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Let $\Sigma$ be a symmetric positive definite matrix. Then the Cholesky decomposition gives us $\Sigma=LL'$ where $L$ is lower triangular and unique.

Under what conditions (if any) does there exist a second symmetric positive definite matrix $\Omega$ which is NOT diagonal that satisfies $\Sigma=\hat{L} \Omega \hat{L}'$ where $\hat{L}$ is lower triangular and not diagonal?

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  • $\begingroup$ I don't understand the question. Obviously $\Omega=\Sigma$ works, so do you want this $\Omega$ have any particular property? $\endgroup$
    – Igor Rivin
    Jan 16, 2011 at 23:56
  • $\begingroup$ Write $\Omega=RR'$. If $R$ is invertible, then $\hat L=LR^{-1}$. $\endgroup$ Jan 17, 2011 at 0:09
  • $\begingroup$ My fault, I left out the crucial bit: $\hat{L} should be lower triangular and not diagonal (edited to reflect this). Thanks $\endgroup$
    – JMS
    Jan 17, 2011 at 0:25
  • $\begingroup$ $LR^{-1}$ is lower triangular, as $L$ and $R$ are. $\endgroup$ Jan 17, 2011 at 0:35
  • $\begingroup$ Right, my edit was @ Igor's comment. I follow you comments, but then I guess my question becomes when can I further factor the Cholesky factor $L$ of $\Sigma$ into $\hat{L}R$ where $R$ isn't diagonal (and is the cholesky factor of the SPD matrix $\Omega$), if that makes sense. It isn't clear to me that this should always be the case, but I may well be missing something simple. $\endgroup$
    – JMS
    Jan 17, 2011 at 0:49

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It seems to me that if you look at http://en.wikipedia.org/wiki/Cholesky_decomposition the "Cholesky outer product algorithm" writes $L = L_1 \dots L_k,$ so if you write $\Lambda_i = L_i\dots L_k,$ then $\Omega=\Lambda_i \Lambda_i^\prime$ should work for most values of $i.$

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  • $\begingroup$ Yes, that works. An example, for posterity: $\Sigma =\left( \begin{array}{cc} s & v' \\ v & T \end{array} \right) = \left( \begin{array}{cc} \sqrt{s} & 0 \\ v/\sqrt{s} & I_{p-1} \end{array} \right) \left( \begin{array}{cc} 1 & 0 \\ 0 & T - vv'/s \end{array} \right) \left( \begin{array}{cc} \sqrt{s} & v'/\sqrt{s} \\ 0 & I_{p-1} \end{array} \right)$ It just remains to show that $T - vv'/s$ is positive definite, which is not difficult (and I guess also that it isn't diagonal in order to fit my original problem statement). $\endgroup$
    – JMS
    Jan 17, 2011 at 1:53

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