Differentiability of eigenvalues of positive-definite symmetric matrices Let $A\in M(n,\mathbb{R})$ be an invertible matrix. Consider the (real) eigenvalues $\lambda_1,\cdots,\lambda_n$, in increasing order, of the positive-definite symmetric matrix $A^t A$. We shall denote the eigenvalues as $\lambda_i(A)$. 
Question What can be said about the differentiability of the functions $\lambda_i:GL(n,\mathbb{R}) \to \mathbb{R}$? 
[We may assume that the domain is $GL^+(n,\mathbb{R})$ for differentiability/smoothness.]
Any reference for this or relevant results would be appreciated.
 A: For the best known positive results under mild hypotheses, you might want to look at 

Armin Rainer, Perturbation theory for normal operators, Trans. A.M.S., Volume 365, Number 10, October 2013, Pages 5545–5577 

A: In the open subset of $M_n(\mathbb{R})$ where the $\lambda_i$ are distinct, they are $C^{\infty}$ functions: this follows from the implicit function theorem.
On the other hand, when some eigenvalue has multiplicity $>1$ you don't get more than continuity. For example if $A=\begin{pmatrix} 0 & 1\\ 1 & t
\end{pmatrix}$ the largest $\lambda_i$ is $\dfrac{1}{2}\left(t^2+2 +|t|\sqrt{t^2+4}\right)$, which is not differentiable (as a function of $t$) at $t=0$.
A: The keyword is the Cartan decomposition in the theory of symmetric spaces.
In short, when an eigenvalue is simple (its multiplicity is $1$) it is locally an analytic function.  But when the eigenspace is degenerate (the multiplicity is greater than $1$), the eigenvalue function is not differentiable.  The problem is essentially one of choosing branches: if you try to deform the identity matrix, there is no consistent way to say which of the resulting distinct eigenvalues after deformation is the eigenvalue that you should have kept track of.
Let $K = \mathrm{O}(n)$, and let $A$ be the group of diagonal matrices with positive entries. You then have $G=KAK$ and if $g=k_1 a k_2$ then the eigenvalues of $g^\dagger g$ are exactly the squares of the eigenvalues of $a$.  The problem is that the decomposition is not unique: you can conjugate $a$ by a permutation matrix, and there will be problems when $a$ is fixed by a permutation matrix.
A: As mentionned by other answers, simple eigenvalues are $C^\infty$, while non-simple ones are not. Let me add however two important properties which you can find in Kato's book Perturbation theory of linear operator.
The first one is that each $\lambda_j$ is a Lipschitz function. This statement is still valid if you replace  ${\bf Sym}_n({\mathbb R})$ by a subspace $E\subset{\bf M}_n({\mathbb R})$ with the property that the eigenvalues are always real.
The second one is that if $t\mapsto A(t)$ is a smooth curve in ${\bf Sym}_n({\mathbb R})$, then there is a labelling of the eigenvalues $t\in{\cal V}\rightarrow(\mu_1(t),\ldots,\mu_n(t))$ such that each $\mu_j$ is smooth. Mind that this labelling does not respect the order between eigenvalues when the multiplicities vary. Mind also that this becomes false if we replace a curve by a surface.
A: Let us consider functions $A$ from (an open interval in) $\mathbb{R}$ into the set of symmetric real $n\times n$ matrices (Hermitian complex $n\times n$ matrices behave analogously).
If $A$ is given by $A(t) = diag(1+t,1-t)$, then the eigenvalue functions $\lambda_1,\lambda_2$ of $A$ with $\lambda_1\leq\lambda_2$ are $\lambda_1(t) = 1-|t|$ and $\lambda_2(t) = 1+|t|$, hence are not differentiable. Instead of differentiability of the ordered tuple of eigenvalues, we should therefore discuss the question whether there is a differentiable function $(\lambda_1,\dots,\lambda_n):\mathbb{R}\to\mathbb{R}^n$ that consists pointwise of the eigenvalues of $A$ counted with multiplicities (i.e.: can the eigenvalue functions be chosen differentiably?).
(I chose the $2\times2$ example $A$ to be pointwise positive definite for $t$ close to $0$, because this was asked for in the original question. But this is not relevant: every differentiability problem that can occur for any eigenvalue $\leq0$ can also occur for positive eigenvalues. Moreover, considering $A^tA$ instead of $A$ does not change any differentiability issue: If, for some pointwise positive definite $A$, the eigenvalues of $A^tA = A^2$ are not [resp. cannot be chosen] as regular as $A^tA$ is, then they are not [resp. cannot be chosen] as regular as $A$, because $A^tA$ and $A=\sqrt{A^tA}$ have the same regularity, due to the real-analyticity of $B\mapsto\sqrt{B}$.)
Some of the results of Alekseevsky/Kriegl/Losik/Michor: Choosing roots of polynomials smoothly and Kriegl/Michor: Differentiable perturbation of unbounded operators, or older results cited therein, are the following:


*

*If $A$ is $C^1$, then the eigenvalue functions $\lambda_1,\dots,\lambda_n$ can be chosen $C^1$ (cf. Kato: Perturbation theory for linear operators, §II.6.3, Theorem 6.8).

*If $A$ is real-analytic, then the eigenvalue functions (and also eigenvector functions) can be chosen real-analytically.

*If $A$ is $C^\infty$, then the eigenvalue functions can be chosen twice differentiably.

*Even if $A$ is $C^\infty$, the eigenvalue functions cannot always be chosen $C^2$ (example 7.4 in AKLM, first example in KM).

*Let $A$ be $C^\infty$. Consider the eigenvalue functions $\lambda_1,\dots,\lambda_n$ with $\lambda_1\leq\dots\leq\lambda_n$ (they are always continuous). Assume for all $i,j\in\{1,\dots,n\}$ that either $\lambda_i=\lambda_j$ or there is no $t\in\mathbb{R}$ at which the functions $\lambda_i,\lambda_j$ meet of infinite order. Then the eigenvalue functions (and also eigenvector functions) can be chosen $C^\infty$.

A: In this paper by Xuwen Zhu it is shown that, after resolution by radial blow-ups, the eigenvalues can be made to be smooth:
https://arxiv.org/abs/1504.07581.
