From help-octave-request at bevo dot che dot wisc dot edu  Wed Mar 25 06:36:16 1998
Subject: Re: LU decomposition: backsubstitution available? 
From: Mario Storti <mstorti at minerva dot unl dot edu dot ar>
To: help-octave at bevo dot che dot wisc dot edu
cc: mstorti at minerva dot unl dot edu dot ar
Date: Wed, 25 Mar 1998 09:26:45 -0300

>>>>> On Wed, 25 Mar 1998 10:27:40 +0100 (MET), 
>>>>>      Thomas Hoffmann <hoffmann at ehmgs2 dot et dot tu-dresden dot de> said:

> It is a known procedure for the solution of a sequence of
> systems of linear equations Ax=b with only varying rhs to
> make once a LU decompostion of A (function lu() of octave)
> and then backsubstitute for the several b's, e.g.
>   x=backsubs(l,u,b)
> As e.g. RLaB has such an mechanism, I thought of similar
> functionality in Octave, but did not find any.

> Can anybody shed some light on this?

> Thomas.

I faced this same problem  before. It would be   nice to have  someone
writing a 'backsubs' routine. It seems to me that it should be written
in fortran in order to be efficient (do to the triangular structure of
l and u). 

Meanwhile, I think  that you make a  significant save in computational
effort if compute the inverse of A, say

> invA=inv(A);
> x1=invA*b1;
> x2=invA*b2;
> x3=invA*b3;
> .
> .

since  computing the  inverse requires  O(N^3) ops. and  matrix-vector
multiplication requires only O(N^2). Of course, if you know all the
rhs's at once, then you can put them incolumns in a matrix B and then
make a:

X= A \ B;

and solve all the systems at once. 

Mario

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Mario Alberto Storti                           | Fax: (54)(42) 55.09.44 |
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