ExampleConstruction: AI-adaptedVerification: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-11

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Chosen bases exhibit $\mathbf{Mat}_F$ as equivalent to finite-dimensional vector spaces

Example

Let $\mathbf{Mat}_F$ have natural numbers as objects and $m\times n$ matrices as morphisms $n\to m$ . The coordinate functor identifies it, up to equivalence, with the category of finite-dimensional $F$ -vector spaces.

Facts & Assumptions

Given: A field $F$ and, for every finite-dimensional $F$ -vector space, a supplied ordered basis.

[L1]

Finite-dimensional vector spaces form a full subcategory of $\mathbf{Vect}_F$ (Subcategory and full subcategory, Vector spaces over a fixed field and linear maps form the large locally small category $\mathbf{Vect}_F$ ).

[L2]

Matrix multiplication is associative and has identity matrices, including the zero-dimensional cases (Rectangular matrix multiplication and the identity matrix $I_n$ , including zero-sized shapes, Matrix multiplication is associative, unital, distributive, and compatible with scalar multiplication).

[L3]

Linear maps in chosen coordinates correspond bijectively to matrices, and composition corresponds to matrix multiplication (Coordinate columns $[v]_{\mathcal B}$ and matrices $[T]_{\mathcal B}^{\mathcal C}$ of linear maps relative to ordered bases, $T\mapsto[T]_{\mathcal B}^{\mathcal C}$ is a vector-space isomorphism $\mathcal L(V,W)\cong M_{m\times n}(F)$ , $[S\circ T]_{\mathcal B}^{\mathcal D}=[S]_{\mathcal C}^{\mathcal D}[T]_{\mathcal B}^{\mathcal C}$ ).

[L4]

Equal dimension characterizes isomorphism of finite-dimensional spaces (Two finite-dimensional vector spaces over $F$ are linearly isomorphic if and only if they have the same dimension).

[L5]

A fully faithful, split essentially surjective functor is an equivalence (Covariant functor, identity functor, composite functor, and contravariant functor, Faithful, full, fully faithful, essentially surjective, and split essentially surjective functors, Equivalence, quasi-inverse, and adjoint equivalence of categories, A functor is an equivalence exactly when it is fully faithful and split essentially surjective, without Choice).

Verification

technique · direct

1.1

By [L2], matrix multiplication and identity matrices make $\mathbf{Mat}_F$ a category. Define $K:\mathbf{Mat}_F\to\mathbf{FinVect}_F$ by $K(n)=F^n$ and by letting $K(A)$ be multiplication by the matrix $A$ .

L1L2

2.1

Identity and composition are preserved by [L2] and [L3], so $K$ is a functor.

step 1.1L2L3L5

2.2

For every $m,n$ , the map $A\mapsto K(A)$ is the coordinate bijection from $m\times n$ matrices to linear maps $F^n\to F^m$ . Thus $K$ is fully faithful.

step 1.1L3L5

2.3

Suppose an ordered basis has been supplied for each finite-dimensional vector space $V$ . If its length is $n_V$ , the coordinate map $F^{n_V}\to V$ is a specified isomorphism, including when $V=0$ . Hence these choices split essential surjectivity.

step 1.1L3L4L5

3.1

The criterion in [L5] now makes $K$ an equivalence. Thus chosen bases turn arbitrary finite-dimensional spaces into coordinate models without asserting that the two categories are strictly identical.

step 2.1step 2.2step 2.3L5∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 76 results over 19 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

Emily Riehl, Category Theory in Context, Example 1.5.12 (standard reference, not scraped)

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