which we will state without proof., , THEOREM 6.3.3 Projection Theorem, , If W is a finite-dimensional subspace of an inner product space V, then every vector u, in V can be expressed in exactly one way as, u = w1 + w2, , (8), , where w1 is in W and w2 is in W ⊥ ., , The vectors w1 and w2 in Formula (8) are commonly denoted by, w1 = projW u and w2 = projW ⊥ u, , W⊥, , u, , 0, , projW u, , Figure 6.3.1, , (9), , These are called the orthogonal projection of u on W and the orthogonal projection of u, on W ⊥ , respectively. The vector w2 is also called the component of u orthogonal to W ., Using the notation in (9), Formula (8) can be expressed as, u = projW u + projW ⊥ u, , (10), , projW⊥ u, , (Figure 6.3.1). Moreover, since projW ⊥ u = u − projW u, we can also express Formula, (10) as, W, , u = projW u + (u − projW u), , (11)

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6.3 Gram–Schmidt Process; QR -Decomposition, , 371, , Step 4. To determine a vector v4 that is orthogonal to v1 , v2 , and v3 , we compute the, component of u4 orthogonal to the space W3 spanned by v1 , v2 , and v3 . From (12),, v4 = u4 − projW3 u4 = u4 −, , u4 , v1 , v1, , 2, , v1 −, , u4 , v2 , v2, , 2, , v2 −, , u4 , v3 , v3, , 2, , v3, , Continuing in this way we will produce after r steps an orthogonal set of nonzero, vectors {v1 , v2 , . . . , vr }. Since such sets are linearly independent, we will have produced, an orthogonal basis for the r -dimensional space W . By normalizing these basis vectors, we can obtain an orthonormal basis., The step-by-step construction of an orthogonal (or orthonormal) basis given in, the foregoing proof is called the Gram–Schmidt process. For reference, we provide the, following summary of the steps., , The Gram–Schmidt Process, To convert a basis {u1 , u2 , . . . , ur } into an orthogonal basis {v1 , v2 , . . . , vr }, perform, the following computations:, Step 1. v1 = u1, Step 2. v2 = u2 −, Step 3. v3 = u3 −, Step 4. v4 = u4 −, , .., ., , u2 , v1 , v1, , 2, , v1, , 2, , v1, , 2, , u3 , v1 , u4 , v1 , , v1, v1 −, v1 −, , u3 , v2 , v2, , 2, , v2, , 2, , u4 , v2 , , v2, v2 −, , u4 , v3 , v3, , 2, , v3, , (continue for r steps), Optional Step. To convert the orthogonal basis into an orthonormal basis, , {q1 , q2 , . . . , qr }, normalize the orthogonal basis vectors., , Historical Note Erhardt Schmidt (1875–1959) was a German mathematician, who studied for his doctoral degree at Göttingen University under David, Hilbert, one of the giants of modern mathematics. For most of his life he taught, at Berlin University where, in addition to making important contributions to, many branches of mathematics, he fashioned some of Hilbert’s ideas into a, general concept, called a Hilbert space—a fundamental structure in the study, of infinite-dimensional vector spaces. He first described the process that bears, his name in a paper on integral equations that he published in 1907., , Jorgen Pederson Gram, (1850–1916), , Historical Note Gram was a Danish actuary whose early education was at village schools supplemented by private tutoring. He obtained a doctorate degree, in mathematics while working for the Hafnia Life Insurance Company, where, he specialized in the mathematics of accident insurance. It was in his dissertation that his contributions to the Gram–Schmidt process were formulated., He eventually became interested in abstract mathematics and received a gold, medal from the Royal Danish Society of Sciences and Letters in recognition of, his work. His lifelong interest in applied mathematics never wavered, however,, and he produced a variety of treatises on Danish forest management., [Image: http://www-history.mcs.st-and.ac.uk/PictDisplay/Gram.html]

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372, , Chapter 6 Inner Product Spaces, , E X A M P L E 8 Using the Gram–Schmidt Process, , Assume that the vector space R 3 has the Euclidean inner product. Apply the Gram–, Schmidt process to transform the basis vectors, u1 = (1, 1, 1), u2 = (0, 1, 1), u3 = (0, 0, 1), into an orthogonal basis {v1 , v2 , v3 }, and then normalize the orthogonal basis vectors to, obtain an orthonormal basis {q1 , q2 , q3 }., Solution, , Step 1. v1 = u1 = (1, 1, 1), Step 2. v2 = u2 − projW1 u2 = u2 −, , u2 , v1 , 2, , v1, , v1, , !, , 2, 3, , 2 1 1, 3 3 3,  u 3 , v1 , u3 , v2 , Step 3. v3 = u3 − projW2 u3 = u3 −, v1 −, v2, 2, v1, v2 2, , = (0, 1, 1) − (1, 1, 1) = − , ,, , 1/3, 2/3, , 1, 3, , = (0, 0, 1) − (1, 1, 1) −, 1 1, 2 2, , !, , 2 1 1, 3 3 3, , !, , − , ,, , = 0, − ,, Thus,, , !, , 2 1 1, , v3 =, 3 3 3, , v1 = (1, 1, 1), v2 =, , − , ,, , 1 1, 0, − ,, 2 2, , !, , form an orthogonal basis for R 3 . The norms of these vectors are, v1 =, , √, , √, , v2 =, , 3,, , 6, ,, 3, , 1, v3 = √, 2, , so an orthonormal basis for R 3 is, v1, =, q1 =, v1, , !, !, v2, 2, 1, 1, = −√ , √ , √ ,, √ , √ , √ , q2 =, 1, , 1, , 3, , 3, , q3 =, , 1, , v2, , 3, , 1, 1, 0, − √ , √, 2, 2, , v3, =, v3, , !, , 6, , 6, , 6, , Remark In the last example we normalized at the end to convert the orthogonal basis into an, orthonormal basis. Alternatively, we could have normalized each orthogonal basis vector as soon, as it was obtained, thereby producing an orthonormal basis step by step. However, that procedure, generally has the disadvantage in hand calculation of producing more square roots to manipulate., A more useful variation is to “scale” the orthogonal basis vectors at each step to eliminate some of, the fractions. For example, after Step 2 above, we could have multiplied by 3 to produce (−2, 1, 1), as the second orthogonal basis vector, thereby simplifying the calculations in Step 3., , CA L C U L U S R E Q U I R E D, , E X A M P L E 9 Legendre Polynomials, , Let the vector space P2 have the inner product, , , , p, q =, , 1, , −1, , p(x)q(x) dx, , Apply the Gram–Schmidt process to transform the standard basis {1, x, x 2 } for P2 into, an orthogonal basis {φ1 (x), φ2 (x), φ3 (x)}.

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6.3 Gram–Schmidt Process; QR -Decomposition, Solution Take u1, , 373, , = 1, u2 = x , and u3 = x 2 ., , Step 1. v1 = u1 = 1, Step 2. We have, , , u2 , v1  =, , so, v 2 = u2 −, Step 3. We have, , 1, , −1, , u2 , v1 , , 1, , v1 = u2 = x, , x dx =, 2, , −1, , , u3 , v2  =, , 2, , v1, , , u3 , v1  =, , x dx = 0, , 1, , −1, , x 3 dx =, , , v1, , 2, , = v1 , v1  =, , so, v3 = u3 −, , u3 , v1 , v1, , 2, , 1, , −1, , v1 −, , x3, , 1, =, , 3, , x, , 4, , −1, , 2, 3, , 1, , 4, , =0, −1, , 1, , 1 dx = x, , =2, −1, , u3 , v2 , v2, , 2, , v2 = x 2 −, , 1, 3, , Thus, we have obtained the orthogonal basis {φ1 (x), φ2 (x), φ3 (x)} in which, 1, φ1 (x) = 1, φ2 (x) = x, φ3 (x) = x 2 −, 3, Remark The orthogonal basis vectors in the last example are often scaled so all three functions, have a value of 1 at x = 1. The resulting polynomials, 1, (3x 2 − 1), 1, x,, 2, which are known as the first three Legendre polynomials, play an important role in a variety of, applications. The scaling does not affect the orthogonality., , Extending Orthonormal, Sets to Orthonormal Bases, , Recall from part (b) of Theorem 4.5.5 that a linearly independent set in a finite-dimensional, vector space can be enlarged to a basis by adding appropriate vectors. The following theorem is an analog of that result for orthogonal and orthonormal sets in finite-dimensional, inner product spaces., THEOREM 6.3.6 If W is a finite-dimensional inner product space, then:, , (a) Every orthogonal set of nonzero vectors in W can be enlarged to an orthogonal, basis for W ., (b) Every orthonormal set in W can be enlarged to an orthonormal basis for W ., We will prove part (b) and leave part (a) as an exercise., , S = {v1 , v2 , . . . , vs } is an orthonormal set of vectors in W ., Part (b) of Theorem 4.5.5 tells us that we can enlarge S to some basis, , Proof (b) Suppose that, , S = {v1 , v2 , . . . , vs , vs+1 , . . . , vk }, for W . If we now apply the Gram–Schmidt process to the set S , then the vectors, v1 , v2 , . . . , vs , will not be affected since they are already orthonormal, and the resulting, set, S = {v1 , v2 , . . . , vs , vs+1 , . . . , vk }, will be an orthonormal basis for W .

