# If $AB-BA=\mu B$, $\mu \ne0$. Determine whether $A$ and $B$ share common eigenvector

$A$ and $B$ are matrices on $\mathbb{C}^{n\times n}$. If $AB-BA=\mu B$, $\mu \ne0$. Determine whether $A$ and $B$ share common eigenvector. Prove or show a counterexample.

I have already read the question about if $AB=BA$ , then they share common eigenvectors. So I believe it's about find some counterexample. But when I tried some small size matrice. They all didn't satisfy this. So I wonder if there is any tricks to find the counterexample.

Suppose $A$ and $B$ did not have any common eigenvectors. Let $\lambda$ be any eigenvalue of $A$ and let $v\neq 0$ such that $Av=\lambda v$. Then $$A(Bv)=AB(v)=(BA+\mu B)v=B(Av+\mu v)=(\lambda+\mu)(Bv).$$ Since $v$ is not an eigenvector of $B$, we must have $Bv \neq 0$, and so $\lambda+\mu$ is an eigenvalue of $A$. Since $\mu \neq 0$, this would mean that $A$ has infinitely many eigenvalues which is a contradiction.
$\mu$ is fixed, if $AB-BA=\mu B$ for some $\mu \neq 0$ then $A$ and $B$ share a common (non zero) eigenvector if and only if $B$ is not invertible.