Is the annihilator of the annihilator of a space the space itself? Let $V$ be an $m$-dimensional subspace of the $n$-dimensional vector space $U$ (of course $m\leq n$). Now, my question is if $V^{00}=(V^0)^0=V$ (where $V^0$ is the annihilator of vector space $V$)?
I just started to learn about dual spaces and annihilators and would appreciate a counterexample or proof to explain why. Thanks.
 A: Yes, this is true. You can prove it concretely by choosing a basis $v_1, \dots v_m$ of $V$ and extending it to a basis $v_1, \dots v_n$ of $U$. If $v_1^{\ast}, \dots v_n^{\ast} \in U^{\ast}$, then you can check that the annihilator, which I'll write
$$V^{\perp} = \{ f \in U^{\ast} : \forall v \in V : f(v) = 0 \}$$
has basis given by the dual vectors $v_{m+1}^{\ast}, \dots v_n^{\ast}$. Applying this result a second time gives that $V^{\perp}$ has basis given by $v_1, \dots v_n$ so $(V^{\perp})^{\perp} = V$ as desired.
Abstractly, let $T : V \to U$ be the inclusion. $V^{\perp}$ is the kernel of the dual map $T^{\ast} : U^{\ast} \to V^{\ast}$ (exercise). Starting from the short exact sequence
$$0 \to V \xrightarrow{T} U \xrightarrow{\text{coker}(T)} U/V \to 0$$
dualizing produces the short exact sequence
$$0 \to (U/V)^{\ast} \xrightarrow{\text{ker}(T^{\ast})} U^{\ast} \xrightarrow{T^{\ast}} V^{\ast} \to 0$$
showing that $V^{\perp} \cong (U/V)^{\ast}$; in other words, the annihilator can be identified (naturally in $U$ and $V$) with the dual of the cokernel of the inclusion $T$. Dualizing a second time gives the original short exact sequence back, which gives
$$(V^{\perp})^{\perp} \cong V$$
although there's some work to do to show that all of the natural isomorphisms we just used reduce to the actual literal equality $(V^{\perp})^{\perp} = V$. This argument shows that the conclusion generalizes to finitely generated projective modules over a ring.
