Trace class


In mathematics, a trace-class operator is a compact operator for which a trace may be defined, such that the trace is finite and independent of the choice of basis.
Trace-class operators are essentially the same as nuclear operators, though many authors reserve the term "trace-class operator" for the special case of nuclear operators on Hilbert spaces and reserve "nuclear operator" for usage in more general topological vector spaces.

Definition

Define the trace, denoted by, of linear operator A to be the sum of the series
where this sum is independent of the choice of the orthonormal basis k of H and where this sum equals if it does not converge.
If H is finite-dimensional then is equal to the usual definition of the trace.
For any bounded linear operator T : H → H over a Hilbert space H, we define its absolute value, denoted by |T|, to be the positive square root of, i.e. is the unique bounded positive operator on H such that.
It may be shown that a bounded linear operator on a Hilbert space is trace class if and only if its absolute value is trace class.
A bounded linear operator T : H → H over a Hilbert space H is said to be in the trace class if any of the following equivalent conditions is satisfied:

  1. T is a nuclear operator.
  2. T is equal to the composition of two Hilbert-Schmidt operators.
  3. is a Hilbert-Schmidt operator.
  4. T is an integral operator.
  5. there exist weakly closed and equicontinuous subsets and of and, respectively, and some positive Radon measure on of total mass ≤ 1 such that for all x ∈ H and :
  6. there exist two orthogonal sequences and in H and a sequence in l1 such that for all x in H,.
    • Here, the infinite sum means that the sequence of partial sums converges to T in H.
  7. T is a compact operator and, where l1, l2,... are the eigenvalues of T with each eigenvalue repeated as often as its multiplicity.
    • Recall that the multiplicity of an eigenvalue r is the dimension of the kernel of T - r IdH, where IdH : H → H is the identity map.
  8. for some orthonormal basis k of H, the sum of positive terms is finite.
  9. the above condition but with the word "some" replaced by "every".
  10. the transpose map is trace class, in which case.
    • Recall that the transpose of T is defined by, for all belonging to the continuous dual space of H. The subscript b indicates that has its usual norm topology.
  11. .
and if T is not already a positive operator then we may add to this list:

  1. the operator is trace class.

Trace-norm

If T is trace class then we define the trace-norm of a trace class operator T to be the common value
and we denote the space of all trace class linear operators on H by B1.
If T is trace class then
When H is finite-dimensional, every operator is trace class and this definition of trace of A coincides with the definition of the trace of a matrix.
By extension, if A is a non-negative self-adjoint operator, we can also define the trace of A as an extended real number by the possibly divergent sum
where this sum is independent of the choice of the orthonormal basis k of H.

Examples

Every bounded linear operator that has a finite-dimensional range is trace class;
furthermore, the space of all finite-rank operators is a dense subspace of B1.
The composition of two Hilbert-Schmidt operators is a trace class operator.
Given any x and y in H, define by = <z, y> x, which is a continuous linear operator of rank 1 and is thus trace class;
moreover, for any bounded linear operator A on H,.

Properties

Lidskii's theorem

Let be a trace-class operator in a separable Hilbert space, and let be the eigenvalues of. Let us assume that are enumerated with algebraic multiplicities taken into account. Lidskii's theorem states that
Note that the series in the left converges absolutely due to Weyl's inequality
between the eigenvalues and the singular values of a compact operator.

Relationship between some classes of operators

One can view certain classes of bounded operators as noncommutative analogue of classical sequence spaces, with trace-class operators as the noncommutative analogue of the sequence space ℓ1.
Indeed, it is possible to apply the spectral theorem to show that every normal trace-class operator on a separable Hilbert space can be realized in a certain way as an ℓ1 sequence with respect to some choice of a pair of Hilbert bases. In the same vein, the bounded operators are noncommutative versions of ℓ∞, the compact operators that of c0, Hilbert–Schmidt operators correspond to ℓ2, and finite-rank operators. To some extent, the relationships between these classes of operators are similar to the relationships between their commutative counterparts.
Recall that every compact operator T on a Hilbert space takes the following canonical form:
for some orthonormal bases and. Making the above heuristic comments more precise, we have that T is trace-class if the series ∑i αi is convergent, T is Hilbert–Schmidt if ∑i αi2 is convergent, and T is finite-rank if the sequence has only finitely many nonzero terms.
The above description allows one to obtain easily some facts that relate these classes of operators. For example, the following inclusions hold and are all proper when H is infinite-dimensional: ⊂ ⊂ ⊂.
The trace-class operators are given the trace norm ||T||1 = Tr = ∑i αi. The norm corresponding to the Hilbert–Schmidt inner product is ||T||2 = ½ = ½. Also, the usual operator norm is ||T|| = supi. By classical inequalities regarding sequences,
for appropriate T.
It is also clear that finite-rank operators are dense in both trace-class and Hilbert–Schmidt in their respective norms.

Trace class as the dual of compact operators

The dual space of c0 is ℓ1. Similarly, we have that the dual of compact operators, denoted by K*, is the trace-class operators, denoted by C1. The argument, which we now sketch, is reminiscent of that for the corresponding sequence spaces. Let f ∈ K*, we identify f with the operator Tf defined by
where Sx,y is the rank-one operator given by
This identification works because the finite-rank operators are norm-dense in K. In the event that Tf is a positive operator, for any orthonormal basis ui, one has
where I is the identity operator:
But this means that Tf is trace-class. An appeal to polar decomposition extend this to the general case, where Tf need not be positive.
A limiting argument using finite-rank operators shows that ||Tf||1 = ||f||. Thus K* is isometrically isomorphic to C1.

As the predual of bounded operators

Recall that the dual of ℓ1 is ℓ∞. In the present context, the dual of trace-class operators C1 is the bounded operators B. More precisely, the set C1 is a two-sided ideal in B. So given any operator T in B, we may define a continuous linear functional φT on by φT = Tr. This correspondence between bounded linear operators and elements φT of the dual space of is an isometric isomorphism. It follows that B is the dual space of. This can be used to define the weak-* topology on B.