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I have the following problem:

$$\begin{bmatrix}A &B \\C& D\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix} = \begin{bmatrix}\lambda I_m & 0 \\ 0& \mu I_n\end{bmatrix}\begin{bmatrix}x \\y\end{bmatrix} $$

where $A$, $B$, $C$, $D$ are matrices, $I_m,I_n$ are identity matrices of sizes $m,n$, $x,y$ are vectors and $\lambda$ and $\mu$ are scalars.

How does one solve such a problem?

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  • 2
    $\begingroup$ That's not an eigenvalue problem. You will want $\lambda$ and $\mu$ to be the same. $\endgroup$ Oct 1, 2018 at 11:44
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    $\begingroup$ This looks like a two-parameter eigenvalue problem in a non-standard form. I would expect that it is possible to transform it to the standard form, under certain conditions. Google: two-parameter eigenvalue problem. See, for example, Philip A. Browne: Numerical Methods for Two Parameter Eigenvalue Problems. $\endgroup$
    – wim
    Oct 1, 2018 at 14:33
  • $\begingroup$ Thanks for the clarification @WolfgangBangerth .. yes, making $\mu,\lambda$ to be the same makes it a trivial eigenvalue problem, but I am particularly looking for the case when they are different. $\endgroup$
    – xadu
    Oct 3, 2018 at 4:06
  • $\begingroup$ @wim Thanks for the reference, though I don't think it is the same type of problem: as per the reference, a two parameter eigenvalue problem is defined as: $\begin{bmatrix} A_1 & 0 \\ 0 & A_2 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} \lambda B_1+\mu C_1 & 0 \\ 0 & \lambda B_2+\mu C_2 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix}$ which is different from the the one I posted (for one, there is no coupling between $x$ and $y$). $\endgroup$
    – xadu
    Oct 3, 2018 at 4:10
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    $\begingroup$ @udax: I see. With a standard eigenvalue problem one requires that $x\neq 0$. I think that in your case $x\neq 0$ and $y\neq 0$ is not sufficient. Otherwise the simple problem $\begin{bmatrix}1 &1 \\1& 1\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix} = \begin{bmatrix}\lambda \cdot 1 & 0 \\ 0& \mu \cdot 1\end{bmatrix}\begin{bmatrix}x \\y\end{bmatrix} $, would have infinitely many solutions for $\mu$ and $\lambda$. Probably $||x|| = ||y|| =1$ is a reasonable requirement here? $\endgroup$
    – wim
    Oct 3, 2018 at 8:40

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