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Here, https://math.stackexchange.com/a/1146455, it is noted that eigenprojections are continuous, but eigenvectors are not. Are there any conditions where the eigenvalues are not simple, but the eigenvectors are continuous?

For example, assume we're using the $\infty$ norm and the eigenvectors are assumed to be normalized to some constant $c$. If $v_P$ and $v_Q$ are the first eigenvectors of $P$ and $Q$, then $||v_P−v_Q||_{\infty}$ is always less than some constant, for example, $c+1$. Can we claim continuity in this case?

EDIT:

To clarify, given that the eigenprojections of $P$ and $Q$, call them $R_P$ and $R_Q$, are continuous functions of $P$ and $Q$, doesn't this imply continuity of the eigenvectors under certain conditions? For example, if we note that $||R_P-R_Q||_{\infty} < \delta$, for some constant $\delta$, then this implies a constant bound on $||v_P-v_Q||_{\infty}$ as well. While continuity may not be the best term to use to describe their relationship with $P$ and $Q$, all the eigenvectors are "close".

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  • $\begingroup$ What do you mean by "the first eigenvectors of $P$ and $Q$"? The point is that the eigenprojections are uniquely determined by the symmetric matrix but unless we assume that the eigenspaces are all one dimensional then the eigenvectors are not, even after normalization. $\endgroup$ Commented Apr 20, 2016 at 22:33
  • $\begingroup$ I mean an eigenvector associated with the largest eigenvalue. In the case where this eigenvalue has algebraic multiplicity >1, you would just pick one of the associated eigenvectors. $\endgroup$
    – billbob
    Commented Apr 20, 2016 at 23:07
  • $\begingroup$ In the case where the multiplicity is bigger than 1, then which one are you going to pick? You are asking if some function is continuous, but you haven't actually defined a function yet. Come to think of it, the question doesn't even make sense if the multiplicity is equal to 1: even if you impose the constraint that the eigenvector have norm 1, there are still two choices (which differ by a minus sign). $\endgroup$ Commented Apr 21, 2016 at 17:52

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