0
$\begingroup$

If Gaussian elimination can be carried out without pivoting for A, then A has an LU decomposition. Is the converse true: if A has an LU decomposition, then Gaussian elimination can be carried out (in exact arithmetic, disregarding stability issues) without pivoting? If yes, can somebody explain carefully why this works in the other direction as well? Im new to numerical methods, thanks.

$\endgroup$
0
$\begingroup$

Look at the construction of LU. You’re basically doing the Gaussian elimination (forward step) and computing the elements of L and U along the way. Thus, having L and U means that you can do the forward step of Gaussian elimination.

$\endgroup$
0
$\begingroup$

Gaussian elimination and LU decomposition are a single method ! The only difference is that LU stores the elimination coefficients in the zeroed triangle of the matrix, so that no information is lost (the original matrix can be reconstructed). But the elimination itself is the same.

If pivoting is used, permutation information must be kept as well to restore the original order of the unknowns.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.