- #1
member 428835
Hi PF!
I am trying to solve the eigenvalue problem ##A v = \lambda B v##. I thought I'd solve this by $$A v = \lambda B v \implies\\
B^{-1} A v = \lambda v\implies\\
(B^{-1} A - \lambda I) v = 0 $$
and then using the built in function Eigenvalues and Eigenvectors on the matrix ##B^{-1}A##. But as you can see from the title, ##B^{-1}## is badly conditioned. Any ideas on how to get around this?
I am trying to solve the eigenvalue problem ##A v = \lambda B v##. I thought I'd solve this by $$A v = \lambda B v \implies\\
B^{-1} A v = \lambda v\implies\\
(B^{-1} A - \lambda I) v = 0 $$
and then using the built in function Eigenvalues and Eigenvectors on the matrix ##B^{-1}A##. But as you can see from the title, ##B^{-1}## is badly conditioned. Any ideas on how to get around this?