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Let $\Sigma$ be a hermitian positive definite matrix and $L$ be it's Cholesky decomposition so that $LL^\ast=\Sigma$. Furthermore, let's diagonalize $\Sigma$ as $\Sigma = P\Lambda P^\ast$. $\Lambda$ is a diagonal matrix containing the real, positive eigenvalues of $\Sigma$, let us denote as $\sqrt{\Lambda}$ the diagonal matrix whose diagonal elements are the square roots of these eigenvalues.

What are some non trivial relationships between $L$ and $(P,\Lambda)$?

The one I have is: Since $(P\sqrt{\Lambda}P^\ast)$ is the unique positive square root of $\Sigma$ then $U = L^\ast P \Lambda^{-1/2} P^\ast$ is unitary.

... and that's about it. Maybe it counts as trivial?

Are there other interesting relations, maybe relations that take into account $L$'s triangular structure? In particular, I'd be interested in algorithm that derive $L$ from $(P,\Lambda)$ or vice-versa (and obviously which aren't merely the trivial composition of two algorithms).

share|cite|improve this question… in the answer to this quest, I tried to demonstrate the following phenomena: if there is small eigenvalue, then there should small element on the diagonal of Cholesky and they should be approximately equal. In that example we get both equal to zero. – Alexander Chervov Apr 4 '12 at 14:17
up vote 3 down vote accepted

First, let me rephrase your remark. Let $L=HU$ be the polar factorization of $L$ ($H$ hermitian positive definite, $U$ unitary). Then $\Sigma=LL^\ast=H^2$ tells you that the Hermitian part of $L$ is $\sqrt\Sigma$. Then $U=L\Sigma^{-1/2}$ is its unitary part.

On the other hand, you have $\sqrt\Sigma=LQ^\ast=QL^\ast$. This is exactly the $QR$-factorization of $\sqrt\Sigma$: $L^\ast$ is the upper triangular part with positive real diagonal in the $QR$-factorization of $\sqrt\Sigma$.

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Just making sure: do you actually mean $L^\ast is the upper triangle part with positive real diagonal in the $QR$-factorization of $\underline{\sqrt(\Sigma)}$? – Arthur B Apr 4 '12 at 17:55
Just making sure: do you actually mean $L^\ast$ is the upper triangle part with positive real diagonal in the $QR$−factorization of $\underline{\sqrt(\Sigma)}$? – Arthur B Apr 4 '12 at 17:56
@Arthur. You're right. I fix it. – Denis Serre Apr 4 '12 at 19:49

Computing a Cholesky factor is a much easier task than computing the eigendecomposition. For instance, you can do Cholesky while staying in the original field (modulo some square roots on the diagonal, which can be avoided if you go for the $LDL^*$ form), while arbitrarily complicated algebraic field extensions can be required for the eigendecomposition. So there is no hope to obtain $(P,\Lambda)$ from $L$ through simple algebraic formulas.

The other direction is typically much less useful, at least from a computational point of view --- on the top of my head, everything you can do with Cholesky, you can do it also with the eigenvalue decomposition, and it's more stable.

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Agreed, it's much more likely to go from $(P,\Lambda)$ to $L$, and also agreed it's probably not useful. Regardless, I'm still interested in finding out if there is anything. – Arthur B Apr 4 '12 at 15:45

In some circumstances, Cholesky factorization is enough, so we don't bother to go through more subtle steps of finding eigenvectors and eigenvalues. For an example, when constructing "correlated Gaussian random variables". By the way, @Federico Poloni, why the Cholesky is less stable?

share|cite|improve this answer
Sorry, that was badly phrased from my side. I don't mean that one is intrinsically more stable than the other (they are both backward stable), but that if you go through an additional step (eigendecomposition -> Cholesky -> whatever you need, correlated variables for instance) then it is less stable than doing it directly skipping a step. – Federico Poloni Jun 12 '15 at 6:47

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