Sylvester equation

Remarks

The map $X \mapsto AX + XB$ is linear, and with the Kronecker product the equation reads $(I_n \otimes A + B^T \otimes I_m) \operatorname{vec}(X) = \operatorname{vec}(C)$. It has a unique solution for every $C$ if and only if $A$ and $-B$ have no common eigenvalue in an algebraic closure of $\mathbb{K}$. Example: $A = \operatorname{diag}(1, 2)$, $B = 1$ and $C = \begin{pmatrix} 2 \\ 3 \end{pmatrix}$ give $x_1 + x_1 = 2$ and $2x_2 + x_2 = 3$, so $X = \begin{pmatrix} 1 \\ 1 \end{pmatrix}$. Non-example: the equation $AXB = C$ is not a Sylvester equation, since $X$ is multiplied on both sides in one term. For $B = -A$, the equation $AX - XA = C$ never has a unique solution, because $A$ and $A$ share their eigenvalues.