Eigenvalues are roots of the characteristic polynomial. In the case the polynomial has simple roots, they smoothly depend on the coefficients. Below is the proof in the real case. The argument below can be easily modified to more general situations.

**Theorem.** For $a=(a_0,a_1,\ldots,a_n)\in\mathbb{R}^{n+1}$, $a_n\neq 0$ let $P_a(x)=a_nx^n+\ldots+a_1x+a_0$. Suppose that for
$a^0=(a_0^0,\ldots,a_n^0)$, $a^0_n\neq 0$ the polynomial $P_{a^0}(x)$
has $n$ distinct real roots. Then, there is $\epsilon>0$ and
$C^\infty$ smooth functions $$ \lambda_1,\ldots,\lambda_n:
B^{n+1}(a^0,\epsilon)\to\mathbb{R} $$ such that for any $a\in
B^{n+1}(a^0,\epsilon)$, $\lambda_1(a),\ldots,\lambda_n(a)$ are
distinct roots of the polynomial $P_a(x)$. In other words, prove that
in a small neighborhood of $a^0$, roots of the polynomial $P_a$
depend smoothly on the coefficients $a_0, a_1,\ldots, a_n$.

**Proof.**
Denote the roots of $P_{a^0}$ by $\lambda_1^0,\ldots,\lambda_n^0$. That is
$P_{a^0}(\lambda)=a_n^0(\lambda-\lambda_1^0)\cdots(\lambda-\lambda_n^0)$ and clearly
$$
\left.\frac{d}{d\lambda}\right|_{\lambda=\lambda_i^0} P_{a^0}(\lambda)\neq 0
\quad
\mbox{for all $i=1,2,\ldots,n$.}
$$
This is where we employ the assumption that the roots are distinct.

Let $F(a,\lambda)=F(a_0,a_1,\ldots,a_n,\lambda)=P_a(\lambda)$.
For each $i=1,2,\ldots,n$ we have
$$
F(a_0^0,a_1^0,\ldots,a_n^0,\lambda_i^0)=P_{a^0}(\lambda_i^0)=0
$$
and
$$
\left.\frac{\partial}{\partial\lambda}\right|_{\lambda=\lambda_i^0} F(a_0^0,a_1^0,\ldots a_n^0,\lambda) =
\left.\frac{d}{d\lambda}\right|_{\lambda=\lambda_i^0} P_{a^0}(\lambda)\neq 0.
$$
Thus according to the Implicit Function Theorem, there is a unique $C^\infty$ smooth function $\lambda(a)$ defined for
$a=(a_0,a_1,\ldots,a_n)$ in a neighborhood of $a^0=(a_0^0,a_1^0,\ldots,a_n^0)$ (neighborhood in $\mathbb{R}^{n+1}$) such that
$\lambda(a^0)=\lambda^0_i$ and $P_a(\lambda(a))=F(a,\lambda(a))=0$. Denote this function by $\lambda_i(a)$. Hence
$\lambda_i(a)$, $i=1,2,\ldots,n$ are roots of the polynomial $P_a$. Since $\lambda_i(a^0)\neq \lambda_j(a^0)$ for $i\neq j$,
we see that these roots are distinct in a neighborhood $B^{n+1}(a^0,\epsilon)$ of $a^0$, provided $\epsilon>0$ is sufficiently small.
$\Box$

Perturbation Theory for Linear Operators$\endgroup$ – Pietro Majer May 11 '18 at 19:52