Alphabeta Math
CorollaryStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-11
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The rank of a matrix equals the rank of the linear map xAxx\mapsto Ax

Statement

For AMm×n(F)A\in M_{m\times n}(F), the matrix rank of AA equals the rank of the linear map LA:FnFmL_A:F^n\to F^m defined by LA(x)=AxL_A(x)=Ax.

Facts & Assumptions

Given: A matrix AMm×n(F)A\in M_{m\times n}(F) and its associated linear map LAL_A.

[L1]

Matrix rank is row rank, which equals column rank (Row space, column space, nullspace, row rank, column rank and matrix rank).

[L3]

The rank of a linear map is the dimension of its image (Rank and nullity of a linear map with finite-dimensional domain).

[L4]

Matrix multiplication is given entrywise by (Ax)i=j<naijxj(Ax)_i=\sum_{j<n}a_{ij}x_j (Rectangular matrix multiplication and the identity matrix InI_n, including zero-sized shapes).

[L5]

Matrix multiplication distributes over addition and is compatible with scalar multiplication (Matrix multiplication is associative, unital, distributive, and compatible with scalar multiplication).

Proof

technique · direct
1.1

Fact [L5] shows that LA(x)=AxL_A(x)=Ax is linear. By [L4], every value AxAx is a linear combination of the columns of AA, with coefficients xjx_j, and every such combination is AxAx for that coefficient column. Hence imLA=Col(A)\operatorname{im}L_A=\operatorname{Col}(A).

L4L5algebra
2.1

Taking dimensions and using [L1]–[L3] gives rankLA=dimCol(A)=rankA\operatorname{rank}L_A=\dim\operatorname{Col}(A)=\operatorname{rank}A.

step 1.1L1L2L3

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 63 results over 18 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