TITLE: Product of complex Hermitian positive semidefinite matrices with trace zero
QUESTION [9 upvotes]: Let $A$ and $B$ are both $n\times n$ complex Hermitian positive semidefinite matrices, then whether $\mathrm{trace}(AB)=0$ implies $AB=0$? When $A$ and $B$ are real  Hermitian positive semidefinite matrices, this is indeed true, is it true for the complex case?

REPLY [7 votes]: By assumption, there exist two matrices $C,D$ sucht that
$$
A=C^*C\quad\mbox{and}\quad B=DD^*.
$$
Now, using the commutativity of the trace,
$$
\mbox{Tr}(AB)=\mbox{Tr}(C^*CDD^*)=\mbox{Tr}(D^*C^*CD)=\mbox{Tr}((CD)^*CD)
$$
appears to be the sum of of all $|(CD)_{i,j}|^2$.
So 
$$
\mbox{Tr}(AB)=0\quad\Rightarrow \quad
CD=0\quad \Rightarrow\quad AB=C^*CDD^*=0.
$$
Note: this proof works in the complex case, like in the real case.