Is there some condition guarantees continuous hermitian matrix $H(t)$ has continuous eigenvectors?

Asked Jul 17 '22 at 09:49

Active Jul 17 '22 at 10:35

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Suppose we have a hermitian matrix $H(t)$ continuously depend on parameter $t$, it's easy to know that the eigenvalue $\lambda_i(t)$ of it is continuous$^1$, but what about eigenvectors of it? Is there some theorem tells us that in what condition of $H(t)$, the continuous eigenvectors will be guaranteed?

This question is duplicated with this one while I think it will be better answered here.

^{1. See this question.}

edited Jul 17 '22 at 10:35

asked Jul 17 '22 at 09:49

narip

I'd like not to score from this question. XD – narip Jul 17 '22 at 10:37
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A sufficient for "continuity of eigenvectors", i.e. the existence of a continuous paremeterization of the eigenvectors, is that $H(t)$ has distinct eigenvalues for all $t$. – Ben Grossmann Jul 17 '22 at 16:49
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I have added another (self contained answer) to this one Perhaps it can be helpful for solving your question. – Ryszard Szwarc Jul 17 '22 at 16:50

Is there some condition guarantees continuous hermitian matrix $H(t)$ has continuous eigenvectors?

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