A. B is a square matrix of order n. It is proved that AB and Ba have the same eigenvalues

A. B is a square matrix of order n. It is proved that AB and Ba have the same eigenvalues

Since a is not necessarily reversible, the solution of AB ~ A ^ {- 1} (AB) a = Ba is defective. Please refer to the figure below for the detailed solution. Note that it is necessary to discuss the classification of whether the eigenvalue is zero

It is proved that if there is a positive integer k such that a ^ k = 0, then the eigenvalue of a can only be 0

Two knowledge points are needed
1. The eigenvalue of zero matrix is only zero
2. If λ is the eigenvalue of a and G (x) is a polynomial of X, then G (λ) is the eigenvalue of G (a)
The purpose of this question is to prove that:
Let λ be the eigenvalue of a, then λ ^ k is the eigenvalue of a ^ K
Because a ^ k = 0, and the eigenvalue of zero matrix is only zero
So λ ^ k = 0
So λ = 0
That is, the eigenvalue of a can only be 0#

The diagonal matrix obtained by diagonalizing the symmetric matrix is composed of the eigenvalues of the original symmetric matrix
Well, teacher, I would like to ask, are the arrangement of these eigenvalues on the diagonal regular? If regular, then there is no need to find the orthogonal matrix, or is there no regular? Must PAP-1 find the diagonal matrix?

It can be arranged arbitrarily, but it must correspond to the column of P