Let $\left( E _ { 1 } , \ldots , E _ { n } \right)$ be the canonical basis of $\mathcal { M } _ { n , 1 } ( \mathbb { R } )$. We denote $V = \sum _ { k = 1 } ^ { n } E _ { k }$.
For $i \in \llbracket 1 , n \rrbracket$, express $E _ { i }$ in terms of $V$ and of $V - 2 E _ { i }$. Deduce that $\mathcal { M } _ { n , 1 } ( \mathbb { R } ) = \operatorname { Vect } \left( \mathcal { V } _ { n , 1 } \right)$.