Suppose ${latex.inlinev_{1}, ..., v_{n}} is a basis of V and ${latex.inlinew_{1}, ..., w_{n} \in W}. Then there exists a unique linear map T: V -> W such that ${latex.inlineTv_{k} = w_{k}} for each k from 1 to n.
First we show that there exists a linear map T with the desired property. We can define T by ${latex.inlineT(c_{1}v_{1} + ... + c_{n}v_{n}) = c_{1}w_{1} + ... + c_{n}w_{n}} where the c’s are arbitrary in F. The list of v’s is a basis of V. Thus the linear map is indeed a function from V to W because each element of V can be uniquely written as a linear combination of the vectors ${latex.inlinev_{1}, ..., v_{n}}.
For each k, taking ${latex.inlinec_{k} } and the other c’s equal to 0, we get that ${latex.inlineTv_{k} = w_{k}}.
Now we show that this function is indeed a linear map. We must show additivity and homogeneity.
additivity Let ${latex.inlineu, v \in V} with ${latex.inlineu = a_{1}v_{1} + ... + a_{n}v_{n}} and ${latex.inlinev = c_{1}v_{1} + ... + c_{n}v_{n}}. Then we need to show ${latex.inlineT(u + v) = Tu + Tv}. Well, ${latex.inlineT(u + v) = T((a_{1} + c_{1})v_{1} + ... + (a_{n} + c_{n})v_{n})}. But, by definition this means ${latex.inlineT(u + v) = (a_{1} + c_{1})w_{1} + ... + (a_{n} + c_{n})w_{n}}. Using the distributive property we get that this equals ${latex.inlinea_{1}w_{1} + ... + a_{n}w_{n} + c_{1}w_{1} + ... + c_{n}w_{n}} which is equivalent to ${latex.inlineTu + Tv}.
homogeneity Suppose ${latex.inline\lambda \in F} and ${latex.inlinev = c_{1}v_{1} + ... c_{n}v_{n}}. Then ${latex.inlineT(\lambda v) = T(\lambda c_{1}v_{1} + ... + \lambda c_{n}v_{n})}. But with our linear map we get ${latex.inlineT(\lambda v) = \lambda c_{1}w_{1} + ... + \lambda c_{n}w_{n} = \lambda(c_{1}w_{1} + ... + c_{n}w_{n})}. But this is equal to ${latex.inline\lambda T(v)}.
Now we need to prove uniqueness. Suppose ${latex.inlineT \in L(V, W)}. and ${latex.inlineTv_{k} = w_{k}} for each k in 1 to n. Let ${latex.inlinec_{1}, ..., c_{k} \in F}. Then the homogeneity of T implies that ${latex.inlineT(c_{k}v_{k}) = c_{k}w_{k}}. The additivity of T implies that ${latex.inlineT(c_{1}v_{1} + ... + c_{n}v_{n}) = c_{1}w_{1} + ... c_{n}w_{n}}. Thus T is uniquely determined on span(v1, ..., vn) by the equation above. Because the list of v’s is a basis of V, this implies that T is uniquely determined on V, as desired.
We need the set of v’s to be a basis because we need to show that the function T is from V to W, which means it must map every element of V to something in W. If we had no basis, we could not claim that each element of V can be wirtten as a linear combination of the set of v’s, which would make T being a function not true.
If we had a spanning set but not a basis, we know by 2.30 that the spanning list would be able to be reduced to a basis anyway without changing the span. So the proof would be the same, youd’d just have the extra step required of reducing the spanning set to a basis.