Prove the nonzero eigenvalues of $M$ are roots of unity. Suppose $A$ and M are $n \times n$ matrices over $\mathbb{C}%$, $A$ is invertible
and $AMA^{-1} = M^2$. Prove the nonzero eigenvalues of $M$ are roots of
unity.
I get that you can rearrange this as $MA^{-1} = A^{-1}M^2$. Then for some eigenvector-eigenvalue pair of $M$ ($Mv=\lambda v$), then $M(A^{-1}v) = A^{-1}M^2v = \lambda^2(A^{-1}v)$ making $\lambda^2$ an eigenvalue. 
Why does it follow that $\lambda^{2^k}$ is an eigenvalue? For reference this is 7.5.9 in the Book "Berkeley Problems In Mathematics". Thanks.
 A: The matrices $AMA^{-1}$ and $M^2$ have the same eigenvalues. In particular, they have the same non-null eigenvalues. Therefore, if $\lambda$ is a non-null eigenvalue of $M$, then so is $\lambda^2$. But then so is $\lambda^4$ and so on. On the other hand, $M$ has only a finite number of eigenvalues. Therefore $\lambda^{2^k}=\lambda^{2^l}$ for some $k,l\in\mathbb{N}$ with $k>l$. So, $\lambda^{2^l}(\lambda^{2^k-2^l}-1)=0$. But $\lambda^{2^l}\neq0$ and therefore $\lambda^{2^k-2^l}=1$.
A: You have shown that if $\lambda$ is an eigenvalue, then so is $\lambda^2$. Note that by the same argument, if $\lambda^2$ is an eigenvalue, then so is $\lambda^4$, therefore also $\lambda^8$, etc.
A: We have arrived a step that $\lambda,\lambda^{2},\ldots,\lambda^{2^{k}},\ldots$
are non-zero eigenvalues. Since there are finitely many eigenvalues,
there must exist $k_{1}<k_{2}$ such that $\lambda^{2^{k_{1}}}=\lambda^{2^{k_{2}}}$.
That is, $\lambda^{2^{k_{2}}-2^{k_{1}}}=1$. It follows that $\lambda$
is a root of unity (i.e., root of the equation $z^{n}=1$, where $n=2^{k_{2}}-2^{k_{1}}$.)
A: By induction: $\lambda, \lambda^2, (\lambda^2)^2, \dots$ are all eigenvalues, so if $\lambda^{2^k}$ is an eigenvalue then $(\lambda^{2^k})^2$ is too, i.e. $\lambda^{2 \cdot 2^k}$, which is $\lambda^{2^{k+1}}$. So by induction, try for all $k$. If $|\lambda|>1$, then taking larger powers of $\lambda$ makes for infinitelty many eigenvalues, ever larger in absolute value. But there are at most $n$ distinct eigenvalues. So $|\lambda|\le 1$. But the same argument, replacing $|\lambda|>1$ with $0 < |\lambda|<1$, we find the same contradiction. Therefore $|\lambda|=1$ or $\lambda=0$.
