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# Maths

CTO and Cofounder at RenaiSense um RenaiSense
23. Sep 2013
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### Maths

1. J.Baskar Babujee Department of Mathematics Anna University, Chennai-600 025. MA6151MA6151
2. CONTENT 1.1 INTRODUCTION 1.2 TYPES OF MATRICES 1.3 CHARACTERISTIC EQUATION 1.4 EIGEN VECTORS 1.5 PROBLEMS 1.6 CAYLEY HAMILTON THEOREM 2
3. 1.7 DIAGONALISATION OF A MATRIX 1.8 REDUCTION OF A MATRIX TO DIAGONAL FORM 1.9 ORTHOGONAL TRANSFORMATION OF A SYMMETRIC MATRIX TO DIAGONAL FORM 1.10 QUADRATIC FORMS 1.11 NATURE OF A QUADRATIC FORM 3
4. 1.12 REDUCTION OF QUADRATIC FORM TO CANONICAL FORM 1.13 INDEX AND SIGNATURE OF THE QUADRATIC FORM 1.14 LINEAR TRANSFORMATION OF A QUADRATIC FORM. 1.15 REDUCTION OF QUADRATIC FORM TO CANANONICAL FORM BY ORTHOGONAL TRANSFORMATION 4
5. 1.1 INTRODUCTION:- A matrix is defined as a rectangular array (or arrangement in rows or columns) of scalar subject to certain rules of operations. If mn numbers (real or complex) or functions are arranged in the columns (vertical lines) then A is called an m n matrix. Each of the mn numbers is called an element of the matrix. × Unit 1 MATRICES
6. An m n matrix is usually written as A matrix is usually denoted by a single capital letter A, B or C etc. 11 12 13 1 21 22 23 2 31 32 33 3 1 2 3 .... .... .... .... .... .... .... .... .... n n n m m m mn a a a a a a a a a a a a a a a a                ×
7. 7 Thus, an m n matrix A may be written as A = , where i = 1, 2, 3, … , m ; j = 1, 2, 3, … , n In Algebra, the matrices have their largest application in the theory of simultaneous equations and linear transformations. × ija  
8. E.g., The set of simultaneous equations 3333232131 2323222121 1313212111 bxaxaxa bxaxaxa bxaxaxa =++ =++ =++
9. may be symbolically represented by the equation Where A = , X = , B =           333231 232221 131211 aaa aaa aaa A X = B           3 2 1 x x x           3 2 1 b b b
10.       − − 720 135 1.2 TYPES OF MATRICES (1) Real Matrix :- A matrix is said to be real if all its elements are real numbers. E.g., is a real matrix.
11. (2) Square Matrix:- A matrix in which the number of rows is equal to the number of columns is called a square matrix, otherwise, it is said to be rectangular matrix. i.e., a matrix A = is a square matrix if m = n a rectangular matrix if m n ij m n a ×    ≠
12. A square matrix having n rows and n columns is called “ n – rowed square matrix”, is a 3 – rowed square matrix The elements of a square matrix are called its diagonal elements and the diagonal along with these elements are called principal or leading diagonal. 11 12 13 21 22 23 31 32 33 a a a a a a a a a          33,2211 aa,a
13. The sum of the diagonal elements of a square matrix is called its trace or spur. Thus, trace of the n rowed square matrix A= isija   ∑ = =++++ n 1i ijnn332211 aa....aaa
14. (3) Row Matrix :- A matrix having only one row and any number of columns, i.e., a matrix of order 1 n is called a row matrix. [2 5 -3 0] is a row matrix. × Example:-
15. (4) Column Matrix:- A matrix having only one column and any number of rows, ii.e., a matrix of order m 1 is called a column matrix. is a column matrix.           − 1 0 2 × Example:-
16. (5) Null Matrix:- A matrix in which each element is zero is called a null matrix or void matrix or zero matrix. A null matrix of order m n is denoted by = 16 × nmO × Example :-       0000 0000 42O ×
17. (6) Sub - matrix :- A matrix obtained from a given matrix A by deleting some of its rows or columns or both is called a sub – matrix of A. is a sub – matrix of       41 03           − − 2461 7053 5210 Example:-
18. (7) Diagonal Matrix :- A square matrix in which all non – diagonal elements are zero is called a diagonal matrix. i.e., A = [a ] is a diagonal matrix if a = 0 for i j. is a diagonal matrix. ij nn× ij ≠ Example:-           − 000 010 002
19. (8) Scalar Matrix:- A diagonal matrix in which all the diagonal elements are equal to a scalar, say k, is called a scalar matrix. i.e., A = [a ] is a scalar matrix if is a scalar matrix.           200 020 002 ij nn×    = ≠ = jiwhenk jiwhen0 aij Example :-
20. (9) Unit Matrix or Identity Matrix:- A scalar matrix in which each diagonal element is 1 is called a unit or identity matrix. It is denoted by . i.e., A = [a ] is a unit matrix if is a unit matrix. nI ij nn×    = ≠ = jiwhen1 jiwhen0 aij Example       10 01
21. (10) Upper Triangular Matrix. A square matrix in which all the elements below the principal diagonal are zero is called an upper triangular matrix. i.e., A = [a ] is an upper triangular matrix if a = 0 for i > j is an upper triangular matrix          − 300 510 432 ij nn× ij Example:-
22. (11) Lower Triangular Matrix. A square matrix in which all the elements above the principal diagonal are zero is called a lower triangular matrix. i.e., A = [a ] is a lower triangular matrix if a = 0 for i < j is a lower triangular matrix. ij nn× ij Example:-          − 023 065 001
23. (12) Triangular Matrix:- A triangular matrix is either upper triangular or lower triangular. (13) Single Element Matrix:- A matrix having only one element is called a single element matrix. i.e., any matrix [3] is a single element matrix.
