TITLE: Proof for infinity matrix norm
QUESTION [5 upvotes]: I need to prove the infity norm of a matrix, which is here: $$||A||_\infty = \max_{x \neq 0}\left\{\frac{\max_{1\le i\le n}(|(Ax)_i|)}{\max_{1 \le i \le n}(|x_i|)}\right\}.$$ I can't find the the proof anywhere and would appreciate if someone could explain every step of the proof. Thanks!

REPLY [13 votes]: It is obviously sufficient to consider vectors $x$ with $\|x\|_{\infty}=1$. Then we have
$$
\|A\|_{\infty} = \max_{\|x\|_{\infty}=1} \|Ax\|_{\infty}
=\max_{\|x\|_{\infty}=1} \max_{i=1,\ldots,n}\left|\sum_{j=1}^{n}a_{ij}x_j\right| \\
=\max_{i=1,\ldots,n} \max_{\|x\|_{\infty}=1}\left|\sum_{j=1}^{n}a_{ij}x_j\right|
=\max_{i=1,\ldots,n} \sum_{j=1}^{n}\left|a_{ij}\right|
$$
The first equality is the definition of the matrix norm. The second equality applies the definition of the vector norm $\|\cdot\|$ to the vector $Ax$. The third equality simply swaps the $\max$s, which is always allowed. For the fourth equality, we can argue that the inner maximum is obviously obtained when we choose the absolute value of $x_j$ as large as possible, i.e. $x_j\in\{-1,1\}$, and when $x_j$ has the same sign as $a_{ij}$.