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 $det \begin{pmatrix} A & B \\ C & D \end{pmatrix} = det\left ( A -BD^{-1}C \right )det(D)$