Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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
(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
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 $v\in\Ann(S^\circ)\subseteq\ker\phi$.

share|cite|improve this answer
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.)

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.