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.

Hint: the map $f(x)=\|A(x)\|$ defined on $S_n=\{x:\|x\|= 1\}$ is continuous, since $S_n$ is compact, there exists $x\in B_n$ such that $f(x)=sup_{y\in S_n}f(y)$.
