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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.

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    $\begingroup$ To explain $\sigma$ vs $\sigma^{-1}$, you could, for instance, consider the fact that $(AP)v=A(Pv)$. The way $P$ permutes the columns of $A$ on the left-hand side is not the same as the way it permutes the rows of $v$ on the right. You would need to tweak this a bit too make eigenvectors of $A$ appear, but it isn't too hard. $\endgroup$ – Arthur May 4 '18 at 21:41
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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 $- s^5+s^3+s^2-1$.

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    $\begingroup$ Yes, you are right, what if A is also a permutation matrix? $\endgroup$ – Leafar May 4 '18 at 22:02

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