Trace class

Suppose ${\displaystyle H}$ is a separable Hilbert space and ${\displaystyle A:H\to H}$ a bounded linear operator on ${\displaystyle H}$ which is non-negative (I.e., semi—positive-definite) and self-adjoint. The trace of ${\displaystyle A}$, denoted by ${\displaystyle \operatorname {Tr} A,}$ is the sum of the series[1]

$${\displaystyle \operatorname {Tr} A=\sum _{k}\left\langle Ae_{k},e_{k}\right\rangle ,}$$

where ${\displaystyle \left(e_{k}\right)_{k}}$ is an orthonormal basis of ${\displaystyle H}$. The trace is a sum on non-negative reals and is therefore a non-negative real or infinity. It can be shown that the trace does not depend on the choice of orthonormal basis.

For an arbitrary bounded linear operator ${\displaystyle T:H\to H}$ on ${\displaystyle H,}$ we define its absolute value, denoted by ${\displaystyle |T|,}$ to be the positive square root of ${\displaystyle T^{*}T,}$ that is, ${\displaystyle |T|:={\sqrt {T^{*}T}}}$ is the unique bounded positive operator on ${\displaystyle H}$ such that ${\displaystyle |T|\circ |T|=T^{*}\circ T.}$ The operator ${\displaystyle T:H\to H}$ is said to be in the trace class if ${\displaystyle \operatorname {Tr} (|T|)<\infty .}$ We denote the space of all trace class linear operators on H by ${\displaystyle B_{1}(H).}$ (One can show that this is indeed a vector space.)

If ${\displaystyle T}$ is in the trace class, we define the trace of ${\displaystyle T}$ by

$${\displaystyle \operatorname {Tr} T=\sum _{k}\left\langle Te_{k},e_{k}\right\rangle ,}$$

where ${\displaystyle \left(e_{k}\right)_{k}}$ is an arbitrary orthonormal basis of ${\displaystyle H}$. It can be shown that this is an absolutely convergent series of complex numbers whose sum does not depend on the choice of orthonormal basis.

When H is finite-dimensional, every operator is trace class and this definition of trace of T coincides with the definition of the trace of a matrix.

Given a bounded linear operator ${\displaystyle T:H\to H}$, each of the following statements is equivalent to ${\displaystyle T}$ being in the trace class:

• ${\displaystyle \operatorname {Tr} (|T|)<\infty .}$[1]
• For some orthonormal basis ${\displaystyle \left(e_{k}\right)_{k}}$ of H, the sum of positive terms ${\textstyle \sum _{k}\left\langle |T|\,e_{k},e_{k}\right\rangle }$ is finite.
• For every orthonormal basis ${\displaystyle \left(e_{k}\right)_{k}}$ of H, the sum of positive terms ${\textstyle \sum _{k}\left\langle |T|\,e_{k},e_{k}\right\rangle }$ is finite.
• T is a compact operator and ${\textstyle \sum _{i=1}^{\infty }s_{i}<\infty ,}$ where ${\displaystyle s_{1},s_{2},\ldots }$ are the eigenvalues of ${\displaystyle |T|}$ (also known as the singular values of T) with each eigenvalue repeated as often as its multiplicity.[1]
• There exist two orthogonal sequences ${\displaystyle \left(x_{i}\right)_{i=1}^{\infty }}$ and ${\displaystyle \left(y_{i}\right)_{i=1}^{\infty }}$ in ${\displaystyle H}$ and a sequence ${\displaystyle \left(\lambda _{i}\right)_{i=1}^{\infty }}$ in ${\displaystyle \ell ^{1}}$ such that for all ${\displaystyle x\in H,}$ ${\textstyle T(x)=\sum _{i=1}^{\infty }\lambda _{i}\left\langle x,x_{i}\right\rangle y_{i}.}$[2] Here, the infinite sum means that the sequence of partial sums ${\textstyle \left(\sum _{i=1}^{N}\lambda _{i}\left\langle x,x_{i}\right\rangle y_{i}\right)_{N=1}^{\infty }}$ converges to ${\displaystyle T(x)}$ in H.
• T is a nuclear operator.
• T is equal to the composition of two Hilbert-Schmidt operators.[1]
• ${\textstyle {\sqrt {|T|}}}$ is a Hilbert-Schmidt operator.[1]
• T is an integral operator.[3]
• There exist weakly closed and equicontinuous (and thus weakly compact) subsets ${\displaystyle A^{\prime }}$ and ${\displaystyle B^{\prime \prime }}$ of ${\displaystyle H^{\prime }}$ and ${\displaystyle H^{\prime \prime },}$ respectively, and some positive Radon measure ${\displaystyle \mu }$ on ${\displaystyle A^{\prime }\times B^{\prime \prime }}$ of total mass ${\displaystyle \leq 1}$ such that for all ${\displaystyle x\in H}$ and ${\displaystyle y^{\prime }\in H^{\prime }}$:
  $${\displaystyle y^{\prime }(T(x))=\int _{A^{\prime }\times B^{\prime \prime }}x^{\prime }(x)\;y^{\prime \prime }\left(y^{\prime }\right)\,\mathrm {d} \mu \left(x^{\prime },y^{\prime \prime }\right).}$$

  

Trace-norm

We define the trace-norm of a trace class operator T to be the value

$${\displaystyle \|T\|_{1}:=\operatorname {Tr} (|T|).}$$

One can show that the trace-norm is a norm on the space of all trace class operators ${\displaystyle B_{1}(H)}$ and that ${\displaystyle B_{1}(H)}$, with the trace-norm, becomes a Banach space.

