Eigen value , eigen vectors, caley hamilton theorem
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Eigen value , eigen vectors, caley hamilton theorem
- 1. Topic: DIAGONALIZATION, EIGEN VALUE, EIGEN VECTOR, ORTHOGONAL, CAYLEY HAMILTON THEORM By rajesh goswami
- 2. DIAGONALIZATION EIGEN VALUE EIGEN VECTOR ORTHOGONAL CAYLEY HAMILTON THEOREM kshumENGG2013 2
- 4. A square matrix M is called diagonalizable if we can find an invertible matrix, say P, such that the product P–1 M P is a diagonal matrix. A diagonalizable matrix can be raised to a high power easily. › Suppose that P–1 M P = D, D diagonal. › M = PD P–1 . › Mn = (PD P–1 ) (PD P–1 ) (PD P–1 ) … (PD P–1 ) = PDn P–1 . 4
- 5. Let A is diagonalizable because we can find a matrix such that 5
- 6. By definition a matrix M is diagonalizable if P–1 M P = D for some invertible matrix P, and diagonal matrix D. or equivalently, 6
- 8. Given a square matrix A, a non-zero vector v is called an eigenvector of A, if we an find a real number λ (which may be zero), such that This number λ is called an eigen value of A, corresponding to the eigenvector v. 8 Matrix-vector product Scalar product of a vector
- 9. If v is an eigenvector of A with eigen value λ, then any non-zero scalar multiple of v also satisfies the definition of eigenvector. 9 k ≠ 0
- 10. First eigenvalue = 2, with eigenvector where k is any nonzero real number. Second eigenvalue = -5, with eigenvector where k is any nonzero real number. 10
- 12. Matrix addition/subtraction › Matrices must be of same size. Matrix multiplication Condition: n = q m x n q x p m x p
- 15. Example:
- 16. 2 x 2 3 x 3 n x n
- 17. diagonal matrix:
- 18. The inverse A-1 of a matrix A has the property: AA-1 =A-1 A=I A-1 exists only if Terminology › Singular matrix: A-1 does not exist › Ill-conditioned matrix: A is close to being singular
- 19. Properties of the inverse:
- 20. The pseudo-inverse A+ of a matrix A (could be non-square, e.g., m x n) is given by: It can be shown that:
- 21. Equal to the dimension of the largest square sub-matrix of A that has a non- zero determinant. Example: has rank 3
- 22. Alternative definition: the maximum number of linearly independent columns (or rows) of A. Therefore, rank is not 4 !Example:
- 24. • A is orthogonal if: • Notation: Example:
- 25. • A is orthonormal if: • Note that if A is orthonormal, it easy to find its inverse: Property:
- 26. Diagonalize 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 −=⇒ =−−−−−⇒ = − − − 26 Example :-
- 27. 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 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 27
- 28. 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 28
- 29. ∴ ∴ ∴ 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 29
- 30. 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 . 30
- 33. 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
- 34. ⇒ 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
- 39. 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 39
Notas do Editor
- \mathbf{A} = \begin{bmatrix} 0.9 & 0.4\\ 0.1 & 0.6 \end{bmatrix}
- \mathbf{P}=\begin{bmatrix} x_1 & x_2 & x_3\\ y_1 & y_2 & y_3\\ z_1 & z_2 & z_3 \end{bmatrix}