necessary and sufficient condition of a trace equation $\operatorname{diag}(AXA^\dagger)=0$

Consider the following matrix equation:

$$\begin{equation} \operatorname{diag}(AXA^\dagger)=0 \end{equation}$$ where $$\operatorname{diag}(.)$$ represent the diagonal elements. with $$X$$ being the variable matrix and $$A$$ being an arbitrary matrix with constant elements and $$A^\dagger=(A^*)^T$$. Assuming they both have complex entries, what is the sufficient and necessary condition that the answer to the above equation be $$\begin{equation} X=0 \end{equation}$$ Update: if there is not a necessary and sufficient condition, is there a necessary condition?

meaning is there a condition on entries of $$A$$ such that $$\operatorname{diag}(AXA^\dagger)=0$$ implies $$X=0$$ ?

• Not an answer, but considering that the trace is cyclic, if $A$ is Unitary, then the expression reduces to $Tr(AXA^{\dagger}) =Tr(A^{\dagger}AX) = Tr(X) =0$. Oct 16 '20 at 14:58
• on one hand, this is not a general case, on the other hand, you cant conclude that X=0 if Tr(X)=0 @RyanK Oct 16 '20 at 15:01
• @Jason I'm quite sure the answer is that no specific value of $A$ can imply that $X = 0$. I'm currently busy at the moment, but the basic idea is to note that the desired quantity is equal to $Tr(A^\dagger A X)$, and that $A^\dagger A = Q D Q^\dagger$ for some diagonal $D$ and orthogonal $Q$. Then you just simply have to set $X = QE Q^\dagger$ for an $E$ where $Tr(DE) = 0$, which is always possible without forcing $E = 0$. I'll be back to write a more complete answer if no one does in the meantime. Oct 16 '20 at 15:05
• @paulinho thank you for your comment I have updated the question a little bit regarding your comment, I have also noticed a mistake that I made so the question is a bit changed now Oct 16 '20 at 15:41

Your notation is very confusing. I suppose that $$A^\dagger$$ means $$A^\ast=\overline{A}^{\,T}$$ (the conjugate transpose of $$A$$) rather than $$(A^\ast)^T=\overline{A}$$ (the complex conjugate of $$A$$). In this case, the statement $$\forall X,\ \operatorname{diag}(AXA^\dagger)=0\Rightarrow X=0\tag{1}$$ holds if and only if $$A$$ is a nonzero column vector.
Let $$A$$ be $$m\times n$$. When $$n>1$$, $$f:X\mapsto \operatorname{diag}(AXA^\dagger)$$ is a linear map from $$M_n(\mathbb C)$$ to $$\mathbb C^n$$. Since $$\dim M_n(\mathbb C)=n^2>n=\dim\mathbb C^n$$, $$\ker f$$ is always nonzero regardless of the value of $$A$$.
When $$n=1$$, $$X$$ is a scalar and $$\operatorname{diag}(AXA^\dagger)=X(|a_1|^2,|a_2|^2,\ldots,|a_m|^2)^\top$$. Therefore $$(1)$$ holds if and only if $$A\ne0$$.
• @Jason In the second paragraph, I've shown that if $n>1$ (i.e. if $A$ is not a column vector), statement $(1)$ cannot possibly hold because there always exists a nonzero $X$ such that $AXA^\ast$ has a zero diagonal. In the third paragraph, I've shown that if $n=1$ (i.e. if $A$ is a column vector), then $(1)$ holds iff $A\ne0$ (i.e. iff $A$ is a nonzero column vector). Oct 16 '20 at 17:35
• @Jason Not exactly. It is true that $A$ cannot be a row vector with two or more elements, but $A$ can be $1\times1$ (i.e. a scalar). In this case $A$ is both a column vector and a row vector and statement $(1)$ holds if and only if $A$ is a nonzero scalar. Oct 16 '20 at 18:14