TITLE: Proving that $\|A\|$ is finite.
QUESTION [7 upvotes]: Let $|v|$ be the Euclidean norm on $\mathbb{R^n} $. For  $A\in \mathrm{Mat}_{n\times n}(\mathbb{R})$ we define $\displaystyle \|A\|:= \sup_{\large v\in \mathbb{R^n},\,v \neq 0}\frac{|Av|}{|v|}$. How to show that $\|A\|$ is finite for every $A$?
It would be very helpful if someone could give hints. I think I should show that $\|-\|$ is bounded,but I don't know how...

REPLY [6 votes]: Assume $A\in \Bbb R^{m\times n}$ and write $Ax=: y$, where $x\in\Bbb R^n$, $\ y\in\Bbb R^m$. Then by Schwarz' inequality
$$y_i^2=\left(\sum_{k=1}^n a_{ik}x_k\right)^2\leq \sum_{k=1}^n a_{ik}^2\ \sum_{k=1}^n x_k^2=|a_{i\cdot}|^2\ |x|^2\qquad(1\leq i\leq m)$$
and therefore
$$|y|^2=\sum_{i=0}^m y_i^2\leq C|x|^2$$
with
$$C:=\sum_{i=1}^m |a_{i\cdot}|^2=\sum_{i,k} a_{ik}^2\ .$$
It follows that
$$\|A\|\ \leq\ \left(\sum_{i,k} a_{ik}^2\right)^{1/2}\ .$$

REPLY [2 votes]: Hint: 
Let $S=\{v\in\mathbb{R}^n\;|\;|v| = 1\}, N = \{\frac{|Av|}{|v|}\;|\;v\in\mathbb{R}^n.\;v\ne 0\}, N' = \{|Av|\;|\;v\in\mathbb{R}^n.\;|v|=1\}$  
Step 1: $||A|| = \sup N = \sup N'$
Step 2: Show that $x\to|Ax|$ is a continuous map. $S$ is closed and bounded in $\mathbb{R}^n$ therefore compact,  |Ax| attains max on $S$. 
Done
Do you happen to be in Linear Algebra II class?