TheoremStatement: AI-adaptedProof: AI-generatedPipeline-generatedjudge pass (gpt-6-luna)

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Dimension and double-dual identity for finite-field code duals

Statement

Let $F$ be a finite field, $n \in N$ , and let $D \leq F^{n}$ be a linear subspace (Linear subspace of a vector space) of dimension $r$ . Then $D^{⊥}$ is a linear subspace of dimension $n - r$ , and $(D^{⊥})^{⊥} = D$ . In particular this holds for every classical linear code, including $D = {0}$ and $D = F^{n}$ (Classical linear code over a finite field).

Facts & Assumptions

Given: A finite field $F$ , $n \in N$ , a subspace $D \leq F^{n}$ of dimension $r$ , and the bilinear dot product of Bilinear dual of a finite-field linear code.

[L1]

Bilinear dual of a finite-field linear code: $D^{⊥}$ consists of vectors whose dot product with every vector of $D$ is zero.

[F1]

Row space, column space, nullspace, row rank, column rank and matrix rank: The row space is the span of matrix rows, the column space is the span of columns, and the nullspace is ${x : A x = 0}$ ; row and column ranks are their dimensions.

[F2]

Row rank equals column rank, and both equal the number of pivots: For every finite matrix over a field, its row rank equals its column rank.

[F3]

Rectangular matrix multiplication and the identity matrix $I_{n}$ , including zero-sized shapes: Matrix products are the finite coordinate sums, including empty shapes.

[F4]

Linear map between vector spaces over the same field: A map between vector spaces over $F$ is linear when it preserves linear combinations.

[F5]

Finite-dimensional vector space, and its dimension $\dim_{F} V$ ; infinite-dimensional means having no finite basis: A finite-dimensional vector space has a finite basis, and its dimension is the cardinality of such a basis.

[F6]

The standard list $e : n \to F^{n}$ with $e_{i} (i) = 1_{F}$ and $e_{i} (j) = 0_{F}$ for $j \neq i$ is an ordered basis of $F^{n}$ ; hence $\dim_{F} F^{n} = n$ , and $F^{0}$ is the zero space with basis $\emptyset$ and dimension $0$ : $F^{n}$ has dimension $n$ , including $F^{0} = {0}$ .

[F7]

Rank-nullity: $\dim_{F} V = nullity T + rank T$ : For a linear map with finite-dimensional domain, domain dimension is kernel dimension plus image dimension.

[F8]

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 $U = V$ : A subspace of a finite-dimensional space has the same dimension as its ambient space exactly when it equals that space.

[F9]

Linear subspace of a vector space: A linear subspace contains zero and is closed under addition and scalar multiplication.

Proof

1.1L1F1F2F3F4F5F9

Since $\dim_{F} D = r$ , [F5] supplies a basis $d_{0}, \dots, d_{r - 1}$ . Form the $r \times n$ matrix $A$ whose $i$ th row is $d_{i}^{T}$ . Its row space has dimension $r$ by [F1]. The map $T_{A} : F^{n} \to F^{r}$ , $y \mapsto A y$ , is linear by [F3] and [F4]; the product formula writes $A y$ as the linear combination of the columns of $A$ with coefficients $y_{j}$ , so its image is the column space. Hence [F2] gives $\dim im T_{A} = r$ . This also works for $r = 0$ , with an empty row list.

2.1L1F3F4F6F7F9step 1.1

A vector $y$ belongs to $\ker T_{A}$ exactly when $d_{i} \cdot y = 0$ for every basis vector $d_{i}$ , which by bilinearity is equivalent to $d \cdot y = 0$ for every $d \in D$ . Thus $\ker T_{A} = D^{⊥}$ . Since $T_{A}$ is linear, this kernel contains zero and is closed under linear combinations, so it is a subspace. By [F6] and [F7], $\dim D^{⊥} = n - r$ .

3.1L1F5F6F7F8step 1.1step 2.1∎

Applying step 1.1 and step 2.1 with $D^{⊥}$ in place of $D$ gives $\dim (D^{⊥})^{⊥} = n - (n - r) = r$ . Symmetry of the dot product gives $D \subseteq (D^{⊥})^{⊥}$ . Apply [F8] to this inclusion inside the $r$ -dimensional space $(D^{⊥})^{⊥}$ ; equal dimensions force equality. This also covers $r = 0$ and $r = n$ .

Remarks

The coordinate pairing is nondegenerate over every field because the standard coordinate rows form a basis. No complex conjugation is used, and no choice principle is needed in finite dimension.

Depends on

Used by

Dependency tree · two levels

52 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Richard Cleve, Quantum Information Processing course notes, §37.1.1 (standard reference, not scraped)
John Preskill, Quantum Computation, Chapter 7, §7.6 (standard reference, not scraped)