Operator trace ($\mathrm{tr}T$)

For a trace-class operator, the series $\mathrm{tr}T={\sum }_n⟨T{e}_n,e_n⟩$ converges absolutely and is independent of the orthonormal basis. It satisfies $|\mathrm{tr}T|\le \Vert T{\Vert }_1$.

Every bounded operator $S$ defines a functional $T\mapsto \mathrm{tr}(ST)$ on the trace-class operators, andFinite-rank density in ${\mathcal{S}}_1$ makes every continuous functional arise uniquely this way.