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It's a known fact that $\mathrm{Ann}(S^\circ)=S$, where $S$ is a subspace of a finite dimensional vector space $V$. I'll include the definitions for the sake of completeness, since $\mathrm{Ann}(S)$ (sometimes notated $^\circ S$ and called the pre-annihilator or joint kernel of $S$) hasn't got a more or less standard definition (for more on this, check this MO thread).

Given a vector space $V$ and a subspace $S\subseteq V$, we define:

$S^\circ=\{\varphi\in V^\star:\varphi(S)=0\}$

Similarly, given a subspace $S^\star\subseteq V^\star$, we define:

$\mathrm{Ann}(S^*)=\{v\in V:\varphi(v)=0\,\,\forall\varphi\in S^\star\}$

On the other hand, if we're dealing with an infinite dimensional vector space, we can only affirm that $S\subseteq \mathrm{Ann}(S^\circ)$. For instance, the inclusion is strict if we work under a normed vector space and choose $S\neq V$ to be any dense subspace (it follows from continuity that any linear functional that vanishes at $S$ must be identically $0$, and therefore $S\subseteq\mathrm{Ann}(S^\circ)=\mathrm{Ann}(\{0\})=V$).

However, a friend of mine sketched a proof that demonstrates that the equality always holds (regardless of the dimension of $V$), and I can't seem to find a hole in neither his argument or the one showed above. It goes like this:

Given $S$, pick any basis $B$ of $S$ and extend it to a basis of $V$ by adding linearly independent vectors outside of $S$, thus forming a basis $B'$ of V. Then, we can define a linear functional $\varphi$ such that $\varphi(v)=0$ for every $v\in B$ and $\varphi(w)=1$ for every $w\in B'\setminus B$. Then, since $\varphi\in S^\circ$, there's no vector $v\in V\setminus S$ such that $v\in\mathrm{Ann}(S^\circ)$. Therefore $S\supseteq\mathrm{Ann}(S^\circ)$, and the other inclusion is easily proved; thus, the equality holds.

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If $w$ and $w'$ are in $B'\setminus B$, $\phi(w-w')=0$, so $w-w'\in\ker\phi$. Notice that $w-w'\in V\setminus S$. –  Mariano Suárez-Alvarez May 9 '12 at 2:50
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(Your example with a normed space only holds if you only look at continuous functionals $\phi$; if you are doing linear algebra, so looking at arbitrary functionals, then it is not true that a functional vanishing on a dense subspace is zero) –  Mariano Suárez-Alvarez May 9 '12 at 2:53
    
@MarianoSuárez-Alvarez: Indeed, I thought of that but I have absolutely no background dealing with normed vector spaces; though I kinda hoped linear functionals would be continuous (totally informal, but I have a hard time picturing something linear and discontinuous). –  F M May 9 '12 at 3:00
    
Linear functionals are continuous if their domain is finite dimensional. Let $V=C^1[-1,1]$ the be space of continuous real functions on $[-1,1]$ which are differentiable on $(0,1)$ and put on it the $\sup$-norm, turning it into a normed space. Then the function $f\in V\mapsto f'(0)\in\mathbb R$ is linear but not continuous. –  Mariano Suárez-Alvarez May 9 '12 at 3:02

2 Answers 2

up vote 2 down vote accepted

$\newcommand\Ann{\operatorname{Ann}}$Let $V$ be a vector space and let $S\subseteq V$ be a subspace. Suppose there is a vector $v\in\Ann(S^\circ)\setminus S$. There exists a linear map $\phi:V\to\mathbb R$ such that $\phi|_S=0$ and $\phi(v)=1$: to construct it, let $B$ be a basis of $S$; then $B'=B\cup\{v\}$ is linearly independent, and we can find a basis $B''$ of $V$ which contains $B'$. There is a linear map $\phi:V\to\mathbb R$ such that $\phi(b)=0$ for all $b\in B''\setminus\{v\}$ and $\phi(v)=1$. But then $\phi\in S^\circ$ and this is absurd, because then $v\in\Ann(S^\circ)\subseteq\ker\phi$.

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Great answer - thanks! –  F M May 9 '12 at 3:18

Questions to ask yourself about the sentence "Then, since $\varphi \in S^{\circ}$ , there's no vector $v \in V \setminus S$ such that $v \in \operatorname{Ann}(S^{\circ})$":

  • How is the "there's no vector..." part of this sentence specific to the functional $\varphi$ just mentioned? (It seems completely independent of it?)

  • How is the assertion "there's no vector $v \in V \setminus S$ such that $v \in \operatorname{Ann}(S^{\circ})$" any different from the statement you are trying to prove? Why is it true?

It may help your intuition to note that, in general, the kernel of the functional $\varphi$ you have just mentioned need not be $S$. (If $B' \setminus B$ has two distinct elements $w$ and $w'$, then $w - w'$ is in $\ker(\varphi)$ also.)

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