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? 



Is it possible to decompose a symmetric, positive definite matrix in this way?
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 $i.$


