Using the shifted power method I find the eigenvalue (of the matrix A) farthest from a number $\mu$ and the corresponding eigenvector .

In the method I follow the below steps:

I first compute the matrix $A-\mu I$ which has the eigenvalues $\{ \lambda_{j}-\mu \}^{n}_{j=1}$ since $Ax = \lambda_{j} x \Leftrightarrow (A-\mu I)x = (\lambda_{j}-\mu)x$

Using the power method on $A-\mu I$ I obtain the eigenvalue farthest from $\mu$ (i.e $\vert \lambda_{j}-\mu \vert$). My question is: Why? I know that the power method find the biggest eigenvalue $\lambda_{j}-\mu$, so how can I conclude that it finds the biggest absolute eigenvalue?

  • 5
    $\begingroup$ You say you know that the power method finds the "biggest eigenvalue", but what you should know is that this "biggest" means the one with largest absolute value. $\endgroup$ – Robert Israel Oct 27 '11 at 19:44
  • $\begingroup$ @RobertIsrael Isn't he using though the "shift power method", which from what I've read it's not just a way of finding the largest eigenvalue. $\endgroup$ – nbro Oct 3 '16 at 20:48
  • $\begingroup$ Actually if you choose $\mu$ wisely you can find any eigenvalue. $\endgroup$ – Leo Lerena Oct 26 '17 at 20:41

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