# Trace Class Operators and Compactness

According to many sources I've looked through, part of the definition of trace-class operators is that they be compact. What's the need for this caveat? Why not just look at all those operators $$T$$ for which $$\sum_{i \in I} \langle T e_i,e_i \rangle$$ converges absolutely for some orthonormal basis $$\{e_i\}_{i \in I}$$? This is the definition of trace-operator with the compactness assumption dropped, right?

As an aside, is there an example of an operator $$T$$ that is not compact yet $$\sum_{i \in I} \langle T e_i,e_i \rangle$$ converges absolutely for some orthonormal basis $$\{e_i\}_{i \in I}$$

• @qbert Oh, so being trace-class implies that the operator is compact? Nov 5, 2019 at 0:35
• @qbert So, do we lose anything if we drop the compactness assumption? That is, is the set of operators $T$ such that $\sum_{i \in I} \langle Te_i,e_i \rangle$ absolutely converges for some ONB $\{e_i\}_{i \in I}$ still of any interest, or does the set of trace-class operators so defined lose nice properties? Nov 5, 2019 at 1:06

Let $$\pi$$ be any permutation of positive integers such that $$\pi (i) \neq i$$ for any $$i$$. Then $$T(e_i)=e_{\pi (i)}$$ gives you an isometric isomorphism with $$\langle Te_i , e_i \rangle =0$$ for all $$i$$. This $$T$$ is surely not compact.
Your definition is not correct. It would be correct (but it is not trivial) if you require that $$\sum_j|\langle Te_j,e_j\rangle|<\infty$$ for all orthonormal bases $$\{e_j\}$$. Here is an example of a compact, not trace-class operator such that all terms in your sum are zero for a certain orthonormal basis.
The usual definition is that $$T$$ is trace class if $$\sum_j\langle |T|e_j,e_j\rangle < \infty$$ for some (and then all!) orthonormal basis $$\{e_j\}$$. Here $$|T|=(T^*T)^{1/2}$$. In particular $$T$$ is trace-class if and only if $$|T|$$ is. As $$|T|$$ is positive, it is easy to see that if it is trace-class then it is compact. Then the polar decomposition and the fact that the trace-class operators form an ideal gives you that $$T$$ is compact if it is trace-class.
• Ah, so compactness $T$ is a consequence of of $T$ being trace-class! Very nice! Nov 5, 2019 at 2:00