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6.3 Gram–Schmidt Process; QR -Decomposition, , 375, , with orthonormal column vectors and an invertible upper triangular matrix R . We call, Equation (15) a QR -decomposition of A. In summary, we have the following theorem., , THEOREM 6.3.7 QR -Decomposition, , It is common in numerical, linear algebra to say that a matrix with linearly independent, columns has full column rank., , If A is an m × n matrix with linearly independent column vectors, then A can be factored as, , A = QR, where Q is an m × n matrix with orthonormal column vectors, and R is an n × n, invertible upper triangular matrix., , Recall from Theorem 5.1.5 (the Equivalence Theorem) that a square matrix has, linearly independent column vectors if and only if it is invertible. Thus, it follows from, Theorem 6.3.7 that every invertible matrix has a QR -decomposition., E X A M P L E 10 QR -Decomposition of a 3 × 3 Matrix, , Find a QR -decomposition of, , ⎡, , 1, ⎢, A = ⎣1, 1, Solution The column vectors of A are, , ⎡ ⎤, , ⎤, , 0, 1, 1, , 0, ⎥, 0⎦, 1, , ⎡ ⎤, , ⎡ ⎤, , 1, 0, 0, ⎢ ⎥, ⎢ ⎥, ⎢ ⎥, u1 = ⎣1⎦, u2 = ⎣1⎦, u3 = ⎣0⎦, 1, 1, 1, Applying the Gram–Schmidt process with normalization to these column vectors yields, the orthonormal vectors (see Example 8), , ⎡, q1 =, , √1, , 3, ⎢ 1, ⎢√, ⎣ 3, √1, 3, , ⎤, , ⎡, , − √26, , ⎥, ⎢, ⎥, q2 = ⎢, ⎦, ⎣, , √1, , 6, √1, 6, , ⎤, , ⎡, , ⎤, , 0, , ⎥, ⎥, ⎢, ⎥, q3 = ⎢− √1 ⎥, ⎦, ⎣ 2 ⎦, √1, , 2, , Thus, it follows from Formula (16) that R is, , ⎡, , u1 , q1 , , ⎢, R=⎣, , 0, 0, , u2 , q1 , u2 , q2 , 0, , ⎤ ⎡ √3, 3, u3 , q1 , ⎥ ⎢, ⎢, 0, u3 , q2 ⎦ = ⎣, u3 , q3 , 0, , √2, , 3, √2, 6, , 0, , √1, , 3, √1, 6, √1, 2, , ⎤, ⎥, ⎥, ⎦, , from which it follows that a QR -decomposition of A is, , ⎡, , 1, ⎢, ⎣1, 1, , 0, 1, 1, , ⎡, , ⎤, , √1, , 3, 0, ⎢ 1, ⎥, √, 0⎦ = ⎢, ⎣ 3, 1, √1, 3, , A, , =, , − √26, √1, , 6, √1, 6, , Q, , ⎤ ⎡, , 0, , √3, , √2, , 0, , 0, , 3, , ⎥ ⎢, ⎢, − √12 ⎥, ⎦ ⎣0, √1, , 2, , 3, √2, 6, , R, , √1, , 3, √1, 6, √1, 2, , ⎤, ⎥, ⎥, ⎦

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376, , Chapter 6 Inner Product Spaces, , Exercise Set 6.3, 1. In each part, determine whether the set of vectors is orthogonal and whether it is orthonormal with respect to the Euclidean inner product on R 2 ., (a) (0, 1), (2, 0), , ', , √1, , form an orthogonal basis for R 3 with respect to the Euclidean, inner product, and then use Theorem 6.3.2(a) to express the, vector u = (−1, 0, 2) as a linear combination of v1 , v2 , and v3 ., , 2, , (d) (0, 0), (0, 1), , 10. Verify that the vectors, , 2. In each part, determine whether the set of vectors is orthogonal and whether it is orthonormal with respect to the Euclidean inner product on R 3 ., , ', , ( ', ( ', (, , √13 , √13 , − √13 , − √12 , 0, √12, ,  ,  , , (b) 23 , − 23 , 13 , 23 , 13 , − 23 , 13 , 23 , 23, ', (, (c) (1, 0, 0), 0, √12 , √12 , (0, 0, 1), ', ( ', (, (d) √16 , √16 , − √26 , √12 , − √12 , 0, (a), , √1, , 2, , , 0,, , 9. Verify that the vectors, v1 = (2, −2, 1), v2 = (2, 1, −2), v3 = (1, 2, 2), , ( ', (, , √12 , √12, ', ( ', (, (c) − √12 , − √12 , √12 , √12, (b) − √12 ,, , 8. Use Theorem 6.3.2(b) to express the vector u = (3, −7, 4) as, a linear combination of the vectors v1 , v2 , and v3 in Exercise 7., , √1, , 2, , v1 = (1, −1, 2, −1),, v3 = (1, 2, 0, −1),, , form an orthogonal basis for R with respect to the Euclidean, inner product, and then use Theorem 6.3.2(a) to express the, vector u = (1, 1, 1, 1) as a linear combination of v1 , v2 , v3 ,, and v4 ., , 11. Exercise 7, , 2, 3, 1, 3, , p3 (x) =, , − 23 x + 13 x 2 , p2 (x) =, , + 13 x − 23 x 2 ,, , 2, 3, , + 23 x + 23 x 2, , (b) p1 (x) = 1, p2 (x) =, , √1, , 2, , x+, , √1, , 2, , x 2 , p3 (x) = x 2, , 4. In each part, determine whether the set of vectors is orthogonal with respect to the standard inner product on M22 (see, Example 6 of Section 6.1)., 1, 0, , 0, ,, 0, , 1, (b), 0, , 0, ,, 0, , (a), , , , 0, , 2, 3, , 1, 3, , − 23, , 0, 0, , 1, ,, 0, , , , , , ,, 0, 1, , 0, , 2, 3, , − 23, , 1, 3, , 0, ,, 1, , , , , , ,, 0, 1, , 0, , 1, 3, , 2, 3, , 2, 3, , , , ⎡1, , ⎥, , ⎢, , 2, , 0, , 5. A = ⎣ 0, , 0, , 5⎦, , −1, , 2, , 0, , ⎢, , 5, , , , , , 14. Exercise 10, 2, , In Exercises 15–18, let R have the Euclidean inner product., (a) Find the orthogonal projection of u onto the line spanned by, the vector v., (b) Find the component of u orthogonal to the line spanned by, the vector v, and confirm that this component is orthogonal, to the line., 15. u = (−1, 6); v =, , 3, 5, , ,, , 4, 5, , , , 16. u = (2, 3); v =, , 5, 13, , ,, , 12, 13, , , , 18. u = (3, −1); v = (3, 4), , In Exercises 19–22, let R have the Euclidean inner product., (a) Find the orthogonal projection of u onto the plane spanned, by the vectors v1 and v2 ., , − 21, , 1, 6. A = ⎢, ⎣5, , 1, 2, , 1, 5, , 0, , 1⎤, 3, , ⎥, , 1⎥, 3⎦, , − 23, , 7. Verify that the vectors, v1 = − 35 , 45 , 0 , v2 =, , 13. Exercise 9, , 3, , 0, −1, , ⎤, , 1, , 12. Exercise 8, , 17. u = (2, 3); v = (1, 1), , In Exercises 5–6, show that the column vectors of A form an, orthogonal basis for the column space of A with respect to the, Euclidean inner product, and then find an orthonormal basis for, that column space., , ⎡, , v4 = (1, 0, 0, 1), 4, , In Exercises 11–14, find the coordinate vector (u)S for the vector u and the basis S that were given in the stated exercise., , 3. In each part, determine whether the set of vectors is orthogonal with respect to the standard inner product on P2 (see, Example 7 of Section 6.1)., (a) p1 (x) =, , v2 = (−2, 2, 3, 2),, , 4, , , , 3 , 0 , v3 = (0, 0, 1), 5 5, , form an orthonormal basis for R 3 with respect to the Euclidean inner product, and then use Theorem 6.3.2(b) to express the vector u = (1, −2, 2) as a linear combination of v1 ,, v2 , and v3 ., , (b) Find the component of u orthogonal to the plane spanned, by the vectors v1 and v2 , and confirm that this component is, orthogonal to the plane., , , , , , 23 , − 23 , v2 = 23 , 13 , 23, ', ', (, 20. u = (3, −1, 2); v1 = √16 , √16 , − √26 , v2 = √13 , √13 ,, 19. u = (4, 2, 1); v1 =, , 1, 3, , √1, , (, , 3, , 21. u = (1, 0, 3); v1 = (1, −2, 1), v2 = (2, 1, 0), 22. u = (1, 0, 2); v1 = (3, 1, 2), v2 = (−1, 1, 1), In Exercises 23–24, the vectors v1 and v2 are orthogonal with, respect to the Euclidean inner product on R 4 . Find the orthogonal projection of b = (1, 2, 0, −2) on the subspace W spanned by, these vectors., 23. v1 = (1, 1, 1, 1), v2 = (1, 1, −1, −1), 24. v1 = (0, 1, −4, −1), v2 = (3, 5, 1, 1)

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6.3 Gram–Schmidt Process; QR -Decomposition, , In Exercises 25–26, the vectors v1 , v2 , and v3 are orthonormal with respect to the Euclidean inner product on R 4 . Find the, orthogonal projection of b = (1, 2, 0, −1) onto the subspace W, spanned by these vectors., , ', , 25. v1 = 0,, v3 =, , ', , √1, , √1, , 18, , 18, , , − √4, , 18, , , − √1, , , 0, √118 , − √418, , (, , 18, , (, , , v2 =, , 1, 2, , ,, , 5 1 1, , ,, 6 6 6, , , , ,, , 377, , 38. Verify that the set of vectors {(1, 0), (0, 1)} is orthogonal with, respect to the inner product u, v = 4u1 v1 + u2 v2 on R 2 ; then, convert it to an orthonormal set by normalizing the vectors., 39. Find vectors x and y in R 2 that are orthonormal with respect, to the inner product u, v = 3u1 v1 + 2u2 v2 but are not orthonormal with respect to the Euclidean inner product., , , , , , 26. v1 = 21 , 21 , 21 , 21 , v2 = 21 , 21 , − 21 , − 21 ,, , 1, v3 = 2 , − 21 , 21 , − 21, , 40. In Example 3 of Section 4.9 we found the orthogonal projection of the vector x = (1, 5) onto the line through the origin, making an angle of π/6 radians with the positive x -axis. Solve, that same problem using Theorem 6.3.4., , In Exercises 27–28, let R 2 have the Euclidean inner product, and use the Gram–Schmidt process to transform the basis {u1 , u2 }, into an orthonormal basis. Draw both sets of basis vectors in the, xy -plane., , 41. This exercise illustrates that the orthogonal projection resulting from Formula (12) in Theorem 6.3.4 does not depend on, which orthogonal basis vectors are used., , 27. u1 = (1, −3), u2 = (2, 2), , 28. u1 = (1, 0), u2 = (3, −5), , (a) Let R 3 have the Euclidean inner product, and let W be the, subspace of R 3 spanned by the orthogonal vectors, v1 = (1, 0, 1) and v2 = (0, 1, 0), , In Exercises 29–30, let R 3 have the Euclidean inner product and, use the Gram–Schmidt process to transform the basis {u1 , u2 , u3 }, into an orthonormal basis., , Show that the orthogonal vectors, , 29. u1 = (1, 1, 1), u2 = (−1, 1, 0), u3 = (1, 2, 1), , span the same subspace W ., , 30. u1 = (1, 0, 0), u2 = (3, 7, −2), u3 = (0, 4, 1), 31. Let R 4 have the Euclidean inner product. Use the Gram–, Schmidt process to transform the basis {u1 , u2 , u3 , u4 } into an, orthonormal basis., u1 = (0, 2, 1, 0),, , u2 = (1, −1, 0, 0),, , u3 = (1, 2, 0, −1),, , u4 = (1, 0, 0, 1), , 32. Let R 3 have the Euclidean inner product. Find an orthonormal basis for the subspace spanned by (0, 1, 2), (−1, 0, 1),, (−1, 1, 3)., , v1 = (1, 1, 1) and v2 = (1, −2, 1), (b) Let u = (−3, 1, 7) and show that the same vector projW u, results regardless of which of the bases in part (a) is used, for its computation., 42. (Calculus required ) Use Theorem 6.3.2(a) to express the following polynomials as linear combinations of the first three, Legendre polynomials (see the Remark following Example 9)., (a) 1 + x + 4x 2, , (b) 2 − 7x 2, , (c) 4 + 3x, , 43. (Calculus required ) Let P2 have the inner product, , , , 1, , p, q =, , p(x)q(x) dx, 0, , 33. Let b and W be as in Exercise 23. Find vectors w1 in W and, w2 in W ⊥ such that b = w1 + w2 ., , Apply the Gram–Schmidt process to transform the standard, basis S = {1, x, x 2 } into an orthonormal basis., , 34. Let b and W be as in Exercise 25. Find vectors w1 in W and, w2 in W ⊥ such that b = w1 + w2 ., , 44. Find an orthogonal basis for the column space of the matrix, , ⎡, , 35. Let R 3 have the Euclidean inner product. The subspace of, R 3 spanned by the vectors u1 = (1, 1, 1) and u2 = (2, 0, −1), is a plane passing through the origin. Express w = (1, 2, 3), in the form w = w1 + w2 , where w1 lies in the plane and w2 is, perpendicular to the plane., 36. Let R 4 have the Euclidean inner product. Express the vector, w = (−1, 2, 6, 0) in the form w = w1 + w2 , where w1 is in the, space W spanned by u1 = (−1, 0, 1, 2) and u2 = (0, 1, 0, 1),, and w2 is orthogonal to W ., 37. Let R 3 have the inner product, , u, v = u1 v1 + 2u2 v2 + 3u3 v3, Use the Gram–Schmidt process to transform u1 = (1, 1, 1),, u2 = (1, 1, 0), u3 = (1, 0, 0) into an orthonormal basis., , 6, , ⎤, −5, 1⎥, ⎥, ⎥, 5⎦, , 1, , ⎢ 2, ⎢, A=⎢, ⎣−2, , −2, , 6, , 8, , 1, , −7, , In Exercises 45–48, we obtained the column vectors of Q by, applying the Gram–Schmidt process to the column vectors of A., Find a QR -decomposition of the matrix A., 1, 45. A =, 2, , ⎡, , 1, 46. A = ⎣0, 1, , −1, 3, , ⎤, , , , , Q=, , √1, , 5, , √2, , 5, , ⎡, , √1, , 2, 2, ⎢, ⎢, ⎦, 1 , Q=⎣0, 4, √1, 2, , − √25, , , , √1, , 5, , − √13, √1, , ⎤, ⎥, ⎥, , 3⎦, , √1, , 3

