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Let $n \in \mathbb{N}$. Let $A,B,C,D$ be non-singular $n \times n$ matrices. If the matrix pencils $A-\lambda C$ and $B-\lambda D$ are regular and have disjoint spectra, then

$$AXB-CXD = 0$$

has a unique solution ($X=0$). A reference for this result is:

What if we are looking for conditions on $A,B,C,D$ such that it implies a non-unique although low-rank solution $X$? Is there any known result in the literature?

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  • $\begingroup$ I am looking for theoretical conditions. Thank you anyway for the suggestion. $\endgroup$
    – baptiste
    Commented Mar 15, 2018 at 7:30

2 Answers 2

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If $0 \ne u \in \ker (A - \lambda C)$ and $0 \ne v \in \ker(D^T - \lambda B^T)$ with $\lambda \ne 0$, then $X = u v^T$ is a rank-one solution.

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  • $\begingroup$ Is it an equivalence? $\endgroup$
    – baptiste
    Commented Mar 15, 2018 at 7:34
  • $\begingroup$ Not quite. You could also have at least one of $Au=0$ and $B^T v = 0$ and at least one of $Cu=0$ and $D^T v=0$. $\endgroup$
    – Robert Israel
    Commented Mar 15, 2018 at 19:59
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You may want to look at the following (not easy to find) preprint: T. Košir, The matrix equation AXD^T − BXC^T = E, Technical Report (Research paper 737), Dept. of Mathematics and Statistics, Univ. of Calgary, 1992.

It contains a full characterization of the solutions of the generalized Sylvester equation depending on the blocks appearing in the Kronecker forms of those two pencils. I'm not sure it is easy to find out from this form when the solutions are low-rank, but it's a natural place to start.

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