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Its characteristic equation can be written as, , where, , ,, , For a 2 x 2 matrix:, Method 1:, The characteristic equation is, Method 2:, Its characteristic equation can be written as, , where, , ,, Problems:, 1. Find the characteristic equation of the matrix (, Solution: Let A = (, , ), , ). Its characteristic equation is, , where, , ,, = 1(2) – 2(0) = 2, Therefore, the characteristic equation is, , 2. Find the characteristic equation of (, , ), , Solution: Its characteristic equation is, , , where, = 8 + 7 + 3 = 18,, |, , ,, , |, , |, , |, , |, , = 8(5)+6(-10)+2(10) = 40 -60 + 20 = 0, , Therefore, the characteristic equation is, 3. Find the characteristic polynomial of (, , ), , |
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Solution: Let A = (, , ), , The characteristic polynomial of A is, , where, = 3(2) – 1(-1) = 7, , = 3 + 2 = 5 and, Therefore, the characteristic polynomial is, CAYLEY-HAMILTON THEOREM:, , Statement: Every square matrix satisfies its own characteristic equation, Uses of Cayley-Hamilton theorem:, (1) To calculate the positive integral powers of A, (2) To calculate the inverse of a square matrix A, Problems:, 1. Show that the matrix [, Solution:Let A = [, , ] satisfies its own characteristic equation, , ] . The characteristic equation of A is, , The characteristic equation is, To prove, [, , ][, [, , ]=[, ], , [, , ], ], , [, , ], , [, , ], , Therefore, the given matrix satisfies its own characteristic equation, 2. If A = [, , ], , Solution:Cayley-Hamilton theorem states that every square matrix satisfies its own, characteristic equation., The characteristic equation of A is, , where, , where
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Therefore, the characteristic equation is, By Cayley-Hamilton theorem,, i.e.,, when A = [, , 3. Verify Cayley-Hamilton theorem, find, Solution: The characteristic equation of A is, , ], where, , Therefore, the characteristic equation is, To prove that:, , ------------- (1), , [, , ][, , ], , [, , ][, , [, [, To find, , [, , ], , ], , ], , [, , [, , ], , ], , [, , ], , ], , :, ------------- (2), , Multiply by A on both sides,, Therefore,, Hence,, , [, , ], , [, , ], , [, , ], , [, , ]
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[, , ], , [, , [, , ], , [, , ], , ], , To find, Multiplying (1) by, , [, , [, , ], , ], , [, , ], , [, , [, , [, , ], , ], , [, , [, , ], , ], , ], , 4. Verify that A = [, , ] satisfies its own characteristic equation and hence find, ]. The characteristic equation of A is, , Solution:Given A =[, , Therefore, the characteristic equation is, To prove:, , where, , i.e.,, , ---------- (1), [, [, , ], , [, , ], , ][, [, , ], , [, , ], , [, , ], , [, , ], , [, , ], , ], , To find, From (1), we get,, Multiplying by, , 5. Find, , on both sides, we get,, , if A = [, , [, , ], , ], using Cayley-Hamilton theorem, , [, , ]
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Solution:The characteristic equation of A is, , where, , The characteristic equation of A is, By Cayley- Hamilton theorem,, , ------------ (1), , To find, Multiplying (1) by, , we get,, ------------- (2), , [, , ][, , ], , [, , ], , [, , From (2),, , 6. If A = [, , [, , ], , [, , ], , [, , ], , [, , [, , ], , ], , Solution:The characteristic equation of A is, , The characteristic equation of A is, , where, , i.e.,, , √, , To find, When, , is divided by, , , let the quotient be, , and the remainder be, , ---------- (1), When, , When, ------------ (2), , ], , ]
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------------ (3), Solving (2) and (3), we get, (2) - (3), (2) – 2 x (3), i.e.,, , Since, , by Cayley-Hamilton theorem, (1), [, , ][, , ], , 7. Use Cayley-Hamilton theorem for the matrix, polynomial in A (i), Solution: Given A = [, , [, , ], ] to express as a linear, , (ii), ]. The characteristic equation of A is, , where, , The characteristic equation is, By Cayley-Hamilton theorem, we get,, , ------------ (1), , (-), , (by (1)) which is a linear polynomial in A, (i)
