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Let $S$ be a set, $F$ a field, and $V(S;F)$ the space of all functions from $S$ into $F$, with the usual function addition and scalar multiplication. Let $W$ be any $n$-dimensional subspace of $V(S;F)$. Show that there exist points $x_1,...,x_n$ in $S$ and functions $f_1,...,f_n$ in $W$ such that $f_i(x_j)=\delta_{ij}$, where $\delta_{ij} = 1$ if $i=j$, and $\delta_{ij} = 0$ otherwise (i.e. the delta function is the Kronecker delta function).

If $S$ were an $n$-dimensional subspace, then the result is true. But I am not sure how to proceed. I was considering the set $(S^0)^0$, which is spanned by $S$, but I am sure where this leads me.

Edit: I found that, since $(S^0)^0$ is a subspace, we can find some $n$ points of a basis $a_1,...,a_n$ of that subspace such that its dual is $f_1,...,f_n$, and each $a$ of which is a linear combination of some elements in $S$. If those $a_i$ elements are in $S$, we are done. But I am having trouble proving the result if at least one $a_i$ is not in $S$.

User New
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  • I haven't thought it through entirely, but have you considered looking at the (algebraic) dual of $W$? Then the elements $x \in S$ form functionals $\hat{x}$ with $\hat{x}(f) = f(x)$, but lie in the finite dimensional space $W^*$. I think then you could construct a biorthogonal sequence that way. – Theo Bendit Jun 26 '17 at 02:33
  • @Theo Bendit thank you, but I am not sure what you mean. Could you explain more in detail? I also do not know whay a bioethogonal sequence is. – User New Jun 26 '17 at 02:54
  • A biorthogonal sequence in a vector space $V$ is a sequence $(v_n, v^_n) \in V \times V^$, where $V^$ is some dual space, algebraic or topological, such that $v_i^(v_j) = \delta_{i,j}$. Since we don't have a norm, I'm talking about the algebraic dual, i.e. the space of linear functions from $V$ to $F$. It's not hard to show that a finite-dimensional space has a biorthogonal sequence. I was thinking, since $\hat{x} \in V^$ for all $x \in S$, you might be able to construct a biorthogonal sequence using only elements of $V^$ of the form $\hat{x}$ for some $x$. Maybe it won't work though. – Theo Bendit Jun 26 '17 at 03:16
  • In fact, any basis in $V$ can be extended to a biorthogonal sequence by choosing appropriate elements of $V^*$ to pair with $V$. – Theo Bendit Jun 26 '17 at 03:18

1 Answers1

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Let's show the result holds for $n=1$.

If $f\in W$, $f\ne0$, then $f(x_1)\ne0$, for some $x_1\in S$. Then the function $$ f_1=(f(x_1))^{-1}f $$ satisfies the requirement.

Suppose the statement holds for $(n-1)$-dimensional subspaces of $V(S;F)$ and choose an $(n-1)$-dimensional subspace $W'$ of $W$.

By inductive hypothesis, we are able to find $x_1,\dots,x_{n-1}\in S$ and $f_1,\dots,f_{n-1}\in W'$ with $f_i(x_j)=\delta_{ij}$.

The set $\{f_1,\dots,f_{n-1}\}$ is linearly independent (easy check). Consider the map $\varphi\colon W\to F^{n-1}$, $\varphi(f)=(f(x_1),\dots,f(x_{n-1}))$. This map is surjective (prove it), so its kernel has dimension $1$.

Can you finish?

egreg
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