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I have heard of infinite-dimensional matrices that have a countably-infinite number of dimension. Is it possible that there could be a matrix with aleph-1, aleph-2, or even aleph-aleph-0 dimensions?

elipson_-1
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    An infinite matrix is basically an operator on a Banach/Hilbert space. And yes, there are Hilbert spaces of uncountable cardinality. – N. S. May 14 '19 at 22:07

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An $m\times n$ matrix $M$ with entries from a field $k$ is really just a representation of a linear transformation - or, more specifically, a way of associating a linear transformation to any appropriate choice of vector spaces over $k$ and bases. We can talk about vector spaces of arbitrarily large (even infinite) dimension as well as bases of such and linear maps between them, so there's no difficulty there.

  • OK fine there's some subtlety here, mainly because once you hit infinite-dimensional spaces you need to be careful about what you mean by "basis" ("Hamel" versus "Schauder"). There are also some subtle set-theoretic issues we normally get to ignore: e.g. the statement "Every vector space has a (Hamel) basis," while uncomplicated by set theory in the finite-dimension case, is in its full generality equivalent to the axiom of choice. But that's sort of a side issue here.

Of course, whether the matrix itself is still a useful idea here isn't as clear. As a tool, matrices are useful because they give compact (intuitively speaking) representations of complicated objects; once the dimension gets infinite, though, it's not clear they provide much advantage.

More abstractly, a matrix is really just a map from a product of two sets (the "vertical" and "horizontal" axes of the matrix) to a third set (= the entries in the matrix). And again these make perfect sense regardless of the cardinalities involved.

So there's absolutely no issue here; the only possible point is that the matrix idea itself may become less convenient as a visualization device as the objects in question get bigger.

Noah Schweber
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  • @MorganRodgers Which "this" do you mean? The real point is that (1) a matrix is really just a map of the form $f:A\times B\rightarrow C$, and (2) if $V,W$ are $k$-vector spaces of dimension $\vert A\vert,\vert B\vert$ then such a map (with $C=k$) determines a way to assign a linear transformation $V\rightarrow W$ to each appropriate choice of basis satisfying all the usual rules. So the same combinatorial gadget (rows + columns), whether we call it a matrix or not, does exactly the same job. – Noah Schweber May 14 '19 at 22:20