Show that invertible matrices with an additional condition are diagonalizable.

Question

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.

Answer 1

Fill in details:

Let $\;\lambda\;$ be an eigenvalue of $\;A\;$ with corresponding eigenvector $\;v\;$ :

$$ABv=A(Bv)=-BAv=-B(\lambda v)=-\lambda Bv$$

and thus also $\;-\lambda\;$ is an eigenvalue of $\;A\;$ with corr. eigenvector $\;Bv\;$ .

Since $\;\lambda\neq0\;$ we get $\;A\;$ is diagonalizable, and by symmetry also $\;B\;$ is.

Answer 2

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 $-\lambda_0$