How to prove that if $f^3 = f$, $f$ is diagonalizable?

Question

Prove that

Let $V$ be a finite dim. vector space over a field of characteristic zero, and $f: V \to V$ be a linear map.Then if $$f^3 = f,$$ then $f$ is diagonalizable.

Since $f$ is zero of the polynomial $$p(x) = x^3 - x$$ the minimal polynomial $m_f$ should divide $p$, hence $m_f$ can be the following polynomials only: $$m(x) = (x-1) \\ = x \\ = x+ 1 \\ =x^2 - x \\ = x^2 + x \\ = x^2 - 1 \\ =x(x-1)(x+1) $$

But, I couldn't show that if $m$ is either of the followings $ x^2 + x $ & $ x^2 - 1 $ & $ x(x-1)(x+1) $, it should be diagonalizable. So for these 3 cases how can we prove that $f$ should be diagonalizable ?

Answer 1

Since the matrix $f$ is a zero of the polynomial $$p(x) = x^3 - x = x (x-1)(x+1),$$ the minimal polynomial of $f$ has to divide the polynomial $p$, but this means that $m$ is some combinations of the factors $x$, $(x-1)$, and $(x+1)$ with each having the multiplicity 1, but this implies that $f$ is diagonalisable.

QED.

Answer 2

The polynomial $x^3-x$ aplits as a product of distinct linear factors, hence the minimal polynomial of $f$, which is a divisor of $x^3-x$, is a product of distinct linear factors too. This is the criterion for a linear map to be diagonalisable.