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What is the result of the method for multiple eigenvalues? Is there any case for which this method will not work altogether?

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    $\begingroup$ When you say "inverse iteration", are you referring to the "power method" applied to the inverse of the matrix to estimate the smallest eigenvalue by magnitude? $\endgroup$ – Wolfgang Bangerth Jun 29 at 22:26
  • $\begingroup$ I am referring to the "power method" applied to the matrix (A - \lambda * I)^-1, where \lambda is an approximation of some eigenvalue. The goal is to find eigenvector corresponding to the eigenvalue, which \lambda is an approximation of. $\endgroup$ – Maristo-Tero Jun 30 at 14:36
  • $\begingroup$ The concerns you have to have in this case are exactly the same as those you have for the power iteration applied to $A$ itself. That is, if the largest eigenvalue of $(A-\lambda I)^{1}$ is multiple (or there are multiple eigenvalues of the same magnitude), then you will only converge to some vector in the space spanned by the corresponding eigenvectors. $\endgroup$ – Wolfgang Bangerth Jun 30 at 21:49
  • $\begingroup$ @WolfgangBangerth thank you! I was assuming this was the case but I wasn't sure. $\endgroup$ – Maristo-Tero Jul 2 at 16:11
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I am going to share the answer I've got from my professor.

In the case of multiple eigenvalues, the exact analytical solution for the eigenvector contains a much greater degree of uncertainty than in the case of a non-multiple one.

For a non-multiple eigenvalue, the eigenvector is defined up to its length, while its direction is determined uniquely, and the normalization condition makes the solution unambiguous.

For multiple eigenvalues, there is uncertainty already in the direction of the eigenvector, and the higher the multiplicity, the greater the degree of freedom in the direction (in the limiting case of an n-fold eigenvalue — for example, an n-fold zero for a null matrix-the eigenvector is formally any vector). For a method of inverse iterations with intermediate normalization, there is no single limit vector that the method can converge to.

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  • $\begingroup$ To me this does not address the questions posed by the OP. $\endgroup$ – nicoguaro Jul 2 at 20:10

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