Another sufficient condition for two matrices having a common eigenvector ?

Question

Let $A , B$ be complex $n \times n$ matrices such that $AB = BA^2$, and assume $A$ has no eigenvalue of absolute value $1$ , then is it true that $A$ and $B$ have a common (nonzero) eigenvector ?

Answer 1

Suppose that $\lambda$ is an eigenvalue of $A$ and $v$ an eigenvector. It follows that $$ABv=\lambda^2 Bv.$$ Repeating this argument, $$A(B^nv)=\lambda^{2^n}(B^nv).$$ It follows that $\lambda^{2^n}$ are all eigenvalues of $A$. Thus, $\lambda^{2^k}=\lambda^{2^l}$ for some $k<l$.

If $\lambda=0$, then $Bv$ is also an eigenvector of $\lambda=0$. If $V$ is the subspace of eigenvectors of $0$ of $A$, then $B$ can be viewed as a linear operation on $V$. It implies that $B$ has an eigenvector in $V$. Thus $A$ and $B$ has a common eigenvector.

If $\lambda\neq0$, then $\lambda^{2^{l}-2^k}=1$ contradicting $|\lambda|\ne1$.