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I have a tricky problem concerning a covariance matrix cholesky decomposition.

What I need is to obtain the cholesky decomposition of the estimated variance matrix of the set of samples stored in a matrix $P$, i.e.:

$$ \begin{align} Input: & & P \in R^{D \times n} & & (\text{$D$ ... sample dimensionality, $n$ - number of samples)} \\ Output: & & L \in R^{D \times 1} & & var(P) = LL' \end{align} $$

The tricky part is that in my problem $D$ is extremely large thus the covariance matrix $cov(P)$ does not fit into the memory. The neat thing is that the only output I need in the end is the cholesky decomposition $L$ vector.

I imagine that the solution to this problem could be a direct estimate of the $L$ vector without the need to compute the actual covariance matrix ... If yes could anybody please point me to a solution/reference?

Many thanks

David

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  • $\begingroup$ Can you do some kind of PCA on your matrix? Or at least have a sense as to whether or not it's sparse? $\endgroup$
    – Alex R.
    Commented Jul 11, 2014 at 17:55
  • $\begingroup$ What $L$ vector? Cholesky decomposition should produce a lower triangular $D \times D$ matrix, not just a vector. The only case where the covariance matrix can be written as $L L'$ where $L$ is a vector is if that covariance matrix has rank 1. $\endgroup$
    – Robert Israel
    Commented Jul 11, 2014 at 21:34
  • $\begingroup$ ... and the covariance matrix has rank 1 if and only if all the samples are collinear. $\endgroup$
    – Robert Israel
    Commented Jul 11, 2014 at 22:55
  • $\begingroup$ Of course I am sorry, at first point I thought that L is a vector but since it is an upper triangular matrix, there is no way how to obtain it concerning the fact that the covariance matrix which is two times bigger does not fit into the memory by far (59 milion dimensions) ... $\endgroup$
    – David Novotny
    Commented Jul 17, 2014 at 14:49
  • $\begingroup$ Thanks for the clarification by the way ... PCA would be nice, however you still need the covariance matrix to do it and projecting each 59 mil dimensional vector would be pretty slow. $\endgroup$
    – David Novotny
    Commented Jul 17, 2014 at 14:51

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