Trace class operator

Remarks

If $T$ is trace class, then for every orthonormal basis $(e_i)$ the sum $\operatorname{tr} T = \sum_i \ip{T e_i}{e_i}$ converges absolutely, and its value does not depend on the basis. On $\C^n$, this is the usual trace of a matrix. Equivalently, $T$ is trace class if and only if $T$ is compact and its singular values satisfy $\sum_n s_n(T) < \infty$, so the trace class is the Schatten class $S_1$. Example: on $\ell^2$, let $T e_n = e_n / n^2$; then $\abs{T} = T$ and $\norm{T}_1 = \sum 1/n^2 < \infty$, so $T$ is trace class with $\operatorname{tr} T = \pi^2 / 6$. Non-example: $D e_n = e_n / n$ gives $\sum_n \ip{\abs{D} e_n}{e_n} = \sum 1/n = \infty$, so $D$ is compact but not trace class. The trace class operators form a two-sided ideal of the bounded operators, and $\operatorname{tr}(AT) = \operatorname{tr}(TA)$ for $T$ trace class and $A$ bounded.