dividing on 0 in your denominator.

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hamid shakeri
hamid shakeri on 27 Aug 2014
Commented: Image Analyst on 28 Aug 2014
Dear MATLAB helpers,
I'm trying to run a project for 2 month but it got stuck this warning every time and this influence my results too much.
Warning: Matrix is close to singular or badly scaled.
Results may be inaccurate. RCOND = 1.345970e-19.
The warning line:
s=K(fdof,fdof)\(f(fdof)-K(fdof,pdof)*dp);
Also I attached variables value such as K, it may help.
I will be thankful for any suggestion.
All the best

Answers (1)

Adam
Adam on 27 Aug 2014
Your matrix K has a rank of just 7, it needs to be 15 (equal to its size) in order to be considered well conditioned.
There are a number of ways to condition a singular (or close to) matrix and the best probably depends on what you will be using it for so I can't really say what would be best in this case.
What I have done in the past, which is a fairly standard method, is to boost the leading diagonal slightly.
I just tried this on your K matrix and multiplying the leading diagonal by 1.01 resulted in an invertible matrix of full rank (15).
I'm not suggested you should just go ahead and multiply by 1.01 as adding 1% to your leading diagonal may not suit your purpose, but if it does then that could be one solution for you.
  5 Comments
Adam
Adam on 27 Aug 2014
Yeah, that one-liner will definitely be more efficient for large k.
How I wish Matlab had operators like *= to make code like that look so much nicer!!
Adam
Adam on 27 Aug 2014
Sorry I didn't run the project as I don't have the time to at the moment.
For the example K matrix you gave in the original question adding 1% to the leading diagonal was enough to make the matrix non-singular, but it may not always be. In some cases you may need to boost it more so try playing around with higher multipliers than 1.01, bearing in mind that you don't want to go too high as you will start to skew the data in your matrix too much.
I use this technique for conditioning a matrix to be used for inversion of a covariance matrix in a Gaussian classifier, but it should work for other cases too with maybe a larger multiplier.

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