Cayley-Hamilton Theorem, Eigenvalues, Eigenvectors and Eigenspace.
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The document discusses the Cayley-Hamilton theorem, which states that every square matrix satisfies its own characteristic equation. It provides a detailed example of finding the eigenvalues, eigenvectors, and eigenspaces for a specific matrix using this theorem. The findings include characteristic equations, traces, and determinants associated with the given matrix.
Cayley-Hamilton Theorem, Eigenvalues, Eigenvectors and Eigenspace.
- 1. Cayley–Hamilton theorem - Eigenvalues, Eigenvectors and Eigenspaces Isaac Amornortey Yowetu NIMS-GHANA March 17, 2021
- 2. Cayley–Hamilton theorem Finding Eigenvalues EigenVectors and EigenSpaces Content 1 Cayley–Hamilton theorem 2 Finding Eigenvalues 3 EigenVectors and EigenSpaces
- 3. Cayley–Hamilton theorem Finding Eigenvalues EigenVectors and EigenSpaces Cayley–Hamilton theorem Every square matrix satisfies its own characteristic equation. Thus: p(λ) = det(λI − A) Substituting the matrix A for λ in the polynomial, p(λ) = det(λI − A) results in a zero matrix. Thus: p(A) = 0
- 4. Cayley–Hamilton theorem Finding Eigenvalues EigenVectors and EigenSpaces Example : Using Non-singular Matrix Consider the given matrix: A = 2 1 −1 1 2 −1 − −1 2 Find the characteristic equation of the square matrix and hence find the eigenvalues, eigenvectors and eigenspaces.
- 5. Cayley–Hamilton theorem Finding Eigenvalues EigenVectors and EigenSpaces Solution p(λ) = det(λI − A) (1) =
- 10. λ 1 0 0 0 1 0 0 0 1 − 2 1 −1 1 2 −1 −1 −1 2
- 15. (2) =
- 20. λ − 2 −1 1 −1 λ − 2 1 1 1 λ − 2
- 25. (3) = (λ − 2)
- 29. λ − 2 1 1 λ − 2
- 33. − (−1)
- 37. −1 1 1 λ − 2
- 41. (4) +1
- 45. −1 λ − 2 1 1
- 50. Cayley–Hamilton theorem Finding Eigenvalues EigenVectors and EigenSpaces Solution Continue p(λ) = (λ − 2)[(λ − 2)2 − 1] + [(−1)(λ − 2) − 1] (5) +[(−1)(1) − 1(λ − 2)] = (λ − 2)(λ2 − 4λ + 3) − 2(λ − 1) (6) = (λ − 2)(λ − 3)(λ − 1) − 2(λ − 1) (7) = (λ − 1)[(λ − 3)(λ − 2) − 2] (8) = (λ − 1)[(λ2 − 5λ + 4] (9) = (λ − 1)(λ − 1)(λ − 4) (10) p(λ) = λ3 − 6λ2 + 9λ − 4 (11)
- 51. Cayley–Hamilton theorem Finding Eigenvalues EigenVectors and EigenSpaces Considering our matrix: A = 2 1 −1 1 2 −1 −1 −1 2 Finding the Eigenvalues λ1 = 1, λ2 = 1 and λ3 = 4 Finding the trace of a matrix A tr(A) = 3 X i=1 λi = 3 X i=1 aii = 6
- 52. Cayley–Hamilton theorem Finding Eigenvalues EigenVectors and EigenSpaces Determinant of the of a matrix A det(A) = 3 Y i=1 λi = 4
- 53. Cayley–Hamilton theorem Finding Eigenvalues EigenVectors and EigenSpaces Eigenvectors Considering the eigenvalues, λ1 = 1 and λ2 = 1 and λ3 = 4 and the matrix. λ − 2 −1 1 −1 λ − 2 1 1 1 λ − 2 When λ = 1: −1 −1 1 0 −1 −1 1 0 1 1 −1 0 Using row reduction method, we have −1 −1 1 0 We consider expressing the argument matrix in terms of equation as;
- 54. Cayley–Hamilton theorem Finding Eigenvalues EigenVectors and EigenSpaces −x − y + z = 0 (12) x = −y + z (13) The eigenspace Λ1 for λ = 1 becomes Λ1 = x y z = −s + t s t (14) = s −1 1 0 + t 1 0 1 (15) Let z = t and y = s and where t 6= 0 and s 6= 0
- 55. Cayley–Hamilton theorem Finding Eigenvalues EigenVectors and EigenSpaces Eigenvectors . λ − 2 −1 1 −1 λ − 2 1 1 1 λ − 2 When λ = 4: 2 −1 1 0 −1 2 1 0 1 1 2 0 Using row reduction method, we have 2 −1 1 0 3 3 0 −3 −3 0
- 56. Cayley–Hamilton theorem Finding Eigenvalues EigenVectors and EigenSpaces 2 −1 1 0 1 1 0 We consider expressing the argument matrix in terms of equation as; 2x − y + z = 0 (16) y + z = 0 (17) y = −z (18) y = −t (19) Let z = t x = −t (20)
- 57. Cayley–Hamilton theorem Finding Eigenvalues EigenVectors and EigenSpaces The eigenspace Λ3 for λ = 4 becomes Λ3 = x y z = −t −t t = t −1 −1 1 Where t 6= 0
- 58. Cayley–Hamilton theorem Finding Eigenvalues EigenVectors and EigenSpaces Conclusion Eigenvectors P = −1 1 −1 1 0 −1 0 1 1 Eigenvalues D = 1 0 0 0 1 0 0 0 4 A = PDP−1
- 59. Cayley–Hamilton theorem Finding Eigenvalues EigenVectors and EigenSpaces End THANK YOU