7

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?

nicoguaro
  • 8,500
  • 6
  • 23
  • 49
xadu
  • 171
  • 2
  • 2
    That's not an eigenvalue problem. You will want $\lambda$ and $\mu$ to be the same. – Wolfgang Bangerth Oct 01 '18 at 11:44
  • 3
    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. – wim Oct 01 '18 at 14:33
  • 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. – xadu Oct 03 '18 at 04:06
  • @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 $y$). – xadu Oct 03 '18 at 04:10
  • 2
    @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? – wim Oct 03 '18 at 08:40
  • @udax The problem can be transformed into a generalized eigenvalue problem for one of the parameters fixed ($\lambda$ or $\mu$) and there is an inverse for either B or C. – Bort Oct 04 '18 at 16:51
  • @Bort Yes, that is possible. The only downside is then one needs to do a parameter sweep over the fixed parameter to get the full range of allowed $\lambda,\mu$. – xadu Oct 06 '18 at 16:16
  • @wim that's an interesting observation. I don't have a good response so far. – xadu Oct 06 '18 at 16:17
  • Are you sure there's no transformation you can perform that would separate the equations? Something like in this paper: https://journals.aps.org/prb/abstract/10.1103/PhysRevB.43.9649 – Yuriy S Oct 10 '18 at 14:56
  • How many solutions would you like to find? Because there would be infinity many, namely you find $n+m$ eigenvalues for every constraint that $\mu=\alpha,\lambda$ where $\alpha$ can be any scalar. Not sure if $\alpha$ has to be non-zero, but either way there are infinite possibilities for $\alpha$. So without more constraints the number of solutions unbounded. – fibonatic Oct 10 '18 at 18:29

0 Answers0