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(-), =, , = 0 + A + 2I = A +2I, , (by (1)) which is a, , linear polynomial in A, when A = [, , 8. Using Cayley-Hamilton theorem, find, Solution:The characteristic equation of A is, , ], where, , The characteristic equation is, By Cayley-Hamilton theorem,, Pre-multiplying by, , [, , we get,, , ][, , [, , ], , ];, , ([, , [, , ], , [, , ], , ], , [, , ], , [, , [, , ], , ]), , 9. Verify Cayley-Hamilton theorem for the matrix, , Solution: Given A = [, , [, , [, , ], , [, , ], , The Characteristic equation of A is, Sum of the main diagonal elements = 1+2+1 = 4, , where, , ], , ]
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The characteristic equation is, To prove that:, [, , ][, , ], , [, , ][, , =[, , [, , ], , ], , [, , ], , [, , ], , ], , [, , ], , =[, , ], , =[, , [, , ], , [, , ], , [, , ], , [, , ], , [, , ], , [, , ], , ], , Therefore, Cayley-Hamilton theorem is verified., 10. Verify Cayley-Hamilton theorem for the matrix (i) A = [, Solution:(i) Given A = [, , ], , The characteristic equation of A is, , where, , The characteristic equation is, To prove that:, [, , ][, , ], , [, , ], , [, , ], , ], , [, , ]
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[, , ], [, , ], , [, , ], , [, , ], , [, , ], , [, , ], , [, , ], , [, , ], , Hence Cayley-Hamilton theorem is verified., (ii) Given A = [, , ], , The characteristic equation of A is, , where, , The characteristic equation is, To prove that:, [, [, , ], , [, , ];, , [, , ], , ][, , ], , [, [, , ], ], , [, , [, , ], , [, , [, , ], , ], , ], , [, , ], , Hence Cayley-Hamilton theorem is verified., , EIGEN VALUES AND EIGEN VECTORS OF A REAL MATRIX:, Working rule to find eigen values and eigen vectors:, 1. Find the characteristic equation, 2. Solve the characteristic equation to get characteristic roots. They are called eigen values, 3. To find the eigen vectors, solve, , ], , for different values of, , Note:, 1. Corresponding to n distinct eigen values, we get n independent eigen vectors, 2. If 2 or more eigen values are equal, it may or may not be possible to get linearly, independent eigen vectors corresponding to the repeated eigen values
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3. If, , is a solution for an eigen value, , , then c, , is also a solution, where c is an arbitrary, , constant. Thus, the eigen vector corresponding to an eigen value is not unique but may, be any one of the vectors c, 4. Algebraic multiplicity of an eigen value, characteristic polynomial (i.e., if, 5. Geometric multiplicity of, , is the order of the eigen value as a root of the, , is a double root, then algebraic multiplicity is 2), , is the number of linearly independent eigen vectors, , corresponding to, Non-symmetric matrix:, If a square matrix A is non-symmetric, then A ≠, Note:, 1. In a non-symmetric matrix, if the eigen values are non-repeated then we get a linearly, independent set of eigen vectors, 2. In a non-symmetric matrix, if the eigen values are repeated, then it may or may not be, possible to get linearly independent eigen vectors., If we form a linearly independent set of eigen vectors, then diagonalization is possible, through similarity transformation, Symmetric matrix:, If a square matrix A is symmetric, then A =, Note:, 1. In a symmetric matrix, if the eigen values are non-repeated, then we get a linearly, independent and pair wise orthogonal set of eigen vectors, 2. In a symmetric matrix, if the eigen values are repeated, then it may or may not be, possible to get linearly independent and pair wise orthogonal set of eigen vectors, If we form a linearly independent and pair wise orthogonal set of eigen vectors, then, diagonalization is possible through orthogonal transformation, Problems:, 1. Find the eigen values and eigen vectors of the matrix (, , )
