Unsure of meaing of $T'(\lambda)$ in definition of simple eigenvalue? Let $T:\Omega \subset \mathbb{C} \to \mathbb{C}^{m \times m}$ be a matrix-valued function. The number $\lambda \in \Omega$ is an eigenvector of $T$ if $T(\lambda)v = 0$, and $v\in\mathbb{C}^n$ is its corresponding eigenvector.
An eigenvalue $\lambda \in \Omega$ is called simple if
$$\text{Ker}(T(\lambda)) = \text{span}\{v\}, \quad v \neq 0, \quad T'(\lambda)v \notin R(T(\lambda).$$
It follows that, for the adjoint $T^H(z)$ we have that
$$\text{Ker}(T^H(\lambda)) = \text{span}\{w\}, \quad v \neq 0,$$
and
$$w^H T'(\lambda)v \neq 0.$$
Question


*

*I'm not sure what's going on here as I don't know the notation $T'(\lambda)$ means. Is it the derivative? If so, how do we know that $T'(\lambda)v \notin R(T(\lambda)$?

*Most importantly, where is the relation $w^H T'(\lambda)v \neq 0$ coming from?


Edit
This comes from page 4 of this paper.

My guess would be indeed $T'(\lambda)$ is the derivative. You don't know that $T'(\lambda)v\notin R(T(\lambda))$, this is a condition.
Note that $x\mapsto w^H$x has co-dimension one kernel. On the other hand $T(\lambda)$ has co-dimension one image, and $(T(\lambda)^Hw)^Hx=w^HT(\lambda)x=0$ for all $x$, so the kernel of $w^H$ must be the range of $T(\lambda)$. Since $T'(\lambda)v\notin R(T(\lambda))$ you have that $w^HT'(\lambda)v\neq0$.
