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From: Nico Galoppo (nico_at_[hidden])
Date: 2007-03-03 09:58:04
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 ())
>
>
-- nico galoppo von borries | address: 119 fidelity st., carrboro comp. graphics phd. student | north carolina, 27510 USA UNC, chapel hill | phone: +1 (919) 962-1898 (office) | +1 (919) 942-4388 (home) | email: nico at cs dot unc dot edu | homepage: http://www.ngaloppo.org