$ A^2 - B^2 = I_{2n+1} \implies det(AB-BA)=0 $ where A,B are complex matrices of odd size

Question

Let $A, B$ be square matrices (with complex entries) of size $2n+1$, where $n$ is a positive integer.
I need help proving the following: $$A^2 - B^2 = I_{2n+1} \implies det(AB-BA)=0 $$

I've tried using characteristic polynomials, properties of eigenvalues, however to no avail. I feel like this kind of problem needs a little bit of experience in working with ranks of matrices (I've tried using Sylvester in more ways) and I would appreciate some help.

Answer 1

Let $P=A+B$ and $Q=A-B$. Then $A=\frac 12(P+Q)$ and $B=\frac 12(P-Q)$. Hence the hypothesis reads as $PQ+QP=2I$ and we have to prove that $\det(QP-PQ)=0$.

We have $I-PQ=-(I-QP)$. Therefore $\det(I-PQ)=(-1)^{2n+1}\det(I-QP)=-\det(I-QP)$. On the other hand, by Sylvester we have $\det(I-PQ)=\det(I-QP)$. It follows that $\det(I-PQ)=0$.

Since $\det(I-PQ)=0$, there is a nonzero vector $v$ with $(I-PQ)v=0$ i.e. $PQv=v$. It follows from $PQ+QP=2I$ that $QPv=v$ as well. Therefore $(QP-PQ)v=0$. Since $v$ is nonzero, the last equality gives $\det(QP-PQ)=0$, as desired.

Answer 2

A partial answer: suppose $A$ is invertible. Since $A^2=I+B^2$, $A^2$ commutes with $B$. Therefore $$ A(AB-BA)=A^2B-ABA=BA^2-ABA=-(AB-BA)A $$ Take determinants on both sides, we get $\det(A)\det(AB-BA)=(-1)^{2n+1}\det(AB-BA)\det(A)$. Hence $\det(AB-BA)=0$.