Let $T: V \to V$ be a linear map such that $T^2-3T+2I=0$. Let $T: V \to V$ be a linear map such that $T^2-3T+2I=0$, where $I$ is the identity map. 

question:

a) Prove that $V=\ker(T-2I) \oplus\ker(T-I)$

b) let $A$ be an $n \times n$ matrix such that $A^2-3A+2I_n=0$ Where $I_n$ is the $n\times n$ identity matrix. True or false: $A$ is diagonalizable 

I attempted $\ker(T-2I)=(T-2I)V=0$

$\ker(T-I)=(T-I)V=0$ 

I know the direct sum should be the join of $\ker(T-2T)$ and $\ker(T-I)$ is $0$, and I am not sure how to prove it 

I am a first year student from McGill U. I am doing linear mapping on linear algebra. The textbook I am using is Linear Algebre edition sixth by SEYMOUR LIPAXHUTZ
Some hints: To show $V= \ker(T-2I) \oplus \ker(T-I)$, you need to show that

(i) $\ker(T - 2I) \cap \ker(T - I) = \{0\}$ and (ii) $\ker(T - 2I) + \ker(T - I) = V$.

(i) Given $v \in \ker(T - 2I) \cap \ker(T - I)$, what can you say about $T(v)$? What does $v \in \ker(T - I)$ imply? What does $v \in \ker(T - 2I)$ imply?

(ii) Given $v \in V$, then
$$
0 = (T^2 - 3T + 2I)(v) = (T - 2I)((T - I)(v)) = (T-2I)(T(v) - v) \, .
$$
What does this say about $T(v) - v$? Similarly,
$$
0 = (T^2 - 3T + 2I)(v) = (T - I)((T - 2I)(v)) = (T-I)(T(v) - 2v) \, .
$$
What does this say about $T(v) - 2v$?
Spanning part is already proved. Now, say $v\in ker (T-2I)\cap ker (T-I)$. Then $(T-2I)v=0$

$\Rightarrow (T-I)v=v $

$\Rightarrow v=0$