TITLE: If $\lambda^n $ is an eigenvalue of $A^n$ then $\lambda $ is an eigenvalue of $A$?
QUESTION [5 upvotes]: I'm trying to figure out whether the following statement is correct:

If $\lambda^n $ is an eigenvalue of $A^n$ associated to a certain
  eigenvector $v$ then $\lambda $ is an eigenvalue of $A$ associated to
  the same eigenvector $v$.

I know that the opposite statement is correct, that is, if $\lambda $ is an eigenvalue of $A$ associated to a certain eigenvector $v$ then $\lambda^n $ is an eigenvalue of $A^n$ associated to the same eigenvector $v$.
I tried to guess some matrices to find out a counter example but I couldn't find anything. It seems that if we take $A$ to be diagonal matrix then the statement is always correct. In particular if we take $A$ such that $A^n$=$\lambda^n I $ then it must be that $A=\lambda I$.

REPLY [4 votes]: In particular, if we take $A$ such that $A^n = \lambda^n I$, then it must be that $A = \lambda I$.

This is not the case.  Two examples:
$$
A = \pmatrix{1&0\\0&-1}
$$
Note that $v = (1,1)$ is an eigenvector of $A^2 = I$, but not of $A$.
$$
A = \pmatrix{0&1\\0&0}
$$
Note that $v = (0,1)$ is an eigenvector of $A^2 = 0$, but not of $A$.