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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?

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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 at 18:19
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I suggest you to change "uncountable matrices" by "square matrices with uncountable size" on the title. –  Matemáticos Chibchas Aug 10 at 20:30
    
I will, thank you for your advise. –  Koenraad van Duin Aug 10 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 at 20:51

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up vote 3 down vote accepted

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.

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But this kind of ring would not have a $1$, would it? –  Koenraad van Duin Aug 10 at 21:14
    
Well, think about it: is the identity transformation a linear transformation? Is the "infinite identity matrix" row finite? ;) –  rschwieb Aug 10 at 21:18
    
No it isn't, I guess. There should be an endless line of ones on the "diagonal" of that matrix, right? –  Koenraad van Duin Aug 10 at 21:27
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Row finite means that each row is finite (i.e. almost all entries are zero). In the identity matrix, every row has only one non-zero entry. –  Martin Brandenburg Aug 10 at 21:40
    
I always get confused here. You want your vectors to have only finitely many nonzero entries as well, don’t you? –  Lubin Aug 10 at 22:11

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