Can we choose continuously the eigenvector corresponding to the maximal eigenvalue? Is there a continuous function $f:\text{Sym}_n \to \mathbb{R}^n$ ($\text{Sym}_n$ is the space of real $n \times n$ symmetric matrices),
such that for every $A \in \text{sym}_n$ : $f(A)$ is an eigenvector of $A$ corresponding to the maximal  eigenvalue of $A$. (They are all real since $A$ is symmetric).
If such a function cannot exist, can we do this if restrict ourselves to the (open) submanifold of symmetric positive definite matrices?
 A: No, there isn't (at least assuming $f(A)$ is supposed to always be nonzero and $n>1$).  For instance, let $n=2$ and consider the matrices $A_t=\begin{pmatrix} 1+t & 0 \\ 0 & 1\end{pmatrix}$.  For $t>0$, $f(A_t)$ must be a multiple of $(1,0)$, and for $t<0$, $f(A_t)$ must be a multiple of $(0,1)$.  By continuity of $f$, we get that $f(A_0)$ must be a multiple of both $(1,0)$ and $(0,1)$.  But the only way this can happen is if $f(A_0)=(0,0)$.
In fact, even if you restrict to symmetric matrices without repeated eigenvalues, there is still no such $f$.  Let us again take $n=2$, identify $\mathbb{R}^2$ with $\mathbb{C}$, and let $A_\theta$ be the matrix which has $e^{i\theta}$ and $e^{i(\theta+\pi/2)}$ as eigenvectors with eigenvalues $2$ and $1$, respectively.  Since $f(A_\theta)$ is continuous as a function of $\theta$ and must always be a nonzero multiple of $e^{i\theta}$, the function $g(\theta)= f(A_\theta)/e^{i\theta}$ is a continuous map $\mathbb{R}\to\mathbb{R}\setminus\{0\}$.  In particular, $g(\theta)$ must always have the same sign.  But $A_{\theta+\pi}=A_\theta$ so $g(\theta+\pi)=-g(\theta)$, so $g$ must change sign.  This is a contradiction.
