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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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    @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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    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