Which of the following is/are true for the trace of the matrices $P$ and $Q$?

Question

If $P$ and $Q$ are invertible matrices such that $PQ=-QP$ than which of the following is/are true?

1. Tr$(P)$$=$Tr$(Q)$$=$$0$
2. Tr$(P)$$=$Tr$(Q)$$=$$1$
3. Tr$(P)$$=-$Tr$(Q)$
4. Tr$(P)$$\not=$Tr$(Q)$

I took the matrices $P=\begin{bmatrix}-1&-1\\0&1\end{bmatrix}$ and $Q=\begin{bmatrix}-1&0\\2&1\end{bmatrix}$ such that $PQ=-QP$, then option 1 is correct. But I am unable to verify the other options. How can I do this? Any help would be great. Thanks.

Answer 1

We have $\textrm{Tr}(P) = \textrm{Tr}(PQQ^{-1})$.

For square matrices $A$ and $B$, it holds that $\textrm{Tr}(AB) = \textrm{Tr}(BA)$. Therefore $\textrm{Tr}(PQQ^{-1})=\textrm{Tr}(Q^{-1}PQ)$. Using the fact that $PQ=-QP$, and the linearity of trace, we obtain $\textrm{Tr}(Q^{-1}PQ)=\textrm{Tr}(-Q^{-1}QP)=-\textrm{Tr}(P)$.

Thus $\textrm{Tr}(P) = -\textrm{Tr}(P)$. So $\textrm{Tr}(P) = 0$. You can also show, the same way, that $\textrm{Tr}(Q) = 0$.

So 1. and 3. are true, 2. and 4. are false.

Answer 2

Note that $$ PQ = -QP \implies PQP^{-1} = -Q $$ However, $Tr(PQP^{-1}) = Tr(Q)$, and $Tr(-Q) = -Tr(Q)$.

Answer 3

Fill in details and justify the following:

Observe that if the matrices are of order $\;n\;$ , then

$$\det (PQ)=\det(-QP)=(-1)^n\det(QP)=(-1)^n\det(PQ)\implies n\;\;\text{ is even}$$ and also

$$PQ=-QP\implies Q^{-1}PQ=-P\;,\;\;\text{but also}\;\;Tr.(Q^{-1}PQ)=Tr.(PQQ^{-1})=Tr. (P)$$

and deduce that $\;Tr.(P)=0\;$, and deduce more taking into account that $\;Tr.(PQ)=Tr.(QP)\;$