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$\dim_F M_{m\times n}(F)=mn$ and $\dim_F\mathcal L(V,W)=(\dim_FV)(\dim_FW)$ for finite-dimensional $V,W$

Statement

For a field $F$ and naturals $m,n$ , $\dim_FM_{m\times n}(F)=mn$ . Consequently, for finite-dimensional $F$ -vector spaces $V,W$ ,

$\dim_F\mathcal L(V,W)=(\dim_FV)(\dim_FW).$

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 $E_{ij}$ have one entry equal to $1$ and all other entries equal to $0$ (Matrix units $E_{ij}$ and the Kronecker delta).

[L2]

A Cartesian product of finite sets has cardinality equal to the product of their cardinalities (The product rule: $\lvert A \times B\rvert = \lvert A\rvert\,\lvert B\rvert$ , and $\big\lvert\prod_{i<m} A_i\big\rvert = \prod_{i<m}\lvert A_i\rvert$ ).

[L3]

Relative to ordered bases, matrix representation is a vector-space isomorphism $\mathcal L(V,W)\cong M_{m\times n}(F)$ ( $T\mapsto[T]_{\mathcal B}^{\mathcal C}$ is a vector-space isomorphism $\mathcal L(V,W)\cong M_{m\times n}(F)$ ).

Proof

technique · direct

1.1

Every matrix $A=(a_{ij})$ has the expansion $A=\sum_{(i,j)\in m\times n}a_{ij}E_{ij}$ , and a linear relation among the $E_{ij}$ 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 $m\times n$ 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 $\dim_FM_{m\times n}(F)=mn$ .

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∎

Depends on

Used by

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Sources

S. Axler, Linear Algebra Done Right, 4th ed., §3C, Dimension of matrix spaces (standard reference, not scraped)
S. Schiavone, MIT 18.700 Day 9, Corollary 27 (standard reference, not scraped)

dim⁡FMm×n(F)=mn\dim_F M_{m\times n}(F)=mndimF​Mm×n​(F)=mn and dim⁡FL(V,W)=(dim⁡FV)(dim⁡FW)\dim_F\mathcal L(V,W)=(\dim_FV)(\dim_FW)dimF​L(V,W)=(dimF​V)(dimF​W) for finite-dimensional V,WV,WV,W