# Maths-->>Eigenvalues and eigenvectors

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21-Aug-2014Category

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It is the topic of maths-2 in the second semester of engineering !! I hope it is useful and satisfactory !!

### 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 (A-I)x=0 The matrix (A-I ) is called the characteristic matrix of a where I is the Unit matrix. The equation det (A-I )= 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 A-1 . 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 A-1. 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(A-I)=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 .
- 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
- 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 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 p +p +p +...+p = 0 p A +p A +p A +...+p I= 0 ...(1) In-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
- 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.
- Verify Cayley Hamilton theorem for the matrix A = . Hence compute A-1 . 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 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
- 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 1-21 i.e.,0IA 23 Example 2:-
- 17 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-
- 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 non-singular matrix P such that, B= P-1AP 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 P-1x is an eigenvector of B corresponding to the eigenvalue where B= P-1AP.
- 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 P-1AP=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 B-1AB 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 1-20 211- 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 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
- 26 Corresponding to = 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
- 27 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
- 28 Now, since D = M-1AM => A = MDM-1 A2 = (MDM-1) (MDM-1) = MD2M-1 [since M-1M = I] 300 020 001 200 21-0 311 300 120 211 2 1 00 11-0 2 1 11 AMM 1
- 29 Similarly, A3 = MD3M-1 = A3 = 2700 19-80 327-1 2 1 00 11-0 2 1 11 2700 080 001 200 21-0 311
- 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= P-1AP Since P is orthogonal, P-1=PT B= P-1AP=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

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