24. (14) Equal Matrices:- Two matrices A and B are said to be equal iff they have the same order and their corresponding elements are equal. i.e., if A = and B = , then A = B iff a) m = p and n = q b) a = b for all i and j. qpij]b[ ×nmij]a[ × ij ij
25. (15) Singular and Non – Singular Matrices:- A square matrix A is said to be singular if | A| = 0 and non – singular if |A| 0. A = is a singular matrix since |A| = 0.           − − − 110 432 432 ≠ Example :-
26. 1.3 CHARACTERISTIC EQUATION If A is any square matrix of order n, we can form the matrix , where is the nth order unit matrix. The determinant of this matrix equated to zero, i.e., IλA − I 0 λa...aa ............ a...λaa a...aλa λA nnn2n1 2n2221 1n1211 = − − − =− I
27. is called the characteristic equation of A. On expanding the determinant, we get where k’s are expressible in terms of the elements a The roots of this equation are called Characteristic roots or latent roots or eigen values of the matrix A. 0k...λkλkλ1)( n 2n 2 1n 1 nn =++++− −− ij
28. 1.4 EIGEN VECTORS Consider the linear transformation Y = AX ...(1) which transforms the column vector X into the column vector Y. We often required to find those vectors X which transform into scalar multiples of themselves. Let X be such a vector which transforms into X by the transformation (1). λ
29. Then Y = X ... (2) From (1) and (2), AX = X AX- X = 0 (A - )X = 0 ...(3) This matrix equation gives n homogeneous linear equations ... (4) I⇒ λ λ λ λ        =−+++ =++−+ =+++− 0λ)x(a...xaxa ................ 0xa...λ)x(axa 0xa...xaλ)x(a nnn2n21n1 n2n222121 n1n212111 ⇒ I
30. These equations will have a non-trivial solution only if the co-efficient matrix A - is singular i.e., if |A - | = 0 ... (5) Corresponding to each root of (5), the homogeneous system (3) has a non-zero solution X = is called an eigen vector or latent vector λI Iλ             4 2 1 x ... x x
31. 31 Properties of Eigen Values:-Properties of Eigen Values:- 1.The sum of the eigen values of a matrix is the sum of the elements of the principal diagonal. 2.The product of the eigen values of a matrix A is equal to its determinant. 3.If is an eigen value of a matrix A, then 1/ is the eigen value of A-1 . 4.If is an eigen value of an orthogonal matrix, then 1/ is also its eigen value. λ λ λ λ
32. 32 PROPERTY 1:-PROPERTY 1:- If λ1, λ2,…, λn are the eigen values of A, then i. k λ1, k λ2,…,k λn are the eigen values of the matrix kA, where k is a non – zero scalar. ii. are the eigen values of the inverse matrix A-1 . iii. are the eigen values of Ap , where p is any positive integer. n21 λ 1 ,..., λ 1 , λ 1 p n p 2 p 1 λ...,,λ,λ
33. 33 Proof:- i. Let λr be an eigen value of A and Xr the corresponding eigen vector. Then, by definition, Multiplying both sides by k, (kA)Xr = (kλr)Xr Then k λr is an eigen value of kA and the corresponding eigen vector is the same as that of λr, namely Xr. AXr = λrXr
34. 34 ii. Pre multiplying both sides of AXr = λrXr by A-1 A-1 (A Xr) = A-1 (λr Xr ) Xr = λr (A-1 Xr ) => A-1 Xr = (Xr) Hence is an eigen value of A-1 and the corresponding eigen vector is the same as that of λr , namely Xr. rλ 1 rλ 1
35. 35 iii. Pre multiplying both sides of AXr = λrXr by A A(A Xr) = A(λr Xr ) A2 Xr = λr (AXr ) = λr (λr xr) = λr 2 Xr. Similarly, we can prove that A3 Xr = λr 3 Xr, …, Ap Xr = λr p Xr, where p is any positive integer. Hence λr p is any eigen value of Ap and the corresponding eigen vector is the same as that of λr, namely Xr.
36. 36 THEOREM :- A matrix A is singular if and only if 0 is an eigen value of A. 1.5 PROBLEMS 1. Find the sum and product of the eigen values of the matrix without finding the eigen values.           −− − −− = 021 612 322 A
37. 37 Solution:- Sum of the eigen values of A = sum of its diagonal elements. = - 2 + 1 + 0 = -1. Product of the eigen values of A = | A | = = 45. 021 612 322 −− − −−
38. 38 2. Two eigen values of the matrix are equal to 1 each. Find the third eigen value. Solution:- Let a be the third eigen value of A. Since sum of the eigen values = sum of the diagonal elements, 1 + 1 + a = 2 + 3 + 2 a = 5 Therefore, the third eigen value of A is 5.           = 221 131 122 A
39. 39 3. The product of two eigen values of the matrix is 16. Find the third eigen value. Solution:- Let a be the third eigen value of A. Since product of the eigen values = | A | 16a = Therefore, a = 2.           − −− − = 312 132 226 A 312 132 226 − −− −
40. 40 4. Find the sum of the eigen values of the inverse of Solution:- The eigen values of the lower triangular matrix A is 1, -3, 2. Then the eigen values of A-1 are Sum of the eigen values of A-1 = =           −= 250 032 001 A . 2 1 , 3 1 ,1 − . 2 1 3 1 1 + − + 6 7
41. 41 5. If , find the eigen values of 3A, A-1 and – 2A-1 . Solution:- The eigen values of A are 2, -1, 4. The eigen values of 3A are 3×2, 3×(-1), 3×4 i.e., 6, -3, 12           −= 400 210 572 A
42. 42 The eigen values of A-1 are 4 1 1,, 2 1 i.e., 4 1 , 1 1 , 2 1 − − 2 1 2,1,i.e., 4 1 21),2)((, 2 1 2 are2AofvalueseigenThe 1 −−       −−−      − − −
43. 43 6. Find the eigen values and eigen vectors of the matrix A = Solution:- The characteristic equation of the given matrix is       − − 45 21 0|IA| =λ−
44. or Thus, Corresponding to =6, the eigen vectors are given by (A – 6 ) X1 = 0 1,6 0652 −=λ=> =−λ−λ=> 0 45 21 = λ−− −λ− λ I 0 x x 25 25 or0 x x 645 261 or 2 1 2 1 =            −− −− =            −− −− the eigen values of A are 6, -1.
