# Hilbert-Schmidt operator gives the same infinite sum acting on any Hilbert basis

If we let $E$ and $F$ be Hilbert spaces and $A: E \rightarrow F$ a bounded linear operator between them, then $A$ is a Hilbert-Schmidt operator if there exists a Hilbert basis for $E$ (i.e. a complete, orthonormal sequence in $E$), $(e_n)_{n=1}^\infty$, such that $$\displaystyle\sum_{n=1}^\infty ||A(e_n)||_F^2<\infty.$$ What I'd like to show is that if $(f_n)_{n=1}^\infty$ is any other complete orthonormal sequence in $E$, then $$\displaystyle\sum_{n=1}^\infty ||A(e_n)||_F^2 = \displaystyle\sum_{n=1}^\infty ||A(f_n)||_F^2.$$ It seems like the easiest way of doing this is using the fact that we can represent each $e_n$ and $f_n$ in terms of each other, since for any $x\in E$, $x = \sum_{n=1}^\infty \langle x, e_n \rangle e_n = \sum_{n=1}^\infty \langle x, f_n \rangle f_n$, and we may allow that $x$ to be any of the $e_n$ or $f_n$.

But I'm getting a bit lost in the details. Is this a sensible way of proving the statement, and if so, am I just a tweak away from finishing the proof?

$$\sum_{i\in I}|\langle x,e_i\rangle|^2=\Vert x\Vert^2$$
for any $x\in\mathcal H$ and $(e_i)_{i\in I}$ any orthonormal basis of $\mathcal H$. Using this for orthonormal bases $(e_i)_{i\in I}$, $(f_i)_{i\in I}$, $(g_i)_{i\in I}$ of $\mathcal H$ one gets
$$\sum_{i\in I}\Vert Ae_i\Vert^2=\sum_{i\in I}\sum_{j\in I}|\langle Ae_i,g_j\rangle|^2=\sum_{j\in I}\sum_{i\in I}|\langle A^\dagger g_j,e_i\rangle|^2=\sum_{j\in I}\Vert A^\dagger g_j\Vert^2\\ \hphantom{\sum_{i\in I}\Vert Ae_i\Vert^2}=\sum_{j\in I}\sum_{i\in I}|\langle A^\dagger g_j,f_i\rangle|^2=\sum_{i\in I}\sum_{j\in I}|\langle Af_i,g_j\rangle|^2=\sum_{i\in I}\Vert Af_i\Vert^2.$$
Note that interchanging the sums over $i,j$ is fine as all the summands are non-negative.