# Could one make a ring of matrices of uncountable size?

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?

• You'd have to restrict yourself to integrable $f,g$. There's no identity in these matrices. This stuff comes up in the field of "functional analysis." (Also: If $x\in[0,1]$ then $-x\notin[0,1]$, so maybe you mean $1-x$ and $1-y$?) Presumably $f(x,y)=A(x,y)$... – Thomas Andrews Aug 10 '14 at 18:19
• I suggest you to change "uncountable matrices" by "square matrices with uncountable size" on the title. – Matemáticos Chibchas Aug 10 '14 at 20:30
• I will, thank you for your advise. – Koenraad van Duin Aug 10 '14 at 20:40
• O no! I've messed up the question again. I'm sorry, this is not what I ment to do. I'll readjust it right now. – Koenraad van Duin Aug 10 '14 at 20:51

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.)
• But this kind of ring would not have a $1$, would it? – Koenraad van Duin Aug 10 '14 at 21:14