= Solution
{c}
For $\phi(\alpha)=\sum_{i=1}^k\alpha_i u_i$, its <image of a linear map> is exactly $\langle u_1,\ldots,u_k\rangle$. Consequently $\phi$ is <surjective> if and only if $u_1,\ldots,u_k$ are a <spanning set> for $V$.
If the <vectors> are <linearly independent> and $\phi(\alpha)=\phi(\beta)$, then $\sum_i(\alpha_i-\beta_i)u_i=0$, so $\alpha_i=\beta_i$ for all $i$. Hence $\phi$ is <injective>. Conversely, if $\phi$ is <injective>, any relation $\sum_i\alpha_i u_i=0$ says $\phi(\alpha)=\phi(0)$ and thus $\alpha=0$, proving <linear independence>.
Combining both equivalences gives
$$
\boxed{(u_1,\ldots,u_k)\text{ is a basis of }V\quad\Longleftrightarrow\quad\phi\text{ is a bijection}.}
$$
A <basis> is therefore exactly a choice of <vectors> giving existence and uniqueness of coordinates, expressed by a <bijection>.
Back to article page
