1
$\begingroup$

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 $X=0$ ?

$\endgroup$
4
  • $\begingroup$ 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
1
$\begingroup$

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 $A\ne0$.

$\endgroup$
6
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.