# 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?

• $\langle x,x\rangle = \langle U^TUx,x\rangle = \langle Ux,Ux\rangle$ – Surb May 24 '17 at 12:28

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

$$\|Ux\|_2^2 = \langle Ux, Ux\rangle = \sum_{i=1}^n |x_i|^2 = \|x\|_2^2$$

• Good proof. Personal opinion, seems like a little bit of overkill on the details. – ADA May 25 '17 at 1:14
• -1: this makes an arbitrary assumption about what properties of a unitary to use. For that matter, one could have started with $\|Ux\|=\|x\|$. – Martin Argerami May 25 '17 at 2:40

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

• i don't see how this is an answer to OP's question. – 5xum May 24 '17 at 12:39
• @5xum: Your answer assumes that a unitary matrix has orthonormal columns. One can take that as a definition of unitary. Or one can define (and this is more general, as it extends to infinite dimension) a unitary as a (surjective) isometry, that is a matrix that satisfies $\|Ux\|=\|x\|$ for all $x$. The point is that all possible definitons are equivalent, which is precisely what my answer states. – Martin Argerami May 25 '17 at 2:28

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.

Assume that $U$ is a complex $n \times n$ matrix such that $U^* U = I$. If $x \in \mathbb C^n$ then $$\|Ux \|_2^2 = (Ux)^* Ux = x^* U^* U x = x^* x = \|x\|_2^2.$$

• Hmmmm the OP said unitary... not orthogonal. – ADA May 25 '17 at 1:00
• @AlecDiaz-Arias Good point, I adjusted my answer. – littleO May 25 '17 at 1:11