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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...

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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.

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  • $\begingroup$ Hmm. I'm not seeing it. Suppose $v_1, ..., v_n$ are linearly dependent; say $v_1 = a_2v_2 + \cdots + a_n v_n$, for some $a_i \in F$. Then $T(v_1) = T(a_2 v_2 + \cdots a_n v_n)$; and so $T(v_1) = a_2 T(v_2) + \cdots a_n T(v_n)$... I'm failing to spot a contradiction here. Also, suppose the $v_1, ..., v_n$ do not span $V$. Then there exists $x \in V$ such that $x \ne a_1v_1 + \dots a_nv_n$, for any $a_i \in F$. This just seems like an awkward premise to argue from... $\endgroup$ May 30, 2019 at 15:50
  • $\begingroup$ In the linearly dependent case, if $w_1 \ne a_2 w_2 + \ldots + a_n w_n$ you have your contradiction. In the not-spanning case, $T(x)$ could be anything so $T$ is not unique. $\endgroup$ May 31, 2019 at 1:21
  • $\begingroup$ But why can't $w_1 = a_2 w_2 + \cdots + a_n w_n$? After all, we don't know that the $w_i$ are linearly independent. $\endgroup$ May 31, 2019 at 16:09
  • $\begingroup$ The assumption is that a unique $T$ exists for any choice of the $w$'s. $\endgroup$ May 31, 2019 at 18:49
  • $\begingroup$ Ah! So, we can choose the $w's$ to be basis elements of $W$, right? (Moreover, we can choose $W$ to be of dimension greater than or equal to $n$.) $\endgroup$ Jun 4, 2019 at 1:26

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