Need help to understand the uniqueness of linear maps on basis of domain.

Question

I am reading Linear Algebra Done Right. Now, I understand that the meaning of this statement is that if we know the basis of $V$ and how $T$ acts on the basis ($Tv_j = w_j$), we will know how it acts on all the vectors in $V$ ($T: V \mapsto W$).

I follow the proof of existence part where it proves the property of additivity and homogeneity of linear map of the following $$T(c_1v_1 +…+ c_nv_n) = c_1w_1 +…+ c_nw_n$$

However, I don't understand the uniqueness part.

To prove uniqueness, now suppose that $T \in \mathcal{L}(V,W)$ and that $Tv_j = w_j$, for $j = 1,…,n$. Let $c_1,…,c_n \in \mathbb{F}$. The homogeneity of $T$ implies that $T(c_jv_j) = c_jT(v_j) = c_jw_j$ for $j = 1,…,n$. 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 $V$.

• What is meant by $T$ is uniquely determined on $span(v_1,…,v_n)$?
• What is meant by Because $v_1,…,v_n$ is a basis of $V$, this implies that $T$ is uniquely determined on $V$?
• What is meant by unique? Unique means no other form of $T$?
• Why uniqueness is important in the proof?

Answer 1

• It means that that map $T$ is the only linear map defined on $\operatorname{span}(v_1,\ldots,v_n)$ for which the given condition ($\forall j\in\{1,2,\ldots,n\}):Tv_j=w_j$) holds.
• Since $\{v_1,\ldots,v_n\}$ is a basis, $\operatorname{span}(v_1,\ldots,v_n)=V$. Therefore, asserting that $T$ is uniquely determind in $\operatorname{span}(v_1,\ldots,v_n)$ is the same thing as asserting that it is uniquely determind on $V$.
• Yes. The map $T$ is the only linear map from $V$ into $W$ for which the condition $\forall j\in\{1,2,\ldots,n\}):Tv_j=w_j$ holds.
• Because the statment that you are proving states that $T$ is unique. Therefore, you have to prove it.