Let $F$ be a field and let $n$ be a positive integer. Let $A,B ∈ M_{n×n}(F)$ be matrices satisfying $A^2 + B^2 = I$ and $AB + BA = O$. Show that $tr(A) = tr(B) = 0$.
I know that $tr(A^2)+tr(B^2)=n$ and that $(A+B)^2=I$. I'm having a hard time showing this. I've been playing around for a bit with no success. Any solutions/hints are greatly appreciated.
Assume that $char(F)\not= 2$ and $A$ is invertible. Then $ABA^{-1}+B=0$, $2tr(B)=0$ and finally $tr(B)=0$. Moreover $spectrum(B)$ is in the form $\{0,\cdots,0,\lambda_1,-\lambda_1,\cdots,\lambda_r,-\lambda_r\}$.
If moreover $B$ is invertible, then $n=2p$ ($n$ is even) and $tr(A)=tr(B)=0$.
If we consider also the equality $(A+B)^2=I$ (again under the previous hypothesis), then $A+B$ is similar to $diag(I_p,-I_p)$.