# Is it possible to decompose a symmetric, positive definite matrix in this way?

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?

• I don't understand the question. Obviously $\Omega=\Sigma$ works, so do you want this $\Omega$ have any particular property? Jan 16, 2011 at 23:56
• Write $\Omega=RR'$. If $R$ is invertible, then $\hat L=LR^{-1}$. Jan 17, 2011 at 0:09
• My fault, I left out the crucial bit: $\hat{L} should be lower triangular and not diagonal (edited to reflect this). Thanks – JMS Jan 17, 2011 at 0:25 •$LR^{-1}$is lower triangular, as$L$and$R$are. Jan 17, 2011 at 0:35 • Right, my edit was @ Igor's comment. I follow you comments, but then I guess my question becomes when can I further factor the Cholesky factor$L$of$\Sigma$into$\hat{L}R$where$R$isn't diagonal (and is the cholesky factor of the SPD matrix$\Omega$), if that makes sense. It isn't clear to me that this should always be the case, but I may well be missing something simple. – JMS Jan 17, 2011 at 0:49 ## 1 Answer It seems to me that if you look at http://en.wikipedia.org/wiki/Cholesky_decomposition the "Cholesky outer product algorithm" writes$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.$• Yes, that works. An example, for posterity:$\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).
– JMS
Jan 17, 2011 at 1:53