45. We get only one independent equation - 5x1 - 2x2 = 0       − =∴ −= = = − =⇒ 5 2 kXarevectorseigenThe 5kx 2kx (say)k 5 x 2 x 11 12 11 1 21
46. Corresponding to = -1, the eigen vectos are given by ( A + ) X2 = 0 λ I       =∴ == ==⇒ =−⇒       =            − − ⇒ 1 1 kXarevectorseigenThe kx,kx (say)k 1 x 1 x 0xx 0 0 x x 55 22 22 2221 2 21 21 2 1
47. 7. Find the eigen values and eigen vectors of the matrix A = Solution:- The characteristic equation of the given matrix is           −− − −− 021 612 322 0|IA| =λ− 0 21 612 322 = λ−−− −λ− −λ−−
48. Thus, 53,3,λ 05)3)(λ3)(λ(λ 015)2λ3)(λ(λ it.satisfies3λtrial,By 04521λλλ 0λ)]1(143[6]2λ2[12]λ)λ(1λ)[2( 2 23 −−=⇒ =−++⇒ =−−+∴ −= =−−+⇒ =−+−−−−−−−−−−⇒ the eigen values of A are -3, -3, 5.
49. Corresponding to = - 3, the eigen vectors are given by λ ( )          − +           =           − = ∴ −=== =−+ =                     −− − −− =+ 0 1 2 k 1 0 3 k k k k2k3 0x3x2 0 x x x 321 642 321 or0XI 21 1 2 21 32 3 2 1 1 1 2112213 1 X bygivenarevectorseigenThe 2k3kxthenkx,kxLet xequationtindependenoneonlygetWe 3A
50. Corresponding to λ = 5, the eigen vectors are given by (A – 5 I )X2 = 0. 052 032 0327 0 0 0 521 642 327 321 321 321 3 2 1 =−−− =−− =−+−⇒           =                     −−− −− −− ⇒ xxx xxx xxx x x x
51. 51           − = −===∴ = − ==⇒ −− = + = 1 2 1 kX kx2kx,kx (say)k 1 x 2 x 1 x 22 x 53 x 6-10 x 32 333,231 3 321 321 bygivenarevectorseigentheHence ,equations.twofirstFrom
52. 8. Find the eigen values and eigen vectors of the matrix Solution:- The characteristic equation is           − −− − = 342 476 268 A 0|IA| =λ− .arevalueseigenHence 3,150,λ 045λ18λλ 0 λ342 4λ76 26λ8 23 15,3,0 =⇒ =−+−⇒ = −− −−− −− ⇒
53. 53 Eigen vector X1 corresponding to λ1= 0 is given by (3)...03xx42x (2)...04x7x6x (1)...02x6x8x 0 0 0 x x x 342 476 268 x x x Xwhere0,I)Xλ(A 321 321 321 3 2 1 3 2 1 111 =+− =−+− =+−⇒           =           − −− − ⇒           ==−
54. 54 Solving equations (1) and (2) by cross-multiplication, we get which satisfy equation (3) also. Required eigen vector corresponding to λ1 = 0 is 131211 11 321 321 2kx,2kx,kx 0kwhere(say)k 2 x 2 x 1 x 3656 x 3212 x 1424 x === ≠===⇒ − = +− = −           =           = 2 2 1 2 2 1 1 1 1 1 k k k k X ∴
55. 55 Eigen vector X2 corresponding to λ2= 3 is given by (6)...0x42x (5)...04x4x6x (4)...02x6x5x 0 0 0 x x x 042 446 265 x x x Xwhere0,I)Xλ(A 21 321 321 3 2 1 3 2 1 222 =− =−+− =+−⇒           =           − −− − ⇒           ==−
56. 56 Solving equations (4) and (5) by cross- multiplication, we get which satisfy equation (6) also. Required eigen vector corresponding to λ2 = 3 is 232221 22 321 321 2kx,kx,kx 0kwhere(say)k 2- x 1 x 2 x 3620 x 2012 x 824 x −=== ≠===⇒ − = +− = − 2           − =           − = 2 1 2 2 2 2 2 2 2 2 k k k k X ∴
57. 57 Eigen vector X3 corresponding to λ3= 15 is given by (9)...012xx42x (8)...04x8x-6x (7)...02x6x7x- 0 0 0 x x x 12-42 48-6 267- x x x Xwhere0,I)Xλ(A 321 321 321 3 2 1 3 2 1 333 =−− =−− =+−⇒           =           − −− − ⇒           ==−
58. 58 Solving equations (7) and (8) by cross-multiplication, we get which satisfy equation (9) also. Required eigen vector corresponding to λ3 = 15 is 333231 33 321 321 kx,kx,kx 0kwhere(say)k 1 x 2- x 2 x 20 x 40- x 40 x =−== ≠===⇒ == 22           −=           −= 1 2 2 2 2 3 3 3 3 3 k k k k X
59. 59 9. Find the eigen values and eigen vectors of the matrix Solution:- The characteristic equation is           − −− − = 312 132 226 A 0|IA| =λ− 8.2,2,arevalueseigenHence 82,2,λ 03236λ12λλ 0 λ312 1λ32 22λ6 23 =⇒ =+−+−⇒ = −− −−− −− ⇒
60. 60 Eigen vector corresponding to is given by 221 == λλ 0xx2x 0xx2x 02x2x4x 0 0 0 x x x 112 112 224 x x x Xwhere0,I)Xλ(A 321 321 321 3 2 1 3 2 1 111 =+− =−+− =+−⇒           =           − −− − ⇒           ==−
61. These equations are equivalent to a single equation … (1) Let x3 = 2 k3 and x2 = 2 k2 then from (1) 2x1 – 2 k2 + 2 k3 = 0 => x1 = k2 – k3 Required eigen vector corresponding to is 02 321 =+− xxx ∴ 21 2 λλ ==          − +           =           − = 2 0 1 0 2 1 2 2 32 3 2 32 1 kk k k kk X
