7
$\begingroup$

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?

Improve this question
$\endgroup$
10
  • 2
    $\begingroup$ That's not an eigenvalue problem. You will want $\lambda$ and $\mu$ to be the same. $\endgroup$
    – Wolfgang Bangerth
    Commented Oct 1, 2018 at 11:44
  • 3
    $\begingroup$ 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
    Commented 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
    Commented 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
    Commented Oct 3, 2018 at 4:10
  • 2
    $\begingroup$ @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
    Commented Oct 3, 2018 at 8:40

0

Reset to default

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Browse other questions tagged or ask your own question.