Let $U \le V^{\ast}$. If $f(u) = 0$ for all $f \in U$ implies $u = 0$, then $U = V^{\ast}$.

Question

Let $V$ be a finite-dimensional vector space over $K$, and let $V^* := \{ f : V \to K : \mbox{ f is linear } \}$ be the space of linear functionals, the so called dual space of $V$. Also let $U \le V^*$ be a subspace of the dual space, set $$ W = \{ v \in V : f(v) = 0 \mbox{ for all } f \in U \}. $$ Is there an easy way to show that $W = 0$ implies $U = V^*$?

Answer 1

First , dual space $V^*$ is also a finite linear space , so is subspace $U$.

To be precise , $V^* \cong V$ . Let's assume that $U \not= V^*$

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Lemma: For every basis $\{f_1 , ...,f_n\}$ of dual space $V^*$ , there is a corresponding basis $\{v_1 , ...,v_n\} $ in $V$ satisfies $f_i({v_j})=\delta_{ij} \text{ where } i,j\in \{1,..,n\}$

Proof: Consider $V^{**}$ , the dual space of $V^*$. We can find the dual basis of $\{f_1 , ...,f_n\}$ in $V^{**}$, namely , $\{\bar v_1 , ...,\bar v_n\} $ with $ \bar v_i (f_j)=\delta_{ij}$ . By the equivalence $V^{**} \cong V^* \cong V$ , there is a isomorphism $\pi : V^{**} \rightarrow V$ . Let $v_i = \pi (\bar v_i)$ , the set $\{v_1 , ...,v_n\} $ is actually the basis of $V$ and satisfies the condition $f_i (v_j)=\delta_{ij}$

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$U$ have a basis $\{f_1 , ...,f_n\}$ corresponding to the elements $\{v_1 , ...,v_n\} $ in $V$

$A=Span\{v_1 , ...,v_n\}$ is a subspace of $V$

Extent $\{f_1 , ...,f_n\}$ to get the basis of $V^*$ , namely , $\{f_1,...,f_n,f_{n+1},...,f_{n+m}\}$

By lemma, we know that the corresponding elements $\{v_1,...,v_n,v_{n+1},...v_{n+m}\}$ is the basis of $V$ and $f_i({v_j})=\delta_{ij} \text{ where } i,j\in \{1,..,n+m\}$

Therefore, $span\{v_{n+1},...v_{n+m}\}\subset Ker(f_i) $ for $i\in \{1,...,n\}$

This result induces that $span\{v_{n+1},...v_{n+m}\}\subset W$ , contradicting to the condition $W=0$