let P and Q are two invertible matrices . and PQ= -QP . then which of the following is true
a) trace(P)=trace(Q)=0
c)trace(P) is not equal to trace(Q)
c) none of the above.
i can show that trace(PQ)=0.
but after that i can't proceed. could anyone help?
Since $P$ and $Q$ are invertible, it follows that $P$ and $-P$ are similar: $$ PQ = -QP \quad \Rightarrow \quad Q = P^{-1}(-Q)P. $$ Hence, if $\lambda$ is an eigenvalue of $Q$, so is $-\lambda$. If we compare the Jordan forms of $Q$ and $-Q$, then for each block of size $k$ to eigenvalue $\lambda$ of $Q$, there is a block of size $k$ to eigenvalue $-\lambda$. Hence the trace of the Jordan form of $Q$ is zero, which implies that the trace of $Q$ is zero.
The same argument applies to $P$ as well. Hence (a) is true.
Here is an argument that works without recurring to Jordan (or related) normal forms. As above, we obtain that $P$ and $-P$ are similar. Hence their characteristic polynomials are equal. In particular, the coefficients of the first-order monomials are equal, which is the trace. Hence $tr(P)=tr(-P)$. Now, trace is a linear mapping, $tr(-P)=-tr(P)$. This amounts to $tr(P)=0$.