In mathematics, specifically functional analysis, a trace-class operator is a linear operator for which a trace may be defined, such that the trace is a finite number independent of the choice of basis used to compute the trace. This trace of trace-class operators generalizes the trace of matrices studied in linear algebra. All trace-class operators are compact operators.

In quantum mechanics, mixed states are described by density matrices, which are certain trace class operators.

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 use the term "nuclear operator" in more general topological vector spaces (such as Banach spaces).

Note that the trace operator studied in partial differential equations is an unrelated concept.


Suppose   is a Hilbert space and   a bounded linear operator on   which is non-negative (I.e., semi—positive-definite) and self-adjoint. The trace of  , denoted by   is the sum of the series[1]

where   is an orthonormal basis of  . 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   on   we define its absolute value, denoted by   to be the positive square root of   that is,   is the unique bounded positive operator on   such that   The operator   is said to be in the trace class if  We denote the space of all trace class linear operators on H by   (One can show that this is indeed a vector space.)

If   is in the trace class, we define the trace of   by

where   is an arbitrary orthonormal basis of  . 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.

Equivalent formulationsEdit

Given a bounded linear operator  , each of the following statements is equivalent to   being in the trace class:

  •  [1]
  • For some orthonormal basis   of H, the sum of positive terms   is finite.
  • For every orthonormal basis   of H, the sum of positive terms   is finite.
  • T is a compact operator and   where   are the eigenvalues of   (also known as the singular values of T) with each eigenvalue repeated as often as its multiplicity.[1]
  • There exist two orthogonal sequences   and   in   and a sequence   in   such that for all    [2] Here, the infinite sum means that the sequence of partial sums   converges to   in H.
  • T is a nuclear operator.
  • T is equal to the composition of two Hilbert-Schmidt operators.[1]
  •   is a Hilbert-Schmidt operator.[1]
  • T is an integral operator.[3]
  • There exist weakly closed and equicontinuous (and thus weakly compact) subsets   and   of   and   respectively, and some positive Radon measure   on   of total mass   such that for all   and  :


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

One can show that the trace-norm is a norm on the space of all trace class operators   and that  , with the trace-norm, becomes a Banach space.

If T is trace class then[4]



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   (when endowed with the   norm).[4] The composition of two Hilbert-Schmidt operators is a trace class operator.[1]

Given any   define the operator   by   Then   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),  [4]


  1. If   is a non-negative self-adjoint operator, then   is trace-class if and only if   Therefore, a self-adjoint operator   is trace-class if and only if its positive part   and negative part   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,
    The bilinear map
    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.   is a positive linear functional such that if   is a trace class operator satisfying   then  [1]
  4. If   is trace-class then so is   and  [1]
  5. If   is bounded, and   is trace-class, then   and   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]
    Furthermore, under the same hypothesis,[1]
    and   The last assertion also holds under the weaker hypothesis that A and T are Hilbert–Schmidt.
  6. If   and   are two orthonormal bases of H and if T is trace class then  [4]
  7. If A is trace-class, then one can define the Fredholm determinant of  :
    where   is the spectrum of   The trace class condition on   guarantees that the infinite product is finite: indeed,
    It also implies that   if and only if   is invertible.
  8. If   is trace class then for any orthonormal basis   of   the sum of positive terms   is finite.[1]
  9. If   for some Hilbert-Schmidt operators   and   then for any normal vector     holds.[1]

Lidskii's theoremEdit

Let   be a trace-class operator in a separable Hilbert space   and let     be the eigenvalues of   Let us assume that   are enumerated with algebraic multiplicities taken into account (that is, if the algebraic multiplicity of   is   then   is repeated   times in the list  ). Lidskii's theorem (named after Victor Borisovich Lidskii) states that


Note that the series on the right converges absolutely due to Weyl's inequality

between the eigenvalues   and the singular values   of the compact operator  [6]

Relationship between some classes of operatorsEdit

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  

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   sequence with respect to some choice of a pair of Hilbert bases. In the same vein, the bounded operators are noncommutative versions of   the compact operators that of   (the sequences convergent to 0), Hilbert–Schmidt operators correspond to   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   on a Hilbert space takes the following canonical form: there exist orthonormal bases   and   and a sequence   of non-negative numbers with   such that

Making the above heuristic comments more precise, we have that   is trace-class iff the series   is convergent,   is Hilbert–Schmidt iff   is convergent, and   is finite-rank iff the sequence   has only finitely many nonzero terms. This allows to relate these classes of operators. The following inclusions hold and are all proper when   is infinite-dimensional:

The trace-class operators are given the trace norm   The norm corresponding to the Hilbert–Schmidt inner product is

Also, the usual operator norm is   By classical inequalities regarding sequences,
for appropriate  

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 operatorsEdit

The dual space of   is   Similarly, we have that the dual of compact operators, denoted by   is the trace-class operators, denoted by   The argument, which we now sketch, is reminiscent of that for the corresponding sequence spaces. Let   we identify   with the operator   defined by

where   is the rank-one operator given by

This identification works because the finite-rank operators are norm-dense in   In the event that   is a positive operator, for any orthonormal basis   one has

where   is the identity operator:

But this means that   is trace-class. An appeal to polar decomposition extend this to the general case, where   need not be positive.

A limiting argument using finite-rank operators shows that   Thus   is isometrically isomorphic to  

As the predual of bounded operatorsEdit

Recall that the dual of   is   In the present context, the dual of trace-class operators   is the bounded operators   More precisely, the set   is a two-sided ideal in   So given any operator   we may define a continuous linear functional   on   by   This correspondence between bounded linear operators and elements   of the dual space of   is an isometric isomorphism. It follows that   is the dual space of   This can be used to define the weak-* topology on  

See alsoEdit


  1. ^ a b c d e f g h i j k l m n Conway 1990, p. 267.
  2. ^ Trèves 2006, p. 494.
  3. ^ Trèves 2006, pp. 502–508.
  4. ^ a b c d Conway 1990, p. 268.
  5. ^ M. Reed and B. Simon, Functional Analysis, Exercises 27, 28, page 218.
  6. ^ Simon, B. (2005) Trace ideals and their applications, Second Edition, American Mathematical Society.


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  • Dixmier, J. (1969). Les Algebres d'Operateurs dans l'Espace Hilbertien. Gauthier-Villars.
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