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Dear all, Let's assume that I have a symmetric matrix $\Sigma$ and a permutation matrix $A$. Is there a relationship between the Cholesky decompositions of $\Sigma$ and of $A^T \Sigma A$ ? Many thanks, Samuel

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The Cholesky decomposition allows you to write (for real symmetric positive definite matrix $\Sigma$) $$\Sigma = U^T U$$ with $U$ an upper triangular matrix. Therefore, $$A^T \Sigma A = A^T U^T U A = B^T B$$ with $B= UA$. Note however that $B$ is (in general) not upper triangular anymore such that $B^T B$ is not the standard Cholesky decomposition (but of course very similar to a Cholesky decomposition).

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In short: not much of a relationship. If the OP were considering the eigendecomposition instead of the Cholesky decomposition, however... – J. M. May 5 '11 at 9:34
@J. M.: of course you are right. However depending on what you want to do afterwards, the decomposition $B^T B$ might be enough (and you don't need the additional property that B is upper triagonal). – Fabian May 5 '11 at 9:41
But maybe the OP has a good reason to consider this, since Cholesky with symmetric pivoting is sometimes done for matrices that are (numerically) symmetric positive semidefinite; see e.g. this. – J. M. May 5 '11 at 9:49
@J. M.: we will never know ;-) – Fabian May 5 '11 at 9:50
You guys are right. Let's say $\Sigma=U^T U$ and $A^T \Sigma A=V^T V$ where $U$ and $V$ are upper triangular. I wanted to know if there was any realtionship between $U$ and $V$. – S4M May 5 '11 at 10:13

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