Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I'm learning linear algebra on Kunze Hoffman book and stuck for a long time by this problem.Please help me solve this. Thanks:

Let $S$ be a set, $F$ a field, and $V(S,F)$ the space of all functions from $S$ into $F$: $$ (f + g)(x) = f(x) +g(x)$$ $$(cf)(x) = cf(x)$$ Let $W$ be any $n$-dimensional subspace of $V(S,F)$ . Show that there exist points $x_1, x_2, ... x_n$ in $S$ and functions $f_1, f_2, ..., f_n$ in $W$ such that $f_i(x_j) = \delta_{ij}$. Here $$\delta_{ij} = \begin{cases} 1 , & \text {if $i = j$} \\ 0 , & \text {if $i \neq j$} \\ \end{cases} $$

share|cite|improve this question
Have you tried defining functions that satisfy linearity and have this delta-property? Also, what assumptions are you given on $S$ and $F$? Presumably $n\leq\vert S\vert$? – icurays1 Dec 4 '12 at 18:10
@icurays1: the subspace $W$ is given, so you cannot really define things in it. – Martin Argerami Dec 4 '12 at 18:14
Ah yes, good point. – icurays1 Dec 4 '12 at 18:17
up vote 1 down vote accepted

This can be shown by induction on $n$. For $n=1$ choose a nonzero function $f$ in $W$; since $f$ is not the zero function there is $x_1 \in S$ for which $f(x_1)$ is nonzero, now define $f_1=f/f(x_1)$ and you have what is claimed.

Suppose now the result holds at $n$, and let $W$ be an $n+1$ dimensional subspace. We can choose $n+1$ linearly independent functions in $W$, and using the first $n$ of them gives a set of $n$ linearly independent functions which span a space $V$ of dimension $n$.

By the inductive hypothesis there are $x_1,...,x_n$ in $S$ and functions $f_1,...,f_n$ in $V$ for which $f_i(x_j)=\delta_{i,j}$. Now since $W$ is of dimension $n+1$ the above $f_i$ do not span $W$, so that we may choose a function $g$ in $W$ which is not a linear combination of the $f_i$. Define, for $1 \le i \le n$, the numbers $k_i=g(x_i)$. Next define

$$h=g-k_1f_1-...-k_nf_n.$$ Note that $h$ is not a linear combination of the $f_i$ either, since $g$ isn't.

Then $h(x_i)=0$ for $1 \le i \le n$. Now if $h(x)=0$ for all other $x$ in $S$ then $h$ would be the zero function, against $h$ not being a combination of the $f_i$. So we may choose some $x_{n+1}$ for which $h(x_{n+1})$ is nonzero. Now define $$f_{n+1}=\frac{1}{h(x_{n+1})}\cdot h.$$

Then we have now $f_i(x_j)=\delta_{i,j}$ for $1 \le i,j \le n+1$ to finish the induction.

EDIT: A slight adjustment is needed, but it still works. That is, for $1 \le i \le n$ we need to have $f_i(x_{n+1})=0$, which the above construction did not guarantee. However we may go back and subtract an appropriate muliple of the present $f_{n+1}$ from each of the first $n$ of the $f_i$, so as to ensure (for the redefined $f_i$) that $f_i(x_{n+1})=0$ as desired.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.