Could one make a ring of matrices of uncountable size?

Question

I've seen several kinds of matrices. You could see a real square matrix as a mapping: $$ A \quad : \quad \{1, 2,\cdots, n \}^2 \ \longrightarrow \ \mathbb{R} $$ I've seen that there were also infinite matrices like these: $$ A \quad : \quad \mathbb{Z} \times \mathbb{Z} \ \longrightarrow \ \mathbb{R} $$ So I wondered if we could make an uncountable matrix like this one: $$ A \quad : \quad[0,1] \times [0,1] \ \longrightarrow \ \mathbb{R} $$ Equipped with the following matrixproduct: $$ f \circ g \quad : \quad (x_0,y_0) \ \longmapsto \ \int_0^1f(t,y_0)g(x_0,1-t) dt $$ The function that maps everything to $0$ could be seen as the $0$-element. I wondered if this could become a ring with with some neutral matrix. I'd say that the following map described should somehow be this map, but the required features of a unitary element don't hold. $$ f(x,y) = 0 \ \text{ if } \ x \neq 1-y \qquad \text{ and } \qquad f(x,1-x) = 1 \ $$ Do you think that we can find another unitary element to make this a ring?

Answer 1

Suppose you have a vector space of dimension $\kappa$ for whatever cardinality you want. Then you can model its linear transformations as row-finite matrices with entries indexed y $\kappa\times\kappa$ operating on the right of row vectors.(Row finite means each row has only finitely many nonzero entries.)

It is virtually the same construction you do for finite dimensional spaces, and it still works for all vector spaces. This is the 'obvious' use of this type of matrix ring. You can try to invent another product as you are attempting, but this is the most direct way.

Of course the column finite matrices are also a ring, as is the intersection of these two rings.