Let \(S\) be a set, \(F\) a field, and \(V(S; F)\) the space of all functions from \(S\) into \(F\): \[ (f + g)(x) = f(x) + g(x), \qquad (cf)(x) = cf(x). \] Let \(W\) be any \(n\)-dimensional subspace of \(V(S; F)\). Show that there exist points \(x_1, \ldots, x_n\) in \(S\) and functions \(f_1, \ldots, f_n\) in \(W\) such that \(f_i(x_j) = \delta_{ij}\).
Solution
For each \(x\in S\), let \[ e_x:W\to F,\qquad e_x(f)=f(x) \] be the evaluation functional. We claim that the family \(\{e_x:x\in S\}\) spans \(W^*\).
If its span were a proper subspace of \(W^*\), then, because \(W\) is finite-dimensional, there would exist a nonzero \(f\in W\) annihilated by every \(e_x\). But then \[ f(x)=e_x(f)=0 \] for every \(x\in S\), so \(f\) would be the zero function, a contradiction. Hence the evaluation functionals span \(W^*\).
Since \(\dim W^*=n\), choose points \(x_1,\ldots,x_n\in S\) such that \[ e_{x_1},\ldots,e_{x_n} \] form a basis of \(W^*\). Let \(f_1,\ldots,f_n\) be the basis of \(W\) dual to this basis of \(W^*\). Then \[ f_i(x_j)=e_{x_j}(f_i)=\delta_{ij}, \] which is the desired conclusion.