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Transcript of Maths>>Eigenvalues and eigenvectors
 Eigenvalues and Eigenvectors
 Eigenvalues and Eigenvectors If A is an n x n matrix and is a scalar for which Ax = x has a nontrivial solution x , then is an eigenvalue of A and x is a corresponding eigenvector of A. Ax=x=Ix (AI)x=0 The matrix (AI ) is called the characteristic matrix of a where I is the Unit matrix. The equation det (AI )= 0 is called characteristic equationof A and the roots of this equation are called the eigenvalues of the matrix A. The set of all eigenvectors is called the eigenspace of A corresponding to . The set of all eigenvalues of a is called spectrum of A.
 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., is called the characteristic equation of A. IA 0 a...aa ............ a...aa a...aa A nnn2n1 2n2221 1n1211 I
 On expanding the determinant, we get where ks 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. X = is called an eigen vector or latent vector 0k...kk1)( n 2n 2 1n 1 nn ij 4 2 1 x ... x x
 5 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 A1 . 4. If is an eigen value of an orthogonal matrix, then 1/ is also its eigen value.
 6 PROPERTY 1: If 1, 2,, n are the eigen values of A, then i. k 1, k 2,,k n are the eigen values of the matrix kA, where k is a non zero scalar. ii. are the eigen values of the inverse matrix A1. iii. are the eigen values of Ap, where p is any positive integer. n21 1 ,..., 1 , 1 p n p 2 p 1 ...,,,
 Algebraic & Geometric Multiplicity If the eigenvalue of the equation det(AI)=0 is repeated n times then n is called the algebraic multiplicity of . The number of linearly independent eigenvectors is the difference between the number of unknowns and the rank of the corresponding matrix A I and is known as geometric multiplicity of eigenvalue .
 CayleyHamilton 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
 Let the characteristic polynomial of A be () Then, The characteristic equation is 11 12 1n 21 22 2n n1 n2 nn () = A  I a  a ... a a a  ... a = ... ... ... ... a a ... a   A  I=0
 Note 1: Premultiplying equation (1) by A1 , we have n n1 n2 0 1 2 n n n1 n2 0 1 2 n We are to prove that p +p +p +...+p = 0 p A +p A +p A +...+p I= 0 ...(1) In1 n2 n3 1 0 1 2 n1 n 1 n1 n2 n3 0 1 2 n1 n 0 =p A +p A +p A +...+p +p A 1 A = [p A +p A +p A +...+p I] p
 This result gives the inverse of A in terms of (n1) 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.
 Verify Cayley Hamilton theorem for the matrix A = . Hence compute A1 . Solution: The characteristic equation of A is 211 121 112 tion)simplifica(on0496or 0 211 121 112 i.e.,0IA 23 Example 1:
 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
 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
 15 Now, pre multiplying both sides of (1) by A1 , we have A2 6A +9I 4 A1 = 0 => 4 A1 = 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
 16 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(on0353or 0 113 110 121 i.e.,0IA 23 Example 2:
 17 By Cayley Hamilton theorem, A should satisfy A3 3A2 + 5A + 3I = 0 Pre multiplying by A1 , we get A2 3A +5I +3A1 = 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
 18 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
 19 173 143 110 AAdj. 173 143 110 3 1 3)(AAdj. 3 113 110 121 ANow,
 Similarity of Matrix If A & B are two square matrices of order n then B is said to be similar to A, if there exists a nonsingular matrix P such that, B= P1AP 1. Similarity matrices is an equivalence relation. 2. Similarity matrices have the same determinant. 3. Similar matrices have the same characteristic polynomial and hence the same eigenvalues. If x is an eigenvector corresponding to the eigenvalue , then P1x is an eigenvector of B corresponding to the eigenvalue where B= P1AP.
 Diagonalization A matrix A is said to be diagonalizable if it is similar to diagonal matrix. A matrix A is diagonalizable if there exists an invertible matrix P such that P1AP=D where D is a diagonal matrix, also known as spectral matrix. The matrix P is then said to diagonalize A of transform A to diagonal form and is known as modal matrix.
