Eigenvalues of symmetric matrices

Suppose $A=A(x)$ is a smooth (or continuous) family of symmetric $n\times n$ (real) matrices on an open set $\Omega \subset \mathbb{R}^n$, is it true that locally there exists a smooth (continuous) family $\lambda_1 (x), \cdots, \lambda_n(x)$ which are the eigenvalues of $A(x)$?

Answer: For the continuous case, true perhaps? YES!

Proof. (Sketch, still can’t do the smooth case)

As the eigenvalues of $A$ are the roots of the characteristic polynomial $p(t)=\det(A-t I)$, we only have to show the continous dependence of the roots of polynomials on their coefficients. To show this, let us work in $\mathbb{C}$ first.  Suppose $p(z)=a_0+a_1 z+\cdots a_nz^n\in \mathbb{C}[z]$, $a_n\neq 0$, be a degree n polynomial. Suppose $\lambda_i$ are the roots of $p(z)$ (which may not be distinct). Consider one of the roots, say, $\lambda_1$, and choose a circle $\gamma$ around $\lambda_1$, so small that it does not encloses other roots which are distinct from $\lambda_1$, and does not contain any other roots of $p(z)$ on it. For another degree $n$ polynomial $q(z)$ close enough to $p(z)$ (depending on $\gamma$ and $p(z)$), we must have, on $\gamma$,

$\left|\frac{q(z)}{p(z)}-1\right|<1.$

So by Rouche’s theorem, if $q(z)$ is close enough to $p(z)$, then there must be a root of $q(z)$ around each root of $p(z)$ (counting multiplicity). As the number of roots of $q(z)$ must be exactly $n$, this shows that for a polynomial $p(z)$, for any $\epsilon >0$, there exists $\delta$ such that if $q(z)=b_0+\cdots +b_n z^n$ with $|b_i-a_i|<\delta$, for any root $\lambda$ of $p(z)$, there exists a root $\tilde \lambda$ of $q(z)$ such that $|\tilde \lambda-\lambda|<\epsilon$.

Note that for any symmetric $n\times n$ matrix $B$, its characteristic must have $n$ real roots. So setting the polynomial $q(z)$ above to be the characteristic polynomial of $B$ (and $p(z)$ to be the real characteristic polynomial of A), we conclude that if $B$ is close enough to $A$, the eigenvalues of $B$ are real and are also close to $A$.

Now, given a continuous family $A(x)$, if we define $\lambda_1(x)\leq \lambda_2(x)\leq \cdots \leq \lambda_n(x)$ to be the eigenvalues of $A(x)$, by the above analysis, $\lambda_i(x)$ would be continuous at each point $x$.

Remark: $\lambda_i$ defined in this way may not be smooth, even $A(x)$ is smooth, e.g.

$A(t)=\begin{pmatrix}&t &0\\&0 &1-t\end{pmatrix}$ for $0.

What is the meaning of smooth? For example, what is $A'(x)$ . Is it if $A(x)=(a_{ij}(x))$, then $A'(x)=(a'_{ij}(x))$?
It means its entries are smooth functions of $x$, as you just wrote.