Do non-square infinite matrices exist?

Question

Sorry, I tried to wrap my head more around this, but I failed.

Given non-square matrix $A$ that has dimension $kn \times n$. Now let $n$ goto infinity.

Is the matrix finally square?

Answer 1

For any two cardinal numbers $\alpha$, $\beta$, we can talk about $\alpha \times \beta$ matrices. In particular, we can have infinite row vectors of dimension $1 \times \aleph_0$, or even longer infinite row vectors of dimension $1 \times \aleph_1$ and so forth.

(if we so desired, we could use any set for the dimensions rather than simply using cardinal numbers. I only mention the cardinal numbers to retain the familiar notion of there being a row $0$, a row $1$, a row $2$, and so forth -- e.g. $\{0,1,2\}$ is the cardinal number $3$. But then again, if you start indexing at row $1$ and don't have a row $0$, I suppose you're already using the generalization of using arbitrary sets, since $\{ 1, 2, 3 \}$ isn't a cardinal number)

We can even have matrices of dimension $\aleph_0 \times \aleph_1$ which are infinite in both dimensions, but still aren't "square" in the sense that the dimensions are different.

Of course, if you want to do algebra with matrices, you'll have to restrict to some subset of these; e.g. the one that works best is the "column finite" matrices where every column only has a finite number of nonzero entries, since the column finite $\alpha \times \beta$ matrices with entries in the field $\mathbf{K}$ correspond to linear transformations $\mathbf{K}^\beta \to \mathbf{K}^\alpha$.

Another popular choice for doing algebra is to put an ordering on the rows and columns so the dimensions are ordinal numbers rather than cardinal numbers. By using convergence, we can define some products of matrices over the reals or complexes where each column only has $\aleph_0$. (we don't need the orderings on the rows and columns if we limit ourselves to only defining matrix produces when there is absolute convergence)

Answer 2

I don't know a lot of formal mathematics, but think about this: There are an infinite number of 1 x n matrices, an infinite number of 2 x n matrices... etc. Therefore, an infinite number of non-square matrices do exist.

I don't know if you are asking if there are an infinite number of matrices that aren't squares or if there are matrices with an infinite number of elements that exist. I will try and answer both.

The question in the title is a bit different than the question in the description, so I will try and answer that now: look at an equation for the ratio of the two sides if, for example, the matrix is 2n x n. This equation would be $f(n) = \frac{2n}{n}$. The limit as n approaches infinity is 2, so the shape would not approach a square.

This can be applied to any matrix. For all integers k, the limit as n approaches infinity of $f(n) = \frac{kn}{n}$ = k, which is just a constant. This goes to show that no matter how big the matrix gets, there will always be more elements in the $kn$ row/column than the $n$ row/column.

Answer 3

I might be taking some creative license on what you meant by "square matrix", but in a way the answer is yes. A finite matrix is really just a linear transformation between finite dimensional vector spaces, and a square matrix is then just a linear transformation between finite dimensional vector spaces of equal dimension (hence, between isomorphic vector spaces). And it's of course true that non-isomorphic infinite dimensional vector spaces with dimensions of different cardinality exist, i.e., countable and uncountable. Linear transformations between two such spaces would be the analogue of non-square infinite dimensional matrices here.

Answer 4

In the sense of matrices that are written down entry by entry, and if you require that both dimensions be infinite, then no they don't exist.

But their cousins, that is linear maps from one vector space to another, can exist.

Take a vector space $V$ with uncountable dimension, and $W$ to have countable dimension. Then any linear map $T:V\to W$ could be thought of as being associated with an infinite-dimensional non-square matrix....

But you couldn't write down the entries (even if there was a pattern).

Answer 5

Look into $\mathcal{l}^p$ spaces. These are sequences $(x_1,x_2,...)$ such that $\sum_{k=1}^\infty |x_k|^p<\infty$ We can think of these are vectors and of course vectors are matrices with one row and in in this case a countable number of columns.