I have recently come across this system of three matrix equations in three unknown with three matrix unknowns (all matrices are real):
We define the matrix $ X $ a square positive definite matrix and the matrix $Y$ as a rectangular real matrix, and $ \alpha > 0 $ is a positive constant, we have the three equations:
$ -B^tX +B^tBAX + \alpha [C^tCAX-C^tY] = 0 $
$ -XA^t + BAXA^t = 0 $
$ CAXA^t -YA^t = 0 $
Where the variables are $ A,B,C $ (we assume dimensions allow multiplication) and we wish to solve for $ A,B,C $ generally. If we assume all three matrices are square and invertible, we may simply solve directly, and yet generally I have no clue on this. Perhaps the pseudoinverse will work here, but I cannot figure out how, also I figured may it is to do with eigenvalue problems, but again I cannot figure it out, maybe matrix decomposition but still I cannot see how. I certainly appreciate all help on this.
Here the unknowns real matrices are $A\in M_{p,n},B\in M_{n,p},C\in M_{q,p}$ and the known real matrices are $X\in M_n,Y\in M_{q,n}$.
There is the trivial solution $CAX=Y,BAX=X$, that implies $CA=YX^{-1},BA=I_n$.
Necessarily $n\leq p,rank(A)=rank(B)=n$, otherwise this trivial solution does not exist. Let $A\in M_{p,n}$ be any matrix with $rank(A)=n$.
Then $A^+=(A^TA)^{-1}A^T$ and we can write $B,C$ as functions of $A$:
$B=A^++W(I_p-AA^+)$ where $W$ is any matrix in $M_{n,p}$.
$C=YX^{-1}A^++Z(I_p-AA^+)$ where $Z$ is any matrix in $M_{q,p}$.
EDIT. If $rank(A)=n$, then $A^T$ is a surjection and, necessarily, $BAX=X,CAX=Y$.
If $rank(A)<n$, then it is more difficult. For example, if $A=0$, then $B^TX+\alpha C^TY=0$ and we obtain $B$ as a function of $C$.