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378, , Chapter 6 Inner Product Spaces, , ⎡, , 1, 47. A = ⎣0, 1, , ⎡, , 1, , 48. A = ⎣1, 0, , 0, 1, 2, , 2, 1, 3, , ⎤, , ⎡, , √1, , 2, 2, ⎢, ⎢, ⎦, 1 , Q=⎣0, 0, √1, , ⎤, , ⎡, , − √13, , 1, ⎢, ⎢, 1⎦ , Q = ⎢ √12, ⎣, 1, 0, , ⎥, ⎥, 6⎦, , √2, , 3, , − √16, , √1, 3, , 2, , True-False Exercises, , 6, , √1, , √, √2, , √1, 2, , ⎤, , √1, , 2 19, , √, , − 2√219, √, , 3√ 2, 19, , − √319, , TF. In parts (a)–(f ) determine whether the statement is true or, false, and justify your answer., , ⎤, , ⎥, ⎥, ⎥, 19 ⎦, , √3, , √1, 19, , 49. Find a QR -decomposition of the matrix, , ⎡, , 1, ⎢−1, A=⎢, ⎣ 1, −1, , 0, 1, 0, 1, , ⎤, , 1, 1⎥, ⎥, 1⎦, 1, , 50. In the Remark following Example 8 we discussed two alternative ways to perform the calculations in the Gram–Schmidt, process: normalizing each orthogonal basis vector as soon as, it is calculated and scaling the orthogonal basis vectors at each, step to eliminate fractions. Try these methods in Example 8., , (a) Every linearly independent set of vectors in an inner product, space is orthogonal., (b) Every orthogonal set of vectors in an inner product space is, linearly independent., (c) Every nontrivial subspace of R 3 has an orthonormal basis with, respect to the Euclidean inner product., (d) Every nonzero finite-dimensional inner product space has an, orthonormal basis., (e) projW x is orthogonal to every vector of W ., (f ) If A is an n × n matrix with a nonzero determinant, then A, has a QR-decomposition., , Working withTechnology, T1. (a) Use the Gram–Schmidt process to find an orthonormal, basis relative to the Euclidean inner product for the column, space of, ⎡, ⎤, 1, 1, 1, 1, , Working with Proofs, 51. Prove part (a) of Theorem 6.3.6., 52. In Step 3 of the proof of Theorem 6.3.5, it was stated that “the, linear independence of {u1 , u2 , . . . , un } ensures that v3  = 0.”, Prove this statement., 53. Prove that the diagonal entries of R in Formula (16) are, nonzero., 54. Show that matrix Q in Example 10 has the property, QQT = I3 , and prove that every m × n matrix Q with orthonormal column vectors has the property QQT = Im ., 55. (a) Prove that if W is a subspace of a finite-dimensional vector space V , then the mapping T : V →W defined by, T (v) = projW v is a linear transformation., , ⎢1, ⎢, A=⎢, ⎣0, , 0, , 0, , 1, , 0, , 2⎦, , 2, , −1, , 1, , 1, , 1⎥, ⎥, , ⎥, , (b) Use the method of Example 9 to find a QR -decomposition, of A., T2. Let P4 have the evaluation inner product at the points, −2, −1, 0, 1, 2. Find an orthogonal basis for P4 relative to this, inner product by applying the Gram–Schmidt process to the vectors, p0 = 1, p1 = x, p2 = x 2 , p3 = x 3 , p4 = x 4, , (b) What are the range and kernel of the transformation in, part (a)?, , 6.4 Best Approximation; Least Squares, There are many applications in which some linear system Ax = b of m equations in n, unknowns should be consistent on physical grounds but fails to be so because of, measurement errors in the entries of A or b. In such cases one looks for vectors that come, as close as possible to being solutions in the sense that they minimize b − Ax with respect, to the Euclidean inner product on R m . In this section we will discuss methods for finding, such minimizing vectors., , Least Squares Solutions of, Linear Systems, , Suppose that Ax = b is an inconsistent linear system of m equations in n unknowns in, which we suspect the inconsistency to be caused by errors in the entries of A or b. Since, no exact solution is possible, we will look for a vector x that comes as “close as possible”, to being a solution in the sense that it minimizes b − Ax with respect to the Euclidean

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CHAPTER, , 7, , Diagonalization and, Quadratic Forms, CHAPTER CONTENTS, , 7.1 Orthogonal Matrices, , 401, , 7.2 Orthogonal Diagonalization, 7.3 Quadratic Forms, , 409, , 417, , 7.4 Optimization Using Quadratic Forms, , 429, , 7.5 Hermitian, Unitary, and Normal Matrices, INTRODUCTION, , 437, , In Section 5.2 we found conditions that guaranteed the diagonalizability of an n × n, matrix, but we did not consider what class or classes of matrices might actually satisfy, those conditions. In this chapter we will show that every symmetric matrix is, diagonalizable. This is an extremely important result because many applications utilize, it in some essential way., , 7.1 Orthogonal Matrices, In this section we will discuss the class of matrices whose inverses can be obtained by, transposition. Such matrices occur in a variety of applications and arise as well as, transition matrices when one orthonormal basis is changed to another., , Orthogonal Matrices, Recall from Theorem 1.6.3, that if either product in (1), holds, then so does the other., Thus, A is orthogonal if either, AAT = I or ATA = I ., , We begin with the following definition., DEFINITION 1 A square matrix A is said to be orthogonal if its transpose is the same, , as its inverse, that is, if, , A−1 = AT, , or, equivalently, if, , AAT = ATA = I, , (1), , E X A M P L E 1 A 3 × 3 Orthogonal Matrix, , ⎡, , The matrix, , A=, is orthogonal since, , ATA =, , ⎡3, 7, ⎢2, ⎢, ⎣7, 6, 7, , − 67, 3, 7, 2, 7, , 3, 7, ⎢ 6, ⎢−, ⎣ 7, 2, 7, , ⎤⎡, , 3, 2, 7, 7, ⎥⎢, 6⎥ ⎢ 6, −, 7⎦ ⎣ 7, 3, 2, −7, 7, , ⎤, , 2, 7, 3, 7, 6, 7, , 6, 7, ⎥, 2⎥, 7⎦, − 37, , 2, 7, 3, 7, 6, 7, , ⎤, , 6, 7, ⎥, 2⎥, 7⎦, − 37, , ⎡, , 1, , 0, , 0, , 0, , 0, , 1, , ⎤, , ⎥, ⎢, = ⎣0 1 0⎦, 401

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402, , Chapter 7 Diagonalization and Quadratic Forms, , E X A M P L E 2 Rotation and Reflection Matrices Are Orthogonal, , Recall from Table 5 of Section 4.9 that the standard matrix for the counterclockwise, rotation of R 2 through an angle θ is, , A=, , cos θ, sin θ, , − sin θ, cos θ, , This matrix is orthogonal for all choices of θ since, , ATA =, , cos θ, − sin θ, , sin θ, cos θ, , cos θ, sin θ, , 1 0, − sin θ, =, 0 1, cos θ, , We leave it for you to verify that the reflection matrices in Tables 1 and 2 and the rotation, matrices in Table 6 of Section 4.9 are all orthogonal., , Observe that for the orthogonal matrices in Examples 1 and 2, both the row vectors and the column vectors form orthonormal sets with respect to the Euclidean inner, product. This is a consequence of the following theorem., , THEOREM 7.1.1 The following are equivalent for an n × n matrix A., , (a) A is orthogonal., (b) The row vectors of A form an orthonormal set in R n with the Euclidean inner, product., (c) The column vectors of A form an orthonormal set in R n with the Euclidean inner, product., , Proof We will prove the equivalence of (a) and (b) and leave the equivalence of (a) and, , (c) as an exercise., (a) ⇔ (b) Let ri be the i th row vector and cj the j th column vector of A. Since transpos-, , ing a matrix converts its columns to rows and rows to columns, it follows that cTj = rj ., Thus, it follows from the row-column rule [Formula (5) of Section 1.3] and the bottom, form listed in Table 1 of Section 3.2 that, , ⎡, , r1 cT1, , ⎢ T, ⎢r2 c1, ⎢, T, AA = ⎢, ⎢ .., ⎢ ., ⎣, rn c1T, , r1 c2T, , ···, , r2 cT2, , ···, , .., ., rn cT2, , ···, , r1 cTn, , ⎤, ⎥, , ⎡, , r1 · r1, , ⎢, ⎢r2 · r1, ⎢, ⎥=⎢, .. ⎥ ⎢ .., ⎢, . ⎥, ⎦ ⎣ ., rn cTn, rn · r1, r2 cTn ⎥, ⎥, , ⎤, , r 1 · r2, , ···, , r1 · rn, , r2 · r2, , ···, , r2 · rn ⎥, ⎥, , .., ., rn · r2, , .., ., , ⎥, ⎥, ⎥, ⎥, ⎦, , · · · rn · r n, , It is evident from this formula that AAT = I if and only if, r1 · r1 = r2 · r2 = · · · = rn · rn = 1, WARNING Note that an or-, , thogonal matrix has orthonormal rows and columns—not, simply orthogonal rows and, columns., , and, ri · rj = 0 when i  = j, which are true if and only if {r1 , r2 , . . . , rn } is an orthonormal set in R n ., The following theorem lists four more fundamental properties of orthogonal matrices. The proofs are all straightforward and are left as exercises.

