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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?

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  • $\begingroup$ I suppose the question is for a positive definite diagonal matrix, could you have say $\pm i$ in two different entries? $\endgroup$ – muzzlator Feb 26 '13 at 13:30
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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.

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  • $\begingroup$ Is there any standard text that can be referenced for this? $\endgroup$ – nuse_li Feb 27 '13 at 8:25
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Yes, it's true in the complex case. Since every Hermitian matrix is unitarily diagonalizable, WLOG we may assume that $A=D\oplus 0$, where $D$ is a positive diagonal matrix. Let $B=\begin{pmatrix}U&W^\ast \\W&V\end{pmatrix}$. Since $B$ is positive semidefinite, so are $U,V$ and hence the diagonal of $U$ is nonnegative. Thus $\operatorname{trace}(AB)=0$ implies that $U$ has a zero diagonal, $\operatorname{trace}(U)=0$ and in turn all eigenvalues of $U$ are zero. Thus $U=0$. But then $W$ must be zero, otherwise $B$ cannot be positive semidefinite. Hence $B=0\oplus V$ and $AB=0$.

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