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I have two matrices $A,B$ given that both are positive semi-definite, complex. Another given is that $B$ has rank $1$ and is hermitian. Additionally $B=ww^{\ast}$ where $w\in \mathbb{C}^{m}$ with $w_k=e^{-2\pi i u k}$.

Now I want to find a condition that implies (or better still is equivalent to) $\operatorname{tr}(A^{\ast}B)=0$. Here $\ast$ stands for the complex cunjugate transpose. This condition should if somehow possible be only dependent on $A$. Originally I thought maybe $\det(A)=0$ would suffice but have not been able to prove this.

Leon Schmitz
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1 Answers1

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Note that $$ \operatorname{tr}(A^*B) = \operatorname{tr}(Aww^*) = \operatorname{tr}(w^*Aw) = w^*Aw. $$ So of course, this trace is zero iff $w^*Aw = 0$. Moreover, because $A$ is positive semidefinite, it can be shown that $w^*Aw = 0 \iff Aw = 0$ (I give some proofs of this result in my earlier post here).

That is, we can say that $\operatorname{tr}(A^*B) = 0$ if and only if $w \in \ker(A)$. That is, this trace is zero iff we have $$ \sum_{k=1}^n a_{jk}e^{- 2 \pi i uk} = 0 \quad j = 1, \dots, n, $$ where $a_{jk}$ denotes the $j,k$ entry of $A$.

Ben Grossmann
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