# Finding the basis of a dual space with respect to a vector space of polynomials of at most degree 3 over a field F

Let $\mathbf{F}$ be a field with at least 4 elements and $\mathbf{V}$ be the vector space over $\mathbf{F}$of all polynomials of degree $\leq$3

For $a\in \mathbf{F}$, define $f_a\colon \mathbf{V}\to \mathbf{F}$ by $f_a(p):=p(a)$.

If $a_1$, $a_2$, $a_3$, $a_4$ are distinct members of $\mathbf{F}$, how would you show that the set $\{f_{a_1}, f_{a_2}, f_{a_3}, f_{a_4}\}$ is a basis for $\mathbf{V}^*$ (the dual space of $\mathbf{V}$)? I have found that the set is linearly independent, however am struggling on proving it spans V*

Any help greatly appreciated.

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Do you know the relation between the dimension of $\mathbf{V}$ and the dimension of $\mathbf{V}^*$? – Arturo Magidin Sep 1 '11 at 16:57
I wanted to let you know a few things about MathSE. We also like to know what you've tried on a problem, so it's best to put that into the question, not a comment. These sort of pleasantries usually result in more and better answers. Finally, I should add that posting questions in the imperative (i.e. Compute all such...) is considered rude by some of the members, so I advise you to change that wording. – Arturo Magidin Sep 1 '11 at 16:59
They are the same (in the finite dimensional case). If you've proven the set is linearly independent, then all you need to do now is show that the dimension of $\mathbf{V}$ is $4$ and you'll be done. – Arturo Magidin Sep 1 '11 at 17:00
@LHS You might want to look into LyX, although learning enough TeX to post something like this might be less work, in the end. As to your generalization, certainly $4$ elements will no longer suffice if $n > 4$ (and it will be too many if $n$ is smaller than $4$). Think of how the number $4$ is used in your proof below and make a conjecture. – Dylan Moreland Sep 1 '11 at 18:12
@LHS: The obvious generalization is that if $V$ is the vector space of polynomials of degree at most $n$ over a field with at least $n+1$ elements, and $a_1,\ldots,a_{n+1}$ are pairwise distinct elements of $F$, then the maps $f_{i}(p) = p(a_i)$ are a basis for $V^*$. – Arturo Magidin Sep 1 '11 at 18:54

## 1 Answer

Define $B:=\{f_{a_1},f_{a_2},f_{a_3},f_{a_4}\}$ and let $$p_i(v)= \left\{\begin{array}{ll} 1 & \text{if }v=a_i\\ 0 & \text{otherwise.} \end{array}\right.$$

Consider $u_1f_{a_1}+u_2f_{a_2}+u_3f_{a_3}+u_4f_{a_4}=0$. By evaluating at $p_i$ for $i=1,2,3,4$ we find $u_i=0$. Therefore, $B$ is linearly independent.

Also, as the dimension of $\mathbf{V}^*$ is 4 we have found a set of 4 linearly independent elements in $\mathbf{V}^*$; therefore $B$ forms a basis for $\mathbf{V}^*$

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Let me ask you this: After you define $p_i$, what do you do with it? I don't see you doing anything with it, which begs the question: why bother defining it in the first place, if you are not going to be using it for anything? – Arturo Magidin Sep 1 '11 at 18:58
I incorrectly said evaluate at $a_i$ instead of $p_i$ as $f_a\colon\mathbf{V}\to\mathbf{F}$ it couldn't be evaluated at $a_i$ (Look i'm learning :P) – Freeman Sep 2 '11 at 12:10