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I have a symmetric positive semidefinite covariance matrix $A$, which is approximately computed as the output of a quadratic regression. I then need to invert $A$, but often it is close to singular. I've reduced the problem by using scaling. That is, I create a diagonal matrix $D$, with elements $D_{ii} = 1/\sqrt{A_{ii}}$. Then

$A^{-1} = D(DAD)^{-1}D$

Where $DAD$ has a lower conditioning number then $A$. Unfortunately in some iterations this is not enough. The size of $A$ is quite small, say maximum $50 \times 50$.

I need the inverse of $A$ because I have to use it in a long calculation, where terms such as $x^TA^{-1}x$ and $A^{-1}B$ appear lots for time. Also: $A^{-1}$ represents a covariance matrix, so it has to be symmetric and positive definite.

Is there some better way to make $A$ invertible?

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  • $\begingroup$ Computing the pseudo-inverse instead is not an option? $\endgroup$
    – Gabriel Landi
    Nov 18, 2012 at 21:29
  • $\begingroup$ I didn't think about that. If I use the pseudo-inverse do I have some measure of the distance between the pseudo-inverse and the "true" inverse? $\endgroup$
    – Jugurtha
    Nov 18, 2012 at 23:14
  • $\begingroup$ When the matrix is invertible, the pseudo-inverse is the inverse. $\endgroup$
    – Wolfgang Bangerth
    Nov 19, 2012 at 0:25

2 Answers 2

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Multiply the diagonal elements by a factor $q>1$ but close to 1. It will usually do the job. (I'd first try $q=1.0001$, but one can experiment with the number of zeros in this expression; e.g., use $q=1.01$ for very noisy data.) This is justified under certain conditions, as it is a specific form of regularization. For more on regularization, see my tutorial
http://mat.univie.ac.at/~neum/ms/regtutorial.pdf

An important exception is when some diagonal element is tiny or negative, in which case your data were insufficent for the attempted estimation.

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  • $\begingroup$ Thanks, could I do this while also using preconditioning? $\endgroup$
    – Jugurtha
    Nov 19, 2012 at 12:30
  • $\begingroup$ @Jugurtha: I didn't understand your two programming lines, so I can't see what you mean. Please rephrase these lines in ordinary mathematical terms. $\endgroup$
    – Arnold Neumaier
    Nov 19, 2012 at 15:19
  • $\begingroup$ I edited the question, I hope it is clearer now. $\endgroup$
    – Jugurtha
    Nov 19, 2012 at 16:15
  • $\begingroup$ @Jugurtha: Yes. But one would call what you do ''scaling'', not preconditioning. You can do the scaling after the diagonal regularization; but it should not need to be necessary, as inverting positive definite matrices is stable even without scaling. $\endgroup$
    – Arnold Neumaier
    Nov 19, 2012 at 18:53
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The question to ask is why you need to invert the matrix. If a matrix is near-singular, it's true that you can define something like a pseudo-inverse in some stable way but it's nevertheless true that for near-singular matrices, solving linear systems in any way is unlikely to result in anything useful because the result is so strongly dependent on small changes in the right hand side.

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  • $\begingroup$ I added some motivation about why I want to invert $A$. $\endgroup$
    – Jugurtha
    Nov 19, 2012 at 12:31
  • $\begingroup$ But you didn't understand what I was saying. If $A$ is ill-conditioned or near singular, then expressions like $x^TA^{-1}x$ or $A^{-1}B$ will not make much sense because the values you get will be very sensitive to small perturbations in $x$ or $B$. The only exception would be if you can show that $x$ or $B$ have no components that lie in the directions of the eigenvectors corresponding to the very small eigenvalues of $A$. $\endgroup$
    – Wolfgang Bangerth
    Nov 19, 2012 at 12:45
  • $\begingroup$ I understand that if $A$ is singular we cannot invert it. In my case $A^{-1}$ is the estimate of a covariance matrix, which I know to exist. So when I get a badly conditioned $A$ I would like to perturb it so that I get an invertible matrix which is not too different from the original one. Which is more or less what Arnold is suggesting. The alternative it to increase my dataset until $A$ becomes invertible, but this is quite expensive. $\endgroup$
    – Jugurtha
    Nov 19, 2012 at 13:27
  • $\begingroup$ I understand what you're trying to do. But what I'm telling you is that if you have a near-singular $A$, then trying to compute $x^TA^{-1}x$ doesn't make any sense. You can paper over that by computing some $x^T\widetilde{A^{-1}}x$ but you're just hiding the problem: Yes, you can compute it; but no, it's not going to be a useful number you compute this way. $\endgroup$
    – Wolfgang Bangerth
    Nov 19, 2012 at 14:28
  • $\begingroup$ Ok, so in your opinion the only sensible thing to do if $A$ is almost singular is to get a larger dataset/repeat the simulation until I get an invertible $A$? $\endgroup$
    – Jugurtha
    Nov 19, 2012 at 14:56

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