Linear mapping/Different eigenvalues/Diagonalizable/Fact

Let ${}K$ denote a field, and let ${}V$ denote a finite-dimensional vector space. Let

$\varphi \colon V\longrightarrow V$

denote a linear mapping. Suppose that there exists ${}n$ different eigenvalues.

Then ${}\varphi$ is diagonalizable.

Proof, Write another proof