Here is a problem that's bugging me:
Let $V$ be a vector space over a field $F$ and let $v_1, ..., v_n \in V$.
Prove the following: If, for any vector space $W$ over $F$ and any elements $w_1, ..., w_n \in W$, there exists a unique linear transformation $T:V \to W$ with $T(v_i) = w_i$ for $i = 1, ..., n$, then $\{v_1, ..., v_n\}$ is a basis for $V$.
I can prove the converse of this statement (as this is a fairly typical proof); but I'm having trouble proving this particular statement. I assume we have to work with some other basis of $V$; say $\mathcal{B}$ is a basis for $V$. Then each $v_i$ can be written (uniquely) as $v_i = \alpha_1b_1 + \cdots + \alpha_1 b_m$, for some $\alpha_j \in F$ and some $b_j \in \mathcal{B}$. Then $T(v_i) = w_i$ implies that $w_i = \alpha_1d_1 + \cdots + \alpha_m d_m$, where the $d_j$ come from some basis $\mathcal{D}$ of $W$. This approach got really messy really quickly, so I couldn't help but think I was on the wrong path...
Hint: try proof by contradiction. If they are not a basis, then they are linearly dependent or they fail to span $V$. In the first case, find $w_i$ such that there is no such $T$. In the second case, show that $T$ is not unique.