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Let $A(x)\in\mathbb{R}^{n\times n}$ be a real symmetric matrix depending on the point $x\in\mathbb{R}^n$, where the eigenvalues are not necessarily simple. Can we say that for all $x$ there exists an associated neighborhood $U$ such that the eigenvalues and eigenvectors of $A$ are differentiable w.r.t each $x_i$ in that neighborhood?

I know there are some similar questions on here (e.g. 1 and 2), which have good responses and point out known results in texts such as Kato's Perturbation Theory for Linear Operators. The conditions for these results are stronger than I would hope are necessary.

I believe my question is different because I am only looking for local differentiability of a simpler operator. In particular, I am interested in the last response to 1, which says the eigenvalues and eigenvectors depend analytically on the entries of $A$ in a region where the multiplicites of the eigenvlaues are constant. Is this true? This seems to imply, perhaps with an extra regularity condition, local differentiability of the eigenstuff.

Update: Theorem 1.1.E in this reference gives the sort of conditions I was looking for, $A(x_i)\in C^{1,\alpha}$, but only in the case of eigenvalues. Their example 8.1 shows that this is in general not enough for the differentiability of the eigenvectors.

However, it is still unclear if differentiability of the eigenvectors holds under similar conditions in a neighborhood where the eigenvalue multiplicities are constant, i.e. 1.

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  • $\begingroup$ There is also this answer that is relevant. $\endgroup$
    – username
    Commented Jun 24, 2022 at 15:55
  • $\begingroup$ The example used in the linked answer is the same as that which I reference in my update, so the question still stands when the multiplicites of the eigenvalues are constant in some region. $\endgroup$
    – RS-Coop
    Commented Jun 24, 2022 at 18:00

1 Answer 1

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Theorem (1.1) of Perturbation theory for normal operators is likely what you are looking for.

See also Differentiable perturbation of unbounded operators.

If you would order the eigenvalues by their magnitude they may well not be differentiable, for example the largest eigenvalue of the matrix $\text{diag}\,(1+x,1-x)$ is $1+|x|$, which is not differentiable at $x=0$. But a differentiable parameterization exists under mild conditions.

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  • $\begingroup$ thanks for the response! Yes, part E of that theorem does seem to be what I am asking for, at least in terms eigenvalues, but what about eigenvectors? In your example it seems the issue at 0 is that there is no neighborhood where the multiplicity is constant. I don't need a specific ordering of the eigenvalues, but I it is unclear to me how that helps the isse at 0. What is meant by this parameterization in general? $\endgroup$
    – RS-Coop
    Commented Jun 22, 2022 at 15:42
  • $\begingroup$ theorem B in the second reference I added addresses both eigenvalues and eigenvectors, is that sufficient? $\endgroup$
    – Carlo Beenakker
    Commented Jun 22, 2022 at 15:50
  • $\begingroup$ I am not familiar with the space $C^{[M]}$, or this quasianalytic condition, so I am not quite sure. I was hoping for a condition like that in part E, i.e. $C^{1,\alpha}$. $\endgroup$
    – RS-Coop
    Commented Jun 22, 2022 at 16:07

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