Find the norm of matrix using the Cauchy-Schwarz inequality

Question

Let $A$ be $n \times n$ matrix and such that all of its entries are uniformly $O(1)$. Using Cauchy-Schwarz inequality, show that the operator norm of matrix $A$, which is

$$\|A\|_{op} := \sup_{x\in R^n: |x|=1}|Ax|$$ is of $O(n)$.

Thank you.

Answer 1

Hint: show $|(Ax)_i| \le K \sqrt{n}$ for each $i$, then $\|Ax\|^2 = \sum_{i=1}^n |(A x)_i|^2 \le K^2 n^2$.

Answer 2

I'm not sure where the application of the Cauchy Schwarz (notice no t) comes in, so what I am about to write may have limited applicability, but it may be of some use:

$||Ax||=||x_{1}Ae_{1}+...+x_{n}Ae_{n}||<|x_{1}| ||Ae_{1}||+...+|x_{n}|||Ae_{n}||$ by the triangle inequality. Let $M=max(Ae_{1},Ae_{2}...,Ae_{n})$. Also $||x||=1$, so $|x_{i}|\leq 1$, so

$||Ax||<nM$