 Solution

For $\varphi (\alpha )={\sum }_{i=1}^k{\alpha }_i{u}_i$, its image of a linear map is exactly $⟨u_1,\dots ,u_k⟩$. Consequently $\varphi$ is surjective if and only if $u_1,\dots ,u_k$ are a spanning set for $V$.

If the vectors are linearly independent and $\varphi (\alpha )=\varphi (\beta )$, then ${\sum }_i({\alpha }_i-{\beta }_i)u_i=0$, so ${\alpha }_i={\beta }_i$ for all $i$. Hence $\varphi$ is injective. Conversely, if $\varphi$ is injective, any relation ${\sum }_i{\alpha }_i{u}_i=0$ says $\varphi (\alpha )=\varphi (0)$ and thus $\alpha =0$, proving linear independence.

Combining both equivalences givesA basis is therefore exactly a choice of vectors giving existence and uniqueness of coordinates, expressed by a bijection.