On Mon, Feb 9, 2009 at 12:32 AM, Xiaoyu Chu <[email protected]> wrote:

> Hey all,
>      I am currently working on a large matrix, and I already have a
> specific eigen value that I want to use in order to find out its
> corresponding eigen vector. Is there an easy way to do so?
>
>     I have tried with linalg.solve(a, b), where I put a as the
> Matrix A - eigen value* unit matrix, and b as the zero matrix. But the
> solution returned is a zero matrix, which I really find disappointing.
>
>     I have also tried with eig(A), which finds out all the eigen
> vectors of matrix A, but it takes too long to run especially the order
> of my matrix is like 10,000.
>
>
>    So right now, I really find myself stuck. Is there anyone who can help
> me?
>

The usual trick for this is inverse iteration,
http://en.wikipedia.org/wiki/Inverse_iteration, which involves solving the
almost singular equation A - eig*I = rhs for an appropriately chosen rhs.
The required eigenvector will dominate the solution and will grow rapidly in
amplitude over a few iterations.

Chuck
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