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?
 A: 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
$$\|Ux\|_2^2 = \langle Ux, Ux\rangle = \sum_{i=1}^n |x_i|^2 = \|x\|_2^2$$
A: Unitaries are precisely the matrices that satisfy  $\|Ux\|_2=\|x\|_2$ for all $x $. It is not hard to show that the following statements are equivalent:


*

*$\|Ux\|_2=\|x\|_2$ for all $x $

*$\langle Ux,Uy\rangle =\langle x,y\rangle $ for all $x,y $

*$U^*U=I_n $

*$UU^*=I_n $

*The rows of $U $ are orthonormal

*The columns of $U $ are orthonormal
A: To prove that $||Ux||_2=||x||_2$ consider the square of the left side then you get that $||Ux||_2^2=\langle Ux, Ux \rangle = \langle UU^*x,x \rangle = \langle x,x \rangle = ||x||_2^2$ taking square roots you have the result. The second equality follows from the fact that U has an adjoint and it is a left/right inverse.
A: Assume that $U$ is a complex $n \times n$ matrix such that $U^* U = I$.  If $x \in \mathbb C^n$ then
$$
\|Ux \|_2^2 = (Ux)^* Ux = x^* U^* U x = x^* x = \|x\|_2^2.
$$