62. Similarly, the eigen vector corresponding to = 8 is given by, 3λ ....(4)05xx2x ....(3)0x5x2x ....(2)02x2x2x 0 0 0 x x x 512 152 222 0)X8-(A x x x Xwhere0I)Xλ-(A 321 321 321 3 2 1 3 3 2 1 333 =−− =−−− =+−−⇒           =                     −− −−− −− ⇒ =⇒           ==
63. Solving equations (2) and (3) by cross-multiplication, we get which satisfy equation (4) also. Required eigen vector corresponding to λ3 = 8 is 131211 1 321 321 ,,2 )( 112 6612 kxkxkx sayk xxx xxx =−== == − =⇒ = − = ∴           −=           −= 1 1 22 1 1 1 1 3 k k k k X
64. 64 Show that if λ1,λ2, … λn are the latent roots of the matrix A, then A3 has the latent roots Solution:- Let λ be a latent root of the matrix A. Then there exists a non – zero vector X such that ....,,, 33 2 3 1 nλλλ Example 1:-
65. 65 A X = λ X … (1) => A2 (AX) = A2 (λ X) => A3 X = λ (A2 X) [using (1)] But A2 X = A ( A X) = A (λ X) = λ (AX) = λ (λX) = λ2 X Therefore, A3 X = λ (λ2 X) = λ3 X => λ3 is a latent root of A3.
66. 66 Therefore, If λ1,λ2, … λn are the latent roots of the matrix A, then are the latent roots of A3 . If λ1,λ2, … λn are eigen values of A then find eigen values of the matrix (A – λI)2 . Solution:- (A – λI)2 = A2 – 2 λAI + λ2 I2 = A2 – 2 λA + λ2 I 33 2 3 1 ...,,, nλλλ Example 2:-
67. 67 Eigen values of A2 are Eigen values of 2 λA are 2 λ λ1,2 λ λ2, …, 2 λ λn. Eigen values of λ2 I are λ2 Eigen values of ( A – λI)2 are .λ...,λ,λ 2 n 2 2 2 1 .λ)(λ,...,λ)(λ,λ)(λor .λ2λλλ,...,λ2λλλ,λ2λλλ 2 n 2 2 2 1 2 n 2 n 2 2 2 2 2 1 2 1 −−− +−+−+− ∴
68. 68 Find the eigen values and eigen vectors of the matrix Solution:- The characteristic equation is           = 500 620 413 A 5.2,3,arevalueseigenHence 52,3,λ 0λ)λ)(5λ)(2 0 λ500 6λ20 41λ3 =⇒ =−−−⇒ = − − − ⇒ 3( 0|IA| =λ− Example 3:-
69. 69 Eigen vector X1 corresponding to λ1= 3 is given by 0x0,x 02x 06xx 04xx 0 0 0 x x x 200 61-0 410 x x x Xwhere0,)X3(A 23 3 32 32 3 2 1 3 2 1 11 ==⇒ = =+− =+⇒           =           ⇒           ==− I
70. The characteristic vector corresponding to eigen value λ1 = 3 is given by When λ2 = 2, let X2 be the eigen vector then (A – 2I) X2 = 0 where X2 = [x1 x2 x3]’ 0k,kXwhere 0 0 k x x x X 111 1 3 2 1 1 ≠=           =           =           =                     ⇒ 0 0 0 300 600 411 3 2 1 x x x
71. 71           −=           −=∴ = −= =∴ ≠== − = −==+∴ =⇒= = =++⇒ 0 1 1 k 0 k k XisvectoreigenRequired 0x kx kx 0kwhere(say)k 0 x 1 x 1 x xxor0xx 0x03x 06x 04xxx 22 2 2 3 22 21 22 321 2121 33 3 321
72. 72 When λ3 = 5, let X3 be the eigen vector then (A – 5I) X3 = 0 where X3 = [x1 x2 x3]’ Solving above equations by cross – multiplication, we get 06x3x 04xx2x 0 0 0 x x x 000 630 412 32 321 3 2 1 =+− =++−⇒           =                     − − ⇒
73. 73 Required eigen vector is 333231 33 321 321 kx,2kx,3kx 0kwhere(say)k 1 x 2 x 3 x 6 x 12 x 126 x ===⇒ ≠===⇒ == +           =           = 1 2 3 k k 2k 3k X 3 3 3 3 3 ∴
74. 1.6 CAYLEY HAMILTON THEOREM Every square matrix satisfies its own characteristic equation. Let A = [aij]n×n be a square matrix then, nnnn2n1n n22221 n11211 a...aa ................ a...aa a...aa A ×             =
75. Let the characteristic polynomial of A be φ (λ) Then, The characteristic equation is             11 12 1n 21 22 2n n1 n2 nn φ(λ) = A - λI a -λ a ... a a a -λ ... a = ... ... ... ... a a ... a -λ | A -λI|= 0
76. Note 1:- Premultiplying equation (1) by A-1 , we have ⇒ n n-1 n-2 0 1 2 n n n-1 n-2 0 1 2 n We are to prove that pλ +p λ +p λ +...+p = 0 p A +p A +p A +...+p I= 0 ...(1) I ⇒ n-1 n-2 n-3 -1 0 1 2 n-1 n -1 n-1 n-2 n-3 0 1 2 n-1 n 0 =p A +p A +p A +...+p +p A 1 A =- [p A +p A +p A +...+p I] p
77. This result gives the inverse of A in terms of (n-1) powers of A and is considered as a practical method for the computation of the inverse of the large matrices. Note 2:- If m is a positive integer such that m > n then any positive integral power Am of A is linearly expressible in terms of those of lower degree.
