Re: eigenvalues on the unit circle
- From: israel@xxxxxxxxxxx (Robert Israel)
- Date: 17 Mar 2006 22:38:25 GMT
In article <1142603926.360117.134770@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
pkg <Patric.Gloede@xxxxxxxxxxx> wrote:
I have a function K( . , . ): S x S --> IR, where S is a countable
abelian group. One can see K as an operator acting on l^2(S) and
mapping again into l^2(S), where l^2 is the Hilbert space of real
square summable sequences, in the following way:
K m : l^2 --> l^2, K m = sum{K( . , x) m(x), x
\in S} \in l^2 .
(probabilists may think of the Green function of a random walk!).
I know that it is selfadjoint, bounded, and positive, and for K( . , .)
seen as a function on S x S we have
sup_x sup_y sum{ K(x,z) K(z,y), z \in S} <
\infty.
That's implied by the fact K is a bounded operator.
If e_x is the standard unit vector (e_x(z) = 1 for z = x, 0 otherwise),
we have K(x,z) = <e_x, K e_z> and
sum_z K(x,z) K(z,y) = sum_z <K* e_x, e_z> <e_z, K e_y> = <K* e_x, K e_y>
so your double sup is at most ||K||^2.
In addition K( . , . ) is symmetric.
Must be, if the operator is self-adjoint.
My question is now: What would be sensible criteria in order to check
whether K has eigenvalues on the unit circle (and then because of the
other properties || K || would be an eigenvalue)?
If K is self-adjoint, the only possible eigenvalues on the unit
circle are -1 and 1.
Robert Israel israel@xxxxxxxxxxx
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada
.
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