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7.1 Orthogonal Matrices, , 403, , THEOREM 7.1.2, , (a) The transpose of an orthogonal matrix is orthogonal., (b) The inverse of an orthogonal matrix is orthogonal., (c) A product of orthogonal matrices is orthogonal., (d ) If A is orthogonal, then det(A) = 1 or det(A) = −1., , E X A M P L E 3 det(A) = ±1 for an Orthogonal Matrix A, , The matrix, , , , √1, , √1, , 2, − √12, , A=, , , , 2, √1, 2, , is orthogonal since its row (and column) vectors form orthonormal sets in R 2 with, the Euclidean inner product. We leave it for you to verify that det(A) = 1 and that, interchanging the rows produces an orthogonal matrix whose determinant is −1., Orthogonal Matrices as, Linear Operators, , We observed in Example 2 that the standard matrices for the basic reflection and rotation, operators on R 2 and R 3 are orthogonal. The next theorem will explain why this is so., THEOREM 7.1.3 If A is an n × n matrix, then the following are equivalent., , (a) A is orthogonal., , Ax = x for all x in R n ., (c) Ax · Ay = x · y for all x and y in R n ., (b), , Proof We will prove the sequence of implications (a), (a) ⇒ (b) Assume that A is orthogonal, so that ATA, , ⇒ (b) ⇒ (c) ⇒ (a)., , = I . It follows from Formula (26), , of Section 3.2 that, , Ax = (Ax · Ax)1/2 = (x · ATAx)1/2 = (x · x)1/2 = x, (b) ⇒ (c) Assume that, , Ax = x for all x in R n . From Theorem 3.2.7 we have, , Ax · Ay =, , 1, 4, , Ax + Ay, , =, , 1, 4, , x+y, , 2, , 2, , −, , 1, 4, , (c) ⇒ (a) Assume that Ax · Ay, , −, , 1, 4, , Ax − Ay, , x−y, , 2, , 2, , =, , 1, 4, , A(x + y), , 2, , −, , 1, 4, , A(x − y), , 2, , =x·y, , = x · y for all x and y in R n . It follows from Formula (26), , of Section 3.2 that, x · y = x · ATAy, which can be rewritten as x · (ATAy − y) = 0 or as, x · (ATA − I )y = 0, Since this equation holds for all x in R n , it holds in particular if x = (ATA − I )y, so, , (ATA − I )y · (ATA − I )y = 0, Thus, it follows from the positivity axiom for inner products that, , (ATA − I )y = 0

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404, , Chapter 7 Diagonalization and Quadratic Forms, , Since this equation is satisfied by every vector y in R n , it must be that ATA − I is the, zero matrix (why?) and hence that ATA = I . Thus, A is orthogonal., , TA (u), , TA (v), , β, α, , v, u, , 0, ||TA(u)|| = ||u||, TA (v)|| = ||v||, α = β, d (TA (u), TA (v)) = d(u, v), , Figure 7.1.1, , Change of Orthonormal, Basis, , Theorem 7.1.3 has a useful geometric interpretation when considered from the viewpoint of matrix transformations: If A is an orthogonal matrix and TA : R n →R n is multiplication by A, then we will call TA an orthogonal operator on R n . It follows from parts, (a) and (b) of Theorem 7.1.3 that the orthogonal operators on R n are precisely those, operators that leave the lengths (norms) of vectors unchanged. However, as illustrated, in Figure 7.1.1, this implies that orthogonal operators also leave angles and distances, between vectors in R n unchanged since these can be expressed in terms of norms [see, Definition 2 and Formula (20) of Section 3.2]., Orthonormal bases for inner product spaces are convenient because, as the following, theorem shows, many familiar formulas hold for such bases. We leave the proof as an, exercise., THEOREM 7.1.4 If S is an orthonormal basis for an n-dimensional inner product space, , V, and if, (u)S = (u1 , u2 , . . . , un ) and (v)S = (v1 , v2 , . . . , vn ), then:, , #, u21 + u22 + · · · + un2, ", (b) d(u, v) = (u1 − v1 )2 + (u2 − v2 )2 + · · · + (un − vn )2, (c) u, v = u1 v1 + u2 v2 + · · · + un vn, (a), , u =, , Remark Note that the three parts of Theorem 7.1.4 can be expressed as, u = (u)S, , , , d(u, v) = d (u)S , (v)S, , /, 0, u, v = (u)S , (v)S, , where the norm, distance, and inner product on the left sides are relative to the inner product on, , V and on the right sides are relative to the Euclidean inner product on R n ., , Transitions between orthonormal bases for an inner product space are of special, importance in geometry and various applications. The following theorem, whose proof, is deferred to the end of this section, is concerned with transitions of this type., THEOREM 7.1.5 Let, , V be a finite-dimensional inner product space. If P is the transition matrix from one orthonormal basis for V to another orthonormal basis for V,, then P is an orthogonal matrix., , E X A M P L E 4 Rotation of Axes in 2-Space, , In many problems a rectangular xy-coordinate system is given, and a new x y -coordinate, system is obtained by rotating the xy-system counterclockwise about the origin through, an angle θ . When this is done, each point Q in the plane has two sets of coordinates—, coordinates (x, y) relative to the xy-system and coordinates (x , y ) relative to the x y system (Figure 7.1.2a)., By introducing unit vectors u1 and u2 along the positive x - and y -axes and unit vectors u1 and u2 along the positive x - and y -axes, we can regard this rotation as a change

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7.1 Orthogonal Matrices, , 405, , from an old basis B = {u1 , u2 } to a new basis B = {u1 , u2 } (Figure 7.1.2b). Thus, the, new coordinates (x , y ) and the old coordinates (x, y) of a point Q will be related by, , x, y, , x, y, , = P −1, , (2), , where P is the transition from B to B . To find P we must determine the coordinate, matrices of the new basis vectors u1 and u2 relative to the old basis. As indicated in, Figure 7.1.2c, the components of u1 in the old basis are cos θ and sin θ , so, cos θ, sin θ, , [u1 ]B =, , Similarly, from Figure 7.1.2d we see that the components of u2 in the old basis are, cos(θ + π/2) = − sin θ and sin(θ + π/2) = cos θ , so, , − sin θ, cos θ, , [u2 ]B =, Thus the transition matrix from B to B is, cos θ, P =, sin θ, , − sin θ, cos θ, , (3), , Observe that P is an orthogonal matrix, as expected, since B and B are orthonormal, bases. Thus, cos θ sin θ, P −1 = P T =, − sin θ cos θ, so (2) yields, cos θ sin θ, x, x, =, (4), y, − sin θ cos θ y, or, equivalently,, , x =, , x cos θ + y sin θ, , (5), , y = −x sin θ + y cos θ, These are sometimes called the rotation equations for R 2 ., y´, , y, , y´, Q, , (x, y), u´, (x´, y´) 2, , y´, , θ, , cos θ, , u1, , (a), , x´, , u1´, , u1´, , x, , y, , u2´, , u2, , x´, θ, , y, , θ, , sin θ, , π, sin θ + 2, , (, , π, cos θ + 2, , (c), , x´, θ, , x, , (, , (b), , π, θ +2, , ), , x, , ), (d ), , Figure 7.1.2, , E X A M P L E 5 Rotation of Axes in 2-Space, , Use form (4) of the rotation equations for R 2 to find the new coordinates of the point, Q(2, 1) if the coordinate axes of a rectangular coordinate system are rotated through an, angle of θ = π/4., Solution Since, , sin, , π, 4, , = cos, , π, 4, , 1, , =√, , 2

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406, , Chapter 7 Diagonalization and Quadratic Forms, , the equation in (4) becomes, , ⎡, x, y, , √1, , √1, , =⎣, − √12, 2, , 2, , √1, 2, , ⎤, ⎦, , x, y, , Thus, if the old coordinates of a point Q are (x, y) = (2, −1), then, , ⎡, , x, y, , √1, , =⎣, − √12, 2, , √1, , 2, , √1, , so the new coordinates of Q are (x , y ) =, , 2, , ⎡, , ⎤, , √1, , ⎤, , 2, 2, ⎦, ⎦, =⎣, 3, −1, −√, , ', , √1, , 2, , 2, , (, , , − √32 ., , Remark Observe that the coefficient matrix in (4) is the same as the standard matrix for the, linear operator that rotates the vectors of R 2 through the angle −θ (see margin note for Table 5, of Section 4.9). This is to be expected since rotating the coordinate axes through the angle θ with, the vectors of R 2 kept fixed has the same effect as rotating the vectors in R 2 through the angle −θ, with the axes kept fixed., z, u3, , z´, , E X A M P L E 6 Application to Rotation of Axes in 3-Space, , u3´, u2´, , y´, y, , u1, x, , u2, , u1´, θ, , Suppose that a rectangular xyz-coordinate system is rotated around its z-axis counterclockwise (looking down the positive z-axis) through an angle θ (Figure 7.1.3). If we, introduce unit vectors u1 , u2 , and u3 along the positive x -, y -, and z-axes and unit vectors u1 , u2 , and u3 along the positive x -, y -, and z -axes, we can regard the rotation as, a change from the old basis B = {u1 , u2 , u3 } to the new basis B = {u1 , u2 , u3 }. In light, of Example 4, it should be evident that, , ⎡, , x´, , Figure 7.1.3, , ⎤, , ⎡, , ⎤, , cos θ, − sin θ, ⎢, ⎥, ⎢, ⎥, [u1 ]B = ⎣ sin θ ⎦ and [u2 ]B = ⎣ cos θ ⎦, 0, 0, Moreover, since u3 extends 1 unit up the positive z -axis,, , ⎡ ⎤, 0, , ⎢ ⎥, [u3 ]B = ⎣0⎦, 1, It follows that the transition matrix from B to B is, ⎡, cos θ − sin θ, ⎢, cos θ, P = ⎣ sin θ, 0, 0, and the transition matrix from B to B is, ⎡, cos θ, ⎢, P −1 = ⎣ − sin θ, 0, , sin θ, cos θ, 0, , ⎤, , 0, ⎥, 0⎦, 1, , ⎤, , 0, ⎥, 0⎦, 1, , (verify). Thus, the new coordinates (x , y , z ) of a point Q can be computed from its old, coordinates (x, y, z) by, , ⎡ ⎤ ⎡, cos θ, x, ⎢ ⎥ ⎢, ⎣y ⎦ = ⎣ − sin θ, 0, z, , O PT I O N A L, , sin θ, cos θ, 0, , ⎤⎡ ⎤, , 0, x, ⎥⎢ ⎥, 0⎦ ⎣y ⎦, 1, z, , We conclude this section with an optional proof of Theorem 7.1.5.

