As @Joe Silverman said in his comment, the key here is that $\lambda$ is simple.

I copy here a few Theorems on the subject that might be helpful.

From the book Linear Algebra and Its Applications (by Peter D. Lax):

**Theorem 7, p.130:**
Let $A(t)$ be a differentiable square matrix-valued function of the real variable $t$. Suppose that $A(0)$ has an eigenvalue $a_0$ of multiplicity one, in the sense that $a_0$ is a simple root of the characteristic polynomial of $A(0)$. Then for $t$ small enough, $A(t)$ has an eigenvalue $a(t)$ that depends differentiably on $t$, and which equals $a_0$ at zero, that is $a(0)=a_0$.

**Theorem 8, p.130:**
Let $A(t)$ be a differentiable matrix-valued function of $t$, $a(t)$ an eigenvalue of $A(t)$ of multiplicity one. Then we can choose an eigenvector $h(t)$ of $A(t)$ pertaining to the eigenvalue $a(t)$ to depend differentiably on $t$.

From the Book Matrix Analysis (by Roger A. Horn & Charles R. Johnson)

**Corollary 6.3.8, p.407:** Let $A,E\in \Bbb C^{n\times n}$. Assume that $A$ is Hermitian and $A+E$ is normal, let $\lambda_1,\ldots,\lambda_n$ be the eigenvalues of $A$ arranged in increasing order $\lambda_1\leq \ldots\leq \lambda_n$ and let $\hat \lambda_1,\ldots,\hat \lambda_n$ be the eigenvalues of $A+E$, ordered so that $\Re(\hat\lambda_1)\leq\ldots\leq\Re(\hat\lambda_n) $. Then
$$ \sum_{i=1}^n |\hat \lambda_i-\lambda_i|^2 \leq \|E\|_F^2,$$
where $\|\cdot\|_F$ is the Frobenius norm.

This is a somehow refined version of Theorem 7 above:

**Theorem 6.3.12, p.409:**
Let $A,E\in \Bbb C^{n\times n}$ and suppose that $\lambda$ is a simple eigenvalue of $A$. Let $x$ and $y$ be, respectively, right and left eigenvectors of $A$ corresponding to $\lambda$. Then

a) for each given $\epsilon>0$ there exists a $\delta>0$ such that, for all $t\in\Bbb C$ such that $|t|<\delta$, there is a unique eigenvalue $\lambda(t)$ of $A+tE$ such that $|\lambda(t)-\lambda-ty^*Ex/y^*x|\leq |t| \epsilon$

b) $\lambda(t)$ is continuous at $t=0$, and $\lim_{t\to 0}\lambda(t)=\lambda$

c) $\lambda(t)$ is differentiable at $t=0$, and
$$ \left.\frac{\operatorname{d}\lambda(t)}{\operatorname{d}t}\right|_{t=0}=\frac{y^*Ex}{y^*x}$$

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