On 6/1/24 14:49, Najwa Alshehri wrote:
I decided to solve the exact problem directly, namely AA x = \lambda M x. To
achieve this, I computed the inverse of the matrix AA= Bt * A^inv * B using a
Conjugate Gradient (CG) solver. Subsequently, I solved for the exact
eigenvalues, and to my satisfaction, I obtained results that aligned with
those obtained from MATLAB.
If anyone has insights into why solving for the reciprocal yields different
results, I would greatly appreciate your input on this matter.
Najwa,
I'm glad you figured it out because I don't really have a good idea about what
ARPACK does. Typically,
AA x = \lambda M x
is solved in exactly this way because M may be a matrix that is singular. In
many applications (not sure whether yours is one of those), M is a matrix that
can have a large null-space -- for example, if you are computing eigenvalues
of the Stokes operator, in some applications you have that M = [M_u 0 ; 0 0],
that is, it has a large null block on all pressures. This matrix is not
invertible, whereas AA is. By convention, the invertible matrix is kept on the
left side of the eigenvalue equation.
Best
W.
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