Today I came up with this proof for the following lemma:
$\| QA \| _F^2 = \|A\|_F^2, A \in E^{m * n}, Q \in E^{m, m}$ with $Q$ orthogonal and $E$ any field.
Proof: $\|QA\|_F^2 = \| (Qa_1 | Qa_2|...|Qa_n)\|_F^2 = \sum_{i = 1}^{n} \| Qa_i\|_2^2 = \sum_{i = 1}^{n} \| a_i \|_2^2 = \| A \|_F^2$.
We used that the 2-norm is invariant to orthogonal matrices. $a_1, a_2, ... , a_n$ denote the columns of $A$.
What do you think? Is this valid? Appreciate any comments.
Yes, your proof is valid. A proof that I prefer is as follows: $$ \|QA\|_F^2 = \operatorname{tr}([QA]^T[QA]) = \operatorname{tr}(A^TQ^TQA) = \operatorname{tr}(A^TA) = \|A\|_F^2 $$