For positive semi-definite symmetric complex matrices $A$, $B$, $B$ being full rank, for $\operatorname{tr}(A^H B)$ to be $0$, $A=0$ has to be true?

Question

I have two matrices $A$,$B$ given that both are positive semi-definite, symmetric, complex. Another given is that $B$ has full rank. Now I want to find a condition that implies (or is equivalent to) $\operatorname{tr}(A^H B)=0$. My idea is that it should be $A=0$. But I am not sure how to get there.

Answer 1

Writing $B$ in the basis in which it is diagonal we have $$ B = \sum_{i=1}^n \lambda_i v_iv_i^*, $$ where $\lambda_i > 0$ for all $i = 1, \dots, n$ and $\{v_1, \dots, v_n\}$ form an orthonormal basis. Note that $B$ must be positive definite as it has full rank. Writing $A$ with respect to this basis, $$ A = \sum_{ij} a_{ij} v_i v_j^*. $$ Computing the trace we find (assuming as $A$ is PSD it is also Hermitian) $$ \begin{aligned} \operatorname{Tr}[ A^* B] &= \operatorname{Tr}[\sum_{ij} a_{ij} v_i v_j^* \sum_k \lambda_k v_k v_k^*] \\ &= \operatorname{Tr}[\sum_{ij} \lambda_j a_{ij} v_i v_j^*] \\ &= \sum_{j} \lambda_j a_{jj}. \end{aligned} $$ Now as $\lambda_j > 0$ the only way this sum can be zero (given $a_{jj} \geq 0$) is if $a_{jj} = 0$ for all $j$. However this means that $\operatorname{Tr}[A]=0$, together with the fact that $A$ is PSD this implies $A=0$.

Thus in the context of this question $\operatorname{Tr}[A^* B] = 0 \implies A = 0$. The converse is immediate and so if $A$ and $B$ are PSD and $B$ has full rank then $\operatorname{Tr}[A^* B] = 0 \iff A=0$.