Task is to show how $$\sup_{x \neq 0 } \frac{||Ax||}{||x||} = \sup_{||x|| =1} ||Ax||$$
Here is what I have so far:
$$ \sup_{x \neq 0 } \frac{||Ax||}{||x||} = \sup_{x\neq0} \frac{1}{||x||} Ax $$ This is because the operator norm is homogeneous (which is given by definition).
Then we can do: $$ \sup_{x\neq0} \frac{1}{||x||} Ax = \sup_{x\neq0} \left|\left| \frac{x}{||x||} A\right|\right|$$
My textbook argues that (this is what I dont understand) that any $\hat{x} = \frac{x}{||x||} \in [0,1]$ which gives that
$$ \sup_{x\neq0} \left|\left| \frac{x}{||x||} A\right|\right| = \sup_{x=1} ||Ax|| $$
which I also dont understand.
Can anyone shed some light into this identity, ideally with an intuitive answer ?
Hint: ${{\|A(x)\|}\over{\|x\|}}=\|A({x\over{\|x\|}})\|$ since for every $c\geq 0$, $c\|x\|=\|cx\|$ and $\|{x\over{\|x\|}}\|$ is $1$. This implies that $Sup_{x\neq 0}{{\|A(x)\|}\over{\|x\|}}\leq sup_{\|x\|=1}\|A(x)\|$.