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.