How to get only linearly independent rows in a matrix or to remove linear dependency b/w rows in a matrix?
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Say I have a matrix A = [1,1,1;1,2,3;4,4,4]; and I want only the linearly independent rows in my new matrix. The answer might be A_new = [1,1,1;1,2,3] or A_new = [1,2,3;4,4,4]
Since I have a very large matrix so I need to decompose the matrix into smaller linearly independent full rank matrix. Can someone please help?
2 Comments
sixwwwwww
on 5 Dec 2013
what is meaning of linear independent in your case? A_new is linearly independent and A is not linearly dependent?
Accepted Answer
Matt J
on 5 Dec 2013
This extracts linearly independent columns, but you can just pre-transpose the matrix to effectively work on the rows.
function [Xsub,idx]=licols(X,tol)
%Extract a linearly independent set of columns of a given matrix X
%
% [Xsub,idx]=licols(X)
%
%in:
%
% X: The given input matrix
% tol: A rank estimation tolerance. Default=1e-10
%
%out:
%
% Xsub: The extracted columns of X
% idx: The indices (into X) of the extracted columns
if ~nnz(X) %X has no non-zeros and hence no independent columns
Xsub=[]; idx=[];
return
end
if nargin<2, tol=1e-10; end
[Q, R, E] = qr(X,0);
if ~isvector(R)
diagr = abs(diag(R));
else
diagr = R(1);
end
%Rank estimation
r = find(diagr >= tol*diagr(1), 1, 'last'); %rank estimation
idx=sort(E(1:r));
Xsub=X(:,idx);
21 Comments
Matt J
on 21 Sep 2020
Edited: Matt J
on 21 Sep 2020
Matt J.:
Thanks for the code. I'd like to provide a reference for your work in my upcoming paper that would benefit from removal of dependent equations in the system I am studying.
Please indicate how you would like that to read.
Very Respectfully,
Mike Mont-Eton
Matt J
on 21 Sep 2020
@Michael,
You can just reference the File Exchange link.
The algorithm is not really mine. I just posted it as a FAQ.
More Answers (4)
Wayne King
on 5 Dec 2013
Edited: Wayne King
on 5 Dec 2013
A = [1,1,1;1,2,3;4,4,4];
[R,basiccol] = rref(A);
B = A(:,basiccol);
The columns of B are a basis for the range of A. B has the same rank as A.
3 Comments
Matt J
on 5 Dec 2013
Edited: Matt J
on 5 Dec 2013
I have been warned not to trust RREF for this kind of thing. That was my reason for coding the QR-based method in my Answer.
Wayne King
on 5 Dec 2013
As Matt advises below just transpose to work on rows.
B = A';
[R,basiccol] = rref(B);
B = B(:,basiccol)'
Lem
on 27 Nov 2015
Hello,
I want to ask: instead of rank estimation, can we not just use the minpoly function, get the largest non-zero degree (r) from there and use r instead?
1 Comment
Matt J
on 27 Nov 2015
Edited: Matt J
on 27 Nov 2015
There's no way to avoid estimating rank. minpoly sounds like an alternative way to do so, but requires the Symbolic Toolbox. It also appears to be a lot slower than a QR approach, even for rather small matrices:
>> A=rand(10);
>> tic;minpoly(A);toc
Elapsed time is 0.875673 seconds.
>> tic;qr(A);toc
Elapsed time is 0.000063 seconds.
Dave Stanley
on 24 Aug 2017
I wrote a few functions to handle this. They do basically the same as Matt J's function above, with some added bells and whistles. (Namely, it includes an option for ignoring columns that are shifted by a constant; for example, if col2 = 10 - col1. It also returns indices to clusters of originally linearly dependent columns ). Again, you'd have to add a transpose to operate on rows instead of columns. Hope it's useful to someone.
For numerical matrices: https://www.mathworks.com/matlabcentral/fileexchange/64221-getlinearindependent-a-ignore-constant-shift-
For cell arrays: https://www.mathworks.com/matlabcentral/fileexchange/64222-getlinearindependentcell-a-ignore-constant-shift-
On your example above:
A = [1,1,1;1,2,3;4,4,4]'
[Abasis, Abasisi, Asub]= getLinearIndependent(A)
Result:
Input:
A =
1 1 4
1 2 4
1 3 4
Output:
Abasis =
1 1
1 2
1 3
Abasisi =
1 2
Asub =
1×2 cell array
[1,3] [2]
Here, Asub{1} contains the indices of all columns linearly dependent with the first basis vector, and Asub{2} contains that for the second.
0 Comments
Dominique Joubert
on 9 Nov 2018
svd , and looking at the number of non-zero singular values
8 Comments
Bruno Luong
on 9 Nov 2018
Edited: Bruno Luong
on 9 Nov 2018
"If for example the values in that column were [0.4 0 0.2]"
But it's not the case. So you can't conclude anything in the above example.
If you want to convince, write down your algorithm to detect independent columns using SVD, then we can speak.
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