Let $\mathbf{B}$ be a basis of $E$. Show that $$\left\{\nu(u) \mid u \in \mathcal{N}_{\mathbf{B}}\right\} = \{\nu(u) \mid u \in \mathcal{N}(E)\} = \llbracket 1, n \rrbracket$$
Let $\mathbf{B}$ be a basis of $E$. Show that
$$\left\{\nu(u) \mid u \in \mathcal{N}_{\mathbf{B}}\right\} = \{\nu(u) \mid u \in \mathcal{N}(E)\} = \llbracket 1, n \rrbracket$$