For positive semi-definite complex matrices A, B, B symmetric of rank 1,find a condition for $\operatorname{tr}(A^{\ast}B)$ to be $0$?

Question

I have two matrices A,B given that both are positive semi-definite, complex. Another given is that B has rank 1 and is symmetric. Now I want to find a condition that implies (or is equivalent to) $\operatorname{tr}(A^{\ast}B)=0$. Originally I thought maybe $A=0$ would suffice but the rank of B is a problem. More specifically $A_{i,j}=f_{i,j}\geq 0$ for all $i,j$. Additionally $B=ww^{\ast}$ where $w\in \mathbb{C}^{m}$ with $w_k=e^{-2\pi i u k}$.

Answer 1

In general, if $A$ and $B$ are PSD, then $\operatorname{tr}(A^\ast B)=\operatorname{tr}(AB)=\operatorname{tr}(B^{1/2}AB^{1/2})=\|A^{1/2}B^{1/2}\|_F^2$. Therefore $$ \operatorname{tr}(A^\ast B)=0\Rightarrow A^{1/2}B^{1/2}=0\Rightarrow AB=0\Rightarrow\operatorname{tr}(A^\ast B)=\operatorname{tr}(AB)=0. $$ That is, $\operatorname{tr}(A^\ast B)=0$ is equivalent to $AB=0$.

In your case, since $B=ww^\ast$, $\operatorname{tr}(A^\ast B)=0$ is equivalent to $Aw=0$.