The fact that the entries of the matrix are real does seem to help. The state of the art is the following.
- The spectrum is continuous functions of $\xi$. However, it is not always possible to label the eigenvalues so that they individually are continuous functions.
- When the multiplicities $m_1,\ldots,m_r$ do not change as $\xi$ varies (no crossing of eigenvalues), then the eigenvalues are as smooth as the matrix. If the domain is simply connected, the eigenvalues may be labelled so as to be smooth functions.
- When the entries are analytic functions of a single variable ($k=1$) and the eigenvalues remain real, then the eigenvalues may be labelled so as to be analytic functions. However, in case of crossing, this nice labelling is not the obvious one (i.e. not $\lambda_1\le\lambda_1\le\cdots$). This become false for $k\ge 2$, as shown by the example $$\left(\begin{array}{cc} \xi_1 & \xi_2 \\ \xi_2 & -\xi_1 \end{array}\right).$$end{array}\right)\qquad\qquad (1).$$
- The situation is not that good concerning the eigenvectors. The following is called Petrowski's example, $$\left(\begin{array}{ccc} 0 & \xi_1 & \xi_1 \\ 0 & 0 & 0 \\ \xi_1 & 0 & \xi_2 \end{array}\right).$$ The eigenvalues are real for every $\xi$, distinct when $ \xi_1\ne0$. The matrix is diagonalisable for every $\xi$, but two eigenvectors have the same limit when $ \xi_1\rightarrow0$.
If the domain is not simply connected, you may have additional difficulties with eigenvectors. Take example (1) above, with $\xi$ running over the unit circle $S^1$. When you follow continuously a unit eigenvector $V(\xi)$, it is flipped (i.e. multiplied by $-1$) after one loop