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Solution: Let A = (, , ) which is a non-symmetric matrix, , To find the characteristic equation:, The characteristic equation of A is, , where, ,, , = 1(-1) – 1(3) = - 4, Therefore, the characteristic equation is, , i.e.,, , or, , Therefore, the eigen values are 2, -2, A is a non-symmetric matrix with non- repeated eigen values, , To find the eigen vectors:, ], , [, , [(, , ), , ][ ], , [ ]--------------- (1), , Case 1: If, , [, , i.e., [, , [ ], , ][ ], , (, , )] [ ], , [ ], , ][ ], , [(, , i.e., we get only one equation, , Case 2: If, , [, , ], , [, , ][ ], , (, , [ ] From (1)], , i.e.,, , Therefore, , ), , [ ] From (1)], , )] [ ], , [ ]
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i.e., [, , ][ ], , [ ], , i.e.,, , i.e., we get only one equation, , Hence,, , [ ], , 2. Find the eigen values and eigen vectors of [, , Solution: Let A = [, , ], , ] which is a non-symmetric matrix, , To find the characteristic equation:, Its characteristic equation can be written as, , where, ,, |, , |, , |, , |, , ,, = 2(4)-2(1)+1(-1) = 5, Therefore, the characteristic equation of A is, 1, , 1, , -6, , √, , 5, , √, , 0, , |, , |
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Therefore, the eigen values are 1, 1, and 5, A is a non-symmetric matrix with repeated eigen values, To find the eigen vectors:, ], , [, , [, , Case 1: If, , i.e., [, , ][ ], , ][ ], , ][ ], , [ ], , [ ], , [ ], , --------- (1), ------------- (2), ------------ (3), Considering equations (1) and (2) and using method of cross-multiplication, we get,, , 2, , 1, , -3, , 2, , -2, , 1, , 1, , -2, , Therefore,, , Case 2: If, , [ ], , ,[, , ][ ], , [ ]
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i.e., [, , ][ ], , [ ], , All the three equations are one and the same. Therefore,, Put, , Therefore,, , [, , ], , [, , ], , Put, , Therefore,, , 3. Find the eigen values and eigen vectors of [, , Solution: Let A = [, , ], , ] which is a non-symmetric matrix, , To find the characteristic equation:, Its characteristic equation can be written as, , where, ,, |, , |, , ,, = 2(-4)+2(-2)+2(2) = - 8 – 4 + 4 = - 8, Therefore, the characteristic equation of A is, , |, , |, , |, , |
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1, , 0, , -4, , Therefore, the eigen values are 2, 2, and -2, A is a non-symmetric matrix with repeated eigen values, To find the eigen vectors:, ], , [, , [, , Case 1: If, , i.e., [, , ][ ], , ][ ], , ][ ], , [ ], , [ ], , [ ], , --------- (1), ------------- (2), ------------ (3) . Equations (2) and (3) are one and the same., Considering equations (1) and (2) and using method of cross-multiplication, we get,, , -1, , 1, , 2, , -1, , 3, , 1, , 1, , 3, , Therefore,, , [, , ], , 0
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Case 2: If, , ,[, , i.e., [, , ][ ], , ][ ], , [ ], , [ ], , ---------- (1), ---------------- (2), ------------ (3), Considering equations (1) and (2) and using method of cross-multiplication, we get,, , -2, , 2, , 0, , -2, , -1, , 1, , 1, , -1, , Therefore,, , [ ], , We get one eigen vector corresponding to the repeated root, , 4. Find the eigen values and eigen vectors of [, , Solution: Let A =[, , ], , ] which is a symmetric matrix, , To find the characteristic equation:, Its characteristic equation can be written as, , where
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,, |, , |, , |, , |, , |, , ,, = 1(4)-1(-2)+3(-14) = - 4 + 2-42 = - 36, Therefore, the characteristic equation of A is, , 1, , √, , -9, , 18, , √, , Therefore, the eigen values are -2, 3, and 6, A is a symmetric matrix with non- repeated eigen values, To find the eigen vectors:, ], , [, , Case 1: If, , i.e., [, , [, , ][ ], , ][ ], , ][ ], , [ ], , --------- (1), ------------- (2), ------------ (3), , [ ], , [ ], , 0, , |
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Considering equations (1) and (2) and using method of cross-multiplication, we get,, , 1, , 3, , 3, , 1, , 7, , 1, , 1, , 7, , Therefore,, , [, , Case 2: If, , ,[, , i.e., [, , ], , ][ ], , ][ ], , [ ], , [ ], , ---------- (1), ---------------- (2), 3, , ------------ (3), , Considering equations (1) and (2) and using method of cross-multiplication, we get,, , 1, , 3, , -2, , 1, , 2, , 1, , 1, , 2, , Therefore,, , [, , Case 3: If, , ,[, , ], , ][ ], , [ ]
