TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-13

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Assuming choice, $^{\circ} (U^{\circ}) = U$ ; in finite dimension, $\dim U^{\circ} = \dim V - \dim U$

Statement

Assume the axiom of choice. For every subspace $U \leq V$ ,

$^{\circ} (U^{\circ}) = U .$

If $V$ is finite-dimensional, then

$\dim U^{\circ} = \dim V - \dim U .$

Facts & Assumptions

Given: The axiom of choice, an $F$ -vector space $V$ , and a subspace $U \leq V$ .

[L1]

The annihilator $U^{\circ}$ consists of the functionals vanishing on $U$ , and $^{\circ} S$ consists of vectors annihilated by every member of $S$ (The annihilator $U^{\circ} \leq V^{*}$ of $U \leq V$ and the preannihilator $^{\circ} S \leq V$ of $S \leq V^{*}$ ).

[L2]

If $v \notin U$ , some $f \in V^{*}$ vanishes on $U$ and has $f (v) = 1$ (Assuming choice, if $v \notin U \leq V$ , some $f \in V^{*}$ vanishes on $U$ and satisfies $f (v) = 1$ ).

[L3]

The dual of a finite-dimensional space has the same dimension (The dual family of a finite basis is a basis of the dual space, with the same dimension).

[L4]

Rank-nullity gives $\dim X = \dim \ker R + \dim im R$ for a linear map with finite-dimensional domain $X$ (Rank-nullity: $\dim_{F} V = nullity T + rank T$ ).

[L5]

In finite dimension, a basis of a subspace extends without Choice to a basis of the ambient space (If $\dim_{F} V = n$ and $U$ is a linear subspace of $V$ , then $U$ is finite-dimensional, $\dim_{F} U \leq n$ , and $\dim_{F} U = n$ if and only if $U = V$ , clause 3).

Proof

technique · direct

1.1

Every $u \in U$ is killed by every $f \in U^{\circ}$ , so [L1] gives $U \subseteq^{\circ} (U^{\circ})$ . If $v \notin U$ , [L2] gives $f \in U^{\circ}$ with $f (v) = 1$ , so $v \notin^{\circ} (U^{\circ})$ . Hence $^{\circ} (U^{\circ}) = U$ .

L1L2

1.2

Now suppose $V$ is finite-dimensional and consider restriction $R : V^{*} \to U^{*}$ , $R (f) = f ∣_{U}$ . Its kernel is $U^{\circ}$ by [L1]. It is surjective: extend a basis of $U$ to one of $V$ by [L5], prescribe any given functional's values on the former basis, and prescribe zero on the added vectors.

L1L5choose

2.1

Rank-nullity and surjectivity give $\dim V^{*} = \dim U^{\circ} + \dim U^{*}$ . Applying [L3] to $V$ and $U$ , then rearranging, gives $\dim U^{\circ} = \dim V - \dim U$ .

step 1.2L3L4algebra

3.1

Step 1.1 proves the double-annihilator identity in arbitrary dimension under Choice, and step 2.1 proves the finite-dimensional formula, including $U = 0$ and $U = V$ .

step 1.1step 2.1∎

Depends on

Used by

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Sources

H. Pinkham, Linear Algebra, Chapter 6 (standard reference, not scraped)
K. Conrad, Infinite-Dimensional Dual Spaces (standard reference, not scraped)

Assuming choice, ∘(U∘)=U∘(U∘)=U; in finite dimension, dim⁡U∘=dim⁡V−dim⁡UdimU∘=dimV−dimU