Linear mapping/Different eigenvalues/Diagonalizable/Fact

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

${\displaystyle \varphi \colon V\longrightarrow V}$

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

Then ${\displaystyle {}\varphi }$ is diagonalizable.