Example of a non-trace class operator

Consider the right unilateral shift $S : \ell^2 \to \ell^2$. We have $Se_n = e_{n+1}$ so $\sum_{n=1}^\infty \langle Se_n, e_n\rangle = 0$.

On the other hand, we have $S^*S = I$ so $|S| = I$. It follows $$\sum_{n=1}^\infty \langle |S|e_n, e_n\rangle = \sum_{n=1}^\infty \langle e_n, e_n\rangle = \infty$$

so $S$ is not trace class.

Your operator is compact. If you represent it on $\ell^2(\mathbb N)$, your $T$ acts on the canonical basis by $$ Te_n=\tfrac{i}{n^2}e_{n-1}. $$ And this is compact, because it is a limit of finite-rank operators. Namely, if $$ T_kx=\sum_{|n|\leq k}\tfrac{i}{n^2}x_ne_{n-1}, $$ Then $\|T_n-T\|\leq1/n^2$.

In fact, your $T$ is also trace class. You have that $T^*$ is given by $T^*e_n=-\tfrac{i}{n^2}e_{n+1}$. Then $$T^*Te_n=\tfrac1{n^4}e_n.$$ So $|T|$ is the operator given by $$|T|e_n=\tfrac1{n^2}e_n,$$ and $$\operatorname{Tr}(|T|)=\sum_n\frac1{n^2}<\infty.$$

Using the above ideas you can play with the sequence you use in defining $T$, as in $$ T_ae_n=a_n\,e_{n-1}, $$ where $a\in\ell^\infty(\mathbb N)$. This $T_a$ will always have $\operatorname{Tr}(T)=0$, and $$|T|e_n=|a_n|\,e_n.$$ So,

• if $\sum_n|a_n|<\infty$, you will have that $T$ is trace-class;

• if $\lim_n|a_n|=0$, you will have that $T$ is compact;

• if $\limsup_n|a_n| >0$, then $T$ is not compact.