Trace class

The trace, denoted by ${\displaystyle \operatorname {Tr} A,}$ of linear operator ${\displaystyle A}$ is the sum of the series[1]

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

where this sum is independent of the choice of the orthonormal basis ${\displaystyle \left(e_{k}\right)_{k}}$ of ${\displaystyle H}$ and where this sum equals ${\displaystyle \infty }$ if it does not converge.

If ${\displaystyle H}$ is finite-dimensional then ${\displaystyle \operatorname {Tr} A}$ is equal to the usual definition of the trace.

For any bounded linear operator ${\displaystyle T:H\to H}$ over a Hilbert space ${\displaystyle H,}$ 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 ${\displaystyle A}$ such that ${\displaystyle |T|\circ |T|=T^{*}\circ T.}$

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.[1]

A bounded linear operator ${\displaystyle T:H\to H}$ over a Hilbert space ${\displaystyle H}$ is said to be in the trace class if any of the following equivalent conditions is satisfied:

• 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.[2]
• 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 ${\displaystyle y^{\prime }\in H^{\prime }}$:
  $${\displaystyle y^{\prime }(T(x))=\int _{A^{\prime }\times B^{\prime \prime }}x^{\prime }(x)\;y^{\prime \prime }\left(y^{\prime }\right)\,\mathrm {d} \mu \left(x^{\prime },y^{\prime \prime }\right).}$$

  
• 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 ${\displaystyle x\in H,}$ ${\textstyle T(x)=\sum _{i=1}^{\infty }\lambda _{i}\left\langle x,x_{i}\right\rangle y_{i}.}$[3]
• 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 compact operator and ${\textstyle \sum _{i=1}^{\infty }l_{i}<\infty ,}$ where ${\displaystyle l_{1},l_{2},\ldots }$ are the eigenvalues of ${\displaystyle |T|}$ with each eigenvalue repeated as often as its multiplicity.[1]
• The multiplicity of an eigenvalue ${\displaystyle r}$ is the dimension of the kernel of ${\displaystyle T-r\operatorname {Id} _{H},}$ where ${\displaystyle \operatorname {Id} _{H}:H\to H}$ is the identity map.
• 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.
• The above condition but with the word "some" replaced by "every".
• The transpose map ${\displaystyle {}^{t}T:H_{b}^{\prime }\to H_{b}^{\prime }}$ is trace class (according to any defining condition other than this one), in which case ${\displaystyle \left\|{}^{t}T\right\|_{1}=\|T\|_{1}.}$[4]
• The transpose of T is defined by ${\displaystyle \left({}^{t}T\right)\left(y^{\prime }\right):=y^{\prime }\circ T,}$ for all ${\displaystyle y^{\prime }}$ belonging to the continuous dual space ${\displaystyle H^{\prime }}$ of H. The subscript ${\displaystyle b}$ indicates that ${\displaystyle H^{\prime }}$ has its usual norm topology.
• ${\displaystyle \operatorname {Tr} (|T|)<\infty .}$[1]

and if ${\displaystyle T}$ is not already a positive operator then this list can be extended to include:

• The operator ${\displaystyle |T|}$ is trace class (according to any defining condition other than this one).

Trace-norm

If ${\displaystyle T}$ is trace class then we define the trace-norm of a trace class operator T to be the common value

$${\displaystyle \|T\|_{1}:=\sum _{k}\left\langle |T|\,e_{k},e_{k}\right\rangle =\operatorname {Tr} (|T|)}$$

(where it may be shown that the last equality necessarily holds). We denote the space of all trace class linear operators on H by ${\displaystyle B_{1}(H).}$

If T is trace class then[5]

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

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

$${\displaystyle \sum _{k}\left\langle Ae_{k},e_{k}\right\rangle }$$

where this sum is independent of the choice of the orthonormal basis ${\displaystyle \left(e_{k}\right)_{k}}$ of H.

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).[5] The composition of two Hilbert-Schmidt operators is a trace class operator.[1]

Given any ${\displaystyle x,y\in H,}$ define ${\displaystyle {\begin{alignedat}{4}x\otimes y:\,&H&&\to \,&&H\\&z&&\mapsto \,&&\langle z,y\rangle x\\\end{alignedat}}}$ that is, ${\displaystyle (x\otimes y)(z):=\langle z,y\rangle x.}$ 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), ${\displaystyle \operatorname {Tr} (A(x\otimes y))=\langle Ax,y\rangle .}$[5]

1. If ${\displaystyle A:H\to H}$ is a non-negative self-adjoint, 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, ${\displaystyle AT}$ and ${\displaystyle TA}$ are also trace-class, and[1] [6][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. The space of trace-class operators on H is an ideal in the space of bounded linear operators on H.[1]
7. 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 ${\textstyle \sum _{k}\left|\left\langle Te_{k},f_{k}\right\rangle \right|\leq \|T\|_{1}.}$[5]
8. If A is trace-class, then one can define the Fredholm determinant of ${\displaystyle 1+A}$:
   $${\displaystyle \det(I+A):=\prod _{n\geq 1}[1+\lambda _{n}(A)],}$$

   
   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.
9. 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]
10. 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

$${\displaystyle \sum _{n=1}^{N}\lambda _{n}(A)=\operatorname {Tr} (A).}$$

Note that the series in the left 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 a compact operator ${\displaystyle A.}$[7]

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

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 (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, for all ${\displaystyle h\in H,}$:

$${\displaystyle Th=\sum _{i=1}\alpha _{i}\langle h,v_{i}\rangle u_{i},\quad {\text{ where }}\quad \alpha _{i}\geq 0,\ \alpha _{i}\to 0,}$$

for some orthonormal bases ${\displaystyle \left(u_{i}\right)_{i}}$ and ${\displaystyle \left(v_{i}\right)_{i}.}$ Making the above heuristic comments more precise, we have that ${\displaystyle T}$ is trace-class if the series ${\textstyle \sum _{i}\alpha _{i}}$ is convergent, ${\displaystyle T}$ is Hilbert–Schmidt if ${\textstyle \sum _{i}\alpha _{i}^{2}}$ is convergent, and ${\displaystyle T}$ is finite-rank if the sequence ${\displaystyle \left(\alpha _{i}\right)_{i}}$ 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 ${\displaystyle H}$ 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

$${\displaystyle \|T\|_{2}=\left[\operatorname {Tr} \left(T^{*}T\right)\right]^{1/2}=\left(\sum _{i}\alpha _{i}\right)^{1/2}.}$$

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 ${\displaystyle C_{1}.}$ 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

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• Dixmier, J. (1969). Les Algebres d'Operateurs dans l'Espace Hilbertien. Gauthier-Villars.
• Schaefer, Helmut H. (1999). Topological Vector Spaces. GTM. Vol. 3. New York, NY: Springer New York Imprint Springer. ISBN 978-1-4612-7155-0. OCLC 840278135.
• Trèves, François (2006) [1967]. Topological Vector Spaces, Distributions and Kernels. Mineola, N.Y.: Dover Publications. ISBN 978-0-486-45352-1. OCLC 853623322.