A similar question was asked here, but due to the application an alternative solution was given. But I really do want a Cholesky decomposition of the inverse of a matrix.

To be specific, I want to compute a lower triangular matrix $L$ such that

$\Sigma L L^T = I$

where $\Sigma$ is a given positive semi-definite matrix and $I$ is an identity matrix.

In MATLAB I can achieve this by calling chol(inv(sigma), 'lower'), but I'd like to avoid inverting a dense matrix if it can be done by inverting a triangular matrix instead.

I know I can do L = inv(chol(sigma, 'lower')), but then I have

$\Sigma L^T L = I$

which is not what I need.

  • 1
    $\begingroup$ So what if you chol $\Sigma$ first and then inv the resulting triangular $L$? $\endgroup$ – kimchi lover Sep 12 '17 at 23:18
  • $\begingroup$ @kimchilover: That would not be a Cholesky decomposition. $\endgroup$ – copper.hat Sep 12 '17 at 23:21
  • $\begingroup$ See math.stackexchange.com/questions/2425878/…. $\endgroup$ – copper.hat Sep 12 '17 at 23:22
  • $\begingroup$ @kimchilover: Sorry, I had a typo. The second piece of MATLAB code was supposed to (and now does) address that question. $\endgroup$ – user664303 Sep 13 '17 at 15:52

Let $A$ be the anti-diagonal matrix, let $M\,M' = A\Sigma A$ be the Choseky decomposition of $A\Sigma A$. Then $\Sigma = (AMA)(AM'A)$ and $\Sigma^{-1}=(AM'A)^{-1} (AMA)^{-1} = L L'$. Note that $M$ is lower triangular, so $M'$ is upper triangular so $AM'A$ is lower triangular so $L=(AM'A)^{-1}$ is lower triangular.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.