I have matrix equation in the form $y = Ax$, where $y=\begin{pmatrix} y'_1 \\ y'_2 \\ \vdots \\ y'_M \\ y''_1 \\ y''_2 \\ \vdots \\ y''_M \end{pmatrix}; $$ and, $$ x = \begin{pmatrix} x'_1 \\ x'_2 \\ \vdots \\ x'_N \\ x''_1 \\ x''_2 \\ \vdots \\ x''_N \end{pmatrix}$ and $A$ is block diagonal matrix with the block size MxN.
Is there a way to obtain equivalent representation where $y=\begin{pmatrix} y'_1 & y''_1 \\ y'_2 & y''_2 \\ \vdots & \vdots \\ y'_M & y''_M\\ \end{pmatrix} $
and $x = \begin{pmatrix} x'_1 & x''_1 \\ x'_2 & x''_2 \\ \vdots & \vdots \\ x'_N & x''_N \\ \end{pmatrix}$
With a suitable transformation on $A$?
I've seen the operator that transform the "x" vector in that way is a type of generalized rvec or devec (reverse vec operator). And they are intimately related with Kronecker and tensor product.
The original matrix can be written as block matrix $$ A =\left( \begin{matrix} C & D \\ E & F \end{matrix} \right) $$ Then $$ A x = \left( \begin{matrix} C & D \\ E & F \end{matrix} \right) \left( \begin{matrix} x'\\ x'' \end{matrix} \right) = \left( \begin{matrix} y'\\ y'' \end{matrix} \right) = y $$ so $$ (C \, D) x = y' \\ (E \, F) x = y'' $$ We are looking for $B$ with $$ B (x' \, x'') = (y' \, y'') $$ This seems possible for the special case $D = E = 0$ and $B = C = F$.