A similar question is asked here, but for LU decomposition instead of Cholesky, which makes this less trivial. I am using an algorithm (in Matlab) that spends a lot of time computing the Cholesky factor $C$ of an upper triangular matrix $R$. The diagonal of $R$ is composed of ones only, and a lot of its values are near zero, in case that's relevant.
I have observed that each row of the matrix $C$ is a scalar multiple of the same row in $R$, and many rows are unaltered (apart from what I think are floating point variations, i.e., in the order of $10^{-10}$). I suspect that this is an easy result to obtain but I am unable to prove it.
What I would really like, however, is to have a way to compute those scalars that would be faster than performing the Cholesky decomposition, as that might speed up my code significantly.
EDIT: I have also observed that those factors are the eigenvalues of $C$, which is a trivial consequence of this and the fact that the diagonal of $R$ is composed of ones.
EDIT2: As I've said in the comments, the Matlab function chol that I'm using is only using the upper triangular entries of the input matrix, so I am not in fact calculating the Cholesky factorization of an upper triangular matrix, but of a symmetric positive semi-definite matrix. However, I still don't know why the rows of $C$ are multiples of the rows of (the upper triangular part of) $R$.
Best Answer
You have a Cholesky decomposition for symmetric, positive definite matrices... In order for your matrix to be upper triangular and symmetric, it must in fact be diagonal. Regarding your observation about the relation between the rows in the original matrix and the lines/rows in decomposition, consider this example: $$ A = \left( \begin{array}{cccc} 1 & 0.2 & 0 & 0 \\ 0.2 & 1 & 0.2 & 0 \\ 0 & 0.2 & 1 & 0.2 \\ 0 & 0 & 0.2 & 1 \\ \end{array} \right), \qquad L = \left( \begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0.2 & 0.979796 & 0 & 0 \\ 0 & 0.204124 & 0.978945 & 0 \\ 0 & 0. & 0.204302 & 0.978908 \\ \end{array} \right) $$
Although it is true that $A = L L^T$, there is no such relation.