Position of the largest Eigenvector about lambda-lanczos HOT 3 CLOSED

Thank you for your comment.

The run() returns only one eigenvector that corresponds to either the largest eigenvalue or the smallest eigenvalue depending on the third argument find_maximum. So the position of the returned eigenvector should be

• (n-1)th, if the third argument is true
• 0th, if the third argument is false

for an n by n matrix if one could know all n eigenvectors in ascending eigenvalue order.

(I may have misunderstood your question. I welcome your other comments.)

Regards,
mrcdr

from lambda-lanczos.

Hi mcrdr,

Thank you for your answer. My bad, I guess, that I might misunderstand the concept of eigenvalues and how you are computing them. Thank you once again :-)

Cheers

from lambda-lanczos.

Hello Dominik,

Thank you too for being interested in my library!

Best,
mrcdr

from lambda-lanczos.