similar matrices have the same eigenvalues

Question

how do I show similar matrices have the same eigenvalues?

I really have no idea, any detailed explanation would be thoroughly helpful.

thank you

Answer 1

$A\in\mathbb R^{n\times n}$, $x\in\mathbb R^n$, $\lambda\in\mathbb R$, $$ Ax = \lambda x. $$ $B\in\mathbb R^{n\times n}$ is invertible; $y=B^{-1}x$. $$ B^{-1}ABy = B^{-1}Ax=B^{-1}(\lambda x) = \lambda y, $$ so $\lambda$ is an eigenvalue of $B^{-1}AB$.

Answer 2

Hint
The defintion says $A\sim B$ iff there is a $U\in \mathrm{GL}(n)$ such that $$B = U^{-1} A U$$ Now look at an eigenpair of $A$, $(\lambda, v)$ and notice something...

Answer 3

The eigenvalues of the matrix $A$ are the zeros of the characteristic polynomial $\det(A - t I)$. What about the polynomial $$ \det[S(A - tI)S^{-1}]? $$

Answer 4

This is immediate, because eigenvalues are properties of linear operators, not of the matrices that represent them. Similar matrices represent the same linear operator with respect to different bases (this is the motivation for the notion of similarity), and so naturally such matrices must have the same eigenvalues.

By contrast, the characteristic polynomial of a linear operator is not so easy to define without using a matrix, even though ultimately it depends only on the operator. Therefore the fact that similar matrices have the same characteristic polynomials is somewhat less trivial (and proved using properties of the determinant).

See also this answer to a somewhat more detailed question.