On a Hilbert space $\mathcal{H}$, it is well known that trace class operators have a finite trace. However, there are operators which are not trace class but have a finite trace, e.g. the unilateral shift. Then, consider a (bounded) operator $A$ such that $$ c=\sum_{n=1}^{\infty}\langle x_n,A x_n\rangle<\infty, $$ for some orthonormal basis $\{x_1,x_2,\ldots\}$. I'm not sure whether this is enough for the trace of $A$ to be defined in a basis independent way. Namely, $\mathrm{tr}(A)=\sum_{n=1}^{\infty}\langle y_n,A y_n\rangle=c$ for any orthonormal basis $\{y_1,y_2,\ldots\}$.
In the classics von Neumann's book on Mathematical Foundations of Quantum Mecanics, Sec. II.11, he commented this is true (in fact, he did not consider trace class operators) using essentially the following argument: $$ c=\sum_{n=1}^{\infty}\langle x_n,A x_n\rangle=\sum_{n=1}^{\infty}\sum_{m=1}^{\infty}\langle x_n,y_m\rangle \langle y_m, A x_n\rangle=\sum_{m=1}^{\infty}\sum_{n=1}^{\infty}\langle y_m, A x_n\rangle\langle x_n,y_m\rangle=\sum_{m=1}^{\infty}\langle y_m,A y_m\rangle. $$ However, it is not clear to me that the double sum can be switched. Using that $|\langle u,y_m \rangle \langle y_m, v\rangle|\leq \frac12 (|\langle u,y_m \rangle|^2+|\langle v,y_m \rangle|^2)$ it is obtained that $$ \sum_{m=1}^{\infty}\langle x_n,y_n\rangle \langle y_n, A x_n\rangle $$ is absolutely convergent. Is this enough for the summation switch?
Thank you in advance.
EDIT: As Robert Israel points, $\sum_{n=1}^{\infty}\langle x_n,A x_n\rangle$ must be absolutely convergent. Is it enough?
If the sum $\sum_{n=1}^\infty \langle x_n, A x_n \rangle$ is only conditionally convergent, you can take the same basis $x_n$ but rearranged, and get a sum that diverges, or converges to an arbitrary value: see the Riemann rearrangement theorem.