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Suppose $A$ and $C$ are known invertible complex matrices of possibly different orders. If $B$ is an unknown matrix of appropriate order such that $AB = BC$, then how could one solve for $B$?

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  • $\begingroup$ There isn't always just one solution. $B=0$ is always a solution, and if $A=C$ for example then $B=I$ is also a solution. $\endgroup$ – Najib Idrissi Sep 20 '14 at 17:02
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    $\begingroup$ @NajibIdrissi: Agreed. Notice that this a homogeneous linear system on the unknown coefficients of $B$ so the solutions form a vector space. The case $A=C$ usually implies a higher dimensional solution space. For in that case $B$ can be any polynomial on $A$ (those commute with $A=C$). $\endgroup$ – Jyrki Lahtonen Sep 20 '14 at 17:05
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    $\begingroup$ A brute force method: Write down the matrix of the mapping $T:B\mapsto ABC^{-1}$ and find the eigenspace belonging to eigenvalue $\lambda=1$ of $T$ :-) $\endgroup$ – Jyrki Lahtonen Sep 20 '14 at 17:07
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    $\begingroup$ en.wikipedia.org/wiki/Sylvester_equation $\endgroup$ – daw Sep 20 '14 at 19:20
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We can solve the problem theoretically and then compute it for low dimensions.

Consider $\phi: M(n, \mathbb{K}) \to M(n, \mathbb{K})$ mapping $X \mapsto AXC^{-1}-X$ and find the kernel.

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  • $\begingroup$ Should you have $\phi(X)=AXC^{-1}-X$? $\endgroup$ – Jyrki Lahtonen Sep 20 '14 at 17:10
  • $\begingroup$ Oh, ye, scuse me. $\endgroup$ – Ivan Di Liberti Sep 20 '14 at 17:11
  • $\begingroup$ then please correct or delete your answer... $\endgroup$ – daw Sep 22 '14 at 10:16

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