Suppose \({latex.inline[v_{1}, ..., v_{n}](v_{1}, ..., v_{n})} is a basis of V and \){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 \({latex.inline[T(c_{1}v_{1} + ... + c_{n}v_{n}) = c_{1}w_{1} + ... + c_{n}w_{n}](T(c_{1}v_{1} + ... + c_{n}v_{n}) = c_{1}w_{1} + ... + c_{n}w_{n})} where the c’s are arbitrary in F. The list of v’s is a basis of V. Thus the linear map is indeed a function from V to W because each element of V can be uniquely written as a linear combination of the vectors \){latex.inlinev{1}, ..., v{n}}.
For each k, taking \({latex.inline[c_{k} ](c_{k} )} and the other c’s equal to 0, we get that \){latex.inlineTv{k} = w{k}}.
Now we show that this function is indeed a 1753749939 - Axler Chapter 3A: Vector Space of Linear Maps|linear map. We must show additivity and homogeneity.
additivity Let \({latex.inline[u, v \in V](u, v \in V)} with \){latex.inlineu = a{1}v{1} + ... + a{n}v{n}} and \({latex.inline[v = c_{1}v_{1} + ... + c_{n}v_{n}](v = c_{1}v_{1} + ... + c_{n}v_{n})}. Then we need to show \){latex.inlineT(u + v) = Tu + Tv}. Well, \({latex.inline[T(u + v) = T((a_{1} + c_{1})v_{1} + ... + (a_{n} + c_{n})v_{n})](T(u + v) = T((a_{1} + c_{1})v_{1} + ... + (a_{n} + c_{n})v_{n}))}. But, by definition this means \){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 \({latex.inline[a_{1}w_{1} + ... + a_{n}w_{n} + c_{1}w_{1} + ... + c_{n}w_{n}](a_{1}w_{1} + ... + a_{n}w_{n} + c_{1}w_{1} + ... + c_{n}w_{n})} which is equivalent to \){latex.inlineTu + Tv}.
homogeneity Suppose \({latex.inline[\lambda \in F](\lambda \in F)} and \){latex.inlinev = c{1}v{1} + ... c{n}v{n}}. Then \({latex.inline[T(\lambda v) = T(\lambda c_{1}v_{1} + ... + \lambda c_{n}v_{n})](T(\lambda v) = T(\lambda c_{1}v_{1} + ... + \lambda c_{n}v_{n}))}. But with our linear map we get \){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 \({latex.inline[T \in L(V, W)](T \in L(V, W))}. and \){latex.inlineTv{k} = w{k}} for each k in 1 to n. Let \({latex.inline[c_{1}, ..., c_{k} \in F](c_{1}, ..., c_{k} \in F)}. Then the homogeneity of T implies that \){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 1753318207 - Axler 2.30 Every spanning list contains a basis|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.