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?
Best 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 $i.$