Proving 2 matrices have the same trace [closed]

Question

I found a problem in my textbook and I have tried solving it, but I had no succes. The problem is:

Let $A$ and $B$ be $n \times n$ matrices with complex number entries. Given that $AB−BA$ is invertible and $(A−B)^2=I_n$, where $I_n$ is the identity matrix, prove that $tr(A)=tr(B)$ and that $n$ is even. How should I approach this problem? I have tried to actually compute the inverse of $AB−BA$ from $(A−B)^2$ but I had no succes. I can also say that the eignevalues of $A-B$ are $+1$ or $-1$, but how does that help?

Answer 1

Let's call $C=A-B$. Then you have $C^2=I_n$ and $BC-CB$ is invertible. You want to show that $C$ has an equal dimension of $1$ and $(-1)$ eigenspaces, which in turn implies both the equality $tr(A)=tr(B)$ and $n$ being even.

Pick an eigen basis of the space to assume that $$ C=\left( \begin{array}{c}I_m & 0\\ 0&-I_k\end{array} \right), ~~ B=\left( \begin{array}{c}B_1 & B_2\\ B_3&B_4\end{array} \right) $$ with the appropriate sizes of $B_i$. Then $$ BC-CB=\left( \begin{array}{c}0 & *\\ * &0\end{array} \right). $$ If $m>k$, then the first $m$ rows are linearly dependent and if $m<k$ then the last $k$ rows are.