Some related problems:
My problem comes from one step of a certain proof:
$\|Av\|^2=(Av)^T(Av)=v^TA^TAv=v^TIv = v^Tv =\|v\|^2 \Rightarrow \|Av\| = \|v\|$
Note: $A$ is an orthogonal matrix.
More general one:
$\|Av\|^2=\langle Av,Av \rangle=\langle v,A^TAv \rangle=\langle v,Iv \rangle= \langle v,v\rangle=\|v\|^2 \Rightarrow \|Av\| = \|v\|$
They look similar.
It seems the above equality only holds on $l_2$ norm(the first case) and inner product induced norm (the second case).
So such equality cannot hold for general norm, right?
I just want to make my proof more strict.
Assuming $A$ is an orthogonal linear map (i.e. $A^T A = I$), you are correct that $\|Av\| = \|v\|$ does not hold in general for norms other than the $2$-norm. For example, take $$A = \frac{1}{\sqrt{2}}\begin{pmatrix}1 & 1 \\ -1 & 1\end{pmatrix}$$ which you can check satisfies $A^T A = I$, and $$v = \begin{pmatrix}1 \\ 0 \end{pmatrix}$$ and compute the $1$-norm of $Av = \begin{pmatrix}1\sqrt{2} \\ -1/\sqrt{2}\end{pmatrix}$ and $v$: we see that $\|Av\|_1 = 2/\sqrt{2}$ whereas $\|v\|_1 = 1$.