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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 $U = V^*$?

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

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


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 $f_i (v_j)=\delta_{ij}$


$U$ 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 $W=0$

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  • $\begingroup$ "It is obvious that $p \in W$" - sorry but I do not see that. $p \in W$ is equivalent to $f_i(p) = 0$ for $i = 1, \ldots, n$, but all we know is $p \notin A$ and I do not see in what sense the $u_i$'s and the $f_i$'s are related to enforce that this implies $p \in W$? Surely as $V \cong V^{\ast}$ there exists a correspondence, but this is not canonical and depends on the choice of a basis, so maybe you should state your correspondence more explicitly so I can see that it is indeed obvious? Otherwise it seems to me like a lot of "hand-waving"... $\endgroup$ – StefanH Oct 11 '15 at 14:52
  • $\begingroup$ @Stefan I edit my answer. The key is using the corresponding condition. $\endgroup$ – Yiran Oct 11 '15 at 15:14
  • $\begingroup$ But I do not see why they should hold? If I have some $f : V \to K$ linear, why should there exists some vector $u_1 \in V$ and a possible extension to a basis $u_1, \ldots, u_m$ such that $f(u_1) = 1, f(u_i) = 0$, $i = 2,\ldots, m$? $\endgroup$ – StefanH Oct 11 '15 at 15:25
  • $\begingroup$ Oh , indeed we can firstly extent $\{f_1 , ...,f_n\}$ to the basis $\{f_1 , ...,f_n,f_{n+1},...,f_{n+m}\}$ of $V^*$. Then $f_i(v_j) = \delta_{ij}$ holds for $ i,j\in {1,...,n+m}$. The corresponding elements $\{v_1 , ...,v_n,v_{n+1},...,v_{n+m}\}$ is exactly basis of $V$ $\endgroup$ – Yiran Oct 11 '15 at 15:30
  • $\begingroup$ I am not sure that I get your points. What is the basis of dual space? How the basis element look like? Think about it. Then I believe you can find the way. $\endgroup$ – Yiran Oct 11 '15 at 15:40

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