Show that for any $x \in \mathbb{R}^{n} $ there holds $ \|A x\|_{p} \leq\|A\|_{p}\|x\|_{p} $.
I am given the following definitions:
For $ x \in \mathbb{R}^{n} $ and $ 1 \leq p<\infty $, we set $ \|x\|_{p}:=\sqrt[p p]{\sum \limits_{i=1}^{n}\left|x_{i}\right|^{p}}, \quad\|x\|_{\infty}:=\max _{1 \leq i \leq n}\left|x_{i}\right| $.
For a matrix $ A \in \mathbb{R}^{m \times n} $ and $ 1 \leq p \leq \infty $ we define the matrix norm $ \|A\|_{p}:=\sup _{x \neq 0} \frac{\|A x\|_{p}}{\|x\|_{p}}=\sup _{\|x\|_{p}=1}\|A x\|_{p} $.
I'm not really familiar with matrix norms, so I would appreciate any ideas or suggestions on how to show the above statement.