Suppose $ A $ and $ B $ are two complex matrices and $ A $ has no eigenvalues with modulus $ 1 $. Given that $ A B = B A ^ 2 $, show that $ A $ and $ B $ have a common eigenvector.
I have tried to use Schur's triangularization theorem. Also it can be shown that $ A ^ k B = B A ^ { 2 k } $. But after that I have been stuck.
If $x$ is an eigenvector of $A$ with eigenvalue $\lambda$, then from $AB = BA^2$, we have $$ A(Bx) = \lambda^2 (Bx) $$ So: if $A$ has a non-zero eigenvalue, then $\lambda$ is an eigenvalue implies that $\lambda^2$ is an eigenvalue. Since $A$ can have only finitely many eigenvalues and $A$ has no eigenvalues of magnitude $1$, we can conclude that $A$ has $0$ as its only eigenvalue.
So, if $x$ is an eigenvector of $A$, we have $$ A(Bx) = 0 $$ That is: if $x \in \ker(A)$, then so is $Bx$. Consider the restriction $B|_{\ker(A)}:\ker(A) \to \ker(A)$. $B|_{\ker(A)}$ must have an eigenvector $y$. This eigenvector satisfies $By = \mu y$ for some $\mu \in \Bbb C$ and $Ay = 0$. That is, $y$ is a common eigenvector of $B$ and $A$.