Trace class

Spectral theorem

Let ${\displaystyle T}$ be a bounded self-adjoint operator on a Hilbert space. Then ${\displaystyle T^{2}}$ is trace class if and only if ${\displaystyle T}$ has a pure point spectrum with eigenvalues ${\displaystyle \left\{\lambda _{i}(T)\right\}_{i=1}^{\infty }}$ such that

${\displaystyle \operatorname {Tr} (T^{2})=\sum _{i=1}^{\infty }\lambda _{i}(T^{2})<\infty .}$

Mercer's theorem

Mercer's theorem provides another example of a trace class operator. That is, suppose ${\displaystyle K}$ is a continuous symmetric positive-definite kernel on ${\displaystyle L^{2}([a,b])}$, defined as

${\displaystyle K(s,t)=\sum _{j=1}^{\infty }\lambda _{j}\,e_{j}(s)\,e_{j}(t)}$

then the associated Hilbert–Schmidt integral operator ${\displaystyle T_{K}}$ is trace class, i.e.,

${\displaystyle \operatorname {Tr} (T_{K})=\int _{a}^{b}K(t,t)\,dt=\sum _{i}\lambda _{i}.}$

Finite-rank operators

Every finite-rank operator is a trace-class operator. Furthermore, the space of all finite-rank operators is a dense subspace of ${\displaystyle B_{1}(H)}$ (when endowed with the trace norm).

Given any ${\displaystyle x,y\in H,}$ define the operator ${\displaystyle x\otimes y:H\to H}$ by ${\displaystyle (x\otimes y)(z):=\langle z,y\rangle x.}$ Then ${\displaystyle x\otimes y}$ 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), ${\displaystyle \operatorname {Tr} (A(x\otimes y))=\langle Ax,y\rangle .}$

Lidskii's theorem

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

Note that the series on the right converges absolutely due to Weyl's inequality $${\displaystyle \sum _{n=1}^{N}\left|\lambda _{n}(A)\right|\leq \sum _{m=1}^{M}s_{m}(A)}$$ between the eigenvalues ${\displaystyle \{\lambda _{n}(A)\}_{n=1}^{N}}$ and the singular values ${\displaystyle \{s_{m}(A)\}_{m=1}^{M}}$ of the compact operator ${\displaystyle A.}$ See also Grothendieck trace theorem.

Relationship between common 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 ${\displaystyle \ell ^{1}(\mathbb {N} ).}$

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 ${\displaystyle \ell ^{1}}$ sequence with respect to some choice of a pair of Hilbert bases. In the same vein, the bounded operators are noncommutative versions of ${\displaystyle \ell ^{\infty }(\mathbb {N} ),}$ the compact operators that of ${\displaystyle c_{0}}$ (the sequences convergent to 0), Hilbert–Schmidt operators correspond to ${\displaystyle \ell ^{2}(\mathbb {N} ),}$ and finite-rank operators to ${\displaystyle c_{00}}$ (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 ${\displaystyle T}$ on a Hilbert space takes the following canonical form: there exist orthonormal bases ${\displaystyle (u_{i})_{i}}$ and ${\displaystyle (v_{i})_{i}}$ and a sequence ${\displaystyle \left(\alpha _{i}\right)_{i}}$ of non-negative numbers with ${\displaystyle \alpha _{i}\to 0}$ such that $${\displaystyle Tx=\sum _{i}\alpha _{i}\langle x,v_{i}\rangle u_{i}\quad {\text{ for all }}x\in H.}$$ Making the above heuristic comments more precise, we have that ${\displaystyle T}$ is trace-class iff the series ${\textstyle \sum _{i}\alpha _{i}}$ is convergent, ${\displaystyle T}$ is Hilbert–Schmidt iff ${\textstyle \sum _{i}\alpha _{i}^{2}}$ is convergent, and ${\displaystyle T}$ is finite-rank iff the sequence ${\displaystyle \left(\alpha _{i}\right)_{i}}$ has only finitely many nonzero terms. This allows to relate these classes of operators. The following inclusions hold and are all proper when ${\displaystyle H}$ is infinite-dimensional:$${\displaystyle \{{\text{ finite rank }}\}\subseteq \{{\text{ trace class }}\}\subseteq \{{\text{ Hilbert--Schmidt }}\}\subseteq \{{\text{ compact }}\}.}$$

The trace-class operators are given the trace norm ${\textstyle \|T\|_{1}=\operatorname {Tr} \left[\left(T^{*}T\right)^{1/2}\right]=\sum _{i}\alpha _{i}.}$ The norm corresponding to the Hilbert–Schmidt inner product is $${\displaystyle \|T\|_{2}=\left[\operatorname {Tr} \left(T^{*}T\right)\right]^{1/2}=\left(\sum _{i}\alpha _{i}^{2}\right)^{1/2}.}$$ Also, the usual operator norm is ${\textstyle \|T\|=\sup _{i}\left(\alpha _{i}\right).}$ By classical inequalities regarding sequences, $${\displaystyle \|T\|\leq \|T\|_{2}\leq \|T\|_{1}}$$ for appropriate ${\displaystyle T.}$

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 ${\displaystyle c_{0}}$ is ${\displaystyle \ell ^{1}(\mathbb {N} ).}$ Similarly, we have that the dual of compact operators, denoted by ${\displaystyle K(H)^{*},}$ is the trace-class operators, denoted by ${\displaystyle B_{1}.}$ The argument, which we now sketch, is reminiscent of that for the corresponding sequence spaces. Let ${\displaystyle f\in K(H)^{*},}$ we identify ${\displaystyle f}$ with the operator ${\displaystyle T_{f}}$ defined by $${\displaystyle \langle T_{f}x,y\rangle =f\left(S_{x,y}\right),}$$ where ${\displaystyle S_{x,y}}$ is the rank-one operator given by $${\displaystyle S_{x,y}(h)=\langle h,y\rangle x.}$$

This identification works because the finite-rank operators are norm-dense in ${\displaystyle K(H).}$ In the event that ${\displaystyle T_{f}}$ is a positive operator, for any orthonormal basis ${\displaystyle u_{i},}$ one has $${\displaystyle \sum _{i}\langle T_{f}u_{i},u_{i}\rangle =f(I)\leq \|f\|,}$$ where ${\displaystyle I}$ is the identity operator: $${\displaystyle I=\sum _{i}\langle \cdot ,u_{i}\rangle u_{i}.}$$

But this means that ${\displaystyle T_{f}}$ is trace-class. An appeal to polar decomposition extend this to the general case, where ${\displaystyle T_{f}}$ need not be positive.

A limiting argument using finite-rank operators shows that ${\displaystyle \|T_{f}\|_{1}=\|f\|.}$ Thus ${\displaystyle K(H)^{*}}$ is isometrically isomorphic to ${\displaystyle B_{1}.}$

As the predual of bounded operators

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