# Find $B$ if $AB=BC$ and $A,C$ are invertible

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 $B$?

• 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. – Najib Idrissi Sep 20 '14 at 17:02
• @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 $A=C$). – Jyrki Lahtonen Sep 20 '14 at 17:05
• 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 $T$ :-) – Jyrki Lahtonen Sep 20 '14 at 17:07
• en.wikipedia.org/wiki/Sylvester_equation – daw Sep 20 '14 at 19:20

## 1 Answer

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.

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