# 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?

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I don't understand the question. Obviously $\Omega=\Sigma$ works, so do you want this $\Omega$ have any particular property? –  Igor Rivin Jan 16 '11 at 23:56
Write $\Omega=RR'$. If $R$ is invertible, then $\hat L=LR^{-1}$. –  Wadim Zudilin Jan 17 '11 at 0:09