If $AB=-BA$ then do $A$ and $B$ share a common eigenvector?

Question

I know that for two matrices $A$ and $B$ if $AB = BA$ then they share a common eigenvector. Even in general, for any $k$ if $AB - BA = kB$ then they have a common eigenvector.

But what about if $AB=-BA$? Do $A$ and $B$ share a common eigenvector?

Answer 1

For example, $A=diag(1,-1) ,B_u=\begin{pmatrix}0&1\\u&0\end{pmatrix}$ have no common eigenvectors and are invertible when $u\not=0$.

Yet $A,B_0$ have a common eigenvector and $B_0$ is singular.

EDIT. $\textbf{Proposition}$. Let $A,B$ be st $AB+BA=0$. Then $A$ or $B$ is singular iff $A,B$ have a common eigenvector.

$\textbf{Proof}$. $\Rightarrow$ For example, $A$ is singular. $\ker(A)$ is invariant for $B$; then $B$ admits (over an algebraically closed field) an eigenvector in $\ker(A)$.

$\Leftarrow$ If $Au=\lambda u,Bu=\mu u$, then $\lambda\mu=0$ ( if the charateristic of the field is not $2$) and we are done.