Linear Algebra – Product of Complex Hermitian Positive Semidefinite Matrices

linear algebramatrices

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?

Best Answer

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.