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?