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I was wondering if there is some way to solve the matrix equation $$XX^*-X^*X=A,$$ where $A$ is a given (necessarily Hermitian and vanishing trace) matrix and we are to solve for unknown square matrix $X$. We can assume further that $X$ is non-singular, non-normal.

If this is too difficult, can we find an $X$ when we know only the eigenvalues of $XX^*-X^*X$?

Here $X^*$ denotes "Hermitian conjugate" (i.e. complex conjugate+ transpose) of $X$.

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  • $\begingroup$ If $A \neq 0$, of course $X$ must be non-normal $\endgroup$ – Omnomnomnom Jan 21 '16 at 14:11
  • $\begingroup$ Yes,...it's true. $\endgroup$ – JSparrow Jan 21 '16 at 14:14
  • $\begingroup$ Interseting question. A trivial observation is that it is enough to solve the problem for diagonal $A$'s. When $A \in M_2(\mathbb{C})$ this can be done easily, but I'm not sure about the general case. $\endgroup$ – levap Jan 21 '16 at 16:28
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    $\begingroup$ Daniel's answer in another thread deals with the real symmetric case. The idea is easily extensible to the Hermitian case. $\endgroup$ – user1551 Jan 23 '16 at 2:55

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