LemmaStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-11

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.

Row-equivalent matrices have the same row space and the same row rank

Statement

Row-equivalent matrices over a field have the same row space. Consequently they have the same row rank.

Facts & Assumptions

Given: Matrices $A$ and $B$ of one shape, with $B$ obtained from $A$ by one elementary row operation.

[L1]

The row space is the span of the rows and the row rank is its dimension (Row space, column space, nullspace, row rank, column rank and matrix rank).

[L2]

Every elementary row operation has an elementary inverse (Every elementary row operation has an elementary inverse, so row equivalence is an equivalence relation).

[L3]

An elementary row operation replaces rows by the three stated swap, scaling or row-addition formulas (Elementary row operations and row equivalence for finite matrices over a field).

Proof

technique · direct

1.1

Every row of $B$ is a row of $A$ , a nonzero scalar multiple of one, or a linear combination of two rows of $A$ . Hence $\operatorname{Row}(B)\subseteq\operatorname{Row}(A)$ .

L1L3

2.1

Applying step 1.1 to the inverse operation from [L2] gives the reverse inclusion. Thus the row spaces, and therefore their dimensions and row ranks, are equal; iteration covers row equivalence.

step 1.1L1L2∎

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 33 results over 13 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

J. Hefferon, Linear Algebra, 4th ed., Ch. Three, §IV.4 (standard reference, not scraped)