Solve the matrix equation $X = AX^T + B$ for $X$

Question

Consider the matrix equation

\begin{equation} X=AX^T+B, \end{equation}

where $X$ is an unknown square matrix. Is it possible to solve it analytically? Moreover, can a general solution be written down in terms of the matrices $A$ and $B$?

Answer 1

Assume $X$ is an $n \times n$ matrix.

If we rewrite the equation as:

$$ X - AX^T = B $$

then we have a system of $n^2$ linear equations in $n^2$ unknowns (the entries of $X$), so in general the existence of solutions $X$ can be computed by standard methods.

Answer 2

$\newcommand{\fvo}{1} \newcommand{\cdf}{\cdot & \cdot & \cdot & \cdot} \newcommand{\cdt}{\cdot & \cdot & \cdot} \newcommand{\cdo}{\cdot} \newcommand{\cdw}{\cdot} $

$\newcommand{\vec}{\text{vec}}$ $\newcommand{\lb}{\left(}$ $\newcommand{\rb}{\right)}$ $\newcommand{\unity}{\bf\text{1}}$ Use the superoperator formalism to rewrite $X=AX^T+B$ as

$$ \vec(X)=\lb\unity\otimes A \rb\vec (X^T) +\vec(B)\\ \vec(X)-\lb\unity\otimes A \rb\vec (X^T)=\vec(B)\\ \lb \lb\unity\otimes \unity\rb -\lb\unity\otimes A \rb \hat T\rb \vec (X)=\vec(B),\\ $$ where $\hat T $ describes the Transposition Superoperator. For $4\times 4$ matrices this looks like

\begin{equation} \widehat T := \left(\, \begin{smallmatrix} \fvo & \cdt & \cdf & \cdf & \cdf \\ \cdf & \fvo & \cdt & \cdf & \cdf \\ \cdf & \cdf & \fvo & \cdt & \cdf \\ \cdf & \cdf & \cdf & \fvo & \cdt \\ %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \cdo & \fvo & \cdw & \cdf & \cdf & \cdf \\ \cdf & \cdo & \fvo & \cdw & \cdf & \cdf \\ \cdf & \cdf & \cdo & \fvo & \cdw & \cdf \\ \cdf & \cdf & \cdf & \cdo & \fvo & \cdw \\ %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \cdw & \fvo & \cdo & \cdf & \cdf & \cdf \\ \cdf & \cdw & \fvo & \cdo & \cdf & \cdf \\ \cdf & \cdf & \cdw & \fvo & \cdo & \cdf \\ \cdf & \cdf & \cdf & \cdw & \fvo & \cdo \\ %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \cdt & \fvo & \cdf & \cdf & \cdf \\ \cdf & \cdt & \fvo & \cdf & \cdf \\ \cdf & \cdf & \cdt & \fvo & \cdf \\ \cdf & \cdf & \cdf & \cdt & \fvo \\ %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \\ \end{smallmatrix}\, \right) \end{equation} (see for example here eq. $(43)$).

If $\lb \lb\unity\otimes \unity\rb -\lb\unity\otimes A \rb \hat T\rb $ is invertible, you'll get $$ \vec (X)=\lb \lb\unity\otimes \unity\rb -\lb\unity\otimes A \rb \hat T\rb ^{-1}\vec(B) $$

Undo $\vec$ and you're done...