Let H {\displaystyle H} be a separable Hilbert space, { e k } k = 1 ∞ {\displaystyle \left\{e_{k}\right\}_{k=1}^{\infty }} an orthonormal basis and A : H → H {\displaystyle A:H\to H} a positive bounded linear operator on H {\displaystyle H} . The trace of A {\displaystyle A} is denoted by Tr ( A ) {\displaystyle \operatorname {Tr} (A)} and defined as23
independent of the choice of orthonormal basis. A (not necessarily positive) bounded linear operator T : H → H {\displaystyle T:H\rightarrow H} is called trace class if and only if
where | T | := T ∗ T {\displaystyle |T|:={\sqrt {T^{*}T}}} denotes the positive-semidefinite Hermitian square root.4
The trace-norm of a trace class operator T is defined as ‖ T ‖ 1 := Tr ( | T | ) . {\displaystyle \|T\|_{1}:=\operatorname {Tr} (|T|).} One can show that the trace-norm is a norm on the space of all trace class operators B 1 ( H ) {\displaystyle B_{1}(H)} and that B 1 ( H ) {\displaystyle B_{1}(H)} , with the trace-norm, becomes a Banach space.
When H {\displaystyle H} is finite-dimensional, every (positive) operator is trace class. For A {\displaystyle A} this definition coincides with that of the trace of a matrix. If H {\displaystyle H} is complex, then A {\displaystyle A} is always self-adjoint (i.e. A = A ∗ = | A | {\displaystyle A=A^{*}=|A|} ) though the converse is not necessarily true.5
Given a bounded linear operator T : H → H {\displaystyle T:H\to H} , each of the following statements is equivalent to T {\displaystyle T} being in the trace class:
Let T {\displaystyle T} be a bounded self-adjoint operator on a Hilbert space. Then T 2 {\displaystyle T^{2}} is trace class if and only if T {\displaystyle T} has a pure point spectrum with eigenvalues { λ i ( T ) } i = 1 ∞ {\displaystyle \left\{\lambda _{i}(T)\right\}_{i=1}^{\infty }} such that14
Mercer's theorem provides another example of a trace class operator. That is, suppose K {\displaystyle K} is a continuous symmetric positive-definite kernel on L 2 ( [ a , b ] ) {\displaystyle L^{2}([a,b])} , defined as
then the associated Hilbert–Schmidt integral operator T K {\displaystyle T_{K}} is trace class, i.e.,
Every finite-rank operator is a trace-class operator. Furthermore, the space of all finite-rank operators is a dense subspace of B 1 ( H ) {\displaystyle B_{1}(H)} (when endowed with the trace norm).15
Given any x , y ∈ H , {\displaystyle x,y\in H,} define the operator x ⊗ y : H → H {\displaystyle x\otimes y:H\to H} by ( x ⊗ y ) ( z ) := ⟨ z , y ⟩ x . {\displaystyle (x\otimes y)(z):=\langle z,y\rangle x.} Then x ⊗ y {\displaystyle x\otimes y} is a continuous linear operator of rank 1 and is thus trace class; moreover, for any bounded linear operator A on H (and into H), Tr ( A ( x ⊗ y ) ) = ⟨ A x , y ⟩ . {\displaystyle \operatorname {Tr} (A(x\otimes y))=\langle Ax,y\rangle .} 16
Let A {\displaystyle A} be a trace-class operator in a separable Hilbert space H , {\displaystyle H,} and let { λ n ( A ) } n = 1 N ≤ ∞ {\displaystyle \{\lambda _{n}(A)\}_{n=1}^{N\leq \infty }} be the eigenvalues of A . {\displaystyle A.} Let us assume that λ n ( A ) {\displaystyle \lambda _{n}(A)} are enumerated with algebraic multiplicities taken into account (that is, if the algebraic multiplicity of λ {\displaystyle \lambda } is k , {\displaystyle k,} then λ {\displaystyle \lambda } is repeated k {\displaystyle k} times in the list λ 1 ( A ) , λ 2 ( A ) , … {\displaystyle \lambda _{1}(A),\lambda _{2}(A),\dots } ). Lidskii's theorem (named after Victor Borisovich Lidskii) states that Tr ( A ) = ∑ n = 1 N λ n ( A ) {\displaystyle \operatorname {Tr} (A)=\sum _{n=1}^{N}\lambda _{n}(A)}
