Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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..

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

share|improve this question
3  
What you have there is a Sylvester equation for which many solution methods are known. –  Guess who it is. Aug 25 '12 at 3:21
1  
Are in interested in solving the matrix equation to solve the differential equation, or in its own right? –  Sasha Aug 25 '12 at 3:22
1  
I think this question was asked here on math.SE before, but I can't find the link. –  user2468 Aug 25 '12 at 3:29
    
"I don't even understand a single word in Wikipedia" - fine... do you at least know what a Kronecker product is? Or, better yet, what part of the Wiki article do you not understand? –  Guess who it is. Aug 25 '12 at 13:27
    
@JenniferDylan: I think the closest thing is this, but it's not exactly a duplicate, since it has no free term: math.stackexchange.com/questions/39906/… –  tomasz Aug 25 '12 at 15:21

2 Answers 2

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.

share|improve this answer
    
This is incorrect. $AX+XB$ is $[A,I]\begin{bmatrix}X\\ &X\end{bmatrix}\begin{bmatrix}I\\ B\end{bmatrix}$, not $[A,I]X[I,B]^T$. Note that $[A,I]X$ is not a legitimate matrix product, as $[A,I]$ has $2n$ columns but $X$ has $n$ rows. –  user1551 Jan 16 at 12:36
    
Yeah, I made a mistake. Thank you for your pointing out. –  olivia Jan 18 at 12:19

Use the superoperator formalism: $$ AXB=C \mapsto ( {B}^T\otimes A)\text{vec} X = C $$ (see here for a definition of $\text{vec}(X)\;$ and here for more information: Kronecker product)

Rewrite your equation (using $AXB\to (B^T\otimes A) \text{vec}\;X$) to $$ AX+XB=C \to (1\otimes A)\text{vec}\;X + (B^T\otimes 1)\text{vec}\;X=\text{vec}\;C\\ \text{vec}\;X=\left((1\otimes A) + (B^T\otimes 1)\right)^{-1}\text{vec}\;C, $$ assuming that the last inverse exists...

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.