I have an equation that is a bit similar to a Sylvester equation.
The equations is $AXB^T+X=E$, where all variables are matrices.
I could try to inverse $B$ and rewrite the equation as $AX+XB^{-T}=EB^{-T}$, which is then the standard form for the Sylvester equation, but I need to invert $B$, which is usually not a good idea.
Is there a more standard approach to work with this type of equation?
Edit:
I can see by doing simple algebra that $X=E-AEB^T+A^2EB^{2T}-A^3EB^{3T}+\dots$ seems to be a solution of the equation $AXB^T+X=E$, if the term $A^nEB^{nT}\to0$ as $n\to\infty$. Would there be a simple way to compute this infinite sum?
Edit 2:
I worked on some potential factorization and I realized that in my problem, $A$ and $B$ are actually symmetrical and positive-definite matrices. The matrix $E$ is not symmetrical (and I don't expect the solution $X$ to be symmetrical either).
So with this information, I rewrite the matrices as $A=P_AD_AP_A^T$ and $B=P_BD_BP_B^T$, which means: $$P_AD_AP_A^TXP_BD_BP_B^T+X=E$$ $$D_AYD_B+Y=F$$ with $Y=P_A^TXP_B$ and $F=P_A^TEP_B$. So the matrix $Y$ can be computed as: $$y_{ij}(1+a_ib_j)=f_{ij}$$ where $a_i$ and $b_j$ are the diagonal entries of $D_A$ and $D_B$ respectively. This equation has stable solutions as $a_i>0$ and $b_j>0$. Then $X=P_AYP_B^T$. So it requires 2 diagonalisations and 4 matrix multiplications, plus one scalar equation per entry.
Is there another matrix factorization approach that could be more efficient here?