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Ublas : |
From: Sourabh (sourabh_at_[hidden])
Date: 2007-03-05 03:29:17
But, when I use inplace_solve of trisolve.hpp,
slopes = omega; // omega is a vector
inplace_solve(lembda, slopes, unit_lower_tag ());
it does not give me the same result as:
lembda1 = I - lembda;
slopes = solve (lembda1, omega, lower_tag);
On Sat, 3 Mar 2007, Nico Galoppo wrote:
> Hi Sourabh,
>
> I believe Gunther is trying to say that using unit_lower_tag() makes the
> following call solve the system (I+A)Y=B
>
> solve(A, B, unit_lower_tag());
>
> So, for your case, if you fill A with -lembda, you will effectively solve
> (I-lembda)Y=B
>
> minuslembda = -lembda;
> for (;;)
> solve(minuslembda, omega, unit_lower_tag());
>
> It will be equivalent to the statement you gave below.
>
> --nico
>
> Sourabh wrote:
> > Do you mean that I fill lembda with negative values and use this
> > statement:
> >
> > solve (lembda, omega, unit_lower_tag ()) ?
> >
> > will it solve the equation treating lemda as (I+lembda)
> > Will it be equivalent to the following statement ?
> >
> > lembda = I - lembda
> > solve (lembda, omega, lower_tag ())
> >
> >
>
>
-- -- Sourabh