Note that the series on the right converges absolutely due to Weyl's inequality ∑ n = 1 N | λ n ( A ) | ≤ ∑ m = 1 M s m ( A ) {\displaystyle \sum _{n=1}^{N}\left|\lambda _{n}(A)\right|\leq \sum _{m=1}^{M}s_{m}(A)} between the eigenvalues { λ n ( A ) } n = 1 N {\displaystyle \{\lambda _{n}(A)\}_{n=1}^{N}} and the singular values { s m ( A ) } m = 1 M {\displaystyle \{s_{m}(A)\}_{m=1}^{M}} of the compact operator A . {\displaystyle A.} 25
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 ℓ 1 ( N ) . {\displaystyle \ell ^{1}(\mathbb {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 ℓ 1 {\displaystyle \ell ^{1}} sequence with respect to some choice of a pair of Hilbert bases. In the same vein, the bounded operators are noncommutative versions of ℓ ∞ ( N ) , {\displaystyle \ell ^{\infty }(\mathbb {N} ),} the compact operators that of c 0 {\displaystyle c_{0}} (the sequences convergent to 0), Hilbert–Schmidt operators correspond to ℓ 2 ( N ) , {\displaystyle \ell ^{2}(\mathbb {N} ),} and finite-rank operators to c 00 {\displaystyle c_{00}} (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 {\displaystyle T} on a Hilbert space takes the following canonical form: there exist orthonormal bases ( u i ) i {\displaystyle (u_{i})_{i}} and ( v i ) i {\displaystyle (v_{i})_{i}} and a sequence ( α i ) i {\displaystyle \left(\alpha _{i}\right)_{i}} of non-negative numbers with α i → 0 {\displaystyle \alpha _{i}\to 0} such that T x = ∑ i α i ⟨ x , v i ⟩ u i for all x ∈ H . {\displaystyle Tx=\sum _{i}\alpha _{i}\langle x,v_{i}\rangle u_{i}\quad {\text{ for all }}x\in H.} Making the above heuristic comments more precise, we have that T {\displaystyle T} is trace-class iff the series ∑ i α i {\textstyle \sum _{i}\alpha _{i}} is convergent, T {\displaystyle T} is Hilbert–Schmidt iff ∑ i α i 2 {\textstyle \sum _{i}\alpha _{i}^{2}} is convergent, and T {\displaystyle T} is finite-rank iff the sequence ( α i ) i {\displaystyle \left(\alpha _{i}\right)_{i}} has only finitely many nonzero terms. This allows to relate these classes of operators. The following inclusions hold and are all proper when H {\displaystyle H} is infinite-dimensional: { finite rank } ⊆ { trace class } ⊆ { Hilbert--Schmidt } ⊆ { compact } . {\displaystyle \{{\text{ finite rank }}\}\subseteq \{{\text{ trace class }}\}\subseteq \{{\text{ Hilbert--Schmidt }}\}\subseteq \{{\text{ compact }}\}.}
The trace-class operators are given the trace norm ‖ T ‖ 1 = Tr [ ( T ∗ T ) 1 / 2 ] = ∑ i α i . {\textstyle \|T\|_{1}=\operatorname {Tr} \left[\left(T^{*}T\right)^{1/2}\right]=\sum _{i}\alpha _{i}.} The norm corresponding to the Hilbert–Schmidt inner product is ‖ T ‖ 2 = [ Tr ( T ∗ T ) ] 1 / 2 = ( ∑ i α i 2 ) 1 / 2 . {\displaystyle \|T\|_{2}=\left[\operatorname {Tr} \left(T^{*}T\right)\right]^{1/2}=\left(\sum _{i}\alpha _{i}^{2}\right)^{1/2}.} Also, the usual operator norm is ‖ T ‖ = sup i ( α i ) . {\textstyle \|T\|=\sup _{i}\left(\alpha _{i}\right).} By classical inequalities regarding sequences, ‖ T ‖ ≤ ‖ T ‖ 2 ≤ ‖ T ‖ 1 {\displaystyle \|T\|\leq \|T\|_{2}\leq \|T\|_{1}} for appropriate T . {\displaystyle T.}
It is also clear that finite-rank operators are dense in both trace-class and Hilbert–Schmidt in their respective norms.
