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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 $B$.

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    $\begingroup$ 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? $\endgroup$ – Bertrand Jan 15 at 13:32
  • $\begingroup$ @Bertrand Yes each $A_ij$ has the same dimension so the full matrix $A$ does have an even number of rows and columns. $\endgroup$ – eurocoder Jan 15 at 13:54
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    $\begingroup$ 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 $A$. $\endgroup$ – Bertrand Jan 15 at 14:11
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    $\begingroup$ With a permutation similarity, you can make $A$ block-diagonal with $2 \times 2$ blocks on the diagonal $\endgroup$ – Omnomnomnom Jan 15 at 17:19
  • $\begingroup$ 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. $\endgroup$ – piyush_sao Jan 15 at 20:26
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$\bigg(I - A(\omega)^{-1}B(\omega)\bigg)x = 0 \implies (A(\omega)-B(\omega) )x=0$

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

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  • $\begingroup$ 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 $B$. $\endgroup$ – eurocoder Jan 15 at 16:19

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