Core-nilpotent decomposition of $A-\lambda I$ where $\lambda$ is a simple eigenvalue of $A$

Question

If $A\in M_n$ is a square, complex matrix, and if $\lambda$ is a simple eigenvalue of $A$, then I want to prove that there exists a nonsingular matrix $S\in M_n$, such that $$ S^{-1}(A-\lambda I)S = \begin{pmatrix}0_1 & 0\\ 0 & C\end{pmatrix}, $$ is a core-nilpotent decomposition, without Jordan canonical form, or generalized eigenvectors.

My approach in the proof is following these steps:

1. Prove that if $Ax=\lambda x$ and $y^*A = \lambda y^*$, then $y^*x \neq 0$
2. Set $S=(x ~ S_1)$, where $R(S_1) = y^\bot$, and prove that $S^{-1}=\begin{pmatrix}(y^*x)^{-1}y^* \\ Z\end{pmatrix}$
3. Compute $$S^{-1}(A-\lambda I)S = \begin{pmatrix}0_1 & 0\\ 0 & Z(A-\lambda I)S_1\end{pmatrix}.$$

I know how to prove each step individually without Jordan form or generalized eigenvectors, but this approach seems too difficult. I’m looking for simpler solutions. For example, proving that $R(A-\lambda I)\cap N(A-\lambda I)={0}$.

Answer 1

For example, proving that $R(A-\lambda I)\cap N(A-\lambda I)=\big\{0\big\}$

fix $\lambda \in \mathbb K$ and
$B:=A-\lambda I$

observe that $\text{rank}\big(B^2\big)\leq \text{rank}\big(B\big)$
with equality iff $\text{Im}(B)\cap \ker(B)=\big\{0\big\}$

Equivalently:
$\text{Im}(B)\cap \ker(B)\neq\big\{0\big\}\implies \text{rank}\big(B^2\big)\lt \text{rank}\big(B\big)\implies\det\big(C\big)=0$
$\implies $ 0 is not a simple eigenvalue of $B$