Does a square matrix A being nonsingular imply that for A=LU decomposition the diagonal elements of L and U are all nonzero?
I was just wondering because intuitively, it seems that A being nonsingular implies that A is full rank and dim(LU) can't equal dim(A) unless the diagonal elements are all nonzero... Thanks
A (guarded) yes. If $A$ can be factored as $A = LU$, where $L$ is lower and $U$ upper triangular, then $0 \not = \text{det}(A) = \text{det}(L) \text{det}(U)$ implies that $L$ and $U$ are both nonsingular. Since they are triangular, they cannot have zero elements on their main diagonal.