TITLE: A is similar to $A^k$, then each eigenvalue of $A$ is a root of unity
QUESTION [7 upvotes]: Let $A \in \mathbb{C}(n,n)$ and $k \geq 2$ be an integer such that $$A \sim A^k$$. Show that if $A$ is non-singular then each eigenvalue of $A$ is a root of unity.
Attempt: Since $A \sim A^k$, $$PA = A^kP$$ where $P$ is an invertible matrix. Since $A$ is invertible, $0$ cannot be an eigenvalue of $A$. Suppose $$Av = \lambda v \quad v \neq 0$$ then $$PAv = \lambda Pv$$ $$\therefore A^k(Pv) = \lambda (Pv) $$
which  implies that $Pv$ is an eigenvector of $A^k$. But the eigenvalues of $A^k$ are $\lambda^k$ $$\therefore \lambda^k=\lambda$$ which gives the conclusion required.
My questions is: Is the logic correct? If so, Am I missing any details? If not, then how could I approach this?
Thanks!

REPLY [6 votes]: You are correct in noting that similar matrices have the same set of eigenvalues.  However, given this, you cannot conclude that $\lambda = \lambda^k$.  However, you can conclude that $\lambda^k$ is an eigenvalue if $\lambda$ is.
So suppose there are $n$ eigenvalues total (there must be finitely many).  Take the set $\{ \lambda , \lambda^k , \lambda^{k^2} , \lambda^{k^3}, \dots, \lambda^{k^n} \}$.  This has $n+1$ elements, and each element is an eigenvalue, so by the pidgeonhole principle, we must repeat an element twice.

 So $\lambda^{k^m} = \lambda^{k^i}$ for some $1 \le i < m \le n$.