Show that if $\lambda $ is a simple eigenvalue of $\textbf A$ then $\operatorname {index}(\textbf A-\lambda \textbf I)=1$ How to prove that if $\lambda$ is a simple eigenvalue of $\textbf A$ then $\operatorname {index}(\textbf A-λ\textbf I)=1$, for any square matrix $\textbf A_{n\times n}$?
Equivalently, how to prove that $\text {rank} (\textbf A - \lambda \textbf I)=\operatorname{rank} ((\textbf A - \lambda \textbf I)^2$? Where we know that $\text {rank} (\textbf A - \lambda \textbf I)=n-1$ since $\lambda$ is simple.
Also since $\operatorname{rank}(\textbf B\textbf B)=\operatorname{rank}(\textbf B)-\dim(N(\textbf B)\cap R(\textbf B))$, does this mean that for every simple eigenvalue $\lambda$ we must have $\dim(N(\textbf A-\lambda \textbf I)\cap R(\textbf A-\lambda \textbf I))=0$, i.e. $N(\textbf A-\lambda \textbf I)\cap R(\textbf A-\lambda \textbf I)=\textbf 0$?
 A: $\newcommand{\rk}{\operatorname{rank}}$Instead of proving $\rk(A-\lambda I) = \rk(A-\lambda I)^2$, we can equivalently prove (by rank - nullity theorem) that $$\dim\ker(A-\lambda I) = \dim\ker(A-\lambda I)^2.$$
We know that the geometric multiplicity of $\lambda$ is $\gamma =\dim\ker(A-\lambda I) = 1.$ 
Let  $y\in \ker (A-\lambda I)^2.$ We claim that $y\in \ker(A-\lambda I).$ 
Proof: Assume that $y\in \ker(A-\lambda I)^2,$ but $y\notin \ker (A-\lambda I)$. Equivalently, we have that $$(A-\lambda I)^2 y = 0 \text{ and } (A-\lambda I)y\neq 0 .$$
By definition, $y$ is a generalized eigenvector of rank $2$, that corresponds to the eigenvalue $\lambda.$ But this cannot happen, since the eigenvalue $\lambda$ is simple, which means the only eigenvectors are of rank $1.$ 
Thus, we proved that $ \ker(A-\lambda I)^2 \subseteq \ker(A-\lambda I)$. But it is always true that $\ker(A-\lambda I) \subseteq \ker(A-\lambda I)^2.$ Hence:
$$\ker(A-\lambda I) = \ker(A-\lambda I)^2,$$ which implies what we want to prove.

About your second question, this might be helpful.
