# Eigenvectors of $P^{-1}AP$

Let $A\in M_{n}(\mathbb{C})$ and assume that $A$ is diagonalizable, let $P\in M_{n}(\mathbb{C})$ be an invertible matrix.

My question is what are the eigenvectors of $P^{-1}AP$ ?

I think it's probably something like $P$(eigenvectors of $A)$ , but I don't remember...

I appriciate any help.

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"I don't remember." --There is nothing wrong with that. Rather than trying to remember, it is a good idea in such situations to try to figure it out. – Jonas Meyer Jun 20 '12 at 19:29
You shouldn't "try to remember", you should try to think. That is how we do mathematics =) but still your guess was a good one. – Patrick Da Silva Jun 20 '12 at 19:31

Assume $Ax = \lambda x$. Then if you take $v = P^{-1}x$, you have $$P^{-1}AP(v) = P^{-1}A P(P^{-1}x) = P^{-1} Ax = P^{-1} \lambda x = \lambda P^{-1} x = \lambda v.$$ Therefore the vectors you are looking for are $P^{-1}$ times the eigenvectors of $A$. (Because conversely, an eigenvector of $P^{-1}AP$ gives eigenvectors for $A = (P^{-1})^{-1}(P^{-1}AP)(P^{-1})$ by letting $x = (P^{-1})^{-1} v = Pv$. The function $x \mapsto P^{-1}x = v$ is therefore a bijection from the eigenvectors of $A$ to the eigenvectors of $P^{-1}AP$.)