Let A be an n by n matrix with the property that $A^TA = AA^T$ Show that $\left\Vert A^Tx \right\Vert = \left\Vert Ax \right\Vert$ for $x$ in $\mathbb{R}^n$.
So this is what I did. I need someone to check my work.
1) I squared the left hand side of the equation
$$\left\Vert A^Tx \right\Vert^2$$
2) I then applied the rule $\left\Vert x \right\Vert^2 = \left<x,x\right>$
$$\left<A^Tx,A^Tx\right>$$
3) Since this is just the dot product, I removed the brackets
$$A^Tx \cdot A^Tx$$
4) I took the transpose 
$$(A^Tx \cdot  A^Tx)^T = Ax \cdot  Ax$$
5) I then placed the brackets since the inner product is just the dot product
$$\left<Ax ,  Ax\right>$$
6) I then used the rule $\left\Vert x \right\Vert^2 = \left<x,x\right>$ again
$$\left\Vert Ax \right\Vert^2$$
7) I took the square root
$$\left\Vert Ax \right\Vert$$
Is this the correct way to prove that $\left\Vert A^Tx \right\Vert = \left\Vert Ax \right\Vert$?

$\Vert A^Tx \Vert^2 = \langle A^T x, A^T x \rangle = \langle x, (A^T)^T A^T x \rangle = \langle x, AA^T x \rangle$
$= \langle x, A^T Ax \rangle = \langle (A^T)^T x, Ax \rangle = \langle Ax, Ax \rangle = \Vert Ax \Vert^2, \tag 1$
whence
$\Vert A^T x \Vert = \Vert Ax \Vert. \tag 2$
