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

  • Thread starter ssayani87
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  • #1
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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.
 

Answers and Replies

  • #2
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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.
 
  • #3
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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!
 
  • #4
AlephZero
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Google for "Hilbert Hotel paradox". That should explain why the answer is "no".
 
  • #5
Bacle2
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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.
 

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