Alphabeta Math
CorollaryStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-11
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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.
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These labels describe origin, not correctness: citations and verification chips remain separate evidence.

For an m×nm\times n matrix AA, rank⁡(A)+dim⁡N(A)=n\operatorname{rank}(A)+\dim N(A)=n

Statement

For A∈Mm×n(F)A\in M_{m\times n}(F), rank⁡(A)+dim⁡FN(A)=n.\operatorname{rank}(A)+\dim_F N(A)=n.

Facts & Assumptions

Given: The linear map LA:Fn→FmL_A:F^n\to F^m, x↦Axx\mapsto Ax.

[L1]

Matrix rank equals the rank of LAL_A (The rank of a matrix equals the rank of the linear map x↦Axx\mapsto Ax).

[L2]

Rank–nullity gives dim⁡V=rank⁡T+dim⁡ker⁡T\dim V=\operatorname{rank}T+\dim\ker T for a linear map with finite-dimensional domain (Rank-nullity: dim⁡FV=nullity⁡T+rank⁡T\dim_F V=\operatorname{nullity}T+\operatorname{rank}T).

[L3]

N(A)={x:Ax=0}N(A)=\{x:Ax=0\} (Row space, column space, nullspace, row rank, column rank and matrix rank).

Proof

technique · direct
1.1

Apply rank–nullity to LA:Fn→FmL_A:F^n\to F^m to obtain n=rank⁡LA+dim⁡ker⁡LAn=\operatorname{rank}L_A+\dim\ker L_A.

L2
2.1

By [L1] the first term is rank⁡A\operatorname{rank}A, and by [L3] the kernel is N(A)N(A). Substitution proves the formula, including n=0n=0.

step 1.1L1L3∎

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 61 results over 17 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