Show that $f$ is diagonalizable

Question

Given an endomorphism $f$ of a vector space on $\mathbb{R}$ of dimension $n$ such that $f(f(x))=3f(x)-2x$.

Let $E_1=\ker(f-Id)$ and $E_2=\ker(f-2Id)$. Show that:

1. $E_1$ and $E_2$ form a direct sum.

2. $E=E_1 \oplus E_2$. Is $f$ diagonalizable?

Answer 1

Hints for you to understand/justify/work out:

$$x\in\ker(f-Id.)\cap\ker(f-2Id.)\implies\begin{cases}f(x)=x\\{}\\f(x)=2x\end{cases}\implies x=0 $$

$$f^2=3f-2I\implies x^2-3x+2=(x-2)(x-1)\;\text{is the minimal polynomial of}\;\;f\;\text{ (why?)}$$

and thus $\;f\;$ is diagonalizable