Fair to say there are twice as many square matrices as rectangular?

ssayani87
Messages
10
Reaction score
0
Fair to say there are "twice" as many square matrices as rectangular?

Is it fair to say that there are at least twice as many square matrices as there are rectangular?

I was thinking something like this...

Let R be a rectangular matrix with m rows and n columns, and suppose either m < n or m > n. Then, we can associate two square matrices with R, namely RRt, and RtR, with Rt being R Transpose.

In other words, for every rectangular matrix there can be associated (at least) two square matrices.

Google brought up nothing, so I figured I would ask it here. It's not for homework or anything; just out of interest.
 
Physics news on Phys.org
Every square matrix IS a rectangular matrix.

If you consider rectangular matrices which are not square matrices only:
For every square matrix S, I can produce an infinite set of rectangular matrices by writing the columns of S once, twice, three times, ... next to each other (like "SSSS" - not a multiplication!).
No, that argument does not work.

There is an infinite amount of matrices, both for square matrices and rectangular matrices (note that the former are a proper subset of the latter). Therefore, intuitive ways to compare their number break down. As another example: There are as many even integers as there are integers.
 
Whoops, I suppose a better way to have phrased my question was "Are there twice as many matrices whose dimensions are the same as those whose dimensions are different," but that's a great answer, thanks!
 
Google for "Hilbert Hotel paradox". That should explain why the answer is "no".
 
To expand a bit on what I think Aleph_0 was getting at:
You should always specify the coefficients you are working with. If your coefficient set has cardinality |S| , then there will be |S|^(m+n) rectangular mxn matrices, since you can use any of the elements of S for any entry.

If you like that type of problem, try to answer if there are more invertible nxn matrices or more singular/non-invertible nxn matrices. And knock yourself out if you find it interesting by finding different choices for the meaning of "more" , in topology, measure, etc.
 
I asked online questions about Proposition 2.1.1: The answer I got is the following: I have some questions about the answer I got. When the person answering says: ##1.## Is the map ##\mathfrak{q}\mapsto \mathfrak{q} A _\mathfrak{p}## from ##A\setminus \mathfrak{p}\to A_\mathfrak{p}##? But I don't understand what the author meant for the rest of the sentence in mathematical notation: ##2.## In the next statement where the author says: How is ##A\to...
The following are taken from the two sources, 1) from this online page and the book An Introduction to Module Theory by: Ibrahim Assem, Flavio U. Coelho. In the Abelian Categories chapter in the module theory text on page 157, right after presenting IV.2.21 Definition, the authors states "Image and coimage may or may not exist, but if they do, then they are unique up to isomorphism (because so are kernels and cokernels). Also in the reference url page above, the authors present two...
##\textbf{Exercise 10}:## I came across the following solution online: Questions: 1. When the author states in "that ring (not sure if he is referring to ##R## or ##R/\mathfrak{p}##, but I am guessing the later) ##x_n x_{n+1}=0## for all odd $n$ and ##x_{n+1}## is invertible, so that ##x_n=0##" 2. How does ##x_nx_{n+1}=0## implies that ##x_{n+1}## is invertible and ##x_n=0##. I mean if the quotient ring ##R/\mathfrak{p}## is an integral domain, and ##x_{n+1}## is invertible then...
Back
Top