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?
Best Answer
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.