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7.1 Orthogonal Matrices, , 407, , V is an n-dimensional inner product space and, that P is the transition matrix from an orthonormal basis B to an orthonormal basis, B . We will denote the norm relative to the inner product on V by the symbol, V to, distinguish it from the norm relative to the Euclidean inner product on R n , which we, will denote by, ., To prove that P is orthogonal, we will use Theorem 7.1.3 and show that P x = x, for every vector x in R n . As a first step in this direction, recall from Theorem 7.1.4(a), that for any orthonormal basis for V the norm of any vector u in V is the same as the, norm of its coordinate vector with respect to the Euclidean inner product, that is,, , Proof of Theorem 7.1.5 Assume that, , Recall that (u)S denotes a, coordinate vector expressed, in comma-delimited form, whereas [u]S denotes a coordinate vector expressed in, column form., , u, , V, , = [u]B, , = [u]B, , or, (6), = P [u]B, Now let x be any vector in R , and let u be the vector in V whose coordinate vector with, respect to the basis B is x, that is, [u]B = x. Thus, from (6),, u, , V, , = [u]B, , n, , u = x = Px, which proves that P is orthogonal., , Exercise Set 7.1, In each part of Exercises 1–4, determine whether the matrix is, orthogonal, and if so find it inverse., , , 1. (a), , , 2. (a), , 1, , 0, , 0, , −1, , 1, , 0, , 0, , 1, , ⎡, , 0, , ⎢, 3. (a) ⎢, ⎣1, 0, , , , , , ⎤, , √1, , 1, , ⎥, 0 ⎥, ⎦, ⎤, , 1, 2, , − 56, , 1, 6, , 1⎥, 6⎥, , 1, 6, , 1, 6, , 1, 6, , − 56, , 1, 2, , 5, √1, 5, , 2, , 1, 2, , ⎥, , ⎥, − 56 ⎥, ⎦, , ⎡, , 1, , ⎢, ⎢0, ⎢, (b) ⎢, ⎢0, ⎣, , 1, 6, , , , √2, , √1, , 2, , 1, 2, , 2, , √1, , √1, , 0, , 8. Let TA : R 3 →R 3 be multiplication by the orthogonal matrix in, Exercise 6. Find TA (x) for the vector x = (0, 1, 4), and confirm TA (x) = x relative to the Euclidean inner product, on R 3 ., , 2, , ⎡ 1, − √2, ⎢, (b) ⎢, ⎣ 0, , 2, , 0, , √1, , 5, √2, 5, , (b), , , , − √12, , √1, , 2, √1, 2, , (b), , , , ⎡1, ⎢1, ⎢, ⎢2, 4. (a) ⎢, ⎢1, ⎣2, , , , 0, , √1, , √1, , 6, − √26, √1, 6, , 3, , √1, , 3, √1, 3, , 0, , 0, , √1, , − 21, , 3, √1, 3, √1, 3, , ⎤, , 9. Are the standard matrices for the reflections in Tables 1 and 2, of Section 4.9 orthogonal?, , ⎥, ⎥, ⎦, , 10. Are the standard matrices for the orthogonal projections in, Tables 3 and 4 of Section 4.9 orthogonal?, 11. What conditions must a and b satisfy for the matrix, , 0, , 5. A =, , 4, 5, ⎢ 9, ⎣− 25, 12, 25, , 0, 4, 5, 3, 5, , − 35, , ⎡, , ⎤, , ⎥, − 12, 25 ⎦, 16, 25, , 6. A =, , 1, 3, ⎢ 2, ⎣ 3, − 23, , 2, 3, − 23, − 13, , ⎥, , 0⎥, ⎥, , 0, , ⎥, 1⎥, ⎦, , 1, 2, , 0, , 2⎤, 3, 1⎥, 3⎦, 2, 3, , a+b, a−b, , ⎤, , In Exercises 5–6, show that the matrix is orthogonal three ways:, first by calculating ATA, then by using part (b) of Theorem 7.1.1,, and then by using part (c) of Theorem 7.1.1., , ⎡, , 7. Let TA : R 3 →R 3 be multiplication by the orthogonal matrix, in Exercise 5. Find TA (x) for the vector x = (−2, 3, 5), and, confirm that TA (x) = x relative to the Euclidean inner, product on R 3 ., , b−a, b+a, , to be orthogonal?, 12. Under what conditions will a diagonal matrix be orthogonal?, 13. Let a rectangular x y -coordinate system be obtained by rotating a rectangular xy-coordinate system counterclockwise, through the angle θ = π/3., (a) Find the x y -coordinates of the point whose, xy-coordinates are (−2, 6)., (b) Find the xy-coordinates of the point whose, x y -coordinates are (5, 2)., 14. Repeat Exercise 13 with θ = 3π/4.

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408, , Chapter 7 Diagonalization and Quadratic Forms, , 15. Let a rectangular x y z -coordinate system be obtained by rotating a rectangular xyz-coordinate system counterclockwise, about the z-axis (looking down the z-axis) through the angle, θ = π/4., (a) Find the x y z -coordinates of the point whose, xyz-coordinates are (−1, 2, 5)., (b) Find the xyz-coordinates of the point whose, x y z -coordinates are (1, 6, −3)., 16. Repeat Exercise 15 for a rotation of θ = 3π/4 counterclockwise about the x -axis (looking along the positive x -axis toward, the origin)., 17. Repeat Exercise 15 for a rotation of θ = π/3 counterclockwise, about the y -axis (looking along the positive y -axis toward the, origin)., 18. A rectangular x y z -coordinate system is obtained by rotating, an xyz-coordinate system counterclockwise about the y -axis, through an angle θ (looking along the positive y -axis toward, the origin). Find a matrix A such that, , ⎡ ⎤, ⎡ ⎤, x, x, ⎢ ⎥, ⎢ ⎥, ⎣y ⎦ = A ⎣y ⎦, z, z, , where (x, y, z) and (x , y , z ) are the coordinates of the same, point in the xyz- and x y z -systems, respectively., 19. Repeat Exercise 18 for a rotation about the x -axis., 20. A rectangular x y z -coordinate system is obtained by first, rotating a rectangular xyz-coordinate system 60◦ counterclockwise about the z-axis (looking down the positive z-axis), to obtain an x y z -coordinate system, and then rotating the, x y z -coordinate system 45◦ counterclockwise about the y axis (looking along the positive y -axis toward the origin)., Find a matrix A such that, , ⎡ ⎤, ⎡ ⎤, x, x, ⎢ ⎥, ⎢ ⎥, ⎣y ⎦ = A ⎣y ⎦, z, z, , where (x, y, z) and (x , y , z ) are the xyz- and x y z coordinates of the same point., 21. A linear operator on R 2 is called rigid if it does not change the, lengths of vectors, and it is called angle preserving if it does, not change the angle between nonzero vectors., (a) Identify two different types of linear operators that are, rigid., (b) Identify two different types of linear operators that are, angle preserving., (c) Are there any linear operators on R 2 that are rigid and not, angle preserving? Angle preserving and not rigid? Justify, your answer., 22. Can an orthogonal operator TA : R n →R n map nonzero vectors that are not orthogonal into orthogonal vectors? Justify, your answer., , 23. The set S =, , 1, , √1, , 3, , ,, , √1, , #, , 2, , x,, , 3 2, x, 2, , # 2, , −, , 2, 3, , is an orthonormal ba-, , sis for P2 with respect to the evaluation inner product at the, points x0 = −1, x1 = 0, x2 = 1. Let p = p(x) = 1 + x + x 2, and q = q(x) = 2x − x 2 ., (a) Find (p)S and (q)S ., (b) Use Theorem 7.1.4 to compute p , d(p, q) and p, q., 24. The sets S = {1, x} and S =, , 1, , √1, , 2, , (1 + x),, , √1, , 2, , 2, (1 − x) are or-, , thonormal bases for P1 with respect to the standard inner, product. Find the transition matrix P from S to S , and verify that the conclusion of Theorem 7.1.5 holds for P ., , Working with Proofs, 25. Prove that if x is an n × 1 matrix, then the matrix, , A = In −, , 2, xxT, xT x, , is both orthogonal and symmetric., 26. Prove that a 2 × 2 orthogonal matrix A has only one of two, possible forms:, , A=, , cos θ, sin θ, , − sin θ, cos θ, , or A =, , cos θ, sin θ, , sin θ, − cos θ, , where 0 ≤ θ < 2π . [Hint: Start with a general 2 × 2 matrix A,, and use the fact that the column vectors form an orthonormal, set in R 2 .], 27. (a) Use the result in Exercise 26 to prove that multiplication, by a 2 × 2 orthogonal matrix is a rotation if det(A) = 1, and a reflection followed by a rotation if det(A) = −1., (b) In the case where the transformation in part (a) is a reflection followed by a rotation, show that the same transformation can be accomplished by a single reflection about, an appropriate line through the origin. What is that line?, [Hint: See Formula (6) of Section 4.9.], 28. In each part, use the result in Exercise 27(a) to determine, whether multiplication by A is a rotation or a reflection followed by rotation. Find the angle of rotation in both cases,, and in the case where it is a reflection followed by a rotation, find an equation for the line through the origin referenced in, Exercise 27(b)., (a) A =, ,  1, − √2, − √12, , √1, , 2, , − √12, , , , ⎡, , (b) A = ⎣, , − 21, √, , 3, 2, , √, , 3, 2, , ⎤, ⎦, , 1, 2, , 29. The result in Exercise 27(a) has an analog for 3 × 3 orthogonal matrices. It can be proved that multiplication by a 3 × 3, orthogonal matrix A is a rotation about some line through the, origin of R 3 if det(A) = 1 and is a reflection about some coordinate plane followed by a rotation about some line through, the origin if det(A) = −1. Use the first of these facts and Theorem 7.1.2 to prove that any composition of rotations about, lines through the origin in R 3 can be accomplished by a single, rotation about an appropriate line through the origin.

