$A,B,C \in M_n$; $AXB=C$; if $A$ or $B$ singular, a solution $X$ exist iff $\text{rank}\{B^T\otimes A\}=\text{rank}\{B^T\otimes A\,\text{ vec}(C)\}$

Question

Let $A,B,C \in M_n$ and the equation $AXB=C$. If either $A$ or $B$ is singular, then a solution $X$ can exist if and only if $\operatorname{rank} \left\{ B^T \otimes A \right\} = \operatorname{rank} \left\{ B^T \otimes A \operatorname{vec}(C ) \right\}$, where $\otimes$ is a Kronecker product.

Thank you so much in advance.

Answer 1

Hint: the key is to note that $$ (B^T \otimes A) \operatorname{vec}(X) = \operatorname{vec}(AXB) $$