Understanding a part of a proof involving Hilbert-Schmidt norm

Question

I came across a proof I do not seem to understand fully, a screenshot is provided below.

my concerns are the following:

1. Why does the fact that $||T||_2 = ||UT||_2$ for every unitary U, allow us to use a second basis $f_j$ in the definition: $||T||_2 ^2 = \sum{|\langle e_i , Tf_i \rangle|^2}$ of the Hilbert-Schmidt norm?
2. How exactly does one get from line 1 to line 2: $$ \sum \big| \langle e_i, (A^2+B^2+2AB)f_j \rangle \big|^2 = \sum (\alpha_i ^2 + \beta_j ^2 - 2\alpha_i\beta_j)^2 \big |\langle e_i,f_i \rangle \big|^2$$ in particular, I seem to have problems with the indices $i$ and $j$.
3. Why is the basis ${f_j}$ necessary? couldn't one do the proof only using ${e_j}$. Since $f_j$ does not occur in the choosing of: $Ae_j = \alpha_j e_j$ and $Be_j = \beta_j e_j$

Answer 1

1. For any two ONBs there is a unitary mapping one to the other. So, if one knows $$\|T\|_2^2 = \sum|\langle e_i, Te_i\rangle |^2$$ and wishes to have an arbitrary pair of UNBs here, one can $$\|T\|_2^2 = \|UT\|_2^2= \sum|\langle e_i, UTe_i\rangle |^2= \sum|\langle U^*e_i, Te_i\rangle |^2$$ Choose $U^*$ so that $U^*e_j=f_j$, and you have the claimed identity, only with letters $e$ and $f$ reversed.

3. Typo. Read as $Bf_j=\beta_j f_j$

2. Apparently, $A$ and $B$ are assumed self-adjoint. Then $$\begin{split}\langle e_i, (A^2+B^2+AB)f_j\rangle &= \langle e_i, A^2f_j\rangle + \langle e_i, B^2 f_j\rangle + \langle e_i,ABf_j\rangle \\ &= \langle A^2e_i, f_j\rangle + \langle e_i, B^2 f_j\rangle + \langle Ae_i,Bf_j\rangle \\ &= (\alpha_i^2+\beta_i^2 +\alpha_i\beta_i)\langle e_i, f_j\rangle \end{split}$$