Proof (or reference) about $\lambda_i(A+\epsilon e_je_j^*) = \lambda_i(A) + \epsilon |v_{i,j}|^2 + O(\epsilon^2).$ I'm looking for a proof (or a reference in a textbook) about the fact that
$$
\lambda_i(A+\epsilon e_je_j^*) =_{\epsilon \to 0} \lambda_i(A) + \epsilon |v_{i,j}|^2 + O(\epsilon^2),
$$
where $A$ is a hermitian matrix (having distinct eigenvalues), $\lambda_i(A)$ is an eigenvalue of $A$, $e_j \in \bf{R}^n$ is defined by $(e_j)_i = \delta_{i,j}$, $v_{i,j}$ is the $j-$th component of a unit eigenvector of $\lambda_i(A)$.
This theorem is from perturbation theory, a field I'm not very familiar with.
This is used in : Peter B. Denton, Stephen J. Parke, Terence Tao and Xining Zhang. $\textit{Eigenvectors from eigenvalues: A survey of a basic identity in linear algebra}, 2021;$
arXiv:1908.03795 (page $13$).
 A: This is  first order perturbation theory: a perturbation $\delta A$ to a Hermitian matrix $A$ gives to first order a correction $\delta \lambda$ to an eigenvalue $\lambda$ (with corresponding eigenvector $v$) equal to
$$\delta\lambda=\sum_{n,m}\delta A_{nm}\bar{v}_nv_m.$$
In this case $\delta A_{nm}=\epsilon\delta_{nj}\delta_{mj}$, hence
$$\delta\lambda=\epsilon|v_j|^2+{\cal O}(\epsilon^2).$$
The link I gave above is to the derivation in Wikipedia, if you prefer a textbook, see chapter 7 of Introduction to Quantum Mechanics by Griffiths.
A: I think I found what I was looking for:

*

*First, I found a proof using the implicit function theorem showing that if $A(\epsilon) = A + \epsilon e_j e_j^*$ is $\mathscr{C}^1$ (which is the case) and the eigenvalues of $A$ are distinct, then for $\epsilon$ small enough $\epsilon \mapsto \lambda_i(A(\epsilon))$ is $\mathscr{C}^1$ and we can pick an eigenvector $v_i(\epsilon)$ such that $\epsilon \mapsto v_i(\epsilon)$ is also $\mathscr{C}^1$ (see here).


*Thanks to @JeanMarie (see here) I saw that what I was looking for is equation $(5)$ in this article (the above justifies taking the derivative in the article), so I get $\dot{{\lambda}_i}(0) = |v_{i,j}|^2$ $-$ where $\lambda_i(\epsilon) = \lambda_i(A(\epsilon))$ $-$ ie. $\lambda_i(\epsilon)=\lambda_i(A + \epsilon e_j e_j^*)=_{\epsilon \to 0} \lambda_i(A) + \epsilon |v_{i,j}|^2 + o(\epsilon)$.