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7.2 Orthogonal Diagonalization, , 30. Euler’s Axis of Rotation Theorem states that: If A is an orthogonal 3 × 3 matrix for which det(A) = 1, then multiplication by, A is a rotation about a line through the origin in R 3 . Moreover,, if u is a unit vector along this line, then Au = u., (a) Confirm that the following matrix A is orthogonal, that, det(A) = 1, and that there is a unit vector u for which, Au = u., ⎡, ⎤, 2, , ⎢ 73, A=⎢, ⎣7, 6, 7, , 3, 7, − 67, 2, 7, , 6, 7, ⎥, 2⎥, 7⎦, − 37, , (b) Use Formula (3) of Section 4.9 to prove that if A is a 3 × 3, orthogonal matrix for which det(A) = 1, then the angle, of rotation resulting from multiplication by A satisfies the, equation cos θ = 21 [tr(A) − 1]. Use this result to find the, angle of rotation for the rotation matrix in part (a)., 31. Prove the equivalence of statements (a) and (c) that are given, in Theorem 7.1.1., , True-False Exercises, TF. In parts (a)–(h) determine whether the statement is true or, false, and justify your answer., , ⎡, , 1, , ⎢, (a) The matrix ⎣0, 0, 1, (b) The matrix, 2, , ⎤, , 0, ⎥, 1⎦ is orthogonal., 0, , −2, 1, , is orthogonal., , (c) An m × n matrix A is orthogonal if ATA = I ., (d) A square matrix whose columns form an orthogonal set is, orthogonal., (e) Every orthogonal matrix is invertible., (f ) If A is an orthogonal matrix, then A2 is orthogonal and, (det A)2 = 1., , 409, , (g) Every eigenvalue of an orthogonal matrix has absolute value 1., (h) If A is a square matrix and Au = 1 for all unit vectors u,, then A is orthogonal., , Working withTechnology, T1. If a is a nonzero vector in R n , then aaT is called the outer, product of a with itself, the subspace a⊥ is called the hyperplane in, R n orthogonal to a, and the n × n orthogonal matrix, , Ha⊥ = I −, , 2, aaT, aTa, , is called the Householder matrix or the Householder reflection, about a⊥ , named in honor of the American mathematician Alston S. Householder (1904–1993). In R 2 the matrix Ha⊥ represents, a reflection about the line through the origin that is orthogonal to, a, and in R 3 it represents a reflection about the plane through the, origin that is orthogonal to a. In higher dimensions we can view, Ha⊥ as a “reflection” about the hyperplane a⊥ . Householder reflections are important in large-scale implementations of numerical, algorithms, particularly QR -decompositions, because they can be, used to transform a given vector into a vector with specified zero, components while leaving the other components unchanged. This, is a consequence of the following theorem [see Contemporary Linear Algebra, by Howard Anton and Robert C. Busby (Hoboken,, NJ: John Wiley & Sons, 2003, p. 422)]., , Theorem. If v and w are distinct vectors in R n with the same, norm, then the Householder reflection about the hyperplane, (v − w)⊥ maps v into w and conversely., (a) Find a Householder reflection that maps the vector, v = (4, 2, 4) into a vector w that has zeros as its second, and third components. Find w., (b) Find a Householder reflection that maps the vector, v = (3, 4, 2, 4) into the vector whose last two entries are, zero, while leaving the first entry unchanged. Find w., , 7.2 Orthogonal Diagonalization, In this section we will be concerned with the problem of diagonalizing a symmetric matrix, A. As we will see, this problem is closely related to that of finding an orthonormal basis for, R n that consists of eigenvectors of A. Problems of this type are important because many of, the matrices that arise in applications are symmetric., , The Orthogonal, Diagonalization Problem, , In Section 5.2 we defined two square matrices, A and B , to be similar if there is an, invertible matrix P such that P −1AP = B . In this section we will be concerned with, the special case in which it is possible to find an orthogonal matrix P for which this, relationship holds., We begin with the following definition.

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410, , Chapter 7 Diagonalization and Quadratic Forms, DEFINITION 1 If A and B are square matrices, then we say that B is orthogonally, similar to A if there is an orthogonal matrix P such that B = P TAP ., , Note that if B is orthogonally similar to A, then it is also true that A is orthogonally, similar to B since we can express A as A = QTBQ by taking Q = P T (verify). This, being the case we will say that A and B are orthogonally similar matrices if either is, orthogonally similar to the other., If A is orthogonally similar to some diagonal matrix, say, , P TAP = D, then we say that A is orthogonally diagonalizable and that P orthogonally diagonalizes A., Our first goal in this section is to determine what conditions a matrix must satisfy, to be orthogonally diagonalizable. As an initial step, observe that there is no hope of, orthogonally diagonalizing a matrix that is not symmetric. To see why this is so, suppose, that, P TAP = D, (1), where P is an orthogonal matrix and D is a diagonal matrix. Multiplying the left side, of (1) by P , the right side by P T , and then using the fact that PP T = P TP = I , we can, rewrite this equation as, (2), A = PDP T, Now transposing both sides of this equation and using the fact that a diagonal matrix is, the same as its transpose we obtain, , AT = (PDP T )T = (P T )T D TP T = PDP T = A, so A must be symmetric if it is orthogonally diagonalizable., Conditions for Orthogonal, Diagonalizability, , The following theorem shows that every symmetric matrix with real entries is, in fact,, orthogonally diagonalizable. In this theorem, and for the remainder of this section,, orthogonal will mean orthogonal with respect to the Euclidean inner product on R n ., THEOREM 7.2.1 If A is an n × n matrix with real entries, then the following are equiv-, , alent., , (a) A is orthogonally diagonalizable., (b) A has an orthonormal set of n eigenvectors., (c) A is symmetric., Proof (a) ⇒ (b) Since A is orthogonally diagonalizable, there is an orthogonal matrix P, such that P −1AP is diagonal. As shown in Formula (2) in the proof of Theorem 5.2.1,, the n column vectors of P are eigenvectors of A. Since P is orthogonal, these column, vectors are orthonormal, so A has n orthonormal eigenvectors., (b) ⇒ (a) Assume that A has an orthonormal set of n eigenvectors {p1 , p2 , . . . , pn }. As, , shown in the proof of Theorem 5.2.1, the matrix P with these eigenvectors as columns, diagonalizes A. Since these eigenvectors are orthonormal, P is orthogonal and thus, orthogonally diagonalizes A., , (a) ⇒ (b) we showed that an orthogonally diagonalizable, n × n matrix A is orthogonally diagonalized by an n × n matrix P whose columns form, an orthonormal set of eigenvectors of A. Let D be the diagonal matrix, (a) ⇒ (c) In the proof that, , D = P TAP

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7.2 Orthogonal Diagonalization, , 411, , from which it follows that, , A = PDP T, Thus,, , AT = (PDP T )T = PD TP T = PDP T = A, which shows that A is symmetric., (c) ⇒ (a) The proof of this part is beyond the scope of this text. However, because it is, , such an important result we have outlined the structure of its proof in the exercises (see, Exercise 31)., Properties of Symmetric, Matrices, , Our next goal is to devise a procedure for orthogonally diagonalizing a symmetric matrix,, but before we can do so, we need the following critical theorem about eigenvalues and, eigenvectors of symmetric matrices., , THEOREM 7.2.2 If A is a symmetric matrix with real entries, then:, , (a) The eigenvalues of A are all real numbers., (b) Eigenvectors from different eigenspaces are orthogonal., , Part (a ), which requires results about complex vector spaces, will be discussed in, Section 7.5., Proof (b) Let v1 and v2 be eigenvectors corresponding to distinct eigenvalues λ1 and λ2, of the matrix A. We want to show that v1 · v2 = 0. Our proof of this involves the trick, of starting with the expression Av1 · v2 . It follows from Formula (26) of Section 3.2 and, the symmetry of A that, (3), Av1 · v2 = v1 · ATv2 = v1 · Av2, , But v1 is an eigenvector of A corresponding to λ1 , and v2 is an eigenvector of A corresponding to λ2 , so (3) yields the relationship, , λ1 v1 · v2 = v1 · λ2 v2, which can be rewritten as, , (λ1 − λ2 )(v1 · v2 ) = 0, , (4), , But λ1 − λ2  = 0, since λ1 and λ2 were assumed distinct. Thus, it follows from (4) that, v1 · v2 = 0., Theorem 7.2.2 yields the following procedure for orthogonally diagonalizing a symmetric matrix., Orthogonally Diagonalizing an n × n Symmetric Matrix, Step 1. Find a basis for each eigenspace of A., Step 2. Apply the Gram–Schmidt process to each of these bases to obtain an orthonormal basis for each eigenspace., Step 3. Form the matrix P whose columns are the vectors constructed in Step 2. This, matrix will orthogonally diagonalize A, and the eigenvalues on the diagonal, of D = P TAP will be in the same order as their corresponding eigenvectors, in P .

