Is there any work on the matrix equation in unknowns $X, Y \in {\Bbb C}^{n \times n}$ $$(X \otimes Y + Y \otimes X) \operatorname{vec}(A)=0$$ where $\otimes$ is the Kronecker product? Or, in general, is there any work on $X \otimes Y + Y \otimes X$?
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$\begingroup$ If $A$ is not the unknown, why vectorise, then? Why not write in matrix form without $\otimes$? $\endgroup$– Rodrigo de AzevedoCommented Dec 25, 2023 at 12:10
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$\begingroup$ One can write this way too, $(X \otimes Y + Y \otimes X)v=0$ for a fixed $v \in \mathbb{R}^{n^2}$. $\endgroup$– mukhujjeCommented Dec 25, 2023 at 12:22
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$\begingroup$ That's because I suspect the matrix $(X \otimes Y + Y \otimes X)$ is too nice and restrictive. Which is not that apparent in the unvectorised matrix equation. $\endgroup$– mukhujjeCommented Dec 25, 2023 at 12:25
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$\begingroup$ I usually do that in my work. I wanted to find out if there are already some known results. $\endgroup$– mukhujjeCommented Dec 25, 2023 at 12:33
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$\begingroup$ We can rewrite this equation as $XAY^T=-YAX^T$. If $A$ is symmetric, then $(XAY^T)^T=YAX^T=-XAY^T$, so $XAY^T$ is anti-symmetric. If $A$ is anti-symmetric, then $(XAY^T)^T=YA^TX^T=-YAX^T$, so $XAY^T$ is symmetric. But for complex matrices, it usually makes more sense to take the adjoint than the transpose. $\endgroup$– Joseph Van NameCommented Dec 25, 2023 at 13:21
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