Trace operator is basis independent Let $H$ be a Hilbert space and call suppose $A:H\rightarrow H$ is positive. How do you show Tr$(A)=\sum_{n}(Ae_n,e_n)$, does not depends on the orthonormal basis $e_n$. 
I was thinking about using an approximation of $A$ as $A_k$ where $A_k$ is finite rank operator, so $Tr(A_k)$ is the sum of diagonal entries of an $\infty\times n$ matrix, which is basis independent. Then use $Tr(A_k)\rightarrow Tr(A)$ to conclude the proof. Is this the right approach?
 A: Note for positive sums:
$$a_{\alpha\beta}\geq0:\quad\sum_\alpha\sum_\beta a_{\alpha\beta}=\sum_\beta\sum_\alpha a_{\alpha\beta}$$
By Parseval one has:
$$\sum_\sigma\langle|A|\sigma,\sigma\rangle=\sum_\sigma\||A|^{1/2}\sigma\|^2=\sum_\sigma\sum_\tau|\langle|A|^{1/2}\sigma,\tau\rangle|^2\\
=\sum_\tau\sum_\sigma|\langle\sigma,|A|^{1/2}\tau\rangle|^2=\sum_\tau\||A|^{1/2}\tau\|^2=\sum_\tau\langle|A|\tau,\tau\rangle$$
Concluding independence.
A: Suppose that $\{e_n\}$ and $\{f_k\}$ are orthonormal bases of $H$. Then you can write
$$ \tag 1 x = \sum_k (x,f_k) f_k$$
and 
$$ \tag 2 y = \sum_n (y,e_n) e_n$$
for all $x,y \in H$. Evaluate (1) with $x = Ae_n$ to find
$$ Ae_n = \sum_k (Ae_n,f_k) f_k $$
so that
$$ \sum_n (Ae_n,e_n) = \sum_n \sum_k (Ae_n,f_k)(f_k,e_n).$$
Next evaluate (2) with $y = f_k$ and then apply $A$ to find
$$Af_k = \sum_n (f_k,e_n)Ae_n$$
so that
$$ \sum_k(Af_k,f_k) = \sum_k \sum_n (f_k,e_n)(Ae_n,f_k)$$
Thus $\displaystyle \sum_n (Ae_n,e_n) = \sum_k(Af_k,f_k)$ provided that you can switch the order of summation which will require some condition on $A$.
