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Re: Question about LU decomposition


From: Carlo de Falco
Subject: Re: Question about LU decomposition
Date: Mon, 19 Apr 2010 20:12:58 +0200


On 19 Apr 2010, at 15:31, forkandwait wrote:

Hi all,

Let A = [2 4; 4 11].

Then the output from the lu command is

L = [1 0; 1/2 1]
U = [4 11; 0 -3/2]
P = [0 1; 1 0]

This is great... but it is different from what would be generated if the
permutation hadn't been done first, which would have been

L = [1 0; 2 1]
U = [2 4; 0 3]

This difference wouldn't be a problem, except I am just starting to work through Gilbert Strang's "Intro to Applied Mathematics" using the computer as a
crutch (^H^H learning tool), and Octave's answers don't match the book
(generally because he doesn't permute the matrices). (This matrix is on p 20).

I am not sure what my question is exactly, but maybe for starts:

(1) is there a way to run lu() without permuting?
yes but it is a dirty hack and is only (questionably) useful if you want to check theoretical results:

[l, u] = lu(sparse(A), 0);
full(l)
full(u)
full(l*u-A)

(2) what is the long story on permuting and LU decomposition -- is it wrong to not permute? Is there a relationship between all the possible decompositions?
etc etc (a link would be awesome...)
in general it is always a good idea to use pivoting (i.e. "permuting") to improve the numerical stability of the algorithm, and it has an almost negligible cost, so avoiding pivoting never makes sense in numerics.

Thanks again


HTH,
c.



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