Prove that:
Let $V$ and $W$ be vector spaces over a common field, and let $\beta$ be a basis for $V$. Then, for any function $f: \beta \to V$ there exists exactly one linear transformation $T: V \to W$ such that $T(x) = f(x)$ for all $x \in \beta$.
I also have a theorem that:
Let $V$ and $W$ be vector spaces over $F$, and suppose that $\beta = \{v_1, v_2, …, v_n\}$ is a basis for $V$. For $w_1, w_2, …, w_n$ in $W$, there exists exactly one linear transformation $T: V \to W$ such that $T(v_i) = w_i$ for $i = 1, 2, …, n$.
From the solution manual: $\beta = \{v_1, v_2, … , v_n\}$. Let $x \in V$. Then, $x \in \sum_{i=1}^n a_i v_i$ for $a_i \in F$ (field). Then,
$$T(\sum_{i=1}^n a_i v_i) = \sum_{i=1}^n a_i T(v_i) = \sum_{i=1}^n a_i f(v_i). $$
However, I don't understand how the second equality holds and how this proves the uniqueness of $T$. I appreciate if you give some help.
Best Answer
We have $T(v_i) = f(v_i)$ as a hypothesis. What this is trying to show is that, if $T$ verifies that, we can calculate $T(x)$ for any $x \in V$. But if that's the case, there cannot be a different transformation satisfying $S(v_i) = f(v_i)$, since then $S(x) = T(x) \forall x \in V$
This is true because you would get
$$S(x) = S(\sum_{i=1}^n a_i v_i) = \sum_{i=1}^n a_i S(v_i) = \sum_{i=1}^n a_i f(v_i) = T(\sum_{i=1}^n a_i v_i) = T(x)$$