Basic question about $\sup_{x\neq 0}{} \frac{\|Ax\|}{\|x\|} = \sup_{\|x\| = 1}{\|Ax\|} $, $x \in\mathbb{R}^n$ I am having trouble with understanding the following definition while studying some basic things related with matrix norms:
For every matrix $A\in M_n(\mathbb{R})$
$$\sup_{x\neq 0}{} \frac{\|Ax\|}{\|x\|} = \sup_{\|x\| = 1}{\|Ax\|},\; x \in\mathbb{R}^n$$
Why we are taking $\|x\| = 1$?  Is there any proof of above statement?
Added: Why to take sup in the definition of norm?
I need help to understand above mentioned fact. 
Thank you very much. 
 A: As julien pointed out, what follows is not a proof that 
$$\sup_{x\neq0}\frac{||Ax||}{||x||}=\sup_{||y||=1}||Ay||,\quad\quad(1)$$
it is a discussion about the consequences of $(1)$. For a proof of $(1)$ see MichaelNgelo's answer.

$(1)$ implies that if we know that a given vector achieves either of the two supremums in $(1)$, then we can deduce from it a vector that achieves the other supremum. 
Specifically, $x^*$ is such that 
$$\frac{||Ax^*||}{||x^*||}=\sup_{x\neq0}\frac{||Ax||}{||x||}\quad\quad(*)$$
if and only if $y^*:=\frac{x^*}{||x^*||}$ is such that
$$||Ay^*||=\sup_{||y||=1}||Ay||.\quad\quad(**)$$
The above is useful because it allows to compute one of the supremums by computing the other instead, in particular, we can substitute a supremum over $\mathbb{R}\backslash \{0\}$ with one over the unit sphere.

An easy way to argue the above is by contradiction. For example, one direction follows from:
Suppose that $x^*$ satisfies $(*)$, but $y^*:=\frac{x^*}{||x^*||}$ does not satisfy $(**)$. Then, there exists some $\hat{y}\neq y^*$ such that  $||\hat{y}||=1$ and
$$||A\hat{y}||>||Ay^*||\Rightarrow \frac{||A\hat{y}||}{1}=\frac{||A\hat{y}||}{||\hat{y}||}>||Ay^*||=\left|\left|A\frac{x^*}{||x^*||}\right|\right|=\frac{||Ax^*||}{||x^*||}=\sup_{x\neq0}\frac{||Ax||}{||x||}$$ 
which gives a contradiction. 

EDIT: The $\sup$ is taken in order to make the function $\sigma$, that maps from the vector space of real $n\times n$ matrices, $M_n(\mathbb{R})$, to $[0,\infty)$, and is defined as
$$\sigma(A):=\sup_{x\neq 0}\frac{||Ax||}{||x||},$$
be a norm (that is, satisfy the norm axioms) on $M_n(\mathbb{R})$. If we drop the sup and instead use a fixed $x$, then $\sigma$ doesn't satisfy the first axiom ($\sigma(A)=0\Leftrightarrow A=0)$ any more.
A: $$\frac{ \|Ax\| }{\|x\|} = \left|\left| A \frac{x}{\|x\|} \right|\right|$$
Running through all $x \neq 0$ is equivalent to running through all $y:= \frac{x}{\|x\|} $ with $\| y \|=1$.