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i.e., [, , ][ ], , [ ], , ---------- (1), ---------------- (2), 3, , ------------ (3), , Considering equations (1) and (2) and using method of cross-multiplication, we get,, , 1, , 3, , -5, , -1, , 1, , 1, , Therefore,, , 1, -1, , [ ], , ]. Determine the, , 5. Find the eigen values and eigen vectors of the matrix[, algebraic and geometric multiplicity, , ] which is a symmetric matrix, , Solution: Let A =[, , To find the characteristic equation:, Its characteristic equation can be written as, , where, ,, |, , ,, = 0 -1(-1)+ 1(1) = 0 + 1 + 1 = 2, Therefore, the characteristic equation of A is, , |, , |, , |, , |, , |
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1, , √, , -1, , -2, , 0, , √, , Therefore, the eigen values are 2, -1, and -1, A is a symmetric matrix with repeated eigen values. The algebraic multiplicity of, To find the eigen vectors:, ], , [, , [, , Case 1: If, , i.e., [, , ][ ], , ][ ], , ][ ], , [ ], , [ ], , [ ], , --------- (1), ------------- (2), ------------ (3), Considering equations (1) and (2) and using method of cross-multiplication, we get,, , 1, , 1, , -2, , -2, , 1, , 1, , 1, -2, , is 2
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Therefore,, , [ ], , Case 2: If, , ,[, , i.e., [, , ][ ], , ][ ], , [ ], , [ ], , ---------- (1), ---------------- (2), ------------ (3). All the three equations are one and the same., Therefore,, , . Put, , [, , Therefore,, , ], , Since the given matrix is symmetric and the eigen values are repeated, let, orthogonal to, , ][ ], , ------------ (1), , ][ ], , -------- (2), , Solving (1) and (2) by method of cross-multiplication, we get,, l, , m, , 1, , 1, , 1, , 1, , 1, , -1, , 0, , 1, , [ ]., , is
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. Therefore,, , [, , ], , Thus, for the repeated eigen value, vectors, , there corresponds two linearly independent eigen, , . So, the geometric multiplicity of eigen value, , is 2, , Problems under properties of eigen values and eigen vectors., , 1. Find the sum and product of the eigen values of the matrix [, , ], , Solution: Sum of the eigen values = Sum of the main diagonal elements = -3, Product of the eigen values = │A│ = -1 (1 – 1) -1(-1 – 1) + 1(1- (-1)) = 2 + 2 = 4, , 2. Product of two eigen values of the matrix A = [, , ] is 16. Find the third eigen, , value, Solution: Let the eigen values of the matrix be, , ., , Given, We know that, , (Since product of the eigen values is equal to the determinant of, , the matrix), , |, , | = 6(9-1)+2(-6+2) +2(2-6) = 48-8-8 = 32, , 3. Find the sum and product of the eigen values of the matrix A = (, finding the roots of the characteristic equation, Solution:We know that the sum of the eigen values = Trace of A = a + d, Product of the eigen values = │A│ = ad – bc, , ) without
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], find │A│, without, , 4. If 3 and 15 are the two eigen values of A = [, expanding the determinant, Solution:Given, , We know that sum of the eigen values = Sum of the main diagonal elements, , 5. If 2, 2, 3 are the eigen values of A = [, , ], find the eigen values of, , Solution:By the property “A square matrix A and its transpose, , have the same eigen, , values”, the eigen values of, , 6. Find the eigen values of A = [, , Solution:Given A = [, , ], , ] .Clearly, A is a lower triangular matrix. Hence, by the, , property “the characteristic roots of a triangular matrix are just the diagonal elements of the, matrix”, the eigen values of A are 2, 3, 4, , 7. Two of the eigen values of A = [, , ] are 3 and 6. Find the eigen values of, , Solution:Sum of the eigen values = Sum of the main diagonal elements = 3 +5+3 = 11, Given 3,6 are two eigen values of A. Let the third eigen value be k.
