# 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 Banach spaces.

## Definition

A bounded linear operator A over a separable Hilbert space H is said to be in the trace class if for some (and hence all) orthonormal bases {ek}k of H, the sum of positive terms

$\|A\|_{1}=\operatorname {Tr} |A|:=\sum _{k}{\Big \langle }(A^{*}A)^{\frac {1}{2}}\,e_{k},e_{k}{\Big \rangle }$ is finite. In this case, the trace of A, which is given by the sum

$\operatorname {Tr} A:=\sum _{k}\langle Ae_{k},e_{k}\rangle$ is absolutely convergent and is independent of the choice of the orthonormal basis. 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

$\sum _{k}\langle Ae_{k},e_{k}\rangle .$ ## Properties

1. If A is a non-negative self-adjoint, then A is trace-class if and only if Tr(A) < ∞. Therefore, a self-adjoint operator A is trace-class if and only if its positive part A+ and negative part A are both trace-class. (The positive and negative parts of a self-adjoint operator are obtained by the continuous functional calculus.)
2. The trace is a linear functional over the space of trace-class operators, i.e.
$\operatorname {Tr} (aA+bB)=a\operatorname {Tr} (A)+b\operatorname {Tr} (B).$ The bilinear map

$\langle A,B\rangle =\operatorname {Tr} (A^{*}B)$ is an inner product on the trace class; the corresponding norm is called the Hilbert–Schmidt norm. The completion of the trace-class operators in the Hilbert–Schmidt norm are called the Hilbert–Schmidt operators.
3. If $A$ is bounded, and $B$ is trace-class, $AB$ and $BA$ are also trace-class, and 
$\|AB\|_{1}=\operatorname {Tr} (|AB|)\leq \|A\|\|B\|_{1},\quad \|BA\|_{1}=\operatorname {Tr} (|BA|)\leq \|A\|\|B\|_{1}.$ Furthermore, under the same hypothesis,

$\operatorname {Tr} (AB)=\operatorname {Tr} (BA).$ The last assertion also holds under the weaker hypothesis that $A$ and $B$ are Hilbert–Schmidt.
4. If $A$ is trace-class, then one can define the Fredholm determinant of $1+A$ :
$\det(I+A):=\prod _{n\geq 1}[1+\lambda _{n}(A)],$ where $\{\lambda _{n}(A)\}_{n}$ is the spectrum of $A$ . The trace class condition on $A$ guarantees that the infinite product is finite: indeed,

$\det(I+A)\leq e^{\|A\|_{1}}.$ It also implies that $\det(I+A)\neq 0$ if and only if $(I+A)$ is invertible.

### Lidskii's theorem

Let $A$ be a trace-class operator in a separable Hilbert space $H$ , and let $\{\lambda _{n}(A)\}_{n=1}^{N},$ $N\leq \infty$ be the eigenvalues of $A$ . Let us assume that $\lambda _{n}(A)$ are enumerated with algebraic multiplicities taken into account (i.e. if the algebraic multiplicity of $\lambda$ is $k$ , then $\lambda$ is repeated $k$ times in the list $\lambda _{1}(A),\lambda _{2}(A),\dots$ ). Lidskii's theorem (named after Victor Borisovich Lidskii) states that

$\sum _{n=1}^{N}\lambda _{n}(A)=\operatorname {Tr} (A).$ Note that the series in the left converges absolutely due to Weyl's inequality

$\sum _{n=1}^{N}{\big |}\lambda _{n}(A){\big |}\leq \sum _{m=1}^{M}s_{m}(A)$ between the eigenvalues $\{\lambda _{n}(A)\}_{n=1}^{N}$ and the singular values $\{s_{m}(A)\}_{m=1}^{M}$ of a compact operator $A$ .

### 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 1(N).

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 1 sequence with respect to some choice of a pair of Hilbert bases. In the same vein, the bounded operators are noncommutative versions of (N), the compact operators that of c0 (the sequences convergent to 0), Hilbert–Schmidt operators correspond to 2(N), and finite-rank operators (the sequences that have only finitely many non-zero terms). 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:

$\forall h\in H,\;Th=\sum _{i=1}\alpha _{i}\langle h,v_{i}\rangle u_{i},\quad {\text{where}}\quad \alpha _{i}\geq 0,\ \alpha _{i}\to 0,$ for some orthonormal bases {ui} and {vi}. 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 {αi} 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: {finite rank} {trace class} {Hilbert–Schmidt} {compact}.

The trace-class operators are given the trace norm ||T||1 = Tr[(T*T)½] = i αi. The norm corresponding to the Hilbert–Schmidt inner product is ||T||2 = [Tr(T*T)]½ = (iαi2)½. Also, the usual operator norm is ||T|| = supi(αi). By classical inequalities regarding sequences,

$\|T\|\leq \|T\|_{2}\leq \|T\|_{1}$ 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 1(N). Similarly, we have that the dual of compact operators, denoted by K(H)*, 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(H)*, we identify f with the operator Tf defined by

$\langle T_{f}x,y\rangle =f(S_{x,y}),$ where Sx,y is the rank-one operator given by

$S_{x,y}(h)=\langle h,y\rangle x.$ This identification works because the finite-rank operators are norm-dense in K(H). In the event that Tf is a positive operator, for any orthonormal basis ui, one has

$\sum _{i}\langle T_{f}u_{i},u_{i}\rangle =f(I)\leq \|f\|,$ where I is the identity operator:

$I=\sum _{i}\langle \cdot ,u_{i}\rangle u_{i}.$ 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(H)* is isometrically isomorphic to C1.

### As the predual of bounded operators

Recall that the dual of 1(N) is (N). In the present context, the dual of trace-class operators C1 is the bounded operators B(H). More precisely, the set C1 is a two-sided ideal in B(H). So given any operator T in B(H), we may define a continuous linear functional φT on $C_{1}$ by φT(A) = Tr(AT). This correspondence between bounded linear operators and elements φT of the dual space of $C_{1}$ is an isometric isomorphism. It follows that B(H) is the dual space of $C_{1}$ . This can be used to define the weak-* topology on B(H).