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.

A full-column-rank matrix has image dimension equal to its number of columns

Statement

Let $F$ be a field, let $n, k \in N$ , and let $G \in M_{n \times k} (F)$ . Regard $G$ as the linear map $T_{G} : F^{k} \to F^{n}$ , $x \mapsto G x$ , using column vectors. If the columns of $G$ are linearly independent, then $im T_{G}$ is a $k$ -dimensional linear subspace of $F^{n}$ . This includes $k = 0$ and $n = 0$ .

Facts & Assumptions

Given: A field $F$ , natural numbers $n, k$ , a matrix $G \in M_{n \times k} (F)$ with linearly independent columns, and the associated column-vector map $T_{G} (x) = G x$ .

[F1]

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 $0$ : The standard vectors form an ordered basis of $F^{m}$ , including the empty basis when $m = 0$ , and $\dim_{F} F^{m} = m$ .

[F2]

Rectangular matrix multiplication and the identity matrix $I_{n}$ , including zero-sized shapes: Matrix multiplication is defined entrywise by finite sums, including the empty-sum case.

[F3]

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

[F4]

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

[F5]

Rank-nullity: $\dim_{F} V = nullity T + rank T$ : If $T : V \to W$ is linear and $V$ is finite-dimensional, then $\dim_{F} V = \dim_{F} \ker T + \dim_{F} im T$ .

Proof

1.1F2F3F4given

For $x, y \in F^{k}$ and $a, b \in F$ , the entry formula for matrix multiplication gives $G (a x + b y) = a G x + b G y$ , so $T_{G}$ is linear by [F2] and [F3]. It sends zero to zero and its image is closed under linear combinations, hence its image is a subspace by [F4]; these calculations also cover empty coordinate sets.

1.2givenalgebra

If $x \in \ker T_{G}$ , then $\sum_{j < k} x_{j} g_{j} = 0$ , where $g_{j}$ is column $j$ of $G$ . Independence of the columns forces every $x_{j} = 0$ , so $\ker T_{G} = {0}$ . For $k = 0$ , $F^{0} = {0}$ and the same conclusion holds.

2.1F1F5step 1.1step 1.2∎

By [F1], $\dim_{F} F^{k} = k$ . Applying [F5] to [step 1.1] and [step 1.2] gives $\dim_{F} im T_{G} = k$ . Since the image of $x \mapsto G x$ is the span of the columns, it is the asserted subspace.

Remarks

No Choice axiom is used. The zero-column case is an empty matrix map from the zero space.

Depends on

Used by

Dependency tree · two levels

31 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, §37.1.2 (standard reference, not scraped)
John Preskill, Quantum Computation, Chapter 7, §7.6 (standard reference, not scraped)