$\def\norm#1{{\Vert#1\Vert}_{\infty}}$
I do not completely understand the steps of the below proof. The result establishes that $\norm{A}=\max_{1 \le i \le n}\sum_j |a_{ij}|$. Any inputs or comments that could help me follow the proof to the end, would be really helpful.
Theorem 7.11 (Burden and Faires – Numerical Analysis).
Prove that
$$\norm{A}=\max_{1 \le i \le n}\sum_j |a_{ij}|$$
Proof.
We first prove that
$$\norm{A} \le \max_{1 \le i \le n}\sum_j |a_{ij}|$$
This part of the proof is clear to me.
Let $x=(x_1,x_2,\ldots,x_n)$ be an arbitrary vector such that the length of the vector $x$ relative to the norm $l_\infty$ is $1$.
$\begin{align}
\norm{A} &= \max_{||x||=1}||Ax||_\infty
= \max_{1 \le i \le n} |(Ax)_i|
= \max_{1 \le i \le n} |\sum_j a_{ij} x_j|\\
&\le \max_{1 \le i \le n} \sum_j |a_{ij}|\cdot |x_j|\\
&\le \max_{1 \le i \le n} \sum_j |a_{ij}|\cdot \max_{1 \le j \le n} |x_j| = \max_{1 \le i \le n} \sum_j |a_{ij}|\cdot ||x||_\infty = \max_{1 \le i \le n} \sum_j |a_{ij}|
\end{align}$
We now prove the opposite side of the inequality. The arguments that follow from this point on are not clear to me. Either I don't fully understand it, or there is a typo in the book.
Let $p$ be an integer such that $\sum_j |a_{pj}|=\max_{1 \le i \le n} \sum_j |a_{ij}|$.
Let $x$ be a vector with components $x_j=1$, if $a_{pj}\ge 0$ and $x_j=-1$ if $a_{pj}<0$.
Then, $\norm{x} = 1$ and $a_{pj}x_j = |a_{pj}|$ for all $j=1,2,\ldots,n$.
So,
$\begin{align}
\norm{Ax} &= \max_{1 \le i \le n} |\sum_j a_{ij}x_j|\\
&\ge |\sum_j a_{pj}x_j| \space (\text{Don't follow this step})\\
&= |\sum_j |a_{pj}||
= \sum_j |a_{pj}|
= \max_{1 \le i \le n}\sum_j |a_{ij}|
\end{align}$
This completes the proof.
Best Answer
In order to explain to you, I take the proof you provided and add comments
$\begin{align} ||Ax||_\infty &= \max_{1 \le i \le n} |\sum_j a_{ij}x_j|\\ &\ge \max_{1 \le i \le n} |\sum_j a_{pj}x_j| \space (\text{Don't follow this step})\\ &= |\sum_j |a_{pj}|| \\ &= \sum_j |a_{pj}| \\ &= \max_{1 \le i \le n}\sum_j |a_{ij}| \\ \end{align}$
First note on the second line that the sum do not depend on $i$ so its constant relatively to $i$. And because $p$ is among the $i\in[1,n]$, you have
$$||Ax||_\infty = \max_{1 \le i \le n} |\sum_j a_{ij}x_j| \geq \sum_j |a_{pj}x_j| = \\ \max_{1 \le i \le n} \sum_j |a_{pj}x_j|\\(\text{because max of a constant is a constant, actually you don't need a max for the second line)} $$