# Analytic perturbation of the eigenvalues/eigenvectors of non-Hermitian matrix

Consider a matrix function $A(x)$, analytically depending on single parameter $x$.

Consider the eigenvalue/eigenvector pair of $A(0)$, namely $\lambda(0)$ and $w(0)$.

The question is whether we can construct the analytic functions $\lambda(x)$ and $w(x)$ that are the eigenvalue/eigenvector of $A(x)$. (I assume that both $\lambda(x)$ and $w(x)$ could have poles or branch points, hence being multi-valued).

In books I found this statement for the case of Hermitian matrix $A(t)$ only. Is it true for non-Hermitian comlex matrices as well?

This is completely answered in Kato's Perturbation theory for linear operators, chapter 1 (he treats the non-hermitian case). Eigenvalues are generally analytic, eigenvectors not so much (in the neighborhood of values of $A(x)$ with multiple eigenvalues.
This isn't a complete answer but may help a little. One can begin by considering the function of two complex variables $$f(z,w)=\det (A(z)-wI)$$ and its zero variety $$V=\{(z,w):f(z,w)=0\}.$$ So, a point $(z,w)$ belongs to $V$ if and only if $w$ is an eigenvalue of $A(z)$. Now away from singular points of $V$, one should be able to apply the implicit function theorem to get an analytic family of eigenvalues; one can then get the corresponding eigenvectors via Cramer's rule. On the other hand, from this point of view one can see that in some cases branching is inevitable: for example consider $$A(z)=\begin{pmatrix} 0 & z^2 \\ z & 0\end{pmatrix}.$$ Then $f(z,w)=w^2-z^3$ and $(0,0)$ is the only singular point of $V$. Away from $0$ we can then take $w(z)$ to be any branch of $z^{3/2}$, but this also makes clear that one can't have analyticity near $0$, since $|w(z)|=|z|^{3/2}$ near $0$ and no analytic function can do this.
The functions $\lambda(x)$ and $w(x)$ obey a set of coupled first-order differential equations in $x$, derived in Structure of trajectories of complex matrix eigenvalues (Bohigas, De Carvalho, and Pato, 2012). For the case considered in that paper (a simple linear dependence of $A$ on $x$), no singularities in the $x$-dependence of the eigenvalues appear, and I would think that this holds more generally.