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Re: operator trace
Posted:
Aug 7, 2005 4:25 PM
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On Sat, 5 Aug 2005, Edward Green wrote:
> The trace of a matrix is the sum of its diagonal elements.
> I've encountered the idea that operators may have traces. Linear
> operators can often be represented by matrices, so, presumably, in case
> the operator is represented by a matrix, the operator trace reduces to
> the matrix trace.
>
> But is there a more general concept of operator trace which does not
> explicitly depend on a matrix representation?
Most responses (maybe all, I may not received all of them yet) considered
traces of operators on Hilbert space.
All you really need in the infinite-dimensional case is a normed vector
space V (the norm does not have to come from an inner product, and the
space need not be complete, nor does it have to have a topological
basis!). For safety's sake, we stipulate that all operators in the
discussion below are bounded with respect to the norm on V.
Finite rank operators have a well-defined trace:
a rank-one operator A=xf (x from V, f from V*, the dual) has a
well-defined trace tr(xf) = f(x). It takes some tedious bookkeeping
algebra to prove that the trace extends linearly to finite-rank operators,
and is independent of the representation using rank-one summands.
Denote F1 the set of bounded finite-rank operators V->V of operator
norm <=1. Then for every linear operator T:V->V and every X from F1,
TX and XT are of finite rank, and tr(TX)=tr(XT) (routine proof).
To put the norm to work, we define the absolute trace of T (need not be
finite) by
||T||_1 = sup{abs(tr(XT)) : X from F1 }
If this absolute trace of T is finite, we call T a trace class operator on
V, and by further work on norms we can prove that the trace extends
linearly and continuously (with respect to the absolute trace, which is a
norm) from finite rank operators to all trace class operators.
Look Ma, no coordinates.
Remark: This definition, restricted to Hilbert spaces, defines the same
set of trace class operators. (An exercise I saw somewhere.)
Also, on Hilbert spaces, the trace class operators are the pre-dual of the
bounded operators. Has anyone studied the Banach space setup of this fact?
Cheers, ZVK(Slavek).
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