Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I am self-studying Hoffman and Kunze's book Linear Algebra. This is Exercise 3 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 $f_{i}(x_{j})=\delta_{ij}$.

PS:This is the section about the double dual.

share|improve this question
    
You left out a rather important part of the question :-) –  joriki Mar 18 '12 at 14:20
    
@spohreis: What is $W$? –  Rudy the Reindeer Mar 18 '12 at 14:20
    
@MattN. $W$ is any $n$-dimensional subespace of $V(S,\mathbb{F})$. –  spohreis Mar 18 '12 at 14:30
    
@joriki: Thank you for correcting my mistake, but I will blame Sunday morning. :) –  spohreis Mar 18 '12 at 14:31

2 Answers 2

up vote 5 down vote accepted

You can prove this by induction on $n$. Here's a sketch:

Base Case: ($n = 0$ is totally trivial). $n =1$: a one-dimensional subspace of $\mathbb{F}^S$ is the set of scalar multiples of a single not identically zero function $f: S \rightarrow \mathbb{F}$. So there exists some $x \in S$ such that $f(x) \neq 0$, and then by rescaling there exists some $x \in S$ and $\alpha \in \mathbb{F}$ such that $\alpha f(x) = 1$.

Inductive Step: Suppose that the result holds for any $n$-dimensional subspace $W = \langle f_1,\ldots,f_n \rangle$, and now suppose that we add to $W$ one linearly independent function $g$. By induction there is a subset $S_n = \{x_1,\ldots,x_n\}$ of $S$ such that that elements of $W$, when restricted to functions on $S_n$, give all possible functions on $S_n$. Therefore there is some linear combination of the $f_i$'s which induces the same function on $S_n$ as $g$ does, i.e., there are scalars $\alpha_1,\ldots,\alpha_n$ such that $(g - \sum_{i=1}^n \alpha_i f_i)(x_j) = 0$ for all $1 \leq j \leq n$. But since $g$ is linearly independent from $W$, $(g - \sum_{i=1}^n \alpha_i f_i)$ is not the zero function. Can you complete the argument from here?

By the way, I agree that double duality is also relevant. But I think the above approach is more "hands on" -- after proving it this way, one can think about what it means in terms of double dual spaces.

share|improve this answer
    
@Peter L. Clark. Sorry I am slow! Then there is $x_{n+1}\in S$ and $x_{n+1}\notin S_{n}$ such that $(g-\sum_{i=1}^{n}\alpha_{i}f_{i}(x_{i}))=\alpha\neq 0$. Then we put $f_{i+1}=\dfrac{1}{\alpha}(g-\sum_{i=1}^{n}\alpha_{i}f_{i})$. Then $S_{n+1}=\{x_{1},\ldots,x_{n},x_{n+1}\}.$ Am I right? –  spohreis Mar 18 '12 at 17:47
    
@spohreis: yes, essentially. I guess what you wrote as $f_{i+1}$ should be $f_{n+1}$. Then you should go back and modify $f_1,\ldots,f_n$ to make sure that they are all zero at $x_{n+1}$... –  Pete L. Clark Mar 18 '12 at 17:52
    
Then there is $x_{n+1}\in S$ and $x_{n+1}\notin S_{n}$ such that $(g-\sum_{i=1}^{n}\alpha_{i}f_{i})(x_{n+1})=\alpha\neq 0$. Then we put $f_{n+1}=\dfrac{1}{\alpha}(g-\sum_{i=1}^{n}f_{i})$. –  spohreis Mar 18 '12 at 18:03

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.

share|improve this answer

Your Answer

 
discard

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.