Distinguishing between the different eigenvalues

Question

Consider the symmetric matrix $$A=\begin{pmatrix} 2 & t & \cos t-1 \\ t & 2 & 0 \\ \cos t-1 & 0 & 2 \end{pmatrix}. $$ The (real) eigenvalues of $A$ can be found easily using the quadratic formula: they are $$\lambda_1,\lambda_2,\lambda_3=2,\frac{1}{2} \left(4-\sqrt{2} \sqrt{2 t^2-4 \cos (t)+\cos (2 t)+3}\right),\frac{1}{2} \left(4+\sqrt{2} \sqrt{2 t^2-4 \cos (t)+\cos (2 t)+3}\right) $$

Plotting them as functions of the real parameter $t$, I obtain this graph

What bothers me, is that we could have defined the eigenvalues differently as $$\lambda_1,\lambda_2,\lambda_3=2,\frac{1}{2} \left(4-\text{sgn } t\sqrt{2} \sqrt{2 t^2-4 \cos (t)+\cos (2 t)+3}\right),\frac{1}{2} \left(4+ \text{sgn } t\sqrt{2} \sqrt{2 t^2-4 \cos (t)+\cos (2 t)+3}\right), $$ so that the graph would look like this:

This way, the functions $\lambda_1(t),\lambda_2(t),\lambda_3(t)$ are of class $C^\infty$ (smooth).

I have a few questions regarding this:

1. Given a symmetric matrix which depends (smoothly) on a real parameter $t$, is it possible to define its eigenvalues $\lambda_1(t), \dots, \lambda_n(t)$ is such a way that they will be smooth functions of $t$?
2. Is there a natural way of finding these "smooth branches"? (The quadratic formula has failed me.)
3. A little more generally, given the equation $f(x,t)=0$, under what conditions is there a choice for all solutions $x(t)$, so that they are as regular as $f$ is WRT $t$?

Thank you

Answer 1

See a related thread in MO. To quote the answer of Denis Serre,

If the symmetric matrix depends analytically upon one parameter, then you can follow analytically its eigenvalues and its eigenvectors. Notice that this requires sometimes that the eigenvalues cross. When this happens, the largest eigenvalues, as the maximum of smooth functions, is only Lipschitz.

This should answer your first question. The answers by the others are also informative.