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