Solution to $A = BX + YC$ where $A$ is a square matrix of rank $n$, $B,C$ known, rank $m<n$

Question

I hope this isn't too trivial of a problem. I'm really struggling with it and I feel like it shouldn't be that difficult.

As stated in the title:

Given (full rank):
$A\in\mathbb{R}_{nxn}$
$B\in\mathbb{R}_{nxm}$
$C\in\mathbb{R}_{mxn}$
with $n> m\ge n/2$

Find:
$X\in\mathbb{R}_{mxn}$
$Y\in\mathbb{R}_{nxm}$

That satisfies

$A = BX + YC$

If we knew Y instead of C this would be a trivial problem.
What I understand:
Since $BX$ and $YC$ can have at most rank $m$ each, then at the very minimum, $m\ge n/2$.

Could anyone point me in the right direction?

Answer 1

Assume $n=2m$. Then it is not difficult to solve the equation. We rewrite it as follows: $\begin{pmatrix}A_1&A_2\\A_3&A_4\end{pmatrix}=\begin{pmatrix}B_1\\B_2\end{pmatrix}\begin{pmatrix}X_1&X_2\end{pmatrix}+\begin{pmatrix}Y_1\\Y_2\end{pmatrix}\begin{pmatrix}C_1&C_2\end{pmatrix}$, that is $A_1=B_1X_1+Y_1C_1,A_2=B_1X_2+Y_1C_2,A_3=B_2X_1+Y_2C_1,A_4=B_2X_2+Y_2C_2$. We assume that $B_1,B_2,C_1,C_2$ are invertible matrices.

EDIT. Thus $X_1=B_1^{-1}(A_1-Y_1C_1)=B_2^{-1}(A_3-Y_2C_1)$, that implies a relation of the form $Y_1=PY_2+Q$ (cf. the Dar's calculation below). In the same way, from the 2 other equations, we obtain a relation of the form: $Y_1=PY_2+R$, that implies $Q=R$ (otherwise, there are no solutions). This necessary condition is a relation linking $A,B,C$. The sequel is easy.

Answer 2

This is not possible. Take $n=2$, $m=1$, and $$ A=\pmatrix{1&0\\0&1},\ B= \pmatrix{0\\1}, \ C=\pmatrix{0 & 1}. $$ Then you want to find $X=\pmatrix{x_1&x_2},Y=\pmatrix{y_1\\y_2}$ to satisfy $$ \pmatrix{1&0\\0&1} = \pmatrix{0\\1}\pmatrix{x_1&x_2} +\pmatrix{y_1\\y_2}\pmatrix{0 & 1} = \pmatrix{0&y_1\\x_1&x_2+y_2}, $$ which is impossible.

The same is true if instead $C$ is unknown, and $Y$ is given with $Y=B$.