Interesting relationships between Cholesky decomposition and diagonalization

Question

Let $\Sigma$ be a hermitian positive definite matrix and $L$ be its 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).

Answer 1

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 $\sqrt\Sigma$.

Answer 2

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.

Answer 3

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?