Alphabeta Math
CorollaryStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-11
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dim⁡FMm×n(F)=mn and dim⁡FL(V,W)=(dim⁡FV)(dim⁡FW) for finite-dimensional V,W

Statement

For a field F and naturals m,n, dim⁡FMm×n(F)=mn. Consequently, for finite-dimensional F-vector spaces V,W,

dim⁡FL(V,W)=(dim⁡FV)(dim⁡FW).

Facts & Assumptions

Given: A field F, naturals m,n, and finite-dimensional spaces V,W with dim⁡FV=n and dim⁡FW=m.

[L1]

The matrix units Eij have one entry equal to 1 and all other entries equal to 0 (Matrix units Eij and the Kronecker delta).

[L2]

A Cartesian product of finite sets has cardinality equal to the product of their cardinalities (The product rule: ∣A×B∣=∣A∣ ∣B∣, and ∣∏i<mAi∣=∏i<m∣Ai∣).

[L3]

Relative to ordered bases, matrix representation is a vector-space isomorphism L(V,W)≅Mm×n(F) (T↦[T]BC is a vector-space isomorphism L(V,W)≅Mm×n(F)).

Proof

technique · direct
1.1

Every matrix A=(aij) has the expansion A=∑(i,j)∈m×naijEij, and a linear relation among the Eij has each coefficient zero when its corresponding entry is read. Thus the matrix units form a basis.

givenL1
2.1

By [L2], the index set m×n has cardinality mn, so this basis has mn elements; if either dimension is zero, it is the empty basis of the zero matrix space. Hence dim⁡FMm×n(F)=mn.

step 1.1L1L2
3.1

The isomorphism in [L3] transports a basis and preserves dimension; substituting n=dim⁡FV and m=dim⁡FW gives the second formula.

step 2.1L3∎

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