Matrices over a commutative ring as category

In the book Categories for the working mathematician of Mac Lane:

For each commutative ring $K$, the set $\mathbf{Matr_K}$ of all rectangular matrices with entries in $K$ is a category; the objects are all positive integers m,n,..., and each $m\times n$ matrix $A$ is regarded as an arrow $A:n\rightarrow m$, with composition the usual matrix product.

Does this category contain as much information as $\mathbf{Matr_K}$? Given an arrow, how do I know which matrice it represents?

• I'm wondering if there's a bit of misstatement about the objects, which you say "are all positive integers." Should we not regard the arrow $A:K^n \to K^m$ instead? Commented Oct 9, 2017 at 18:57
• @hardmath In this category, $n$ is an object that can be interpreted as $K^n$. Commented Oct 9, 2017 at 19:00
• What are you asking? The arrows are labelled by matrices, so by definition no information is lost. Commented Oct 9, 2017 at 19:10
• In that case, I think the answer would be "no" since you could make an equivalence from $\mathbf{Matr}_{\mathbf{K}}$ to itself which corresponds to "changing the basis of a single $n$", which wouldn't in general restrict to the identity on $\operatorname{Hom}_{\mathbf{Matr}_{\mathbf{K}}}(n, m)$. Commented Oct 9, 2017 at 19:48
• @DanielSchepler But can we hope to retrieve the matrix up to some equivalence? Commented Oct 9, 2017 at 19:54

To make the composition you describe ("the usual matrix product") well-defined, we must assign a canonical ordered basis to each object $n$ or $K^n$. The standard ordered basis $(e_1,\ldots,e_n)$ would work nicely.
• How do you need canonical bases just to multiply matrices? I think the point of the example is that $\mathbf{Matr}_{\mathbf{K}}$ isn't immediately defined in terms of functions (even though as you observe, it's easy to see it's equivalent to a full subcategory of the category $\mathbf{Mod}_{\mathbf{K}}$ of $K$-modules). Commented Oct 9, 2017 at 19:18
• I don't need bases to multiply matrices, but I do need bases to "recover" the matrix from a map (arrow) from $K^n$ to $K^m$. Dropping the objects entirely and working only with arrows amounts to the same thing, where the identity arrows correspond to identity matrices. Commented Oct 9, 2017 at 19:24