You could simply let $S=-L$ and have $A=SS^{T}$, but the resulting factorization wouldn’t be a Cholesky factorization. This element of the definition of the Cholesky factorization makes the factorization unique.
– Brian BorchersMay 09 '19 at 17:35
Aside from the fact that $L_{j,j} = \sqrt{ A_{j,j} - \sum_{k=1}^{j-1} L_{j,k}^{2} } $?
– May 09 '19 at 18:00
@HansEngler, your comment as a one line answer would be the easiest complete solution. The only thing I would add is a reference that says as much.
– DamienJul 05 '23 at 22:33