78. Verify Cayley – Hamilton theorem for the matrix A = . Hence compute A-1 . Solution:- The characteristic equation of A is           − −− − 211 121 112 tion)simplifica(on049λ6λλor 0 λ211 1λ21 11λ2 i.e.,0λIA 23 =−+− = −− −−− −− =− Example 1:-
79. To verify Cayley – Hamilton theorem, we have to show that A3 – 6A2 +9A – 4I = 0 … (1) Now,           − −− −− =           − −− −           − −− − =×=           − −− − =           − −− −           − −− − = 222121 212221 212222 211 121 112 655 565 556 655 565 556 211 121 112 211 121 112 23 2 AAA A
80. A3 -6A2 +9A – 4I = 0 = - 6 + 9 -4 = This verifies Cayley – Hamilton theorem. ∴           − −− −− 222121 212221 212222           − −− − 655 565 556           − −− − 211 121 112           100 010 001 0 000 000 000 =          
81. 81 Now, pre – multiplying both sides of (1) by A-1 , we have A2 – 6A +9I – 4 A-1 = 0 => 4 A-1 = A2 – 6 A +9I           − − =∴           − − =           +           − −− − −           − −− − =⇒ − − 311 131 113 4 1 311 131 113 100 010 001 9 211 121 112 6 655 565 556 4 1 1 A A
82. 82 Given find Adj A by using Cayley – Hamilton theorem. Solution:- The characteristic equation of the given matrix A is           − − − = 113 110 121 A tion)simplifica(on035λ3λλor 0 λ113 1λ10 1-2λ1 i.e.,0λIA 23 =++− = −− −− − =− Example 2:-
83. 83 By Cayley – Hamilton theorem, A should satisfy A3 – 3A2 + 5A + 3I = 0 Pre – multiplying by A-1 , we get A2 – 3A +5I +3A-1 = 0           − − − =           − −− −− =           − − −           − − − == +−−=⇒ 339 330 363 3A 146 223 452 113 110 121 113 110 121 A.AANow, (1)...5I)3A(A 3 1 A 2 21-
84. 84 AAAAdj. A AAdj. Athat,knowWe 173 143 110 3 1 500 050 005 339 330 363 146 223 452 3 1 AFrom(1), 1 1 1 − − − =∴ =           − − −− −=                     +           − − − −           − −− −− −=∴
85. 85           − − −− =           − − −−       −−=∴ −= − − − = 173 143 110 AAdj. 173 143 110 3 1 3)(AAdj. 3 113 110 121 ANow,
86. 86 1.7 DIAGONALISATION OF A MATRIX Diagonalisation of a matrix A is the process of reduction A to a diagonal form. If A is related to D by a similarity transformation, such that D = M-1 AM then A is reduced to the diagonal matrix D through modal matrix M. D is also called spectral matrix of A.
87. 87 1.8 REDUCTION OF A MATRIX TO DIAGONAL FORM If a square matrix A of order n has n linearly independent eigen vectors then a matrix B can be found such that B-1 AB is a diagonal matrix. Note:- The matrix B which diagonalises A is called the modal matrix of A and is obtained by grouping the eigen vectors of A into a square matrix.
88. 88 Similarity of matrices:- A square matrix B of order n is said to be a similar to a square matrix A of order n if B = M-1 AM for some non singular matrix M. This transformation of a matrix A by a non – singular matrix M to B is called a similarity transformation. Note:- If the matrix B is similar to matrix A, then B has the same eigen values as A.