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Then, 3 + 6 + k = 11, Therefore, the eigen values of A are 3, 6, 2, By the property “If the eigen values of A are, are, , ”, the eigen values of, , , then the eigen values of, , are, , 8. Find the eigen values of the matrix[, , ]. Hence, form the matrix whose eigen, , values are, ]. The characteristic equation of the given matrix is, , Solution: Let A =[, where, , and, √, , Therefore, the characteristic equation is, , =, , Therefore, the eigen values of A are 6, -1, Hence, the matrix whose eigen values are, , │A│ = 4 -10 = - 6; adj A = [, [, , Therefore,, , is, , ], , ], , 9. Find the eigen values of the inverse of the matrix A = [, , ], , Solution:We know that A is an upper triangular matrix. Therefore, the eigen values of A are, 2, 3, 4. Hence, by using the property “If the eigen values of A are, values of, , are, , ”, the eigen values of, , 10. Find the eigen values of, , given A = [, , ], , , then the eigen
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Solution:Given A = [, , ]. A is an upper triangular matrix. Hence, the eigen values of, , A are 1, 2, 3, Therefore, the eigen values of, , are, , i.e., 1,8,27, , 11. If 1 and 2 are the eigen values of a 2 x 2 matrix A, what are the eigen values of, ?, Solution:Given 1 and 2 are the eigen values of A., Therefore,, , i.e., 1 and 4 are the eigen values of, , and 1 and are the eigen, , values of, , 12. If 1,1,5 are the eigen values of A = [, , Solution:By the property “If, , ], find the eigen values of 5A, , are the eigen values of A, then, , are the, , eigen values of kA, the eigen values of 5A are 5(1), 5(1), 5(5) ie., 5,5,25, if A = [, , 13. Find the eigen values of A,, , Solution:Given A = [, , ]. A is an upper triangular matrix. Hence, the eigen values of A are, , 2, 5, The eigen values of, , are, , i.e., 4, 25, , The eigen values of, , are, , i.e., 8, 125, , The eigen values of, , are, , i.e., 16, 625, , The eigen values of 3A are 3(2), 3(5) i.e., 6, 15, The eigen values of, A–I=[, , ]-[, , ], , are, ]=[, , ]
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Since A - I is an upper triangular matrix, the eigen values of A- I are its main diagonal, elements i.e., 1,4, Eigen values of, , where, , First eigen value =, = 3(2)3+ 5(2)2 - 6(2) + 2 = 24 + 20 -12 + 2=34, Second eigen value =, = 3(5)3+ 5(5)2- 6(5) + 2, = 375+ 125-30 + 2 = 472, , 14. Find the eigen values of adj A if A = [, , ]. A is an upper triangular matrix. Hence, the eigen values of A, , Solution:Given A =[, are, , ], , 3, 4, 1, , We know that, Adj A = │A│, The eigen values of, , are, , │A│=Product of the eigen values = 12, Therefore, the eigen values of adj A is equal to the eigen values of 12, , i.e.,, , i.e.,, , 4, 3, 12, , Note: A =[, , ], , [, , ], , [, , ]. Here, A is an upper triangular matrix,, , B is a lower triangular matrix and C is a diagonal matrix. In all the cases, the elements in the, main diagonal are the eigen values. Hence, the eigen values of A, B and C are 1, 4, 6
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15. Two eigen values of A = [, , ] are equal and they are times the third. Find, , them, Solution:Let the third eigen value be, We know that, Given, , 2+3+2 = 7, =, , [, Therefore,, , ], , and hence the eigen values of A are 1,1, 5, , 16. If 2, 3 are the eigen values of [, , ], , ]. Let the eigen values of A be 2, 3, k, , Solution:Let A =[, , We know that the sum of the eigen values = sum of the main diagonal elements, Therefore, 2 +3 +k = 2+ 2+2 = 6, We know that product of the eigen values = │A│, 2(3)(k) = │A│, , |, , |, , 17. Prove that the eigen vectors of the real symmetric matrix A = [, orthogonal in pairs, , ] are
