$\sup_{x \in \mathbb{R}, x \neq 0} \frac{\|Ax\|}{\|x\|}$ equivalent to $\sup_{\|x\| = 1} \|Ax\|$

Question

I have seen it mentioned in many places that for some matrix $A \in \mathbb{R}^{n \times n}$

$\displaystyle\sup_{x \in \mathbb{R}^n, x \neq 0} \frac{\|Ax\|}{\|x\|}$ is equivalent to $\displaystyle\sup_{x \in \mathbb{R}^n, \|x\| = 1} \|Ax\|$.

I can see why this makes sense intuitively, but it is possible to prove it formally?

Answer 1

Take

$$0\neq x\in\Bbb R^n\;,\;\;u:=\frac x{||x||}$$

then

$$||Au||=\left\|A\left(\frac x{||x||}\right)\right\|\stackrel{\text{linearity}}=\left\|\frac1{||x||}A x\right\|=\frac{||Ax||}{||x||}$$