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

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!

• It's not at all clear why you'd want to do this? – Brian Borchers Oct 13 '17 at 22:01
• @BrianBorchers its convoluted, but it involves assuming $\mathbf{A} = \mathbf{LL}^{\top}$ in order to apply the Sylvester determinant theorem, and as a result we end up with an $\mathbf{LDL}^{\top}$ term, and the decomposition is not strictly needed anymore, so if we can express it without $L$ we wouldn't need to compute the decomp in the first place. – CBowman Oct 13 '17 at 22:06
• Cholesky is cheap enough to do that your requirement just seems unreasonable. – J. M. is a poor mathematician Oct 14 '17 at 3:19
• @J.M.isnotamathematician well that rather depends on how large the matrix is, in this case its very very large, and sparse, and is part of a code that needs to run many times. It doesn't look like it is possible, but it would have been a nice cost saving if it were. – CBowman Oct 14 '17 at 9:07
• "very very large, and sparse" - this is a thing that should have been mentioned in the question to begin with. In this case, you can't do Cholesky without a pivoting/reordering strategy, which complicates things even further. – J. M. is a poor mathematician Oct 14 '17 at 9:31

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 $\mathbf{L}$.