Please note that here is Gauss_jordan elimination which help us get inverse of A.
I am wondering, is there any condition that it could work without pivoting?
I try to prove this under column diagonally dominant, but I could pass one step in my proof.
Has anyone heard about sth about this topic?
yes, we could naively say for each step $A^{(i)}$ has no 0 elements in the diagonal line but could we directly add some conditions on A rather than $A^{(i)}$?
thanks
I think you are asking whether a given matrix $A$ can be written as $A =L\cdot U$ where $L$ is lower triangular, $U$ is upper triangular and both are invertible. This is called the Gauss decomposition ( you may have an extra diagonal term in the middle - harmless).
The following result is true:
The matrix $A$ can be written as $A = L\cdot D \cdot U$ (all invertible) if and only if all the leading principal minors of $A$ are nonzero.