Proof $\Vert Ux \Vert_2 = \Vert x\Vert_2$ for unitary $U$? The statement
$\Vert Ux \Vert_2 = \Vert x\Vert_2$ for unitary $U$
is part of a proof that the 2-norm of a matrix $A=UDV^\top$ yields its largest singular value. It's not immediately clear to me why this is the case and the whole statement is simply labelled "matrix fact", so I'm not even sure how to search for it.
Why does above statement hold?

It's because the columns of $U$ form an orthonormal basis.
So, if $x=[x_1,x_2,\dots x_n]^T$ and $U=[u_1,u_2,\dots, u_n]$ (here $u_i$ are columns) then
$$\langle Ux, Ux\rangle = \left\langle\sum_{i=1}^n x_i u_i, \sum_{i=1}^n x_i u_i\right\rangle = \sum_{i,j=1}^n\left\langle x_i u_i, x_ju_j\right\rangle$$
Now use the fact that if $i\neq j$, then $\langle u_i, u_j\rangle = 0$ and you get
$$\langle Ux, Ux\rangle = \sum_{i=1}^n\langle x_iu_i, x_iu_i\rangle=\sum_{i=1}^n|x_i|^2\langle u_i, u_i\rangle$$
And of course, since $\langle u_i, u_i\rangle = 1$, you get
$$\|Ux\|_2^2 = \langle Ux, Ux\rangle = \sum_{i=1}^n |x_i|^2 = \|x\|_2^2$$
