# Necessary conditions for the existence of solution of Sylvester equation AX=XB

Let's consider square matrices $$A_{n \times n}$$, $$B_{n \times n}$$ and $$X_{n \times n}$$ with elements from $$\mathbb{R}$$. Could you tell me please, what would be the necessary conditions for the existence of solution (may be not unique) of Sylvester equation: $$AX=XB.$$ As I know, sufficient condition looks like (but probably it is a necessary and sufficient condition) $$\sigma_p(A) \cap \sigma_p(B) \neq \varnothing,$$ here $$\sigma_p(A)$$ and $$\sigma_p(B)$$ are the spectra of matrices $$A$$ and $$B$$.

• $X = 0$ works. What conditions on $X$ do you want? – LSpice Feb 29 at 21:22

This equation always has a solution: $$X = O$$. I'll assume throughout this answer that you're interested in a non-zero solution.
The equation $$AX = XB$$ is equivalent to $$(A \otimes I - I \otimes B^T)\mathbf{x} = \mathbf{0}$$, where $$\otimes$$ denotes the Kronecker product and $$\mathbf{x}$$ is the vectorization of $$X$$. Your question is thus equivalent to asking when the matrix $$A \otimes I - I \otimes B^T$$ is not invertible (i.e., when $$0$$ is not an eigenvalue of $$A \otimes I - I \otimes B^T$$).
Since the eigenvalues of $$A \otimes I - I \otimes B^T$$ are exactly the sums of the eigenvalues of $$A$$ and $$-B$$, the condition that you wrote ($$\sigma_p(A) \cap \sigma_p(B) \neq \varnothing$$) is in fact both necessary and sufficient.
• @Randal'Thor - The letter $O$ seems to be a common notation for the zero matrix (to distinguish from the zero vector, for example). – mr_e_man Mar 1 at 22:58