89. 89 Reduce the matrix A = to diagonal form by similarity transformation. Hence find A3 . Solution:- Characteristic equation is => λ = 1, 2, 3 Hence eigen values of A are 1, 2, 3.           − − 300 120 211 0=           − − λ-300 1λ-20 21λ1- Example:-
90. 90 Corresponding to λ = 1, let X1 = be the eigen vector then           3 2 1 x x x           =∴ ===∴ = =− =+−⇒           =                     − − =− 0 0 1 kX x0x,kx 02x 0xx 02xx 0 0 0 x x x 200 110 210 0X)I(A 11 3211 3 32 32 3 2 1 1
91. 91 Corresponding to λ = 2, let X2 = be the eigen vector then,           3 2 1 x x x           =∴ ===∴ = =− =+−−⇒           =                     − − =− 0 1- 1 kX x-kx,kx 0x 0x 02xxx 0 0 0 x x x 100 100 211- 0X)(A 22 32221 3 3 321 3 2 1 2 0, I2
92. 92 Corresponding to λ = 3, let X3 = be the eigen vector then,           3 2 1 x x x           =∴ − ===∴ =−− =+−−⇒           =                     − − =− 2 2- 3 kX xk-x,kx 0x 02xxx 0 0 0 x x x 000 11-0 212- 0X)(A 33 13332 3 321 3 2 1 3 3 2 2 3 , 2 I3 k x
93. 93 Hence modal matrix is               −==∴           − = −=           −= − 2 1 00 11-0 2 1- 11 M MAdj. M 1-00 220 122- MAdj. 2M 200 21-0 311 M 1
94. 94 Now, since D = M-1 AM => A = MDM-1 A2 = (MDM-1 ) (MDM-1 ) = MD2 M-1 [since M-1 M = I]           =           −           − −               − − =− 300 020 001 200 21-0 311 300 120 211 2 1 00 11-0 2 1 11 AMM 1
95. 95 Similarly, A3 = MD3 M-1 = A3 =                         − −                     − 2700 19-80 327-1 2 1 00 11-0 2 1 11 2700 080 001 200 21-0 311
96. 96 1.9 ORTHOGONAL TRANSFORMATION OF A SYMMETRIC MATRIX TO DIAGONAL FORM A square matrix A with real elements is said to be orthogonal if AA’ = I = A’A. But AA-1 = I = A-1 A, it follows that A is orthogonal if A’ = A-1 . Diagonalisation by orthogonal transformation is possible only for a real symmetric matrix.
97. 97 If A is a real symmetric matrix then eigen vectors of A will be not only linearly independent but also pairwise orthogonal. If we normalise each eigen vector and use them to form the normalised modal matrix N then it can be proved that N is an orthogonal matrix.
98. 98 The similarity transformation M-1 AM = D takes the form N’AN = D since N-1 = N’ by a property of orthogonal matrix. Transforming A into D by means of the transformation N’AN = D is called as orthogonal reduction or othogonal transformation. Note:- To normalise eigen vector Xr, divide each element of Xr, by the square root of the sum of the squares of all the elements of Xr.
99. 99 Diagonalise the matrix A = by means of an orthogonal transformation. Solution:- Characteristic equation of A is 204 060 402 66,2,λ 0λ)16(6λ)λ)(2λ)(6(2 0 λ204 0λ60 40λ2 −=⇒ =−−−−−⇒ = − − − Example :-
100. 100 I                                       ⇒ ∴    ∴      1 1 2 3 1 1 2 3 1 3 2 1 3 1 1 2 3 1 1 1 x whenλ = -2,let X = x be theeigenvector x then (A + 2 )X = 0 4 0 4 x 0 0 8 0 x = 0 4 0 4 x 0 4x + 4x = 0 ...(1) 8x = 0 ...(2) 4x + 4x = 0 ...(3) x = k ,x = 0,x = -k 1 X = k 0 -1
101. 101 2 2I 0                                       ⇒ − ∴ 1 2 3 1 2 3 1 3 1 3 1 3 2 2 2 3 x whenλ = 6,let X = x betheeigenvector x then (A -6 )X = 0 -4 0 4 x 0 0 0 x = 0 4 0 -4 x 0 4x +4x = 0 4x - 4x = 0 x = x and x isarbitrary x must be so chosen that X and X are orthogonal among th .1 emselves and also each is orthogonal with X
102. 102                    ∴ ∴          ∴ 2 3 3 1 3 2 3 1α Let X = 0 and let X =β 1γ Since X is orthogonal to X α - γ = 0 ...(4) X is orthogonal to X α + γ = 0 ...(5) Solving (4)and(5), we getα = γ = 0 and β is arbitra ry. 0 Takingβ =1, X = 1 0 1 1 0 Modal matrix is M = 0 0 1 -1 1         0
103. 103                                                           The normalised modal matrix is 1 1 0 2 2 N = 0 0 1 1 1 - 0 2 2 1 1 0 - 1 1 02 2 2 0 4 2 2 1 1 D =N'AN = 0 0 6 0 0 0 1 2 2 4 0 2 1 1 - 00 1 0 2 2 -2 0 0 D = 0 6 0 which is the required diagonal matrix 0 0 6 .
104. 104 1.10 QUADRATIC FORMS DEFINITION:-DEFINITION:- A homogeneous polynomial of second degree in any number of variables is called a quadratic form. For example, ax2 + 2hxy +by2 ax2 + by2 + cz2 + 2hxy + 2gyz + 2fzx and ax2 + by2 + cz2 + dw2 +2hxy +2gyz + 2fzx + 2lxw + 2myw + 2nzw are quadratic forms in two, three and four variables.
