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PhiNotPi's avatar

Is there a way to factor a matrix into two matrices with this special property? (see details)

Asked by PhiNotPi (12681points) October 10th, 2011

You have a 2×2 matrix M. Is there a way to factor the matrix into two matrices A and B so that M = AB = BA? All matrices should consist of non-negative integers only.

For example, the matrix
[21, 30]
[60, 51]
can be factored into
[1, 4]
[8, 5]
and
[5, 2]
[4, 7]

Any links to relevant sources will be welcomed.

The reason I am interested with this is that if I could factor a matrix into a “prime factorization” that could not be factored any more, then those “prime” factors could be multiplied in any order and I would end up with the original matrix. I think that any matrix would only have one factorization like this, making the factorization unique. Being able to do this is being able to truly factor matrices.

This is NOT homework.

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5 Answers

LostInParadise's avatar

My gut feeling is that this cannot be done. If it could, it would be such an extraordinary result that I am sure that I would have heard of it. I would think that in general integer matrices can not be factored as the product of integer matrices.

ratboy's avatar

If
[w, x] [a, b]      [a, b][w, x]
[y, z] [c, d] and [c, d][y, z]

are identical, then xc = by.

PhiNotPi's avatar

I have written a Perl program to brute-force matrix factorization. I found out that the matrix

[134, 217]
[217, 351]

has a factorization of

[0, 1] ^ 7
[1, 1]
times
[3, 1]
[1, 4]
times
[2, 1]
[1, 3]

It does seem that every matrix has one unique factorization. I also noticed that when you double the difference of a and d, then both b and c are doubled.

LostInParadise's avatar

I have been thinking about your question and I realize that I am not quite sure what you are asking. Suppose for a given M we could find A and B such that M = AB=BA. Further, suppose that A could be similarly factored into A’ and A’’ and B could be factored into B’ and B’’. What then? Just because A’ and A’’ commute does not mean that A’ and B’ commute. If all the submatrices did commute that would imply that for any two matrices M and N, we would have MN = NM, which is clearly not true.

I did a Web search for prime matrices. The only sites that I found that made any mention of them required payment to get additional information. A really good site that I have used to get answers about math is NRICH The site has some really math savvy people. You have to register to ask questions but it does not cost anything and you do not get put on any mailing lists.

BonusQuestion's avatar

Even if you could do the factorization, it wouldn’t be “unique”. One simple example is this: Look at the 2×2 matrix A with [0 1] in the first row and [1 0] in its second row. We have A^2 = 1. So to any factorization you can add even powers of A. You need to somehow eliminate such possibilities.

Generally we only talk about unique factorization when we are dealing with integral domains (which is a commutative ring with identity that has cancellation, but for matrices we do not have cancellation, neither is it commutative). These domains are called UFDs.

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