CorollaryStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-11

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$\dim_{F} M_{m \times n} (F) = m n$ and $\dim_{F} L (V, W) = (\dim_{F} V) (\dim_{F} W)$ for finite-dimensional $V, W$

Statement

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

$\dim_{F} L (V, W) = (\dim_{F} V) (\dim_{F} W) .$

Facts & Assumptions

Given: A field $F$ , naturals $m, n$ , and finite-dimensional spaces $V, W$ with $\dim_{F} V = n$ and $\dim_{F} W = m$ .

[L1]

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

[L2]

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

[L3]

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

Proof

technique · direct

1.1givenL1

Every matrix $A = (a_{i j})$ has the expansion $A = \sum_{(i, j) \in m \times n} a_{i j} E_{i j}$ , and a linear relation among the $E_{i j}$ has each coefficient zero when its corresponding entry is read. Thus the matrix units form a basis.

2.1step 1.1L1L2

By [L2], the index set $m \times n$ has cardinality $m n$ , so this basis has $m n$ elements; if either dimension is zero, it is the empty basis of the zero matrix space. Hence $\dim_{F} M_{m \times n} (F) = m n$ .

3.1step 2.1L3∎

The isomorphism in [L3] transports a basis and preserves dimension; substituting $n = \dim_{F} V$ and $m = \dim_{F} W$ gives the second formula.

$\dim_{F} M_{m \times n} (F) = m n$ and $\dim_{F} L (V, W) = (\dim_{F} V) (\dim_{F} W)$ for finite-dimensional $V, W$

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources

dim⁡FMm×n(F)=mndimF​Mm×n​(F)=mn and dim⁡FL(V,W)=(dim⁡FV)(dim⁡FW)dimF​L(V,W)=(dimF​V)(dimF​W) for finite-dimensional V,WV,W

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources

$\dim_{F} M_{m \times n} (F) = m n$ and $\dim_{F} L (V, W) = (\dim_{F} V) (\dim_{F} W)$ for finite-dimensional $V, W$