Finding accurate inverse of binary circulant matrix

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Corbin
Corbin 2024-6-10
评论： Corbin 2024-6-11
I would like to find the inverse of a binary circulant matrix using MATLAB. I have the 24x24 binary circulant matrix stored in BCM and use the function inv(BCM) and I get a lot of garbage values. I have the inverse matrix that I am expecting to get, which is the multiplicative inverse of BCM. When I multiply them in MATLAB, I get the identity matrix which is correct. However, I'm going to need to calculate new inverses of new matrices and would like to do so with MATLAB instead of guessing over and over. How can I do this?
Both BCM and it's inverse are stored in the attached Excel spreadsheet. Thank you!
Garbage Values:
  2 个评论
Steven Lord
Steven Lord 2024-6-11
Are you trying to take the inverse mod 2 (or in GF(2))?
If so the gf function in Communications Toolbox may be of use to you.
Corbin
Corbin 2024-6-11
I didn't know it then but from @Torsten's answer I know that's exactly what I need now, so thank you very much! You guys are very prompt and quick in your answers!

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采纳的回答

Torsten
Torsten 2024-6-11
Ran in:
BCM=readmatrix('MixingLayer.xlsx');
inv(gf(BCM,2))
ans = GF(2^2) array. Primitive polynomial = D^2+D+1 (7 decimal) Array elements = 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1 0 1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1 0 1 1 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1 0 0 1 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1 1 0 1 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 1 0 1 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 1 0 1 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 1 0 1 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 1 0 1 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 1 0 1 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 1 0 1 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 1 0 1 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 1 0 1 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 1 0 1 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 1 0 1 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 1 0 1 0 0 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 1 0 1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1 0 1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1 0 1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1 0 1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1 0 1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1 0 1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1 0 1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1 0 1 0 0
  1 个评论
Corbin
Corbin 2024-6-11
This is exactly what I was looking for! Thank you very much :)
You're absolutely correct about the 0 and 1. I forgot to mention that.
I didn't know that finite fields were used in the calculation, and I know that now! This is very helpful to know!

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更多回答（2 个）

Matt J
Matt J 2024-6-10
编辑：Matt J 2024-6-10
Ran in:
We don't know what it is that you consider improper about the results of inv(). The test below indicates to me that everything is fine.
BCM=readmatrix('MixingLayer.xlsx');
[low,high]=bounds(BCM*inv(BCM)- eye(size(BCM)) ,'all')
low = -3.6082e-16
high = 2.7756e-16
John D'Errico
John D'Errico 2024-6-10
编辑：John D'Errico 2024-6-11
Ran in:
Gosh. I thought that multiplying a matrix by its inverse gives an identity matrix. :)
A = readmatrix('MixingLayer.xlsx','sheet','BCM');
Ainv = readmatrix('MixingLayer.xlsx','sheet','inverseBCM');
A*Ainv
ans = 24x24
1 0 0 0 0 0 0 0 2 0 2 0 0 0 0 0 0 0 0 0 2 0 2 0 0 1 0 0 0 0 0 0 0 2 0 2 0 0 0 0 0 0 0 0 0 2 0 2 2 0 1 0 0 0 0 0 0 0 2 0 2 0 0 0 0 0 0 0 0 0 2 0 0 2 0 1 0 0 0 0 0 0 0 2 0 2 0 0 0 0 0 0 0 0 0 2 2 0 2 0 1 0 0 0 0 0 0 0 2 0 2 0 0 0 0 0 0 0 0 0 0 2 0 2 0 1 0 0 0 0 0 0 0 2 0 2 0 0 0 0 0 0 0 0 0 0 2 0 2 0 1 0 0 0 0 0 0 0 2 0 2 0 0 0 0 0 0 0 0 0 0 2 0 2 0 1 0 0 0 0 0 0 0 2 0 2 0 0 0 0 0 0 0 0 0 0 2 0 2 0 1 0 0 0 0 0 0 0 2 0 2 0 0 0 0 0 0 0 0 0 0 2 0 2 0 1 0 0 0 0 0 0 0 2 0 2 0 0 0 0
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And, honestly, it does not look like an identity matrix. ;-) But then what do I know?
norm(A*Ainv - eye(24))
ans = 8
Norm tells us the product of the two is not even close to an identity. And yes, the actual inverse matrix is not the same as what you give. But it does have the properties of an inverse matrix, a good thing.
inv(A)
ans = 24x24
0.2540 0 -0.1270 0 0.0635 0 -0.0317 0 0.0159 0 -0.5079 0 0.2540 0 -0.1270 0 0.0635 0 -0.0317 0 0.0159 0 0.4921 0 0 0.2540 0 -0.1270 0 0.0635 0 -0.0317 0 0.0159 0 -0.5079 0 0.2540 0 -0.1270 0 0.0635 0 -0.0317 0 0.0159 0 0.4921 0.4921 0 0.2540 0 -0.1270 0 0.0635 0 -0.0317 0 0.0159 0 -0.5079 0 0.2540 0 -0.1270 0 0.0635 0 -0.0317 0 0.0159 0 0 0.4921 0 0.2540 0 -0.1270 0 0.0635 0 -0.0317 0 0.0159 0 -0.5079 0 0.2540 0 -0.1270 0 0.0635 0 -0.0317 0 0.0159 0.0159 0 0.4921 0 0.2540 0 -0.1270 0 0.0635 0 -0.0317 0 0.0159 0 -0.5079 0 0.2540 0 -0.1270 0 0.0635 0 -0.0317 0 0 0.0159 0 0.4921 0 0.2540 0 -0.1270 0 0.0635 0 -0.0317 0 0.0159 0 -0.5079 0 0.2540 0 -0.1270 0 0.0635 0 -0.0317 -0.0317 0 0.0159 0 0.4921 0 0.2540 0 -0.1270 0 0.0635 0 -0.0317 0 0.0159 0 -0.5079 0 0.2540 0 -0.1270 0 0.0635 0 0 -0.0317 0 0.0159 0 0.4921 0 0.2540 0 -0.1270 0 0.0635 0 -0.0317 0 0.0159 0 -0.5079 0 0.2540 0 -0.1270 0 0.0635 0.0635 0 -0.0317 0 0.0159 0 0.4921 0 0.2540 0 -0.1270 0 0.0635 0 -0.0317 0 0.0159 0 -0.5079 0 0.2540 0 -0.1270 0 0 0.0635 0 -0.0317 0 0.0159 0 0.4921 0 0.2540 0 -0.1270 0 0.0635 0 -0.0317 0 0.0159 0 -0.5079 0 0.2540 0 -0.1270
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A quick check of the condition number of A shows a very small number. And since A is full rank, the inverse exists, and is unique.
cond(A)
ans = 3.0000
norm(A*inv(A) - eye(24))
ans = 9.9603e-16
I'm sorry, but we cannot know why you think that matrix is the inverse of the binary matrix you show. But it is not. Not even close.

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