Cool stuff! I had previously seen that Vect has a biproduct structure, but never even considered that matrices could be appropriately generalized using it.
One thought I had was about handling potentially infinite-dimensional vector spaces. In that case, if V has a basis {vi:i∈I}, we still get a decomposition into a direct sum of copies of R:
V≅⨁i∈IR,
However, when the indexing set I is infinite, the coproducts and products in Vect differ, so the argument from the post doesn't go through verbatim. But it still feels like we have a (perhaps less useful) notion of matrix here: if V has basis {vi:i∈I} and U has basis {uj:j∈J}, then a linear map T:V→U is still fully determined by the collection {ai,j:(i,j)∈I×J}⊆R, where T(vi)=∑j∈Jai,juj and, for any fixed i∈I, ai,j vanishes for all but finitely many j∈J. Conversely, any collection {ai,j:(i,j)∈I×J}⊆R with the aforementioned... (read more)
Cool stuff! I had previously seen that Vect has a biproduct structure, but never even considered that matrices could be appropriately generalized using it.
One thought I had was about handling potentially infinite-dimensional vector spaces. In that case, if V has a basis {vi:i∈I}, we still get a decomposition into a direct sum of copies of R:
V≅⨁i∈IR,However, when the indexing set I is infinite, the coproducts and products in Vect differ, so the argument from the post doesn't go through verbatim. But it still feels like we have a (perhaps less useful) notion of matrix here: if V has basis {vi:i∈I} and U has basis {uj:j∈J}, then a linear map T:V→U is still fully determined by the collection {ai,j:(i,j)∈I×J}⊆R, where T(vi)=∑j∈Jai,juj and, for any fixed i∈I, ai,j vanishes for all but finitely many j∈J. Conversely, any collection {ai,j:(i,j)∈I×J}⊆R with the aforementioned... (read more)