Im trying to prove that $\Vert A-A_k\Vert _2 =\sigma_{k+1}$ for SVD with the usual notation that sigma represents singular values and if $A=\sum_{i=1}^r \sigma_iu_iv_i^T$, then $A_k=\sum_{i=1}^k \sigma_iu_iv_i^T$ for $A=U\sum V^T$, $k<r$.
So in the process, I got $A-A_k=\sum_{i=k+1}^r\sigma_iu_iv_i^T = U\sum'V^T$ where the sigma prime considers only the $k+1$ and above singular values upto $r$ in their appropriate diagonal positions. Now I have already proved that for, in general, $Q$ orthogonal, $\Vert QA\Vert _2=\Vert A\Vert_2$, but I'm having trouble proving $\Vert AQ\Vert_2=\Vert A\Vert_2$ (*).
$$\Vert AQ\Vert_2=\text{sup}\Vert AQx\Vert_2$$
Over all unit vectors $x$. From here I cannot proceed further since trying to tranpose $AQ$ and multiplying it with $AQ$ isnt very fruitful, $Q^TA^TAQ$. So how do I prove (*)?
Since for any $x\in\mathbb{R}^n$, there exists $u$ such that $x = Qu$, it follows that $$\|A\|=\sup_{\|x\| = 1}\|Ax\| = \sup_{\|Qu\|=1}\|AQu\| = \sup_{\|u\|=1}\|AQu\| = \|AQ\|.$$