Nilpotent endomorphism/Trigonalizable/Fact

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

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

be a nilpotent linear mapping.

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

There exists a basis such that ${\displaystyle {}\varphi }$ is described, with respect to this basis, by an upper triangular matrix, in which all diagonal entries are ${\displaystyle {}0}$.