Do eigenvalues depend smoothly on the matrix elements of a diagonalizable matrix?

Suppose I have a matrix $M(t)$ whose matrix elements depend smoothly on a real parameter $t$. I also know that this matrix is diagonalizable for the $t$s I'm interested in. Can I say that its eigenvalues depend smoothly on $t$?

Thanks very much!

The answer is NO; more pecisely, even for the real symmetric matrices, a $C^{\infty}$ parmetrization $M(t)$ gives birth to (only) a continuous paremetrization of the eigenvalues.

Of course, if, for every $t$, the eigenvalues of $M(t)\in M_n$ are simple, then there are smooth local parametrizations of the spectrum: $\lambda_1(t),\cdots,\lambda_n(t)$. More generally, this property stands when the mutiplicy of the eigenvalues are locally constant. When the mutiplicity of some eigenvalue varies, then the storm is threatening...

Now we assume that we work in $S_n$, the set of symmetric real matrices of dimension $n$ and that the eigenvalues may be multiple.

For the $C^{\infty}$ parametrization, the standard counterexample is due to Rellich, but I have not it on hand; you can make one by using the function $\exp(-1/t^2)$.

There exists a precise result when the symmetric matrix depends analytically on one parameter.

$\textbf{Proposition.}$ Assume that $t\in\mathbb{R}\rightarrow M_t\in S_n$ is analytic. Then the eigenvalues and a basis of (unit length) eigenvectors of $M_t$ are globally analytically parametrizable (even if the eigenvalues present some mutiplicities; note that the natural ordering of the eigenvalues is not met).

$\textbf{Remark.}$ That works also when $t\in\mathbb{R}\rightarrow M_t\in S_n$ is smooth; we must add the condition that two continuous curves $(t\rightarrow \lambda_i(t),t\rightarrow \lambda_j(t))$ (where $(\lambda_i(t),\lambda_j(t))$ are any couple of eigenvalues of $M_t$) are the same or intersect only a finite number of times.