If $\mathbf{A} = \mathbf{LL}^{\top}$, can we find $\mathbf{LDL}^{\top}$ for diagonal $\mathbf{D}$ without computing the decompositon?

Question

Given a symmetric positive-definite $\mathbf{A}$ which permits a Cholesky decomposition $\mathbf{A} = \mathbf{LL}^{\top}$, and a positive diagonal matrix $\mathbf{D}$, is it possible to express $\mathbf{LDL}^{\top}$ in terms of $\mathbf{A}$ and $\mathbf{D}$ only, or do we need to compute the decomposition?

Thanks!

Answer 1

It's not possible: if we had a formula to compute $\mathbf{LDL}^{\top}$, it would be straightforward from there to compute $\mathbf{L}$.

Set $\mathbf{D}$ to be the matrix with $\mathbf{D}_{ii}=1$ and all other entries $0$. Then $\mathbf{LDL}^{\top}$ simplifies to $\mathbf{xx}^\top$ where $\mathbf x$ is the $i^{\text{th}}$ column of $\mathbf L$. In particular, the diagonal entries of $\mathbf{xx}^\top$ are the squared entries of $\mathbf x$, so we can read off $\mathbf x$ up to signs from there alone.

(The signs of the entries of $\mathbf x$ relative to each other can be deduced from looking at the off-diagonal entries. We can't deduce the absolute sign, since negating a column of $\mathbf L$ won't change $\mathbf{LL}^\top$.)

Doing this for all $i$ gets us $\mathbf{L}$.

Answer 2

I think, that is not possible, because matrix multiplication is not conmutative