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412, , Chapter 7 Diagonalization and Quadratic Forms, , Remark The justification of this procedure should be clear: Theorem 7.2.2 ensures that eigenvectors from different eigenspaces are orthogonal, and applying the Gram–Schmidt process ensures, that the eigenvectors within the same eigenspace are orthonormal. Thus the entire set of eigenvectors obtained by this procedure will be orthonormal., , E X A M P L E 1 Orthogonally Diagonalizing a Symmetric Matrix, , Find an orthogonal matrix P that diagonalizes, , ⎡, , 4, ⎢, A = ⎣2, 2, , ⎤, , 2, 4, 2, , 2, ⎥, 2⎦, 4, , Solution We leave it for you to verify that the characteristic equation of A is, , ⎡, λ−4, ⎢, det(λI − A) = det ⎣ −2, −2, , ⎤, −2, ⎥, −2 ⎦ = (λ − 2)2 (λ − 8) = 0, λ−4, , −2, λ−4, −2, , Thus, the distinct eigenvalues of A are λ = 2 and λ = 8. By the method used in Example 7, of Section 5.1, it can be shown that, , ⎡, , ⎤, , ⎢, , ⎥, , −1, , ⎡, , u1 = ⎣ 1⎦, 0, , ⎤, −1, ⎢ ⎥, u2 = ⎣ 0⎦, , and, , (5), , 1, , form a basis for the eigenspace corresponding to λ = 2. Applying the Gram–Schmidt, process to {u1 , u2 } yields the following orthonormal eigenvectors (verify):, , ⎡, , − √12, , ⎢, , v1 = ⎢, ⎣, , √1, , 2, , ⎤, ⎥, ⎥, ⎦, , ⎡ 1, − √6, ⎢ 1, √, v2 = ⎢, ⎣− 6, , and, , ⎤, ⎥, ⎥, ⎦, , (6), , √2, 6, , 0, The eigenspace corresponding to λ = 8 has, , ⎡ ⎤, , 1, ⎢ ⎥, u3 = ⎣1⎦, 1, as a basis. Applying the Gram–Schmidt process to {u3 } (i.e., normalizing u3 ) yields, , ⎡, , v3 =, , ⎤, , √1, , 3, ⎢ 1, ⎢√, ⎣ 3, √1, 3, , ⎥, ⎥, ⎦, , Finally, using v1 , v2 , and v3 as column vectors, we obtain, , ⎡, , − √12, , ⎢, P =⎢, ⎣, , √1, , 2, , 0, , − √16, , √1, , 3, √1, 3, √1, 3, , − √16, √2, , 6, , ⎤, ⎥, ⎥, ⎦, , which orthogonally diagonalizes A. As a check, we leave it for you to confirm that, , ⎡, , − √12, , ⎢ 1, √, P TAP = ⎢, ⎣− 6, √1, , 3, , √1, , 2, − √16, √1, 3, , 0, , ⎤, , ⎡, , ⎡, ⎤ − √1, 2, , 4 2 2, ⎥, ⎢, √2 ⎥ ⎣2 4 2⎦ ⎢, 6⎦, ⎣, 2 2 4, √1, 3, , √1, , 2, , 0, , − √16, − √16, √2, , 6, , √1, , ⎤, , 3, ⎥, √1 ⎥, 3⎦, √1, 3, , ⎡, , ⎤, , 2 0 0, = ⎣0 2 0⎦, 0 0 8

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7.2 Orthogonal Diagonalization, , Spectral Decomposition, , 413, , If A is a symmetric matrix that is orthogonally diagonalized by, , P = [u1, , · · · un ], , u2, , and if λ1 , λ2 , . . . , λn are the eigenvalues of A corresponding to the unit eigenvectors, u1 , u2 , . . . , un , then we know that D = P TAP , where D is a diagonal matrix with the, eigenvalues in the diagonal positions. It follows from this that the matrix A can be, expressed as, , ⎡, , λ1, , A = PDP T = [u1, , u2, , ⎢0, ⎢, · · · un ] ⎢ .., ⎣., , λ2, .., ., , 0, , 0, , ⎡, , T, , u1, , = [λ1 u1, , λ2 u2, , 0, , ⎤, , ···, ···, .., ., , ⎤⎡, , ⎤, , u1T, 0, ⎢, T⎥, 0⎥, ⎥ ⎢u2 ⎥, , .. ⎥ ⎢, .⎥, ⎥, . ⎦⎢, ⎣ .. ⎦, · · · λn, uTn, , ⎢ T⎥, ⎢u2 ⎥, ⎥, · · · λn un ] ⎢, ⎢ .. ⎥, ⎣.⎦, uTn, , Multiplying out, we obtain the formula, , A = λ1 u1 u1T + λ2 u2 u2T + · · · + λn un uTn, , (7), , which is called a spectral decomposition of A.*, Note that in each term of the spectral decomposition of A has the form λu uT , where, u is a unit eigenvector of A in column form, and λ is an eigenvalue of A corresponding to, u. Since u has size n × 1, it follows that the product u uT has size n × n. It can be proved, (though we will not do it) that u uT is the standard matrix for the orthogonal projection, of R n on the subspace spanned by the vector u. Accepting this to be so, the spectral, decomposition of A tells that the image of a vector x under multiplication by a symmetric, matrix A can be obtained by projecting x orthogonally on the lines (one-dimensional, subspaces) determined by the eigenvectors of A, then scaling those projections by the, eigenvalues, and then adding the scaled projections. Here is an example., E X A M P L E 2 A Geometric Interpretation of a Spectral Decomposition, , The matrix, , A=, , 1, 2, , 2, , −2, , has eigenvalues λ1 = −3 and λ2 = 2 with corresponding eigenvectors, x1 =, , 1, , −2, , and x2 =, , (verify). Normalizing these basis vectors yields, , ⎡, , u1 =, , √1, , ⎡, , √2, , ⎤, , x1, x2, 5, 5, = ⎣ 2 ⎦ and u2 =, =⎣ 1 ⎦, √, x1, x2, −√, 5, , *, , ⎤, , 2, 1, , 5, , The terminology spectral decomposition is derived from the fact that the set of all eigenvalues of a matrix, , A is sometimes called the spectrum of A. The terminology eigenvalue decomposition is due to Professor Dan, Kalman, who introduced it in an award-winning paper entitled “A Singularly Valuable Decomposition: The, SVD of a Matrix,” The College Mathematics Journal, Vol. 27, No. 1, January 1996.

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414, , Chapter 7 Diagonalization and Quadratic Forms, , so a spectral decomposition of A is, 1, 2, , ⎡, , ⎤, , √, %, 2, 5, = λ1 u1 u1T + λ2 u2 uT2 = (−3) ⎣ 2 ⎦ √15, −2, −√, , ⎡, , 1, , 5, , − 25, , 1, 5, , = (−3) ⎣, − 25, , 4, 5, , ⎡, , &, , − √25 + (2) ⎣, , ⎤, , ⎡, , ⎦ + (2 ) ⎣, , 4, 5, , 2, 5, , 2, 5, , 1, 5, , ⎤, , √2, , ⎤, , 5, ⎦, √1, 5, , %, , ⎦, , √2, , 5, , √1, , &, , 5, , (8), , where, as noted above, the 2 × 2 matrices on the right side of (8) are the standard matrices, for the orthogonal projections onto the eigenspaces corresponding to the eigenvalues, λ1 = −3 and λ2 = 2, respectively., Now let us see what this spectral decomposition tells us about the image of the vector, x = (1, 1) under multiplication by A. Writing x in column form, it follows that, 1, 2, , Ax =, and from (8) that, 1, Ax =, 2, , 2, −2, , 2, −2, , , , = (−3), , =, , 3, 5, − 65, , − 25, , 1, , 1, 5, = (−3), 2, 1, −5, , , , 1, 3, =, 1, 0, , − 15, 2, 5, , , , , , 4, 5, , , , 1, + (2 ), 1, , 6, , + (2 ), ,  12 , , +, , 5, 6, 5, , =, , (9), , 4, 5, 2, 5, , 2, 5, 1, 5, , , , 1, 1, , 5, 3, 5, , 3, 0, , (10), , Formulas (9) and (10) provide two different ways of viewing the image of the vector (1, 1), under multiplication by A: Formula (9) tells us directly that the image of this vector is, (3, 0), whereas Formula (10) tells us that this image can also be obtained by projecting, (1,, onto, corresponding to λ1 = −3 and λ2 = 2 to obtain,  1),  the eigenspaces, ,  the vectors, , − 15 , 25 and 65 , 35 , then scaling by the eigenvalues to obtain 35 , − 65 and 125 , 65 ,, and then adding these vectors (see Figure 7.2.1)., , λ2 = 2, x = (1, 1), , ( 125 , 65 ), , ( 65 , 35 ), , (– 15 , 52 ), , Ax = (3, 0), , ( 35 , – 56 ), Figure 7.2.1, , The Nondiagonalizable, Case, , λ1 = – 3, , If A is an n × n matrix that is not orthogonally diagonalizable, it may still be possible, to achieve considerable simplification in the form of P TAP by choosing the orthogonal, matrix P appropriately. We will consider two theorems (without proof) that illustrate, this. The first, due to the German mathematician Issai Schur, states that every square, matrix A is orthogonally similar to an upper triangular matrix that has the eigenvalues, of A on the main diagonal.

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7.2 Orthogonal Diagonalization, , 415, , THEOREM 7.2.3 Schur’s Theorem, , If A is an n × n matrix with real entries and real eigenvalues, then there is an orthogonal, matrix P such that P TAP is an upper triangular matrix of the form, , ⎡, , λ1, , ⎢0, ⎢, ⎢, P AP = ⎢ 0, ⎢ .., ⎣., T, , 0, , ×, λ2, , ⎤, ×, ×⎥, ⎥, ×⎥, ⎥, .. ⎥, .⎦, , .., ., , ×, ×, λ3, .., ., , ···, ···, ···, .., ., , 0, , 0, , · · · λn, , 0, , (11), , in which λ1 , λ2 , . . . , λn are the eigenvalues of A repeated according to multiplicity., It is common to denote the upper triangular matrix in (11) by S (for Schur), in which, case that equation would be rewritten as, (12), A = PSP T, which is called a Schur decomposition of A., The next theorem, due to the German electrical engineer Karl Hessenberg (1904–, 1959), states that every square matrix with real entries is orthogonally similar to a matrix, in which each entry below the first subdiagonal is zero (Figure 7.2.2). Such a matrix is, said to be in upper Hessenberg form., , First subdiagonal, , Figure 7.2.2, , THEOREM 7.2.4 Hessenberg’s Theorem, , Note that unlike those in (11),, the diagonal entries in (13), are usually not the eigenvalues, of A., , If A is an n × n matrix with real entries, then there is an orthogonal matrix P such that, P TAP is a matrix of the form ⎡, ⎤, × × ··· × × ×, ⎢× × · · · × × ×⎥, ⎢, ⎥, .., ⎢, ⎥, ., ⎢, × × ×⎥, 0 ×, T, ⎢, (13), P AP = ⎢ .., .. . ., ., .., .. ⎥, ⎥, . .., ., ., .⎥, ⎢., ⎢, ⎥, ⎣ 0 0 · · · × × ×⎦, 0 0 ··· 0 × ×, It is common to denote the upper Hessenberg matrix in (13) by H (for Hessenberg),, in which case that equation can be rewritten as, , A = PHP T, , (14), , which is called an upper Hessenberg decomposition of A., , Issai Schur, (1875–1941), , Historical Note The life of the German mathematician Issai Schur is a sad reminder, of the effect that Nazi policies had on Jewish intellectuals during the 1930s. Schur, was a brilliant mathematician and a popular lecturer who attracted many students, and researchers to the University of Berlin, where he worked and taught. His lectures, sometimes attracted so many students that opera glasses were needed to see him, from the back row. Schur’s life became increasingly difficult under Nazi rule, and in, April of 1933 he was forced to “retire” from the university under a law that prohibited, non-Aryans from holding “civil service” positions. There was an outcry from many, of his students and colleagues who respected and liked him, but it did not stave off, his complete dismissal in 1935. Schur, who thought of himself as a loyal German,, never understood the persecution and humiliation he received at Nazi hands. He left, Germany for Palestine in 1939, a broken man. Lacking in financial resources, he had, to sell his beloved mathematics books and lived in poverty until his death in 1941., [Image: Courtesy Electronic Publishing Services, Inc., NewYork City ]

