Behaviour of backslash operator for non-square matrices least-squares fitting

4 views (last 30 days)
I am currently trying to reproduce a set of results from a pre-existing project and can't get to the bottom of the following difference.
Phi = design matrix of input data size(29507x97)
Ref = reference values to fit to, six data sets, size(29507x6)
Phi = rand(29507,97);
Ref = rand(29507,6);
c1 = Phi\Ref;
for ii = 1:6
c2(:,ii) = Phi\Ref(:,ii);
end
all(all(c1==c2))
I would have expected c1 to give identical results to c2. I can't find detailed information on what the "\" algorithm is doing differently between these case.
Any help appreciated!

Accepted Answer

the cyclist
the cyclist on 24 Feb 2020
They are equal, to within floating-point precision. Notice that
max(abs(c1(:)-c2(:)))
is around 1e-16.
  4 Comments
Matt J
Matt J on 24 Feb 2020
Edited: Matt J on 24 Feb 2020
Also, internal parallelization of mldivide and other linear algebra operations is different depending on the size of the inputs. So, when you perform the mldivide column-by-column, the inputs are divided up into parallel chunks in a different way, leading to floating point differences.
Matt O'Donnell
Matt O'Donnell on 24 Feb 2020
@the cyclist Yes. I agree with that as the likely root cause.
I thought it would be doing the same least squares calculation on a column by column basis as each column should be treated independently from each other in terms of doing the expected calculation.i.e. column 1 should never know anything about the other 5 columns's calculations mathematically speaking. But as @Matt J has suggested it is doing something across the columns, or even splitting each column differently, which leads to the difference/calculation order effect as suggested.
Many thanks to both of you for helping to get my head around this.

Sign in to comment.

More Answers (0)

Products


Release

R2018a

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!