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Solution:The characteristic equation of A is, where, , |, , ;, , |, , The characteristic equation of A is, 3, , 1-7, , 0, , 36, , 0, , 3, , -12, , -36, , 1, , -4, , -12, , 0, , √, , Therefore,, Therefore, the eigen values of A are -2, 3, 6, To find the eigen vectors:, , [, , Case 1: When, , ][ ], , [ ], , -------- (1), ---------- (2), --------- (3), Solving (1) and (2) by rule of cross-multiplication, we get,, , 1, , 3, , 3, , 1, , 7, , 1, , 1, , 7, [, , ]
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[, , Case 2:When, , ][ ], , [ ], , --------- (1), --------- (2), --------- (3), Solving (1) and (2) by rule of cross-multiplication, we get,, , 1, , 3, , -2, , 1, , 2, , 1, , 1, , 2, [, , [, , Case 3:When, , ][ ], , ], , [ ], , --------- (1), --------- (2), --------- (3), Solving (1) and (2) by rule of cross-multiplication, we get,, , 1, , 3, , -5, , 1, , -1, , 1, , 1, , -1, [, , Therefore,, , [, , ],, , [, , To prove that:, ][, , ], , ], , [, , ], , ]
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][ ], , ][, , ], , Hence, the eigen vectors are orthogonal in pairs, 18. Find the sum and product of all the eigen values of the matrix A = [, , ].Is the, , matrix singular?, Solution:Sum of the eigen values = Sum of the main diagonal elements =Trace of the matrix, Therefore, the sum of the eigen values = 1+2+7=10, Product of the eigen values = │A│ = 1(14 - 8) -2(14 - 4) + 3(4 - 2) = 6-20+ 6= - 8, │A│≠0. Hence the matrix is non-singular., 19. Find the product of the eigen values of A = [, , Solution:Product of the eigen values of A = │A│=|, , ], , |, , ORTHOGONAL TRANSFORMATION OF A SYMMETRIC MATRIX TODIAGONAL FORM:, Orthogonal matrices:, A square matrix A (with real elements) is said to be orthogonal if, , or, , Problems:, 1. Check whether the matrix B is orthogonal. Justify. B = [, Solution: Condition for orthogonality is, To prove that:, B=[, , ];, , [, , ], , ]
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[, , ][, , ], , [, , ], , [, , ], , Similarly,, [, , ][, , ], , [, , ], , [, , ], , Therefore, B is an orthogonal matrix, 2. Show that the matrix P = [, , ] is orthogonal, , Solution:To prove that:, [, , ];, , [, , ], , [, , ], [, , ][, , [, , ]= I, , ], , [, Therefore, P is an orthogonal matrix, WORKING RULE FOR DIAGONALIZATION, [ORTHOGONAL TRANSFORMATION]:, Step 1: To find the characteristic equation, Step 2: To solve the characteristic equation, Step 3:To find the eigen vectors, Step 4: If the eigen vectors are orthogonal, then form a normalized matrix N, Step 5: Find, Step 6: Calculate AN, Step 7: Calculate D =, , ], , [, , ]
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Problems:, 1. Diagonalize the matrix [, , Solution: Let A = [, , ], , ], , The characteristic equation is, , where, , Therefore, the characteristic equation is, 2, , 1, 0, 1, , -11, , 36, , -36, , 2, , -18, , 36, , 18, , 0, , -9, , √, , √, , =, , Hence, the eigen values of A are 2, 3, 6, To find the eigen vectors:, , [, , Case 1: When, , [, , ][ ], , ][ ], , [ ], , ---------- (1), -------- (2), --------- (3), Solving (1) and (2) by rule of cross-multiplication,, , [ ]
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-1, , 1, , 1, , -1, , 3, , -1, , -1, , 3, , [, , ], , Case 2: When, , [, , ][ ], , [ ], , ---------- (1), -------- (2), --------- (3), Solving (1) and (2) by rule of cross-multiplication,, , -1, , 1, , 0, , -1, , 2, , -1, , -1, , 2, , [ ], , Case 3: When, , [, , ][ ], , [ ], , ---------- (1), -------- (2), --------- (3), Solving (1) and (2) by rule of cross-multiplication,
