I have a (supposedly) easy question. Let $A$ be a square matrix and let $B=AP$, where $P$ is a permutation matrix (it permutes the columns of $A$, so that $B$ consists in the columns of $A$ in a different order). Naturally, we can associate a permutation $\sigma$ to the matrix $P$.
I would like to prove that the eigenvalues of $B$ are the same as the eigenvalues of $A$, and that if $v$ is an eigenvector of $A$, then $\sigma.v$ is an eigenvector of $B$.
Notice that these assertions may be false. In fact, from one example that I've done, instead of $\sigma.v$ we should have $\sigma^{-1}.v$, although I don't know why.
To make it clear, $\sigma.v$ is the action of $\sigma$ on $v$.
I'm asking for help just because I don't find anything useful about this anywhere and I cannot even prove the first part (eigenvalues, there must be any trick).
Thanks in advance.
What you are asking is not true. Consider the eigenvalues of $$ \left[ \begin{matrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3 \\ \end{matrix} \right] $$
and that of
$$ \left[ \begin{matrix} 0 & 0 & 1 \\ 0 & 2 & 0 \\ 3 & 0 & 0 \\ \end{matrix} \right] $$
I just swapped the first and last columns; the eigenvalues are different.
EDIT: In response to comment.
If $A$ is also a permutation matrix and your field is $\mathbb{C}$ the eigenvalues are also not necessarily the same. Consider the $5 \times 5$ identity matrix and
$$ \left[ \begin{matrix} 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 1 & 0 \\ \end{matrix} \right] $$
In one case the characteristic polynomial is $-(s-1)^5$ and in the other $- s^5+s^3+s^2-1$.