# Existence of Unique Linear Transformation implies Basis

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.
• 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... May 30, 2019 at 15:50
• 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. May 31, 2019 at 1:21
• 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. May 31, 2019 at 16:09
• The assumption is that a unique $T$ exists for any choice of the $w$'s. May 31, 2019 at 18:49
• 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 $n$.) Jun 4, 2019 at 1:26