8
$\begingroup$

If I have already accurately known the eigenvalue spectrum (i.e. all eigenvalues) of a matrix, is there any efficient numerical algorithm to compute all the eigenvectors corresponding to these eigenvalues?

I guess with the information about eigenvalues, there should be some quicker way to compute eigenvectors of the matrix compared with simply diagonalize it without any information.

Improve this question
$\endgroup$
1
  • $\begingroup$ I just had a thought, maybe eigen decomposition can be used ($PDP^{-1}=A$ thus $P=APD^{-1}$, where $P$ is a matrix containing an eigen vector in each column) I have no idea if this can be made into a converging method and if so whether it would converge faster than the already mentioned method. $\endgroup$
    – fibonatic
    Commented Mar 28, 2015 at 4:22

1 Answer 1

Reset to default
3
$\begingroup$

If you can invert the matrix, the simplest choice is shifted inverse iteration, which is just power iteration for $(\mu I - A)^{-1}$, where $\mu$ is some estimate of an eigenvalue whose eigenvector you want. Convergence speed depends on how close you set $\mu$ is to your desired eigenvalue and how close other eigenvalues are to $\mu$.

http://en.wikipedia.org/wiki/Inverse_iteration

Improve this answer
$\endgroup$
4
  • 1
    $\begingroup$ Inverse power iteration is a good idea. I just want to point out that you don't even need to invert $A$, just solve the system $(\mu I - A) v^{n+1} = v^{n}$ with $v^0$ being some guess (e.g.: a vector with each component set to one). $\endgroup$
    – Juan M. Bello-Rivas
    Commented Mar 26, 2015 at 2:12
  • 1
    $\begingroup$ Of course - one should (typically) never invert a matrix :). One thought that came to mind is that approximately solving the system with a known spectrum might be done cheaply using a matrix polynomial if the eigenvalues are clustered. $\endgroup$
    – Jesse Chan
    Commented Mar 26, 2015 at 2:53
  • $\begingroup$ If you want all eigenvectors (or even a significant fraction, $O(n)$ of them, let's say), this method costs $O(n^4)$, because it needs $n$ different LU factorizations, while throwing away the eigenvalues and starting from scratch would cost $O(n^3)$. $\endgroup$
    – Federico Poloni
    Commented Mar 28, 2015 at 8:39
  • $\begingroup$ @JuanM.Bello-Rivas Can you please tell me what is the name of the method? I.e., using $(\mu I-A)v^{n+1}=v^n$ to find eigenvectors. $\endgroup$
    – Tan
    Commented Jan 12, 2022 at 17:17

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.