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 $X=0$ ?

  • $\begingroup$ Not an answer, but considering that the trace is cyclic, if $A$ is Unitary, then the expression reduces to $Tr(AXA^{\dagger}) =Tr(A^{\dagger}AX) = Tr(X) =0$. $\endgroup$
    – RyanK
    Oct 16 '20 at 14:58
  • $\begingroup$ on one hand, this is not a general case, on the other hand, you cant conclude that X=0 if Tr(X)=0 @RyanK $\endgroup$
    – Jason
    Oct 16 '20 at 15:01
  • 2
    $\begingroup$ @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. $\endgroup$
    – paulinho
    Oct 16 '20 at 15:05
  • $\begingroup$ @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 $\endgroup$
    – Jason
    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 $A\ne0$.

  • $\begingroup$ Sorry, I don't fully understand it in what part of your statement you conclude A should be a nonzero "column vector"? @user1551 $\endgroup$
    – Jason
    Oct 16 '20 at 17:27
  • $\begingroup$ @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). $\endgroup$
    – user1551
    Oct 16 '20 at 17:35
  • $\begingroup$ Got it, just one question, based on your argument there is no way that A be a row vector instead of a column vector, is that correct? @user1551 $\endgroup$
    – Jason
    Oct 16 '20 at 17:39
  • $\begingroup$ @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. $\endgroup$
    – user1551
    Oct 16 '20 at 18:14
  • $\begingroup$ I see the core of your arguments depend on the fact that the matrices are finite dimension, but what if instead of finite-dimensional matrices the question be asked for infinite-dimensional matrices i.e., operators on some vector space, can we use the same argument? $\endgroup$
    – Jason
    Oct 20 '20 at 16:02

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