Proving a linear transformation is unique In Axler's Linear Algebra Done Right, there is this theorem. (3.5)

Suppose $v_1. . .  v_n $ is a basis of $V$ and $w_ , . . . w_n \in W$. Then there exists
  a unique linear map $T: V \rightarrow  W$ such that
  $$T (v_j) = w_j$$
  for each $j\in   1, . . . , n $.

I understood the first part of the proof proving existence of such a transformation, but didn't understand the uniqueness part of the proof.

To prove uniqueness, now suppose that $T \in \cal L $$(L,V)$;and that $T( v_j)=  w_j$ for each $j\in   1, . . . , n $.
Let $c_1,. . . ,c_n \in F$. 
The homogeneity of $T$ implies that
  $T(c_j v_j) =  c_jw_j$ for each $j\in   1, . . . , n $.
The additivity of T now implies that
  $T(c_1v_1 + . . . + c_nv_n) = c_1w_1 + . . . + c_nw_n$.  
Thus $T$ is uniquely determined on span($v_1, . . . ,v_n$) by the equation above.
  Because $v_1,  . . . v_n$ is a basis of $V$, this implies that $T$ is uniquely determined
  on $V$.

How does "equation above" show that $T$ is uniquely determined on span($v_1, . . . ,v_n$)?
Is it because there is one way to get each of the basis vectors using the equation? How do we know  there isn't another transformation that will work?
 A: By the way, the proof given above does indeed show that $T$ is unique. Perhaps a different phrasing of the argument might help you understand the uniqueness.
You said you understood the existence part, so ok let $T: V \to W$ be the linear map you constructed, which satisfies $T(v_j) = w_j$ for all $j \in \{1, \dots, n\}$. Suppose now that there is a linear map $S: V \to W$ such that $S(v_j) = w_j$ for all $j \in \{1, \dots, n\}$. We have to show that $T=S$; i.e we have to show that for every $x \in V$, $T(x) = S(x)$.
To prove this, pick any $x \in V$. Since $\{v_1, \dots, v_n\}$ is a basis for $V$, there exist (unique) scalars $c_1, \dots, c_n \in F$ such that
\begin{align}
x = \sum_{i=1}^n c_iv_i \tag{*}
\end{align}
Now, we perform a simple computation:
\begin{align}
T(x) &= T\left( \sum_{i=1}^n c_iv_i \right) \tag{by (*)} \\
&= \sum_{i=1}^n c_i T(v_i) \tag{$T$ is linear by construction} \\
&= \sum_{i=1}^n c_i w_i \tag{by definition of $T$} \\
&= \sum_{i=1}^n c_i S(v_i) \tag{by assumption on $S$} \\
&= S\left( \sum_{i=1}^n c_i v_i \right) \tag{$S$ linear by assumption} \\
&= S(x) \tag{by (*)}
\end{align}
So we have shown that for every $x \in V$, $T(x) = S(x)$. Hence, $T=S$, proving uniqueness of the original $T$ you constructed.
This result is often stated as "a linear transformation is specified by its values on a basis" or something to that effect.
A: Because the vectors $\vec{w}_i$ form a basis for $W$, any two different linear combinations of the vectors $\vec{w}_i$ must yield a different result. Let $\vec{w} = c_1 \vec{w}_1 + \cdots + c_n \vec{w}_n$, with all $\vec{c}_i \in \mathbb{R}$. By construction, the transformation $T(c_1 \vec{v}_1 + \cdots + c_n \vec{v}_n) = c_1 \vec{w}_1 + \cdots + c_n \vec{w}_n$. But as mentioned before, any linear combination of the $\vec{w}_i$ must be unique, and thus, there can be no numbers $d_1, d_2, \cdots, d_n$ such that $T(d_1 \vec{v}_1 + \cdots + d_n \vec{v}_n) = d_1 \vec{w}_1 + \cdots + d_n \vec{w}_n = c_1 \vec{w}_1 + \cdots + c_n \vec{w}_n$ without $d_i = c_i$ for all $i$.
That proves that if $T(\vec{a}) = \vec{w} = T(\vec{b})$, $\vec{a} = \vec{b}$. Now comes the part where the $\vec{v}_i$'s form a basis. This means that any $\vec{v} \in V$ must also be a unique linear combination of the $\vec{v}_i$'s. In other words, if $c_1 \vec{v}_1 + \cdots + c_n \vec{v}_n = \vec{v} = d_1 \vec{v}_1 + \cdots + d_n \vec{v}_n$, we also must have $c_i = d_i$ for all $i$. So the transformation $T$ must be unique: in other words, for every $\vec{w} \in W$, there exists one and only one $\vec{v} \in V$ such that $\vec{v} \in V$. Furthermore, this $\vec{v}$ must be unique. So we say that $T:V \to W$ is uniquely determined on $V$.
