I'd like to solve $AX -BX + XC = D$, for the matrix $X$, where all matrices have real entries and $X$ is a rectangular matrix, while $B$ and $C$ are symmetric matrices and $A$ is formed by an outer product matrix (i.e, as $vv^T$ for some real vector $v$) while $D$ is 'not' symmetric. $A,B,C,D$ matrices are fixed while $X$ is the unknown.
How can this equation be solved? Secondly, is there any case, where the solution of this equation has a closed form?
Best Answer
More generally, Sylvester's equation of the form $$AX+XB=C$$ can be put into the form $$M\cdot \textrm{vec}X=L$$ for larger matrices $M$ and $L$.
Here $\textrm{vec}X$ is a stack of all columns of matrix $X$.
How to find the matrix $M$ and $L$, is shown in chapter 4 of this book: http://www.amazon.com/Topics-Matrix-Analysis-Roger-Horn/dp/0521467136
Indeed, $M=(I\otimes A)+(B^T\otimes I)$, and $L=\textrm{vec}C$, where $\otimes$ denotes the Kronecker product.
Special case with $M$ invertible, we have $\textrm{vec}X=M^{-1}L$.