![Eigen-value AND EigenEigen-value AND Eigen
vectorsvectors
Prepared By: Riddhi Patel[160630107092]
Vrunda Purohit [160630107088]
Dhara Soni [160630107107]
Guided By: Prof.Sejal Patel](https://image.slidesharecdn.com/16063010709288107-171227180834/85/eigen-values-and-eigen-vectors-1-320.jpg)
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Eigen values and eigen vectors
- 1. Eigen-value AND EigenEigen-value AND Eigen vectorsvectors Prepared By: Riddhi Patel[160630107092] Vrunda Purohit [160630107088] Dhara Soni [160630107107] Guided By: Prof.Sejal Patel
- 2. INTRODUCTIONINTRODUCTION • If A is an n x n matrix and λ is a scalar for which Ax = λx has a nontrivial solution x ∈ ⁿ, thenℜ λ is an eigenvalue of A and x is a corresponding eigenvector of A. ◦ Ax=λx=λIx ◦ (A-λI)x=0 • The matrix (A-λI ) is called the characteristic matrix of a where I is the Unit matrix. ◦ • The equation det (A-λI )= 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 λ. The set of all eigenvalues of a is called spectrum of A.
- 3. Determining EigenvaluesDetermining Eigenvalues Vector equation ◦ Ax = λx (A-λΙ)x = 0 ◦ A-λΙ is called the characteristic matrix Non-trivial solutions exist if and only if: ◦ This is called the characteristic equation Characteristic polynomial ◦ nth-order polynomial in λ ◦ Roots are the eigenvalues {λ1, λ2, …, λn} 0)det( 21 22221 11211 = − − − =− λ λ λ λ nnnn n n aaa aaa aaa IA
- 4. Eigenvalue ExampleEigenvalue Example Characteristic matrix Characteristic equation Eigenvalues: λ1 = -5, λ2 = 2 −− − = − − =− λ λ λλ 43 21 10 01 43 21 IA 0103)3)(2()4)(1( 2 =−+=−−−−=− λλλλλIA
- 5. Eigenvalue PropertiesEigenvalue Properties Eigenvalues of A and AT are equal Singular matrix has at least one zero eigenvalue Eigenvalues of A-1 : 1/λ1, 1/λ2, …, 1/λn Eigenvalues of diagonal and triangular matrices are equal to the diagonal elements Trace Determinant ∑= = n j jTr 1 )( λA ∏= = n j j 1 λA
- 6. Determining EigenvectorsDetermining Eigenvectors First determine eigenvalues: {λ1, λ2, …, λn} Then determine eigenvector corresponding to each eigenvalue: Eigenvectors determined up to scalar multiple Distinct eigenvalues ◦ Produce linearly independent eigenvectors Repeated eigenvalues ◦ Produce linearly dependent eigenvectors ◦ Procedure to determine eigenvectors more complex (see text) ◦ Will demonstrate in Matlab 0)(0)( =−⇒=− kk xIAxIA λλ
- 7. Eigenvector ExampleEigenvector Example Eigenvalues Determine eigenvectors: Ax = λx Eigenvector for λ1 = -5 Eigenvector for λ1 = 2 − = − =⇒ =+ =+ 3 1 or 9487.0 3162.0 03 026 11 21 21 xx xx xx = =⇒ =− =+− 1 2 or 4472.0 8944.0 063 02 22 21 21 xx xx xx 2 5 43 21 2 1 = −= − = λ λ A 0)4(3 02)1( 43 2 21 21 221 121 =+− =+− ⇒ =− =+ xx xx xxx xxx λ λ λ λ
- 8. Algebraic & GeometricAlgebraic & Geometric MultiplicityMultiplicity If the eigenvalue λ of the equation det(A-λI)=0 is repeated n times then n is called the algebraic multiplicity of λ. The number of linearly independent eigenvectors is the difference between the number of unknowns and the rank of the corresponding matrix A- λI and is known as geometric multiplicity of eigenvalue λ.
- 9. Cayley-Hamilton TheoremCayley-Hamilton Theorem Let the characteristic polynomial of A be φ (λ)then, The characteristic equation 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
- 10. 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 =−+− = −− −−− −− =−
- 11. − −− −− = − −− − − −− − =×= − −− − = − −− − − −− − = 222121 212221 212222 211 121 112 655 565 556 655 565 556 211 121 112 211 121 112 23 2 AAA A To verify Cayley – Hamilton theorem, we have to show that A3 – 6A2 +9A – 4I = 0 … (1)
- 12. ∴ A3 -6A2 +9A – 4I = 0 − −− −− 222121 212221 212222 = − −− − 655 565 556 - 6 + 9 − −− − 211 121 112 -4 100 010 001 = 0 000 000 000 = This verifies Cayley – Hamilton theorem
- 13. − − =∴ − − = + − −− − − − −− − =⇒ − − 311 131 113 4 1 311 131 113 100 010 001 9 211 121 112 6 655 565 556 4 1 1 A A 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