Let $A$ be an $N\times N$ matrix and let $\| A \|_{sum} = \sum_{i,j} |a_{ij}|$ and $\|\vec{r}\|_{sum} = \sum_k |r_k|$.
Let $\| A \| = \sqrt{N}\|A\|_{sum}$ and let $\| \vec{r}\| = \sqrt{r_1^2+\ldots+r_N^2}$; show that: $$\| A\vec{r}\| \le \| A\| \cdot \| \vec{r} \|$$
I think I have seen it before in ODE 1 course, but I cannot remember how to show it, do I need Cauchy Schwartz inequality here?
Can you show me how to prove this inequality?
Thanks.
Denote by $(Ar)_i$ the $i$-th component of $Ar$, and by $A_{i*}$ the $i$-th row of $A$. Now $(Ar)_i = A_{i*} \cdot r$, so by Cauchy–Schwarz $$\left |(Ar)_i \right |^2 = \left |A_{i*} \cdot r \right |^2 \le \|{A_{i*}} \|^2 \|{r}\|^2.$$ Summing over $i$, you find $$\|Ar\|^2 = \sum \left |(Ar)_i \right |^2 \le \left ( \sum \|{A_{i*}} \|^2 \right ) \|r\|^2.$$ But $\sum \|{A_{i*}} \|^2 = \sum |a_{ij}|^2 \le \left (\sum |a_{ij}| \right )^2$, so we get $$\|Ar\|^2 \le \left (\sum |a_{ij}| \right )^2 \|r\|^2 \quad \Rightarrow \quad \|Ar\| \le \left (\sum |a_{ij}| \right ) \|r\| = \|A\|_{\text{sum}} \|r\| \le \|A\| \|r\|.$$