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 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