Re: Stochastic matrices (operators?)

igor.kh_at_gmail.com
Date: 01/30/05 

Date: 30 Jan 2005 14:40:24 -0800

Jannick Asmus wrote:
> On 30.01.2005 01:18, igor.kh@gmail.com wrote:

> > Another interesting question is about the type of cone that's
allowed.
> > For example, a cone generated by non-negative linear combinations
of
> > elements of S has "edges", while I can imagine having a round cone.
> > Obviously one cannot be transformed into another. Is this
difference
> > significant?
>
> It might be of advantage to construct a situation the separation
> theorems of convex sets can apply to. I think it is sufficient to
take C
> open with <C,C> > 0.

Thinking of this some more, I found that there is quite a bit more
structure lurking in the background than I previously thought.

The two clues are (1) probability theory deals with a measure space and
the algebra of measurable functions on it, (2) the Birkoff-von Neuman
theorem describes all doubly stochastic matrices as the convex hull of
the set
of permutation matrices. With the structure of a cone and a linear
functional
positive on its interior we still can't say which endomorphisms are
permutations and which aren't. So at least in the case of doubly
stochastic
matrices, we need more structure.

Consider the following. Let A be a real (complex probably also works)
finite dimensional unital commutative algebra. Let it be equipped with
a mean m:A->R, a linear functional. There is a natural cone structure
on A given by all elements that can be written as squares a = b^2, a,b
in A. We also require that m is positive on the interior of that cone
and that m(1) = 1. For free we also get an inner product (a,b) = m(ab),
its positive definite by positivity of m.

It is well known that we can find a preferred basis in A, consisting of
idempotent elements, e_i^2 = e_i and e_i e_j = 0 if i!=j. At least now
we can say which endomorphisms of A are permutations, the ones that
permute the e_i among each other. But these are precisely the
automorphisms of A. The definition of a stochastic map from A to A
would then be one that preserves the mean m and the positive cone. It
is doubly stochastic if the same is true for its adjoint with respect
to the above inner product.

If I want automorphisms to be permutations, they should also be doubly
stochastic. But that implies that all automorphisms must leave m
invariant. Hmm, I think this actually puts severe restrictions on m. I
think m is then unique, it gives equal weight to each idempotent and
gives weight 1 to the identity element, which is the sum of all the
idempotents. In essence, given a finite dimensional algebra, we get an
automorphism invariant mean for free, as well as an inner product
associated with that mean.

In this new language, the Birkoff-von Neumann theorem takes on a pretty
form. If a linear endomorphism of A as well as its adjoint preserve the
positive cone, then it is a convex linear combination of automorphisms
of A.

It would be nice if the condition "preserves the positive cone" could
be stated as some kind of inequality. An automorphism is multiplicative
S(ab)=S(a)S(b), so perhaps stochastic maps could be submultiplicative
P(ab)<=P(a)P(b) or supermultiplicative P(ab)>=P(a)P(b). I tried playing
around with stochastic matrices, but couldn't see if any such
inequality holds.

This formulation also beggs the question of how classic results about
stochastic matrices generalize to infinite dimension.

Igor

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