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.

  • $\begingroup$ 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 Mar 28 '15 at 4:22

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 $\mu$.


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    $\begingroup$ 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 Mar 26 '15 at 2:12
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    $\begingroup$ 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 Mar 26 '15 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 Mar 28 '15 at 8:39

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