Norm equivalence of a vector norm and its induced matrix norm using compactness argument I have a theorem in my book on matrix computations that states the following: 
A vector norm and its induced matrix norm satisfy the inequality:
$\|Ax\|\leq \|A\|$$\|x\|$ where A $\in R^{nxn}$ and x $\in R^n$. 
The book states that equality holds, "if and only if x is a vector for which the maximum magnification is attained. (That such a vector exists is actually not obvious. It follows from a compactness argument that works because $R^n$ is a finite-dimensional space. We omit the argument.)" 
I am interested in knowing how I would go about showing that equality holds if x is a vector for which max magnification is attained. This is not a homework problem; however, I would like to know for my upcoming exam.
 A: You have to be a little careful here. A norm that satisfies $||Ax|| \leq ||A||||x||$ is called a submultiplicative norm. Not all induced norms are submultiplicative. Take for example the max norm $||A||_{max}=max\{|a_{ij}|\}$. This is not submultiplicative. A good book for all of this is Matrix Analysis by Horn and Johnson. See here http://www.amazon.com/Matrix-Analysis-Roger-A-Horn/dp/0521386322
A: In $\mathbb{R}^n$, the closed unit ball $\overline{B}$ is compact. The function $f(x) = \|Ax\|$ is continuous, and continuous functions attain their maximum on a compact set, ie, there exists $x_0 \in \overline{B}$ such that $f(x_0) = \max_{x \in \overline{B}} f(x)$. In particular, $\|A x_0 \| = \max_{x \in \overline{B}} \|Ax\|$.
The induced norm is defined as $\|A\| = \max_{x \in \overline{B}} \|A x\|$, and the above remark shows that this is attained for some $x_0$.
A: Without loss of generality we can assume $x$ to be in the boundary of  unit sphere $B$, as this inequality is invariant for scaling as the norms are homogene.
The right hand side is 
\begin{align*}
\|A\|\cdot \|x\|&= \sup_{v\in \partial B} \frac{\| Av \|}{\|v\|} \cdot \|x\|\\
&=\sup_{v\in \partial B} \|A \cdot v\|
\end{align*}
It is trivial that 
$$\|A\cdot x \|\leq \sup_{v \in \partial B} \|A\cdot v\|$$
As the supremum is always greater than $\|A\cdot x\|$ if $x$ is not the maximum, the equality hold iff the maximum is attached at $x$
