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From: goodjeff on 7 Feb 2010 11:32
Dear all,

Given a real symmetric matrix A, the eigenvector, V_min, for the
minimum eigenvalue, Lambda_min, need to be calculated.

Surely there are various numerical algorithms to calculate it, but
my impression is that if the minimum eigenvalue is very close to 0,
the solution is not numerically stable (the resulting V_min might be
very different from the ideal one).

I am thinking if there is a way to rearrange the matrix A to be B,
who has the same eigenvectors but A's V_min now becoming V_max,
corresponding to its biggest eigenvalue, whose numerical solution is
more reliable.

Thanks
Jeff
From: Greg Heath on 7 Feb 2010 13:04
On Feb 7, 11:32 am, goodjeff <jiang1f...(a)gmail.com> wrote:
> Dear all,
>
>    Given a real symmetric matrix A, the eigenvector, V_min, for the
> minimum eigenvalue, Lambda_min, need to be calculated.
>
>    Surely there are various numerical algorithms to calculate it, but
> my impression is that if the minimum eigenvalue is very close to 0,
> the solution is not numerically stable (the resulting V_min might be
> very different from the ideal one).
>
>    I am thinking if there is a way to rearrange the matrix A to be B,
> who has the same eigenvectors but A's V_min now becoming V_max,
> corresponding to its biggest eigenvalue, whose numerical solution is
> more reliable.

Since B = inv(A), I don't think that approach will work.

Best bet is finding a stable iterative algorithm for

lambda = min(x'*A*x/(x'*x))

Hope this helps.

Greg
From: Ray Koopman on 7 Feb 2010 14:00
On Feb 7, 8:32 am, goodjeff <jiang1f...(a)gmail.com> wrote:
> Dear all,
>
> Given a real symmetric matrix A, the eigenvector, V_min, for the
> minimum eigenvalue, Lambda_min, need to be calculated.
>
> Surely there are various numerical algorithms to calculate it, but
> my impression is that if the minimum eigenvalue is very close to 0,
> the solution is not numerically stable (the resulting V_min might be
> very different from the ideal one).
>
> I am thinking if there is a way to rearrange the matrix A to be B,
> who has the same eigenvectors but A's V_min now becoming V_max,
> corresponding to its biggest eigenvalue, whose numerical solution is
> more reliable.
>
> Thanks
> Jeff

Look up "inverse iteration".
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