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Let a be a 3-order square matrix with eigenvalues of 1,2, - 3, find the eigenvalues of a ^ 2-3a + A ^ - 1 + 2E, and | a ^ 2-3a + A ^ - 1 + 2e |, hope to give the process
Let a be a third-order square matrix with eigenvalues of 1,2, - 3, find a Λ 2-3a + a Λ (- 1) + 2e and|a Λ 2-3a + a Λ (- 1) + 2E|
The first question is to find eigenvalues
The eigenvalues of a are 1,2, - 3
Then, the eigenvalues of a Λ 2-3a + a Λ (- 1) + 2E are
1－3＋1＋2＝1
4－6＋1/2＋2＝1/2
9＋9－1/3＋2＝59/3
The value of determinant = the product of eigenvalues
|A∧2－3A＋A∧(-1)＋2E|＝1×(1/2)×(59/3)＝59/6
Let λ be the eigenvalue of square matrix A of order n, and prove that the eigenvalue of α + 2E is λ + 2
If λ is the eigenvalue of square matrix A of order n, then:
Ax = λ x, where x is the eigenvector corresponding to λ
Investigation of (a + 2e) x
(A+2E)x = Ax +2Ex
=λx + 2x
=(λ+2)x
So the eigenvalue of a + 2E is λ + 2, and we can see that the corresponding eigenvector remains unchanged
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