105. 105 In n – variables x1,x2,…,xn, the general quadratic form is In the expansion, the co-efficient of xixj = (bij + bji). ∑∑= = ≠ n 1j n 1i jiijjiij bbwhere,xxb ).b(b 2 1 awherexxaxxb baandaawherebb2aSuppose jiijijji n 1j n 1i ijji n 1j n 1i ij iiiijiijijijij +== ==+= ∑∑∑∑ = == =
106. 106 Hence every quadratic form can be written as ( ) ( ) getweform,matrixin formsquadraticofexamplessaidabovethewritingNow .x,...,x,xXandaAwhere symmetric,alwaysisAmatrixthethatso AX,X'xxa n21ij ji n 1j n 1i ij == =∑∑= =             =++ y x bh ha y][xby2hxyax(i) 22
107. 107 [ ] [ ]                         = +++++++++                     =+++++ w z y x dnml ncgf mgbh lfha wzyx 2nzw2myw2lxwzx2f2gyz2hxydw2czbyax(iii) z y x cgf gbh fha zyx2fzx2gyz2hxyczbyax(ii) 222 222
108. 108 1.11 NATURE OF A QUADRATIC FORM A real quadratic form X’AX in n variables is said to be i. Positive definite if all the eigen values of A > 0. ii. Negative definite if all the eigen values of A < 0. iii. Positive semidefinite if all the eigen values of A 0 and at least one eigen value = 0. iv. Negative semidefinite if all the eigen values of A 0 and at least one eigen value = 0. v. Indefinite if some of the eigen values of A are + ve and others – ve. ≥ ≤
109. 109 Find the nature of the following quadratic forms i. x2 + 5y2 + z2 + 2xy + 2yz + 6zx ii. 3x2 + 5y2 + 3z2 – 2yz + 2zx – 2xy Solution:- i. The matrix of the quadratic form is           = 113 151 311 A Example :-
110. 110 The eigen values of A are -2, 3, 6. Two of these eigen values being positive and one being negative, the given quadratric form is indefinite. ii. The matrix of the quadratic form is The eigen values of A are 2, 3, 6. All these eigen values being positive, the given quadratic form is positive definite.           − −− − = 311 151 113 A
111. 111 1.12 REDUCTION OF QUADRATIC FORM TO CANONICAL FORM A homogeneous expression of the second degree in any number of variables is called a quadratic form. For instance, if which is a quadratic form. (i)....2hxy2gzx2fyzczbyaxAXX'then ],zyx[X'and z y x X, cfg fbh gha A 222 +++++= =           =           =
112. 112 Let λ1, λ2, λ3 be the eigen values of the matrix A and be its corresponding eigen vectors in the normalized form (i.e., each element is divided by square root of sum of the squares of all the three elements in the eigen vector).           =           =           = 3 3 3 3 2 2 2 2 1 1 1 1 z y x X, z y x X, z y x X
113. 113 Then B-1 AB = D, a diagonal matrix. Hence the quadratic form (i) is reduced to a sum of squares (i.e., canonical form). λ1x2 + λ2y2 + λ3z2 And B is the matrix of transformation which is an orthogonal matrix. Note:- 1. Here some of λi may be positive or negative or zero 2. If ρ(A) = r, then the quadratic form X’AX will contain only r terms.
114. 114 1.13 INDEX AND SIGNATURE OF THE QUADRATIC FORM The number p of positive terms in the canonical form is called the index of the quadratic form. (The number of positive terms) – ( the number of negative terms) i.e., p – (r – p) = 2p – r is called signature of the quadratic form, where ρ(A) = r.
115. 115 1.14 LINEAR TRANSFORMATION OF A QUADRATIC FORM. Let X’AX be a quadratic form in n- variables and let X = PY ….. (1) where P is a non – singular matrix, be the non – singular transformation. From (1), X’ = (PY)’ = Y’P’ and hence X’AX = Y’P’APY = Y’(P’AP)Y = Y’BY …. (2) where B = P’AP.
116. 116 Therefore, Y’BY is also a quadratic form in n- variables. Hence it is a linear transformation of the quadratic form X’AX under the linear transformation X = PY and B = P’AP. Note. (i) Here B = (P’AP)’ = P’AP = B (ii) ρ(B) = ρ(A) Therefore, A and B are congruent matrices.
117. 117 Reduce 3x2 + 3z2 + 4xy + 8xz + 8yz into canonical form. Or Diagonalise the quadratic form 3x2 + 3z2 + 4xy + 8xz + 8yz by linear transformations and write the linear transformation. Or Reduce the quadratic form 3x2 + 3z2 + 4xy + 8xz + 8yz into the sum of squares. Example:-
118. 118 Solution:- The given quadratic form can be written as X’AX where X = [x, y, z]’ and the symmetric matrix A = Let us reduce A into diagonal matrix. We know tat A = I3AI3.           344 402 423                               =           100 010 001 344 402 423 100 010 001 344 402 423
119. 119 − − −                  − = −                − −     21 31OperatingR ( 2 / 3),R ( 4 / 3) (for A onL.H.S.andpre factor on R.H.S.),weget 3 2 4 1 0 0 1 0 0 4 4 2 0 1 0 A 0 1 0 3 3 3 0 0 1 4 7 4 0 0 1 3 3 3               −−                 − −=                 − − − −− 100 010 3 4 3 2 1 A 10 3 4 01 3 2 001 3 7 3 4 0 3 4 3 4 0 423 getweR.H.S),onfactorpostandL.H.S.onA(for 4/3)(C2/3),(COperating 3121
120. 120               −−           − −=           − − 100 010 3 4 3 2 1 A 112 01 3 2 001 100 3 4 3 4 0 003 getwe(1),ROperating 32 APP'1, 3 4 3,Diagor 100 110 2 3 2 1 A 112 01 3 2 001 100 0 3 4 0 003 getwe(1),COperating 32 =      −−               −−           − −=           − −
121. 121 The canonical form of the given quadratic form is Here ρ(A) = 3, index = 1, signature = 1 – (2) = -1. Note:- In this problem the non-singular transformation which reduces the given quadratic form into the canonical form is X = PY. i.e.,                     − − =           3 2 1 112 01 3 2 001 y y y z y x [ ] 2 3 2 2 2 1 3 2 1 321 yy 3 4 3y y y y 100 0 3 4 0 003 yyyAP)Y(P'Y' −−=                     − − =
122. 122 1.15 REDUCTION OF QUADRATIC FORM TO CANANONICAL FORM BY ORTHOGONAL TRANSFORMATION Let X’AX be a given quadratic form. The modal matrix B of A is the matrix whose columns are characteristic vectors of A. If B represents the orthogonal matrix of A,
123. 123 then X = BY will reduce X’AX to Y’ diag(λ1, λ2,…, λn) Y, where λ1, λ2,…, λn are characteristic values of A. Note. This method works successfully if the characteristic vectors of A are linearly independent which are pairwise orthogonal.
