Question about $\infty$-norms According to my textbook, a matrix $A \in \mathbb{C}^{n \times n}$ has $\infty$-norm equal to the maximum row sum of the matrix. Is there any way of gaining intuition for this fact?
 A: I never really proved it before, so what follows is a solution I just came up with (possibly not the cleverest one). The $ \infty $-norm for a vector $ v = (v_i)_i $ is just $ \max_i{|v_i|} $, and the corresponding matrix norm is
$$
\max_v{\big(\max_i\left|(Av)_i\right|\big)} = \max_v{\left(\max_i\left|\sum_{j}A_{ij}v_j\right|\right)} \quad \mathrm{provided}\quad\max_i\left|v_i\right| = 1. 
$$
Now any $ A_{ij} $ may be written as $ a_{ij}e^{\mathrm i\varphi_{ij}} $, with $ a_{ij} \in \mathbb R^+_0 $ (the modulus), $ \varphi_{ij} \in \mathbb R $ (the phase).
In order to maximize the sum, you should take a $ v $ which suitably rotates any $ A_{ij} $ of a given row onto a fixed direction—say the positive real semiaxis—while only possessing entries of modulus one, explicitly
$$
v_j = e^{-\mathrm i\varphi_{ij}} \quad\mathrm{for\ a\ chosen}\, i;
$$
in this way, the modulus of that sum equals the sum of the moduli, since
$$
\left|\sum A_{ij}v_j \right| = \left|\sum a_{ij}\right| = \sum a_{ij} = \sum\left|A_{ij}\right|,
$$
and that sum gets no bigger due to the triangular inequality. Since $ i $ (the row label, not the imaginary unit!) is arbitrary, you might as well take that which maximises the above sum.
Therefore,
$$ \|A\|_\infty = \max_i{\sum_j{\left|A_{ij}\right|}}. $$
