Can we simplify $ A^{-1}Bx = x$ where $A$ is a block matrix with each block being diagonal and half the blocks of $B$ are zero? I have the following eigenvalue problem involving block matrices $A$ and $B$:
$$
A^{-1}Bx = x. \quad \quad \quad \quad (*)
$$
$A$ and $B$ have special structures. I would like to reduce/simplify this system to a nicer/alternative form.


*

*Structure of $A$:
$$
A =
\begin{bmatrix}
A_{11} & A_{12} \\
A_{21} & A_{22}
\end{bmatrix}
$$
where each block $A_{ij}$ is a diagonal matrix.

*Structure of $B$:
$$
B =
\begin{bmatrix}
B_{11} & 0 \\
B_{21} & 0
\end{bmatrix}
$$
Initial thoughts: As the blocks of $A$ are diagonal, and hence simple to invert, it would be great if we could somehow rearrange the system so that instead of $A^{-1}Bx = x$ we have something like $\hat A^{-1} \hat B x = x$ with
$$
\hat A =
\begin{bmatrix}
A_{11} & 0 & 0 & 0 \\
0 & A_{12} & 0 & 0 \\
0 & 0 & A_{21} & 0 \\
0 & 0 & 0 & A_{22}
\end{bmatrix}.
$$
Questions:


*

*Can we re-arrange the system as proposed above? What would $\hat B$ need to be so that the new system corresponds exactly to the original one $(*)$?

*Are there other was of exploiting the structures of $A$ and $B$ such that we can get nice or alternative representations for the problem $(*)$?


Extra note: I am actually dealing with a non-linear eigenvalue problem: Finding $\omega$ such that $\bigg(I - A(\omega)^{-1}B(\omega)\bigg)x = 0$ has a non-trivial solution. My main concern at the moment is somehow exploiting the structures of $A$ and $B$.

$\bigg(I - A(\omega)^{-1}B(\omega)\bigg)x = 0 \implies (A(\omega)-B(\omega) )x=0$
so $A(\omega)-B(\omega)$ has 0 as an eigenvalue; so find an $\omega$ such that $det(A(\omega)-B(\omega))=0$ 
Now you can use $det  \begin{pmatrix}
A & B \\ 
C & D
\end{pmatrix}   =  det\left ( A -BD^{-1}C \right )det(D)$
