If $P$ and $Q$ are invertible matrices $PQ=-QP$, then which claim about their traces is true?

Question

If $P$, $Q$ are invertible and $PQ=-QP$, then what can we say about traces of $P$ and $Q$.

I faced this question in an exam but according to me this question is wrong as $Q=-P^{-1}QP$, which implies $\det(Q)=0$ and it implies $Q$ is not invertible? But it is given invertible in hypothesis.

Options were both traces $0$, both $1$, $Tr(Q)\neq Tr(P)$ or $Tr(Q)=-Tr(P)$

Answer 1

use the fact that $tr(AB)=tr(BA)$ so $tr(-P^{-1}QP)=-tr(Q)$.

Answer 2

Note that we can only deduce that $Q=0$ (as you did using the determinant) if $P$ and $Q$ have an even size. In particular, note that $\det(-Q)= (-1)^n \det(Q)$, where $Q$ has size $n$. For a specific example, take $$ P = \pmatrix{0&1\\1&0}, \quad Q = \pmatrix{1&0\\0&-1} $$

However, using your equation, we could take the trace of both sides to find that $Q$ has trace $0$. Symmetrically, we may argue that $P$ also has trace $0$.

Answer 3

Pre-multiply by $P^{-1}$ to get $Q=P^{-1}(-Q)P$ which implies that matrices $Q$ and $-Q$ are similar, so $tr(Q)=tr(-Q)\implies tr(Q)=0$. Similarly you can get $tr(P)=0$ also.