Are eigenvector continuous? Lets consider a matrix $A\in \mathbb R^{d\times d}$, where $d=2$ or $3$,
$\ A$ has entries $a_{ij}(x)$, where $a_{ij}\in C^1(\mathbb{R})$. Also, $A$ is positive definite. Given the eigenvalues $\lambda_i$ of $A$ are the values that satisfy
$$Av_i = \lambda_i v_i,$$
where $v_i$ is the corresponding eigenvector. I think the derivative 
$$\frac{\mathrm d\lambda_i}{\mathrm dx}$$
exists, since $\lambda_i$ are the roots of a polynomial with coefficients that are again multivariate polynomials in $a_{ij}$, therefore being continuous in $\lambda_i$.
What I was wondering was if the eigenvectors $v_i$ are as well. Usually the eigenvectors are all vectors which satisfy 
$$ (A-\lambda_iI)v_i=0,$$
but the matrix $(A-\lambda_iI)$ is singular, so I don't know whether the eigenvectors are still continuous. 
 A: Let $s$ denote $\frac{\sqrt{2}}{2}$, and let 
$$
M = \pmatrix{s & -s \\ s & s}
$$
be rotation by $\pi/4$. 
Let 
$$
X(u) = \pmatrix{u & 0 \\ 0 & 1}.
$$
Then for $u \ne 1$, the matrix 
$$
M X(u) M^t
$$
has eigenvectors in the directions $\pmatrix{\pm s\\s}$ (the "diagonal directions") while the matrix $X(u)$ has eigenvectors in the directions $\pmatrix{1\\0}$ and $\pmatrix{0\\1}$ (the "axial directions"). 
Now consider
$$
H(t) = \begin{cases}
M X(1 + t^4) M^t & t < 0 \\
X(1 + t^4) & t \ge 0
\end{cases}
$$
Then $H$ has the form required by your problem (although my variable is $t$ rather than $x$), but for $t > 0$, the eigenvectors are axial, and for $t < 0$, the eigenvectors are diagonal, and no choice of eigenvectors for $t = 0$, where the matrix is the identity and every vector is an eigenvector, will make either eigenvector a continuous (let alone differentiable) function of $t$. 
(I admit I haven't written out every detail here, and maybe I need to put in something like $\exp(-\frac{1}{t^2})$ instead of $t^4$ to make everything smooth....but I don't think so). 
