For any $n\times n$ matrix $A$, there corresponds a vector $x\neq 0$ such that $\|Ax\|=\|A\|\|x\|$ Prove that for any vector norm and its subordinate matrix norm, and for any $n\times n$ matrix $A$, there corresponds a vector $x\neq 0$ such that $\|Ax\|=\|A\|\|x\|$
I know that $\|Ax\|\leq \|A\|\|x\|$ for all $x\in \mathbb{R}^n$, but I do not know how to find $x\neq 0$ in such a way that the other equality is fulfilled, could someone help me please? Thank you very much.

$$\|A\| = \sup_{\|x\| =1}\|Ay\|=\sup_{\|x\| \le1}\|Ay\| $$
Since $\{ x: x \le 1\}$ is compact, and $\|Ax\|$ is continuous, then the maximum is well defined. 
Hence there exists $\|x\|=1$ such that $\|A\|= \|Ax\|$.
$$\|A\|\|x\|=\|Ax\|$$