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416, , Chapter 7 Diagonalization and Quadratic Forms, , Remark In many numerical algorithms the initial matrix is first converted to upper Hessenberg, form to reduce the amount of computation in subsequent parts of the algorithm. Many computer, packages have built-in commands for finding Schur and Hessenberg decompositions., , Exercise Set 7.2, In Exercises 1–6, find the characteristic equation of the given, symmetric matrix, and then by inspection determine the dimensions of the eigenspaces., , ⎡, , 1., , 1, 2, , ⎡, , 1, ⎢, 3. ⎣1, 1, , ⎡, , 4, ⎢4, ⎢, 5. ⎢, ⎣0, 0, , 2, 4, , ⎤, , 1, 1, 1, , 1, ⎥, 1⎦, 1, , 4, 4, 0, 0, , 0, 0, 0, 0, , ⎡, , 4, ⎢, 4. ⎣2, 2, , ⎤, , 2, ⎥, −2⎦, −2, , 1, −2, 2, ⎥, 2⎦, 4, , −1, , 2, ⎢−1, ⎢, 6. ⎢, ⎣ 0, 0, , ⎤, , ⎡ ⎤, , ⎡ ⎤, , ⎡, , ⎤, , ⎡ ⎤, , ⎡ ⎤, , 0, 1, 1, ⎢ ⎥, ⎢ ⎥, ⎢ ⎥, 20. x1 = ⎣ 1⎦, x2 = ⎣0⎦, x3 = ⎣1⎦, 0, 1, −1, , ⎤, , 0, 0, 2, −1, , 2, 0, 0, , ⎡, , 0, 1, 0, ⎢ ⎥, ⎢ ⎥, ⎢ ⎥, 19. x1 = ⎣ 1⎦, x2 = ⎣0⎦, x3 = ⎣1⎦, 0, 1, −1, , ⎤, , 2, 4, 2, , ⎡, , 0, 0⎥, ⎥, ⎥, 0⎦, 0, , ⎤, , −4, , 1, ⎢, 2. ⎣−4, 2, , In Exercises 19–20, determine whether there exists a 3 × 3 symmetric matrix whose eigenvalues are λ1 = −1, λ2 = 3, λ3 = 7 and, for which the corresponding eigenvectors are as stated. If there is, such a matrix, find it, and if there is none, explain why not., , 0, 0⎥, ⎥, ⎥, − 1⎦, 2, , 21. Let A be a diagonalizable matrix with the property that eigenvectors corresponding to distinct eigenvalues are orthogonal., Must A be symmetric? Explain your reasoning., 22. Assuming that b  = 0, find a matrix that orthogonally diagonalizes, , a, b, , In Exercises 7–14, find a matrix P that orthogonally diagonalizes A, and determine P −1AP ., , , 7. A =, , √ , , 6, , √, , 2 3, , 2 3, , 7, , −2, , 0, −3, 0, , ⎡, ⎢, , 9. A = ⎣, , 0, −36, , ⎡, , 2, ⎢, 11. A = ⎣−1, −1, , ⎡, , −7, , ⎢ 24, ⎢, ⎣ 0, , 13. A = ⎢, , 0, , 3, 8. A =, 1, , −36, , ⎤, , ⎤, −1, ⎥, −1⎦, , −1, , ⎡, , 2, , 1, ⎢, 12. A = ⎣1, 0, , 24, 0, 7, 0, 0 −7, 0 24, , 0, 3, ⎢1, 0⎥, ⎥, ⎢, ⎥ 14. A = ⎢, 24⎦, ⎣0, 0, 7, , 2, −1, , 23. Let TA : R 2 →R 2 be multiplication by A. Find two orthogonal unit vectors u1 and u2 such that TA (u1 ) and TA (u2 ) are, orthogonal., , 1, 3, , ⎤, , ⎡, , (a) A =, , −2, , 6, ⎥, 0⎦ 10. A =, −2, −23, , 3, , ⎤, , 1, 1, 0, , 0, ⎥, 0⎦, 0, , 1, 3, 0, 0, , 0, 0, 0, 0, , ⎡, , −3, ⎢, 17. ⎣ 1, 2, , 1, 3, , 6, 16., −2, 1, , −3, 2, , ⎤, , 2, 2⎥, ⎦, 0, , ⎡, ⎢, , 18. ⎣, , , −1, , 1, , 1, , 1, , , , , , (b) A =, , , , 1, , 2, , 2, , 1, , 24. Let TA : R 3 →R 3 be multiplication by A. Find two orthogonal unit vectors u1 and u2 such that TA (u1 ) and TA (u2 ) are, orthogonal., , ⎡, , ⎤, , 0, 0⎥, ⎥, ⎥, 0⎦, 0, , ⎤, , 4, , 2, , 2, , (a) A = ⎣2, , 4, , 2⎦, , 2, , 2, , 4, , ⎢, , In Exercises 15–18, find the spectral decomposition of the, matrix., 3, 15., 1, , b, a, , ⎥, , ⎡, , ⎤, , 1, , 0, , 0, , (b) A = ⎣0, , 1, , 1⎦, , 0, , 1, , 1, , ⎢, , ⎥, , Working with Proofs, 25. Prove that if A is any m × n matrix, then ATA has an orthonormal set of n eigenvectors., 26. Prove: If {u1 , u2 , . . . , un } is an orthonormal basis for R n , and, if A can be expressed as, , A = c1 u1 uT1 + c2 u2 u2T + · · · + cn un uTn, , −2, , then A is symmetric and has eigenvalues c1 , c2 , . . . , cn ., , 3, , −2, , 0, , 0, −36, , −3, 0, , −36, , ⎤, ⎥, , 0⎦, −23, , 27. Use the result in Exercise 29 of Section 5.1 to prove Theorem 7.2.2(a) for 2 × 2 symmetric matrices., 28. (a) Prove that if v is any n × 1 matrix and I is the n × n identity matrix, then I − vvT is orthogonally diagonalizable.

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7.3 Quadratic Forms, , (b) Find a matrix P that orthogonally diagonalizes I − vvT if, , (c) Use Theorem 7.2.2(b) and the fact that A is diagonalizable, to prove that A is orthogonally diagonalizable., , ⎡ ⎤, 1, , ⎢ ⎥, , v = ⎣0⎦, , True-False Exercises, , 1, 29. Prove that if A is a symmetric orthogonal matrix, then 1 and, −1 are the only possible eigenvalues., 30. Is the converse of Theorem 7.2.2(b) true? Justify your answer., 31. In this exercise we will show that a symmetric matrix A is, orthogonally diagonalizable, thereby completing the missing, part of Theorem 7.2.1. We will proceed in two steps: first we, will show that A is diagonalizable, and then we will build on, that result to show that A is orthogonally diagonalizable., (a) Assume that A is a symmetric n × n matrix. One way, to prove that A is diagonalizable is to show that for each, eigenvalue λ0 the geometric multiplicity is equal to the, algebraic multiplicity. For this purpose, assume that the, geometric multiplicity of λ0 is k , let B0 = {u1 , u2 , . . . , uk }, be an orthonormal basis for the eigenspace corresponding to the eigenvalue λ0 , extend this to an orthonormal, basis B0 = {u1 , u2 , . . . , un } for R n , and let P be the matrix having the vectors of B as columns. As shown in Exercise 40(b) of Section 5.2, the product AP can be written, as, , , , AP = P, , TF. In parts (a)–(g) determine whether the statement is true or, false, and justify your answer., (a) If A is a square matrix, then AAT and ATA are orthogonally, diagonalizable., (b) If v1 and v2 are eigenvectors from distinct eigenspaces of a, symmetric matrix with real entries, then, v1 + v2, , 2, , = v1, , 2, , + v2, , 2, , (c) Every orthogonal matrix is orthogonally diagonalizable., (d) If A is both invertible and orthogonally diagonalizable, then, A−1 is orthogonally diagonalizable., (e) Every eigenvalue of an orthogonal matrix has absolute value 1., (f ) If A is an n × n orthogonally diagonalizable matrix, then there, exists an orthonormal basis for R n consisting of eigenvectors, of A., (g) If A is orthogonally diagonalizable, then A has real eigenvalues., , λ0 Ik, , X, , Working withTechnology, , 0, , Y, , T1. If your technology utility has an orthogonal diagonalization, capability, use it to confirm the final result obtained in Example 1., , Use the fact that B is an orthonormal basis to prove that, X = 0 [a zero matrix of size n × (n − k)]., (b) It follows from part (a) and Exercise 40(c) of Section 5.2, that A has the same characteristic polynomial as, , , , C=P, , 417, , λ0 Ik, , 0, , 0, , Y, , , , Use this fact and Exercise 40(d) of Section 5.2 to prove that, the algebraic multiplicity of λ0 is the same as the geometric, multiplicity of λ0 . This establishes that A is diagonalizable., , T2. For the given matrix A, find orthonormal bases for the, eigenspaces of A, and use those basis vectors to construct an orthogonal matrix P for which P TAP is diagonal., , ⎡, , −4, , 2, , ⎢, A=⎣ 2, , −7, , −2, , 4, , −2, , ⎤, ⎥, , 4⎦, , −7, , T3. Find a spectral decomposition of the matrix A in Exercise T2., , 7.3 Quadratic Forms, In this section we will use matrix methods to study real-valued functions of several, variables in which each term is either the square of a variable or the product of two, variables. Such functions arise in a variety of applications, including geometry, vibrations, of mechanical systems, statistics, and electrical engineering., , Definition of a Quadratic, Form, , Expressions of the form, , a1 x1 + a2 x2 + · · · + an xn, occurred in our study of linear equations and linear systems. If a1 , a2 , . . . , an are, treated as fixed constants, then this expression is a real-valued function of the n variables, x1 , x2 , . . . , xn and is called a linear form on R n . All variables in a linear form occur to