If T is trace class then[4]

$${\displaystyle \|T\|_{1}=\sup \left\{|\operatorname {Tr} (CT)|:\|C\|\leq 1{\text{ and }}C:H\to H{\text{ is a compact operator }}\right\}.}$$

Every bounded linear operator that has a finite-dimensional range (i.e. operators of finite-rank) is trace class;[1] furthermore, the space of all finite-rank operators is a dense subspace of ${\displaystyle B_{1}(H)}$ (when endowed with the ${\displaystyle \|\cdot \|_{1}}$ norm).[4] The composition of two Hilbert-Schmidt operators is a trace class operator.[1]

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), ${\displaystyle \operatorname {Tr} (A(x\otimes y))=\langle Ax,y\rangle .}$[4]

1. If ${\displaystyle A:H\to H}$ is a non-negative self-adjoint operator, then ${\displaystyle A}$ is trace-class if and only if ${\displaystyle \operatorname {Tr} A<\infty .}$ Therefore, a self-adjoint operator ${\displaystyle A}$ is trace-class if and only if its positive part ${\displaystyle A^{+}}$ and negative part ${\displaystyle A^{-}}$ 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, that is,
   $${\displaystyle \operatorname {Tr} (aA+bB)=a\operatorname {Tr} (A)+b\operatorname {Tr} (B).}$$

   
   The bilinear map
   $${\displaystyle \langle A,B\rangle =\operatorname {Tr} (A^{*}B)}$$

   
   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. ${\displaystyle \operatorname {Tr} :B_{1}(H)\to \mathbb {C} }$ is a positive linear functional such that if ${\displaystyle T}$ is a trace class operator satisfying ${\displaystyle T\geq 0{\text{ and }}\operatorname {Tr} T=0,}$ then ${\displaystyle T=0.}$[1]
4. If ${\displaystyle T:H\to H}$ is trace-class then so is ${\displaystyle T^{*}}$ and ${\displaystyle \|T\|_{1}=\left\|T^{*}\right\|_{1}.}$[1]
5. If ${\displaystyle A:H\to H}$ is bounded, and ${\displaystyle T:H\to H}$ is trace-class, then ${\displaystyle AT}$ and ${\displaystyle TA}$ are also trace-class (i.e. the space of trace-class operators on H is an ideal in the algebra of bounded linear operators on H), and[1] [5][1]
   $${\displaystyle \|AT\|_{1}=\operatorname {Tr} (|AT|)\leq \|A\|\|T\|_{1},\quad \|TA\|_{1}=\operatorname {Tr} (|TA|)\leq \|A\|\|T\|_{1}.}$$

   
   Furthermore, under the same hypothesis,[1]
   $${\displaystyle \operatorname {Tr} (AT)=\operatorname {Tr} (TA)}$$

   
   and ${\displaystyle |\operatorname {Tr} (AT)|\leq \|A\|\|T\|.}$ The last assertion also holds under the weaker hypothesis that A and T are Hilbert–Schmidt.
6. If ${\displaystyle \left(e_{k}\right)_{k}}$ and ${\displaystyle \left(f_{k}\right)_{k}}$ are two orthonormal bases of H and if T is trace class then ${\textstyle \sum _{k}\left|\left\langle Te_{k},f_{k}\right\rangle \right|\leq \|T\|_{1}.}$[4]
7. If A is trace-class, then one can define the Fredholm determinant of ${\displaystyle I+A}$:
   $${\displaystyle \det(I+A):=\prod _{n\geq 1}[1+\lambda _{n}(A)],}$$

   
   where ${\displaystyle \{\lambda _{n}(A)\}_{n}}$ is the spectrum of ${\displaystyle A.}$ The trace class condition on ${\displaystyle A}$ guarantees that the infinite product is finite: indeed,
   $${\displaystyle \det(I+A)\leq e^{\|A\|_{1}}.}$$

   
   It also implies that ${\displaystyle \det(I+A)\neq 0}$ if and only if ${\displaystyle (I+A)}$ is invertible.
8. If ${\displaystyle A:H\to H}$ is trace class then for any orthonormal basis ${\displaystyle \left(e_{k}\right)_{k}}$ of ${\displaystyle H,}$ the sum of positive terms ${\textstyle \sum _{k}\left|\left\langle A\,e_{k},e_{k}\right\rangle \right|}$ is finite.[1]
9. If ${\displaystyle A=B^{*}C}$ for some Hilbert-Schmidt operators ${\displaystyle B}$ and ${\displaystyle C}$ then for any normal vector ${\displaystyle e\in H,}$ ${\textstyle |\langle Ae,e\rangle |={\frac {1}{2}}\left(\|Be\|^{2}+\|Ce\|^{2}\right)}$ holds.[1]

Lidskii's theorem

Let ${\displaystyle A}$ be a trace-class operator in a separable Hilbert space ${\displaystyle H,}$ and let ${\displaystyle \{\lambda _{n}(A)\}_{n=1}^{N},}$ ${\displaystyle N\leq \infty }$ 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 ${\displaystyle A.}$[6]

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 ${\displaystyle \left(u_{i}\right)_{i}}$ and ${\displaystyle \left(v_{i}\right)_{i}}$ 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 ${\displaystyle C_{1}.}$

• Nuclear operator
• Nuclear operators between Banach spaces
• Trace operator

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