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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.

IsUnder what conditions (if any) does there everexist a second symmetric positive definite matrix $\Omega$ which is NOT diagonal that satisfies $\Sigma=\hat{L} \Omega \hat{L}'$? If so, where $\hat{L}$ is such a decomposition always availablelower triangular and if not under what conditions does such an $\Omega$ existdiagonal?

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

Is there ever a second symmetric positive definite matrix $\Omega$ which is NOT diagonal that satisfies $\Sigma=\hat{L} \Omega \hat{L}'$? If so, is such a decomposition always available and if not under what conditions does such an $\Omega$ exist?

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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JMS
JMS
  • 269
  • 2
  • 11

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.

Is there ever a second symmetric positive definite matrix $\Omega$ which is NOT diagonal that satisfies $\Sigma=\hat{L} \Omega \hat{L}'$? If so, is such a decomposition always available and if not under what conditions does such an $\Omega$ exist?