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Issue with Cholesky decomposition and positive definiteness

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James Barrett
James Barrett on 23 Sep 2013
Edited: Jai on 28 Oct 2015
I'm getting the following error when using chol
Error using chol
Matrix must be positive definite.
My matrix is a squared exponential kernel matrix and by definition should always be positive definite. It's defined as
K(i,j) = exp(-(1/L)*(x_i-x_j)^2) + beta*delta(i,j)
The problem occurs during an attempt to optimise over the parameter L. When the value of L is very small (approx 1e-6) I end up with zero off diagonal elements. I have saved the matrix to a text file before calling chol and the problematic matrix looks like
10.25310436455727 0 0
0 10.25310436455727 0
0 0 10.25310436455727
Using chol on this matrix obviously doesn't lead to an error. Does anyone know why chol is giving me this error? Could there be rounding errors due to taking the exponential of very small numbers that aren't being saved to the text file?
Kind Regards, James

  1 Comment

Matt J
Matt J on 24 Sep 2013
I have saved the matrix to a text file before calling chol and the problematic matrix looks like
Like Alan, I suspect you are not saving the matrix that is actually causing the error, even if you think you are. You should use
>> dbstop if error
to trap the error, if you aren't doing so already. You should also save the matrix K that triggers dbstop to a .mat file (not a text file) and attach it here, so that we can examine K with the full numerical precision that CHOL sees.

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Answers (3)

Alan Weiss
Alan Weiss on 24 Sep 2013
You say you are optimizing. Are you calling fmincon? It is possible that your optimization is stepping out of bounds, taking L to be negative or zero, during a finite differencing step.
If this is the case, perhaps you should set a lower bound on L that is a bit greater than zero, and use the sqp or interior-point algorithms because these algorithms always respect bounds.
If you are using a different optimizer, then perhaps what is happening is when L becomes small, the exponential terms become zero due to underflow, and the matrix is no longer positive definite.
Just my random thoughts.
Alan Weiss
MATLAB mathematical toolbox documentation

  2 Comments

Matt J
Matt J on 24 Sep 2013
If this is the case, perhaps you should set a lower bound on L that is a bit greater than zero, and use the sqp or interior-point algorithms because these algorithms always respect bounds.
Or, perhaps parametrize K instead as
K(i,j) = exp( z *(x_i-x_j)^2) + beta*delta(i,j)
and optimize with respect to z. Then, later, after the optimal z is found, retransform to obtain L=-1/z. This way, K is defined for all parameter values.
James Barrett
James Barrett on 24 Sep 2013
Thanks both for your comments. I have in fact imposed a lower bound on L by writing it as
L = log(1 + LOWER_BOUND + exp(z))
and optimising over z (this is a nice way to impose constraints without using the optimiser to do so).
Alan, the matrix should be positive definite even if the exponential terms go to zero because of the diagonal beta elements (which are non negative).

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Sean de Wolski
Sean de Wolski on 24 Sep 2013
Hi James,
When I started at MathWorks, I asked our MATLAB/Math development team a very similar question. It was the first time I was ever properly scolded by Cleve :)
Here's what's happening: from the documentation for chol
R = chol(A) produces an upper triangular matrix R from the diagonal and upper triangle of matrix A, satisfying the equation R'*R=A. The chol function assumes that A is (complex Hermitian) symmetric. If it is not, chol uses the (complex conjugate) transpose of the upper triangle as the lower triangle. Matrix A must be positive definite.
So A becomes
Ah = triu(A)+triu(A,1)'
If you look at:
eigs(Ah)
There will be a negative one.

  2 Comments

James Barrett
James Barrett on 24 Sep 2013
Thanks Sean, shouldn't the matrix be symmetric (unless there are some rounding errors)? I have even done
K = (K+K')/2
before I call chol(K).
Secondly, it's not clear to me why there would any negative eigenvalues. I've constrained beta to be always greater that 1e-4 to ensure the matrix will always be positive definite.
Jai
Jai on 28 Oct 2015
Hi Sean, It seems you are correct, indeed eigs(Ah) gives negative eigen value. Is there any workaround to make it work using chol ?

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Leah
Leah on 24 Sep 2013
I used Spectral decomposition when this happens. Which is just a fancy name for setting the negative eigenvalues to zero before chol.

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