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-1, , 1, , -3, , -1, , -1, , -1, , -1, , -1, , [, , ], , ][ ], , ][, , ], , =0, , ][, , ], , Hence, the eigen vectors are orthogonal to each other, √, , √, , √, , √, , √, , √, , [√, , √, , √, , The Normalized matrix N =, , AN = [, , i.e.,, , ], , √, , √, , √, , √, , √, , [√, , √, , √, , √, , √, , √, , √, , √, , √, , √, , √, , √, , √, , √, , √, , [√, , √, , √, , [√, , √, , √, , √, , √, , √, , √, , √, , √, , √, , √, , √, , √, , √, , √, , [√, , √, , √ ] [√, , √, , √ ], , ], , [, , ;, , √, , ], , ], , The diagonal elements are the eigen values of A, , ], , ], , √, √, [√, , √, [, , √, √, , ], , ]
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2. Diagonalize the matrix [, , Solution: Let A = [, , ], , ], , The characteristic equation is, , where, , Therefore, the characteristic equation is, , i.e.,, √, , √, , Hence, the eigen values of A are 0, 3, 15, To find the eigen vectors:, , [, , ][ ], , Case 1: When, , [ ], , [, , ][ ], , [ ], , ---------- (1), -------- (2), --------- (3), Solving (1) and (2) by rule of cross-multiplication,, , -6, , 2, , 8, , -6, , 7, , -4, , -6, , 7
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[ ], , Case 2: When, , [, , ][ ], , [ ], , -------- (1), ------- (2), --------- (3), Solving (1) and (2) by rule of cross-multiplication,, , -6, , 2, , 5, , -6, , 4, , -4, , -6, , 4, , [, , ], , Case 3: When, , [, , ][ ], , [ ], , ---------- (1), ---------- (2), ---------- (3), Solving (1) and (2) by rule of cross-multiplication,, , -6, , 2, , -7, , -6, , -8, , -4, , -6, , -8
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[, , ], , ][, , ], , ][, , ], , 4-2-2 = 0, , ][ ], Hence, the eigen vectors are orthogonal to each other, , [, , The Normalized matrix N =, [, , ], , [, [, , ], , ], , ], , AN =, [, , ] [, , ], , [, , [, , ], , [, , ][, [, , i.e.,, , [, , ][, , ], , [, , ], , ], ], , [, , ], , The diagonal elements are the eigen values of A, , [, ], , [, , ], , ]
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QUADRATIC FORM- REDUCTION OF QUADRATIC FORM TO CANONICAL FORM BY, ORTHOGONAL TRANSFORMATION:, Quadratic form:, A homogeneous polynomial of second degree in any number of variables is called a quadratic, form, Example:, , is a quadratic form in three variables, , Note:, The matrix corresponding to the quadratic form is, , [, , ], , Problems:, 1. Write the matrix of the quadratic form, , Solution:Q =, [, Here, , ], ;, , [, , ;, ], , 2. Write the matrix of the quadratic form, , [, , Solution: Q =, [, , ], , ], , 3. Write down the quadratic form corresponding to the following symmetric matrix, [, , Solution:Let [, , ], , ], , [, , ]
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The required quadratic form is, , =, NATURE OF THE QUADRATIC FORM:, Rank of the quadratic form:The number of square terms in the canonical form is the rank (r) of, the quadratic form, Index of the quadratic form: The number of positive square terms in the canonical form is, called the index (s) of the quadratic form, Signature of the quadratic form:The difference between the number of positive and negative, square terms = s – (r-s) = 2s-r, is called the signature of the quadratic form, , The quadratic form is said to be, (1) Positive definite if all the eigen values are positive numbers, (2) Negative definite if all the eigen values are negative numbers, (3) Positive Semi-definite if all the eigen values are greater than or equal to zero and at, least one eigen value is zero, (4) Negative Semi-definite if all the eigen values are less than or equal to zero and at least, one eigen value is zero, (5) Indefinite if A has both positive and negative eigen values, Problems:, 1. Determine the nature of the following quadratic form f(, Solution:The matrix of the quadratic form is Q = [, , ], , The eigen values of the matrix are 1, 2, 0, Therefore, the quadratic form is Positive Semi-definite, 2. Discuss the nature of the quadratic form, canonical form, Solution:The matrix of the quadratic form is Q = [, , without reducing it to, , ]
