# Solution of a Sylvester equation?

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?

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 $\textrm{vec}X=M^{-1}L$.

• Apart from the invertibility, are there other required conditions for this closed form to exist? Also, can I use the pseudoinverse of $M$, and would that be justified for this solution? Apr 7 '13 at 4:37
• As usual for solving the equation $Ax=b$, if $A$ is not invertible then there are either infinitely many solutions or no solutions depending on $b$. Apr 7 '13 at 4:45
• A similar description in wiki here en.wikipedia.org/wiki/Sylvester_equation considers $X$ to be a square matrix. Is that the case, in the suggested book and in your answer? Because, in my question, $X$ is a rectangular matrix. Apr 7 '13 at 4:58
• It doesn't matter that the matrices are square. It only requires that the products of matrices are well-defined. Apr 7 '13 at 7:18

How about a small but concrete example of $AX + XB = C$: $$\pmatrix{a_0 & a_1 \\a_2 & a_3}\pmatrix{x_0 \\ x_1} + \pmatrix{x_0 \\ x_1} b= \pmatrix{c_0 \\ c_1}$$

This gives a system of two equations: \begin{align} a_0x_0 + a_1x_1+bx_0 &= c_0 \\ a_2x_0 + a_3x_1+bx_1 &= c_1 \\ \end{align} Which leads to the vectorized (using the Kroneker product formula for the Sylvester Equation): $$\pmatrix{a_0 +b & a_1 \\a_2 & a_3+b}\pmatrix{x_0 \\ x_1} = \pmatrix{c_0 \\ c_1}$$

And depending on the determinant, it has a single solution. In this particular example, the $b$ as a scalar was transformed into $bI$ the scale of the identity. In higher dimensions, it will not generalize that way, you just have to follow the vectorization, and the dimension of the new equation will be a square of the multiple of the respective dimensions.