How to Prove that if $a^n$ is an eigenvalue of $M^n$, then $a^{1/n}$ is an eigenvalue of $M$

Question

It's trivial to prove that if $a$ is an eigenvalue of $M$, then $a^n$ an eigenvalue of $M^n$

Is the other way hold as well?

Answer 1

This is not true in general : consider the permutation matrix M such that $M_{i,j}=\delta_{i,\sigma(j)}$ where $\sigma$ is the circular permutation ($\sigma(i)=i+1$).

Then $M^n=I$ and every vector is an eigenvector of $M^n$, but this obviously doesn't holds true for $M$.

On the other hand you have conditions under which this is the case, the most trivial one is for example when M is diagonalisable with $M^n$ having distinct eigenvalues (EDIT : This is unclear. $M$ being of size k, you must have $k$ distinct eigenvalues for $M^n$). So if you are in a particular case, you have to be more specific.

Answer 2

The revised version is correct as long as you allow complex eigenvalues, and as long as you interpret $a$ as some $n$th root of $a^n$. The proof is the direction you already know.