Prove that $\mathrm{Ker}(T - \lambda I_V)^n = {0}$

Question

I´m having trouble proving this statement, I already tried induction but I failed miserably.

Let $V$ a $K$-vectorial space and $T: V\to V$ endomorphism. Let $\lambda\in K$ that it's not a eigenvalue of $T$. Prove that $\mathrm{Ker}(T - \lambda I_V)^{n} = \{0\}$ for all $n\in\mathbb{N}$.

Does anyone know how to prove it?

Thanks beforehand

Answer 1

Given, $V$ a $K$-vectorial space and $T: V\to V$ endomorphism. Whenever, $\mathrm{Ker}(T - \lambda I_V)^{n} \neq \{0\}$ for some $n\in\mathbb{N}$, it means that geometric multiplicity of $\lambda $ is $\ge 1 $ [because, $\mathrm{Ker}(T - \lambda I_V)^{n} \neq \{0\}$ for some $n\in\mathbb{N}$ $\implies$ $\mathrm{Ker}(T - \lambda I_V) \neq \{0\}$ $\implies $ $\dim(\mathrm{Ker}(T - \lambda I_V)) \ge 1 $, and Geometric multiplicity of $\lambda = \dim(\mathrm{Ker}(T - \lambda I_V)) $ ]

that means $\lambda $ is an eigenvalue of $A$.

So, whenever $\lambda $ is not an eigenvalue of $A$, $\mathrm{Ker}(T - \lambda I_V)^{n} = \{0\}$ for all $n\in\mathbb{N}$.

Answer 2

Hints:

• If $\lambda \in K$ is not an eigenvalue, what can you say about $T-\lambda I$?
• If two such maps satisfy this property, what can you say about their composition? Can you extend this by induction to the case of $n$ such maps?