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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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    "Duel" is when two people try to kill each other over a matter of "personal honor".2011-09-01
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    Do you know the relation between the dimension of $\mathbf{V}$ and the dimension of $\mathbf{V}^*$?2011-09-01
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    Oh yes.. thanks for pointing that out. They are the same.2011-09-01
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    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.2011-09-01
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    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.2011-09-01
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    Ah yes, I apologize about the later, I completely understand. I have tried to be careful to as it has in the past not let me post things due to 'the potential for them to be discussed not answered' I will correct these mistakes2011-09-01
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    Why did I not see that.. I've been trying all sorts of overly complicated nonsense. Thanks!2011-09-01
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    Might I make a suggestion? Post your complete proof as an answer. Then people can give you pointers if there is something that needs to be cleaned up. It will also prevent this question from going "unanswered."2011-09-01
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    Ok, I will do, but it won't look very pretty as I am unsure as to how to use latex2011-09-01
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    An excellent excuse to practice it, then. Then others can edit, and by right-clicking on a formula and selecting "Show Source", you can see how to write it out yourself....2011-09-01
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    So is latex a language you need to learn or are there visual programs you can use to create the code quickly? Because I do not intentionally want to inconvenience people here.2011-09-01
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    I honestly don't know if there are visual programs that will let you translate easily (I know one that can offer symbol-by-symbol suggestions); I've been using TeX since 1987, so it's even hard to remember how to learn it.2011-09-01
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    Impressive. Unfortunately i'm currently trying to learn another coding language and make a stab at next terms work before October so I might have to leave it for the moment.2011-09-01
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    Do you have any idea how you would generalize this to a vector space of polynomials degree $\leq $ n over F? You can still only take 4 distinct elements in F? Would that not mean we would need to find a new method for showing independence?2011-09-01
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    @LHS You might want to look into [LyX](http://www.lyx.org/), 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.2011-09-01
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    @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 $V^*$.2011-09-01
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    LHS, it might be good for you to use [this](http://www.codecogs.com/latex/eqneditor.php) to aid you in constructing $\LaTeX$ expressions.2011-09-02
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    @Arturo Magidin: Thanks, that was what I was thinking.2011-09-02
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    @J.M.: That looks fantastic! Many thanks.2011-09-02

1 Answers 1

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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?2011-09-01
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    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)2011-09-02