 22 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 B1AB 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.
 23 Reduce the matrix A = to diagonal form by similarity transformation. Hence find A3. Solution: Characteristic equation is => = 1, 2, 3 Hence eigenvalues of A are 1, 2, 3. 300 120 211 0 300 120 211 Example:
 24 Corresponding to = 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
 25 Corresponding to = 2, let X2 = be the eigen vector then, 3 2 1 x x x 0 1 1 kX xkx,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
 26 Corresponding to = 3, let X3 = be the eigen vector then, 3 2 1 x x x 2 2 3 kX xkx,kx 0x 02xxx 0 0 0 x x x 000 110 212 0X)(A 33 13332 3 321 3 2 1 3 3 2 2 3 , 2 I3 k x
 27 Hence modal matrix is 2 1 00 110 2 1 11 M MAdj. M 100 220 122 MAdj. 2M 200 210 311 M 1
 28 Now, since D = M1AM => A = MDM1 A2 = (MDM1) (MDM1) = MD2M1 [since M1M = I] 300 020 001 200 210 311 300 120 211 2 1 00 110 2 1 11 AMM 1
 29 Similarly, A3 = MD3M1 = A3 = 2700 1980 3271 2 1 00 110 2 1 11 2700 080 001 200 210 311
 Orthogonally Similar Matrices If A & B are two square matrices of order n then B is said to be orthogonally similar to A, if there exists orthogonal matrix P such that B= P1AP Since P is orthogonal, P1=PT B= P1AP=PTAP 1. A real symmetric of order n has n mutually orthogonal real eigenvectors. 2. Any two eigenvectors corresponding to two distinct eigenvalues of a real symmetric matrix are orthogonal.
 31 Diagonalise the matrix A = by means of an orthogonal transformation. Solution: Characteristic equation of A is 204 060 402 66,2, 0)16(6))(2)(6(2 0 204 060 402 Example :
 32 I 1 1 2 3 1 1 2 3 1 3 2 1 3 1 1 2 3 1 1 1 x when = 2,let X = x betheeigenvector 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
 33 2 2I 0 1 2 3 1 2 3 1 3 1 3 1 3 2 2 2 3 x when = 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
 34 2 3 3 1 3 2 3 1 Let X = 0 and let X = 1 Since X is orthogonal to X  = 0 ...(4) X is orthogonal to X + = 0 ...(5) Solving (4)and(5), we get = = 0 and is arbitrary. 0 Taking =1, X = 1 0 1 1 0 Modal matrix is M = 0 0 1 1 1 0
 35 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 .
 36 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. Quadratic Forms
 37 In n variables x1,x2,,xn, the general quadratic form is In the expansion, the coefficient 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
 38 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
 39 w z y x dnml ncgf mgbh lfha wzyx 2nzw2myw2lxwzx2f2gyz2hxydw2czbyax(iii) z y x cgf gbh fha zyx2fzx2gyz2hxyczbyax(ii) 222 222
 40 Two Theorems On Quadratic Form Theorem(1): A quadratic form can always be expressed with respect to a given coordinate system as where A is a unique symmetric matrix. Theorem2: Two symmetric matrices A and B represent the same quadratic form if and only if B=PTAP where P is a nonsingular matrix. AxxY T
 Nature of Quadratic Form A real quadratic form XAX 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.
 42 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 :
 43 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
 Linear Transformation of a Quadratic Form 44 Let XAX 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) = YP and hence XAX = YPAPY = Y(PAP)Y = YBY . (2) where B = PAP.
 45 Therefore, YBY is also a quadratic form in n variables. Hence it is a linear transformation of the quadratic form XAX under the linear transformation X = PY and B = PAP. Note. (i) Here B = (PAP) = PAP = B (ii) (B) = (A) Therefore, A and B are congruent matrices.
 46 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:
 47 Solution: The given quadratic form can be written as XAX 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
 48 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
 49 100 010 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
 50 The canonical form of the given quadratic form is Here (A) = 3, index = 1, signature = 1 (2) = 1. Note: In this problem the nonsingular 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'
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