RE: LU decomposition/backsubstitution

[Michael Hanke]

On Mon, 06 Dec 1999, Ted Harding wrote:
Surely setting
>
>  B = A\U
>
>where U is an identity matrix of appropriate dimensions, would give
>you what you are looking for (to within rounding errors, anyway):
>
>  B*b = A\U*b = A\b
>
>Ted.
Hi,

I have the same problem sometimes. Your proposed solution does not seem to be
optimal. B is simply the inverse matrix, and the complexity is O(n^3). The
problem appears if one has to solve a number m << n (=size(A)) systems wher m
is not known in advance. One could use the LU decomposition [L,U]=lu(A), and
then x=U\(L\b). This works efficiently if octave sees that L and U are
triangular (at least after permutation). Does octave see this? According to the
documentation, matlab does.

Regards,
Michael

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|  Michael Hanke                Royal Institute of Technology   |
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