TITLE: Show that invertible matrices with an additional condition are diagonalizable.
QUESTION [6 upvotes]: Let $A$ and $B$ be invertible $2 \times 2$ matrices such that $AB = -BA$ over the complex numbers. Show that $A$ and $B$ are diagonalizable.

REPLY [6 votes]: We have $B=A^{-1}(-B)A$, hence $B$ and $-B$ are similar. Since $B$ is invertible , $0$ is not an eigenvalue of $B$.
Now, if $\lambda_0$ is an eigenvalue of $B$, then $-\lambda_0$ is an eigenvalue of $-B$. By similarity:  $-\lambda_0$  is an eigenvalue of $B$.
The $2 \times 2$ - matrix $B$ has therefore the two distinct eigenvalues  $\lambda_0$ and  $-\lambda_0$