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|, , |, , |, , |, , Therefore, the quadratic form is positive definite, REDUCTION OF QUADRATIC FORM TO CANONICAL FORM THROUGH ORTHOGONAL, TRANSFORMATION [OR SUM OF SQUARES FORM OR PRINCIPAL AXES FORM], Working rule:, Step 1: Write the matrix of the given quadratic form, Step 2: To find the characteristic equation, Step 3: To solve the characteristic equation, Step 4: To find the eigen vectors orthogonal to each other, Step 5: Form the Normalized matrix N, Step 6: Find, Step 7: Find AN, Step 8: Find D =, Step 9: The canonical form is, , ], , ][ ], , Problems:, 1. Reduce the given quadratic form Q to its canonical form using orthogonal, transformation Q =, , Solution: The matrix of the Q.F is A =, [, i.e., A = [, , ], , ], , The characteristic equation of A is, , where
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The characteristic equation of A is, 1, , 1, , -7, , 14, , -8, , 0, , 1, , -6, , 8, , 1, , -6, , 8, , 0, , √, , The eigen values are 1, 2, 4, To find the eigen vectors:, , [, , ][ ], , [, , Case 1: When, , [ ], , ][ ], , [ ], , = 0 -------- (1), -------- (2), -------- (3), Solving (2) and (3) by rule of cross multiplication, we get,, , 2, , -1, , 0, , 2, , -1, , 2, , 0, , -1, , [ ], , √
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[, , Case 2: When, , ][ ], , [ ], , = 0 -------- (1), -------- (2), -------- (3), Solving (1) and (2) by rule of cross multiplication, we get,, , 0, , 0, , -1, , 0, , 1, , -1, , 0, , 1, , [ ], , Case 3: When, , [, , ][ ], , [ ], , = 0 -------- (1), -------- (2), -------- (3), Solving (1) and (2) by rule of cross multiplication, we get,, , 0, , 0, , -3, , 0, , -1, , -1, , 0, , -1, , [, , ]
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The Normalized matrix N =, , AN = [, , ], , [, , √, , √, , √, , √, , √, , √, , [√, , √, , √, , √, , √, , √, , √, , √, , √, , [√, , √, , √, , [, ], , √, , √, , √, , √, , √, , √ ][, , √, , √ ], , [, , ], , i.e.,, , [, ], , √, , √, , √, , √, , √, , √, , √, , √, , ], , ], , [, , ], , The canonical form is, , ][, , ][ ], , canonical form is, , ][, , ][ ], , 1. Reduce the quadratic form to a canonical form by an orthogonal reduction, .Also find its nature., , Solution: The matrix of the Q.F is A =, [, i.e., A, , [, , ], , ], , The characteristic equation of A is, , The characteristic equation of A is, , where
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1, , 1, , 0, , -3, , 0, , 1, , 1, , 1, , 1, , -2, √, , The eigen values are 1, 1, -2, To find the eigen vectors:, , [, , ][ ], , [ ], , [, , Case 1: When, , ][ ], , [ ], , ][ ], , [ ], , ------- (1), ------- (2), ------- (3), , 1, , 1, , 2, , 1, , 2, , -1, , 1, , 2, , [, , ], , Case 2: When, , [, , 2, -2, 0, √
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------- (1), ------ (2), ------ (3), All three equations are one and the same., Put, , ,, [, , . Let, , . Then, , ], , [ ]. Since, , Let, , is orthogonal to, , ][ ], , ,, , ][ ], -------- (1), --------- (2), , l, , m, , n, , 1, , 1, , -1, , 1, , -1, , 1, , 0, , -1, , [ ], , The Normalized matrix N =, , AN = [, , ], , √, , √, , √, , √, , √, , √, , [√, , √, , √, , √, , √, , √, , √, , √, , [√, , √, , √, , ;, ], , √, , √, , √, , √, , √, , √, , √, , √, , √, , √, , √, , √, , [√, , √, , √, , [√, , √, , √, , ], , √, , ], , ]
