A is similar to $A^k$, then each eigenvalue of $A$ is a root of unity

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!

• The phrase "but the eigenvalues of $A^k$ are $\lambda^k$" seems imprecise. Aug 10, 2013 at 5:25
• Your last conclusion is invalid. $A$ may have several eigenvalues, $\lambda, \mu$, and you might have that $\mu^k=\lambda$. Aug 10, 2013 at 5:25

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 $1 \le i < m \le n$.

• How did you get the set of $\lbrace \lambda,...,\lambda^{k^{n}}\rbrace$
– AAP
Aug 10, 2013 at 7:02
• @AAP Since you know that $\lambda$ is an eigenvalue, you know that $\lambda^k$ is as well, and so is $(\lambda^k)^k = \lambda^{k^2}$ and so on...
– A.S
Aug 10, 2013 at 17:17
• @A.S The lower bound of $i$ should be $0$. Jan 10 at 5:05