Editing Trace class (section)

== Examples ==
=== Spectral theorem===
Let <math>T</math> be a bounded self-adjoint operator on a Hilbert space. Then <math>T^2</math> is trace class ''if and only if'' <math>T</math> has a [[Spectrum_(functional_analysis)#Point_spectrum|pure point spectrum]] with eigenvalues <math>\left\{\lambda_i(T)\right\}_{i=1}^{\infty}</math> such that{{sfn|Simon|2010|page=21}} 
:<math>\operatorname{Tr}(T^2) = \sum_{i=1}^{\infty}\lambda_i(T^2) < \infty.</math>
=== Mercer's theorem===
[[Mercer's theorem]] provides another example of a trace class operator. That is, suppose <math>K</math> is a continuous symmetric [[positive-definite kernel]] on <math>L^2([a,b])</math>, defined as
:<math> K(s,t) = \sum_{j=1}^\infty \lambda_j \, e_j(s) \, e_j(t) </math>
then the associated [[Hilbert–Schmidt integral operator]] <math>T_K</math> is trace class, i.e.,
:<math>\operatorname{Tr}(T_K) = \int_a^b K(t,t)\,dt = \sum_i \lambda_i.</math>

=== 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 <math>B_1(H)</math> (when endowed with the trace norm).{{sfn|Conway|1990|p=268}}

Given any <math>x, y \in H,</math> define the operator <math>
x \otimes y : H \to H</math> by <math>(x \otimes y)(z) := \langle z, y \rangle x.</math> 
Then <math>x \otimes y</math> 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''), <math>\operatorname{Tr}(A(x \otimes y)) = \langle A x, y \rangle.</math>{{sfn|Conway|1990|p=268}}