124. 124 Reduce 8x2 + 7y2 + 3z2 – 12xy + 4xz – 8yz into canonical form by orthogonal reduction. Solution:- The matrix of the quadratic form is           − −− − = 342 476 268 A Example 1:-
125. 125 The characteristic roots of A are given by 0|| =− IλA 153,0,λ 015)3)(λλ(λor 0 λ342 4λ76 26λ8 =∴ =−− = −− −−− −− .,.ei
126. 126 Characteristic vector for λ = 0 is given by [A – (0)I] X1 = 0 ' 11 321 321 321 321 2)2,(1,kXvectoreigenthegiving 2 x 2 x 1 x getwe,twofirstSolving 03x4x2x 04x7x6x 02x6x8xi.e., = == =+− =−+− =+−
127. 127 When λ = 3, the corresponding characteristic vector is given by [A – 3I] X2 = 0 i.e., Solving any two equations, we get X2 = k2 (2, 1, -2)’. Similarly characteristic vector corresponding to λ = 15 is X3 = k3 (2, -2, 1)’. 04x2x 04x4x6x 02x6x5x 21 321 321 =− =−+− =+−
128. 128 Now, X1, X2, X3 are pairwise orthogonal i.e., X1 . X2 = X2 . X3 = X3 . X1 = 0. The normalised modal matrix is                 − −= 3 1 3 2 3 2 3 2 3 2 3 1 3 2 3 1 3 2 B ∴
129. 129 Now B is orthogonal matrix and 1B =           =                 − −                 − − === −− 1500 000 003 3 1 3 2 3 2 3 2 3 2 3 1 3 2 3 1 3 2 A 3 1 3 2 3 2 3 2 3 2 3 1 3 2 3 1 3 2 ie., 15}0,{3,diagDABBandBBi.e., 1T1
130. 130 which is the required canonical form. Note. Here the orthogonal transformation is X =BY, rank of the quadratic form = 2; index = 2, signature = 2. It is positive definite. [ ] 2 3 2 2 2 1 3 2 1 321 1 15y0.y3y y y y 1500 000 003 yyy DYY'AB)Y(BY'AXX' ++=                     = == −
131. 131 TEST YOUR KNOWLEDGE 1. If then find the eigen value of 2. Write the matrix of the Quadratic form 3. Obtain the characteristic equation of the matrix whose eigen values are 1,-2 and 3. 4. A = , then find the eigen values of 3A3 +5A2 -6A+2I. 1 2 3 0 3 2 0 0 2 A −   ÷ =  ÷  ÷−  ( ) 2 3A I− 1 2 3 1 2 32 2 2x x x x x x+ − .           − 200 320 061
132. 132 5. Write the quadratic form corresponding to the following symmetric matrix 6. Find the sum of the eigen values of the inverse of. 7. Obtain the latent roots of A4 where A = 8. If A is idempotent matrix then A2 =A. What will be the eigen values of A.           − − 342 411 210 1 0 0 2 3 0 0 5 2    ÷ − ÷  ÷   5 4 1 2    ÷  
133. 133 9. If are the eigen values of the matrix A, whose characteristic equation is Obtain using the property. 10. (i) Using Cayley-Hamilton Theorem find the inverse of the matrix (ii) Find the Characteristic roots and Characteristic vectors of the matrix 1 2 2 3 3 1λ λ λ λ λ λ+ + 3 2 21 45 0λ λ λ+ − − = 1 2 3, andλ λ λ 1 0 3 2 1 1 1 1 1 A    ÷ = − ÷  ÷−  1 1 3 1 5 1 3 1 1 A    ÷ = ÷  ÷  
134. 134 11. Reduce the quadratic form to the canonical form by orthogonal transformation. Also specify the matrix of transformation. Obtain its index, signature and nature of the quadratic form. 12. (i) Find the eigen value and eigen vector of the matrix (ii) Using Cayley Hamiltonian find the inverse of 2 2 2 5 2 2 6x y z xy yz xz+ + + + +           −− −− −− 6410 527 7411           − − 111 112 301
135. 135 13. Discuss the nature, index and signature of the quadratic form 14. Diagonalise the matrix by orthogonal reduction and provide the normalized modal matrix. 15. Reduce the quadratic form 2x1x2+2x1x3-2x2x3 to the canonical form by an Orthogonal transformation. 2 2 2 10 2 5 6 10 4x y z yz zx xy+ + + − − 8 6 2 6 7 4 2 4 3 A −   ÷ = − − ÷  ÷− 
136. 136 16. (i) Find the eigen value and eigen vector of the matrix (ii) For A = , compute the value of , Using Cayley- Hamilton theorem. 17. Reduce the quadratic form 3x1 2 + 5x2 2 + 3x3 2 -2x2x3+2x3x1-2x1x2 to a Canonical form by orthogonal reduction. 2 1 0 0 2 1 0 0 2    ÷  ÷  ÷             − − 111 112 301 6 5 4 3 2 5 8 2 9 31 36A A A A A A I− + − − + −
137. THANK YOU
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