LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-16

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 quotient basis lifts to a basis adapted to $W$

Statement

Let $V$ be finite-dimensional, let $W \leq V$ , let $(w_{1}, \dots, w_{r})$ be an ordered basis of $W$ , and let $(v_{1} + W, \dots, v_{s} + W)$ be an ordered basis of $V / W$ . Then $(w_{1}, \dots, w_{r}, v_{1}, \dots, v_{s})$ is an ordered basis of $V$ . Consequently, $\dim (V / W) = \dim V - \dim W .$

Facts & Assumptions

Given: The spaces, bases, and representatives in the Statement.

[L1]

The canonical projection $π : V \to V / W$ is linear and surjective with kernel $W$ (Coset equality, well-defined quotient operations, and the canonical projection with kernel $W$ ).

[L2]

A basis is a linearly independent spanning family, with the empty family a basis exactly for the zero space (Basis of a vector space: a linearly independent spanning subset; and ordered basis: an injective finite list whose image is a basis).

[L3]

The dimension of a finite-dimensional vector space is the size of any finite basis, and the zero space has dimension $0$ (Finite-dimensional vector space, and its dimension $\dim_{F} V$ ; infinite-dimensional means having no finite basis).

Proof

technique · direct

1.1L1L2

If $\sum_{i} a_{i} w_{i} + \sum_{j} b_{j} v_{j} = 0$ , applying $π$ gives $\sum_{j} b_{j} (v_{j} + W) = 0$ ; independence of the quotient basis forces every $b_{j} = 0$ , and independence of the basis of $W$ then forces every $a_{i} = 0$ .

2.1step 1.1L1L2L3∎

For $v \in V$ , expand $π (v) = \sum_{j} b_{j} (v_{j} + W)$ ; then $v - \sum_{j} b_{j} v_{j} \in \ker π = W$ and is a combination of the $w_{i}$ , so the displayed independent family spans $V$ and is a basis; counting its $r + s$ members gives $\dim V = \dim W + \dim (V / W)$ , including $W = 0$ , $W = V$ , and $V = 0$ .

Depends on

Used by

Dependency tree · two levels

22 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

S. Axler, Linear Algebra Done Right, 4th ed., Result 3.105 (standard reference, not scraped)

A quotient basis lifts to a basis adapted to WW