The dual space of c 0 {\displaystyle c_{0}} is ℓ 1 ( N ) . {\displaystyle \ell ^{1}(\mathbb {N} ).} Similarly, we have that the dual of compact operators, denoted by K ( H ) ∗ , {\displaystyle K(H)^{*},} is the trace-class operators, denoted by B 1 . {\displaystyle B_{1}.} The argument, which we now sketch, is reminiscent of that for the corresponding sequence spaces. Let f ∈ K ( H ) ∗ , {\displaystyle f\in K(H)^{*},} we identify f {\displaystyle f} with the operator T f {\displaystyle T_{f}} defined by ⟨ T f x , y ⟩ = f ( S x , y ) , {\displaystyle \langle T_{f}x,y\rangle =f\left(S_{x,y}\right),} where S x , y {\displaystyle S_{x,y}} is the rank-one operator given by S x , y ( h ) = ⟨ h , y ⟩ x . {\displaystyle S_{x,y}(h)=\langle h,y\rangle x.}
This identification works because the finite-rank operators are norm-dense in K ( H ) . {\displaystyle K(H).} In the event that T f {\displaystyle T_{f}} is a positive operator, for any orthonormal basis u i , {\displaystyle u_{i},} one has ∑ i ⟨ T f u i , u i ⟩ = f ( I ) ≤ ‖ f ‖ , {\displaystyle \sum _{i}\langle T_{f}u_{i},u_{i}\rangle =f(I)\leq \|f\|,} where I {\displaystyle I} is the identity operator: I = ∑ i ⟨ ⋅ , u i ⟩ u i . {\displaystyle I=\sum _{i}\langle \cdot ,u_{i}\rangle u_{i}.}
But this means that T f {\displaystyle T_{f}} is trace-class. An appeal to polar decomposition extend this to the general case, where T f {\displaystyle T_{f}} need not be positive.
A limiting argument using finite-rank operators shows that ‖ T f ‖ 1 = ‖ f ‖ . {\displaystyle \|T_{f}\|_{1}=\|f\|.} Thus K ( H ) ∗ {\displaystyle K(H)^{*}} is isometrically isomorphic to B 1 . {\displaystyle B_{1}.}
Recall that the dual of ℓ 1 ( N ) {\displaystyle \ell ^{1}(\mathbb {N} )} is ℓ ∞ ( N ) . {\displaystyle \ell ^{\infty }(\mathbb {N} ).} In the present context, the dual of trace-class operators B 1 {\displaystyle B_{1}} is the bounded operators B ( H ) . {\displaystyle B(H).} More precisely, the set B 1 {\displaystyle B_{1}} is a two-sided ideal in B ( H ) . {\displaystyle B(H).} So given any operator T ∈ B ( H ) , {\displaystyle T\in B(H),} we may define a continuous linear functional φ T {\displaystyle \varphi _{T}} on B 1 {\displaystyle B_{1}} by φ T ( A ) = Tr ( A T ) . {\displaystyle \varphi _{T}(A)=\operatorname {Tr} (AT).} This correspondence between bounded linear operators and elements φ T {\displaystyle \varphi _{T}} of the dual space of B 1 {\displaystyle B_{1}} is an isometric isomorphism. It follows that B ( H ) {\displaystyle B(H)} is the dual space of B 1 . {\displaystyle B_{1}.} This can be used to define the weak-* topology on B ( H ) . {\displaystyle B(H).}
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