Differentiability of matrix Let $B(0,R)$ be an open ball in $\mathbb R^d$ and let $A:B(0,R)\to\mathscr{M}_{d\times d}(\mathbb R).$ Assume that for any $x\in B(0,R),$ $A(x)$ is symmetric with $0<\lambda_1(x)<\ldots<\lambda_d(x),$ where $\sigma(A(x))=\{\lambda_1,\ldots,\lambda_d\}.$
If $A\in C^1(B(0,R)),$ is there a selection of $P:B(0,R)\to\mathscr{M}_{d\times d}(\mathbb R),$ where for any $x\in B(0,R),$ $A(x)=P(x)D(x)P^t(x),$ such that $P,P^t\in C^1(B(0,R))?$ Does $D$ belong to $C^1(B(0,R))?$
 A: The implicit function theorem tells you that the eigenvalues, as long as they remain simple, are real analytic functions of the matrix. The implicit function theorem also tells you that there is a local real analytic choice of basis of unit length eigenvectors, unique up to signs, depending on the matrix, i.e the locus of symmetric matrices with simple eigenspaces has a real analytic covering space by the product of the orthogonal group and the choices of distinct eigenvalues. The sheets of the covering are the choices of signs on the eigenvectors. Any $C^k$ map $x \mapsto A(x)$ from any simply connected manifold lifts to a $C^k$ map to the covering space, with any initial condition at a single point.
Since there is some doubt, let's do the calculation. To each orthogonal matrix $P$ and diagonal matrix $D$ with all entries distinct, associate the matrix $A=f(P,D)=PDP^t$. Take some $\dot{P}$ in the Lie algebra of the orthogonal group, i.e. antisymmetric, and so $P\dot{P} \in T_P O(n)$ lies in the tangent space. Take some $\dot{D}$ diagonal matrix. So $$f'(P,D)(P\dot{P},\dot{D})=P\dot{P}DP^t + PD\dot{P}^tP^t+P\dot{D}P^t=P(\dot{P}D+D\dot{P}+\dot{D})P^t.$$
This vanishes if and only if $$\dot{P}D+D\dot{P}+\dot{D}=0.$$
Take off-diagonal entries: $0=\dot{P}_{ij}(D_j-D_i)$, and diagonal entries $\dot{D}_i=0$. Hence $f'(P,D)$ is injective, so $f$ is local real analytic diffeomorphism, so has real analytic local inverses. Each local inverse is defined in the same open set, corresponding to an ordering of eigenvalue $D_i$ and a direction of each eigenvector in its eigenspace. So the map $f$ is a covering map.
A: In general this is wrong. If $x$ is 1-dimensional there are positive results. 
See the following papers and references therein for this:


*

*Andreas Kriegl, Peter W. Michor, Armin Rainer: Denjoy-Carleman differentiable perturbation of polynomials and unbounded operators. Integral Equations and Operator Theory 71,3 (2011), 407-416. (pdf

*Andreas Kriegl, Peter W. Michor, Armin Rainer: Many parameter Hölder perturbation of unbounded operators. Math. Ann. 353, 2 (2012), 519-522. (pdf)
Example (due to Rellich):
\begin{align}
x_+(t) :&= \begin{pmatrix} \cos\frac1t \\ \sin\frac1t \end{pmatrix}, \quad
x_-(t) := \begin{pmatrix} -\sin\frac1t \\ \cos\frac1t \end{pmatrix}, \quad
\lambda_\pm(t) = \pm e^{-\frac1{t^2}},\\
A(t) :&= (x_+(t),x_-(t))
     \begin{pmatrix} \lambda_+(t) & 0 \\
                 0 & \lambda_-(t) \end{pmatrix}
      (x_+(t),x_-(t))^{-1} \\
&= e^{-\frac1{t^2}}\begin{pmatrix} \cos\frac2t & \sin\frac2t \\
                         \sin\frac2t & -\cos\frac2t 
                                \end{pmatrix}.
\end{align}
Here $t\mapsto A(t)$ and $t\mapsto \lambda_\pm(t)$ are smooth, whereas 
the eigenvectors cannot be chosen continuously.
Added:
Sorry, I overlooked that the eigenvalues are all simple. Then everything is true as shown in the answer of Ben McKay. For the eigenvectors one can also use the resolvent integral:
$$
P(\gamma,x) = \int_{\gamma} (A(x) - \lambda)^{-1} d\lambda
$$
which is real analytic in $x$ and a projection onto the sum of all eigenspaces for eigenvalues in the interior of the curve $\gamma$. If there is only one eigenvalue in the interior, then this is true locally around $x$, and $P(\gamma,x)v/\|P(\gamma,x)v\|$ is a smooth eigenvector, locally around $x$. 
