A direct result from the definition of operator norm: $\|Av\|\leq \|A\|_{op} \|v\|$

Question

Although Wikipedia says this result comes from the definition of Operator Norm directly, I am not quite sure how to understand it:

Let $\|\cdot\|$ denote Euclidean norm. Given a $n\times n$ matrix $A$, $\forall v\in R^n$, we have $\|Av\|\leq \|A\|_{op} \|v\|$.

The definition of operator norm is $\|A\|_{op}=\sup_{\|x\|\leq 1, \|x\|\in R^n} \|Ax\|$. Probably, we need to apply Cauchy-Schwartz inequality here, but then we need $\|A\|\leq \|A\|_{op}$, which may not be true. Any explanation will be appreciated.

Answer 1

Suppose that $\| v \| \neq 0$, because otherwise there is nothing to prove.

By definition, we then have that $$\frac{\| Av \|}{\| v \|} = \left\|A \frac{v}{\| v \|} \right\| \le \| A \|_{op}. $$

The inequality follows from the fact that $$ \left\| \frac{v}{\|v\|} \right\| =1. $$