I am self-studying Hoffman and Kunze's book Linear Algebra. This is Exercise 3.6.3(Linear Transformation-The Double Dual) from page 111.
Let $S$ be a set, $\mathbb{F}$ a field, and $V(S,\mathbb{F})$ the
space of all functions from $S$ into $\mathbb{F}:$
$$(f+g)(x)=f(x)+g(x)\hspace{0.5cm}(\alpha f)(x)=\alpha f(x).$$ Let $W$ be any $n$-dimensional subspace of $V(S,\mathbb F)$. Show
that there exist points $x_{1},\ldots,x_{n}\in S$ and functions
$f_{1},\ldots, f_{n}\in W$ such that $f_{i}(x_{j})=\delta_{ij}$.
Since $W$ is an $n$-dimensional subspace of $V(S,\mathbb{F})$ we can say find a basis $\mathcal{B}=\{f_{1},\ldots, f_{n}\}$. But I got stuck here. I don't know what to do from now on. I mean, what should I do in order to find those points $x_{1},\ldots,x_{n}\in S$ such that $f_{i}(x_{j})=\delta_{ij}$.
Best Answer
For your basis $\mathcal B$, for each $x$ consider the $n$-dimensional vector with components $f_i(x)$. There is a linearly independent set of $n$ of these vectors. If $S$ is finite, this follows directly because the matrix formed by all these vectors has rank $n$ because $\mathcal B$ is a basis. It also holds for infinite $S$, however, for if not, some $n-1$ of these vectors would have to span all of them; the matrix formed by those $n-1$ vectors would have rank at most $n-1$, so it would be possible to express one of the functions as a linear combination of the others at the corresponding $n-1$ points, but then since the vectors corresponding to the remaining points are spanned by the $n-1$ vectors, the one function would in fact be identical to the linear combination of the others at all points, contrary to the fact that $\mathcal B$ is a basis.
So we have $n$ points $x_1,\dotsc,x_n$ such that the corresponding $n$ vectors $f_i(x_j)$ are linearly independent. But then their entries $A_{ij}=f_i(x_j)$ form an invertible matrix, and since
$$\delta_{ij}=\sum_kA^{-1}_{ik}A_{kj}=\sum_kA^{-1}_{ik}f_k(x_j)=\left(\sum_kA^{-1}_{ik}f_k\right)(x_j)$$
the points $x_1,\dotso,x_n\in S$ and the functions $\sum_kA^{-1}_{ij}f_k\in W$ have the desired property.