Let $C$ be an $n\times n$ Hermitian matrix. Show that $$\text{Tr}(C)=0 \iff \exists P,Q \text{ Hermitian s.t.} \,\, PQ-QP=iC.$$
Ideas: The right-to-left direction I have no problem with. For the left-to-right direction, I'm going to start with the fact that $$\text{Tr}(C)=0 \implies \exists A,B \in M_n \text{ s.t. } C = AB-BA.$$
I'm having trouble showing we can choose $A,B$ Hermitian. One method would be to try to show $A,B$ must be Hermitian, but that seems too strong. Along this road...
$\text{Tr}(iC)=0$, so we can write $iC=AB-BA$. $iC$ is diagonalizable with all imaginary eigenvalues, so it is skew-Hermitian. Thus $$BA-AB=-(AB-BA)=(AB-BA)^*\\=(AB)^* - (BA)^* = B^*A^*-A^*B^*.$$
But this hasn't gotten me anywhere.
