[Trilinos-Users] Trilinos-Users Digest, Vol 57, Issue 18

[Mark Hoemmen]

On May 24, 2010, at 12:00 PM, trilinos-users-request at software.sandia.gov wrote:
> You might also try solving the (nongeneralized) eigenvalue problem $B^{-1} A x = \lambda x$, if your B matrix is indeed diagonal and positive.

Please disregard this formula and use the formula in Chris Baker's post today ($B^{-1/2} A B^{-1/2} x = \lambda x$) -- I forgot that A is symmetric and needs to stay that way.

mfh