How to solve the Sylvester matrix equation $AX+XB=C$ for $X$?

Question

How does one solve the matrix equation $AX+XB=C$ for $X$? It doesn't seem too difficult. I tried many times but failed.

I'm an adult student... I am now vexed about Gilbert Strang - An Introduction to Linear Algebra. I don't even understand a single word in Wikipedia: Sylvester equation. If you have ever use some nice workable materials or lecture notes? You can generously upload and share the links of the lecture notes and assignments. Different subjects/ topics are welcome, as long as you deem they are nice and workable.

The problem origins from a system of diff equation, using undetermined coefficients (matrix) to find the particular solution. Try $y_p=X\begin{pmatrix} e^{\alpha t} \\ e^{\beta t} \end{pmatrix}$

$\dot{y}+Ay=C\begin{pmatrix} e^{\alpha t} \\ e^{\beta t}\end{pmatrix}$

$\dot{y_p}=X\begin{pmatrix} \alpha & 0 \\ 0 & \beta \end{pmatrix} \begin{pmatrix}e^{\alpha t} \\ e^{\beta t}\end{pmatrix}$

substitute $\dot{y_p}$ and $y_p$ into the original differential equation..

$X\begin{pmatrix} \alpha & 0 \\ 0 & \beta \end{pmatrix}+AX=C$

Answer 1

We need to solve $AX + XB = C$ given that $A(n \times n)$, $B(m \times m)$ and $X(n \times m)$.

$vec(AXB) = ({B}^T\otimes A)\text{vec} X$

(see here for a definition of $\text{vec}(X)\;$ and here for more information: Kronecker product)

Note that $AX = AXI_m$ where $I_{m}$ is the m x m identity matrix.

$vec(AXI_m) = {(I_m}\otimes A)\text{vec} X$

Similarly, note that $XB = I_nXB$.

$vec(I_nXB) = (B^{T}\otimes I_n)\text{vec}(X)$

Taking the vec of on both sides of $AX + XB = C$ we get $\text{vec}(AX + XB) = \text{vec}(C)$.

Therefore $$ vec(AX + XB) \\ = vec(AX) + vec(XB) \\ = (I_{m}\otimes A)\text{vec} X + (B^{T}\otimes I_n)\text{vec}(X)\\ = ((I_{m}\otimes A) + (B^{T}\otimes I_n))\text{vec}(X) $$

Hence $((I_{m}\otimes A) + (B^{T}\otimes I_n))\text{vec}(X) = \text{vec}(C)$ Taking the inverse we get: $$ \text{vec}\;X=\left((I_{m}\otimes A) + (B^T\otimes I_{n})\right)^{-1}\text{vec}\;C, $$ assuming that the last inverse exists...

Answer 2

The matrix equation that prompted this post, $$X\begin{pmatrix} \alpha & 0 \\ 0 & \beta \end{pmatrix}+AX=C,$$ actually has a very easy solution.

Let $X$ and $C$ have columns $X_1,X_2$ and $C_1,C_2$, respectively. Then $$X_1=(A+\alpha I)^{-1}C_1,X_2=(A+\beta I)^{-1}C_2.$$

Footnote

A nice feature of this solution is that the condition for the existence and uniqueness of a solution to a Sylvester Equation is self evident in this simple case :-

$A$ and $-B$ must have no common eigenvalue.

Answer 3

It seems $AX+XB=[A,I_1]X[I_2,B]^T$, where $I_1$ and $I_2$ are identity matrices and subscript 'T' is transpose of a matrix. Thus, the problem becomes how to solve $[A,I_1]X[I_2,B]^T = C $. Obviously, $X=[A,I_1]^\dagger C {[I_2,B]^T}^\dagger$, where subscript '$\dagger$' means the Pseudo-inverse.