Proving that $\|A\|$ is finite. 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...
 A: 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
$$\|A\|\ \leq\ \left(\sum_{i,k} a_{ik}^2\right)^{1/2}\ .$$
A: Ok seems like I should make it a bit more explicit:
At first we can scale the problem using the homogeneity of norms:
$$ \sup_{v\in \mathbb{R}^n, v\neq 0} \frac{|Av|}{|v|}=\sup_{v\in \mathbb{R}^n, v\neq 0} 
|A\frac{v}{|v|}| = \sup_{|v|=1, v\in \mathbb{R}^n} |Av|$$ 
Now we write the Matrix $A$ like this one 
$$\begin{pmatrix} 
a & b \\
c & d\\
\end{pmatrix} = \begin{pmatrix} a& b \\0 & 0 \\ \end{pmatrix} + \begin{pmatrix} 0 & 0 \\ c & d \\ \end{pmatrix}$$
We decompose the Matrix $A$ in a sum of matrices $A_i$ where only 1 column has not only $0$ entries. So we get
$$\sup_{|v|=1,v\in \mathbb{R}^n } |Av|=\sup_{|v|=1,v\in \mathbb{R}^n} |\sum_{i=1}^n A_i v|$$
Using the triangle inequality we get 
$$\sup_{|v|=1,v\in \mathbb{R}^n} |Av|\leq \sup_{|v|=1,v\in \mathbb{R}^n} \sum_{i=1}^n |A_i v|$$
We can identify $A_i\in \mathbb{R}^{n\times n}$ with a vector $b_i\in \mathbb{R}^n$, $b_i$ does have the non zero entries of $A_i$. Note that $A_i v$ is a vector with only 1 non $0$ entry. So $$A_i \cdot v=\langle b_i,v \rangle \cdot e_i$$ where $e_i=(0,\dots,1,0,\dots,0)$
Let $\langle \cdot , \cdot \rangle$ be the euclidean scalar product we get:
$$\sup_{|v|=1,v\in \mathbb{R}^n} \sum_{i=1}^n |A_i v| =\sup_{|v|=1,v\in \mathbb{R}^n} \sum_{i=1}^n \operatorname{abs}(\langle b_i, v\rangle) \cdot |e_i|  $$
Using Cauchy Schwarz we get
$$\sup_{|v|=1,v\in \mathbb{R}^n}\sum_{i=1}^n \operatorname{abs}(\langle b_i, v\rangle) \cdot |e_i| \leq \sup_{|v|=1, v\in \mathbb{R}^n} \sum_{i=1}^n |b_i|\cdot |e_i|$$
Since this is (finally) independent of $v$
$$ \sup_{|v|=1, v\in \mathbb{R}^n} \sum_{i=1}^n |b_i|\cdot |e_i|=\sum_{i=1}^n |b_i|\cdot |e_i|$$
A: Let $||\cdot ||$ be any norm on a finite dimensional vector space $X$. Then define the norm $N$ on the space of endomorphisms $\mathcal{L}(X)$ by $N(\varphi)= \sup\limits_{x \in X \backslash \{0\}} ||\varphi(x)||/||x||$. It is straightforward that $N(\varphi)= \sup\limits_{x \in X, ||x||=1} ||\varphi(x)||$. You can deduce that $N(\varphi)<+ \infty$ from the continuity of $\varphi$ (any endomorphism of a finite dimensional normed spaces is continuous) and the compacity of the sphere $\{x \in X : ||x||=1\}$ (a finite dimensional normed space is locally compact).
A: 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?
