From: israel@math.ubc.ca (Robert Israel)
Subject: Re: Generalized eigenvalue problem
Date: 17 Mar 1999 17:35:43 GMT
Newsgroups: sci.math,sci.math.num-analysis

In article <36EE2AD0.E5A88895@ind-cr.uclm.es>, Juan Jose Garcia Ripoll <jjgarcia@ind-cr.uclm.es> writes:

|> perhaps people here know the answer to this problem. I know that every
|> hermitian operator "A" has complete set of eigenvectors which are solutions of
|> 	Ax = \lambda x
|> Now let "B" be also hermitian and nonsingular. Does "A" have a complete set of
|> generalized eigenvectors that are solutions of
|> 	Ax = \lambda Bx

(If you're talking about hermitian operators in an infinite-dimensional
Hilbert space, I'll interpret "complete set of eigenvectors" as physicists' 
terminology for the Spectral Theorem.)

If B is positive definite, yes.  For then B has a positive definite square root,
and the generalized eigenvector problem is equivalent to the ordinary eigenvector
problem

   B^(-1/2) A B^(-1/2) y = lambda y

where B^(-1/2) A B^(-1/2) is hermitian.  Note that the generalized eigenvectors
x are related to the eigenvectors y of this operator by x = B^(-1/2) y, so
these are not orthogonal in the ordinary inner product.  They are orthogonal
in the inner product u* B v.

If B is not positive definite, the answer can be no.  Example:
    [ 1 1 ]      [ 1 0 ]
A = [ 1 1 ], B = [ 0 -1].  Then A x = lambda B x is equivalent to 
                          [ 1 1 ]
BA x = lambda x, but BA = [-1 -1] does not have a complete set of 
eigenvectors (its only eigenvalue is 0, which only has a one-dimensional
eigenspace).

Robert Israel                                israel@math.ubc.ca
Department of Mathematics        http://www.math.ubc.ca/~israel 
University of British Columbia            
Vancouver, BC, Canada V6T 1Z2