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 $L$? $\endgroup$ Commented Sep 12, 2017 at 23:18
  • $\begingroup$ @kimchilover: That would not be a Cholesky decomposition. $\endgroup$
    – copper.hat
    Commented Sep 12, 2017 at 23:21
  • $\begingroup$ See math.stackexchange.com/questions/2425878/…. $\endgroup$
    – copper.hat
    Commented Sep 12, 2017 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
    Commented Sep 13, 2017 at 15:52

1 Answer 1


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.


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