# 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.

1. 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.
2. 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 $$B$$.

• So each block $A_{ij}$ is assumed to have the same dimension? (Consequence of assuming $A_{12}$ diagonal.) And the matrix $A$ necessarily has an even number of rows and columns? – Bertrand Jan 15 at 13:32
• @Bertrand Yes each $A_ij$ has the same dimension so the full matrix $A$ does have an even number of rows and columns. – eurocoder Jan 15 at 13:54
• OK, yes in this case, the matrix $A^{-1}$ is block-diagonal as a result of matrix block-inversion. So $A^{-1}$ inherits the same structure as $A$. – Bertrand Jan 15 at 14:11
• With a permutation similarity, you can make $A$ block-diagonal with $2 \times 2$ blocks on the diagonal – Omnomnomnom Jan 15 at 17:19
• I don't understand. $A$ is already easily invertible, so how does having $\hat{A}$ helps? $A$ can be permuted to almost diagonal matrix, i.e. matrix with $dim(A_{12})$ block diagonals of size 2x2. – piyush_sao Jan 15 at 20:26

$$\bigg(I - A(\omega)^{-1}B(\omega)\bigg)x = 0 \implies (A(\omega)-B(\omega) )x=0$$
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 $$det \begin{pmatrix} A & B \\ C & D \end{pmatrix} = det\left ( A -BD^{-1}C \right )det(D)$$
• I had already considered the determinant formula for block matrices but it doesn't seem to lead anywhere, i.e. it doesn't give us an opportunity to exploit the structure of $A$ and $B$. – eurocoder Jan 15 at 16:19