LemmaStatement: AI-adaptedProof: AI-adaptedprecheck passaudited 2026-08-26

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.

For a subspace $U \leq F^{n}$ , $\dim_{F} U^{⊥} = n - \dim_{F} U$ , where $U^{⊥} = {x : ⟨ x, u ⟩ = 0 for all u \in U}$

Statement

Let $F$ be a field, let $U \leq F^{n}$ be a subspace, and define

$U^{⊥} : = {x \in F^{n} : ⟨ x, u ⟩ = 0 for every u \in U} .$

Then

$\dim_{F} U^{⊥} = n - \dim_{F} U .$

Facts & Assumptions

Given: a field $F$ , a natural number $n$ , and a subspace $U \leq F^{n}$ .

[F1]

A finite-dimensional subspace has a basis, and its dimension is at most $n$ (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 $U = V$ ).

[F2]

Row rank equals column rank (Row rank equals column rank, and both equal the number of pivots).

[F3]

Rank-nullity gives $\dim_{F} F^{n} = nullity Φ + rank Φ$ for a linear map $Φ : F^{n} \to F^{d}$ (Rank-nullity: $\dim_{F} V = nullity T + rank T$ ).

[F4]

The standard form is $⟨ x, y ⟩ = \sum_{i < n} x_{i} y_{i}$ (The standard bilinear form $⟨ x, y ⟩ = \sum_{i < n} x_{i} y_{i}$ on $F^{n}$ ).

Proof

technique · direct

1.1F1F4construct

Let $u_{1}, \dots, u_{d}$ be a basis of $U$ , where $d = \dim_{F} U$ , and define the linear map $Φ : F^{n} \to F^{d}$ by $Φ (x) = (⟨ x, u_{1} ⟩, \dots, ⟨ x, u_{d} ⟩)$ .

2.1F1F2step 1.1

By definition, $\ker Φ = U^{⊥}$ . The matrix of $Φ$ has the vectors $u_{1}, \dots, u_{d}$ as its rows, so its row rank is $d$ because those rows are independent; hence its column rank is also $d$ by [F2], and therefore $rank Φ = d$ .

3.1F3step 2.1∎

Rank-nullity [F3] now gives $n = \dim_{F} F^{n} = nullity Φ + rank Φ = \dim_{F} U^{⊥} + d$ , so $\dim_{F} U^{⊥} = n - d = n - \dim_{F} U$ .

Remarks

The cases $U = 0$ and $U = F^{n}$ are included automatically: then $U^{⊥} = F^{n}$ and $U^{⊥} = 0$ respectively.

Depends on

Used by

Dependency tree · two levels

48 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

L. Babai and P. Frankl, Linear Algebra Methods in Combinatorics, §2.3.2 (standard reference, not scraped)

For a subspace U≤FnU≤Fn, dim⁡FU⊥=n−dim⁡FUdimF​U⊥=n−dimF​U, where U⊥={x:⟨x,u⟩=0 for all u∈U}U⊥={x:⟨x,u⟩=0 for all u∈U}