By definition, we have
$$
\|V\|_p
:= \sqrt[p]{\displaystyle \sum_{i=1}^{n}|v_i|^p}
\qquad \text{and} \qquad
\|A\|_p
:= \sup_{x\not=0}\frac{||Ax||_p}{||x||_p}
$$
and if $A$ is finite, we change sup to max.
However I don't really get how we get to the definition of $||A||_1$ as the maximum absolute column sum of the matrix as stated in Wikipedia
For example, assume $A=\begin{bmatrix} a_{11} & a_{12} \\ a_{21} & a_{22}\end{bmatrix}$.
Then
$$
||A||_1
= \max_{x\not=0} \frac{\left\|\begin{bmatrix} a_{11} & a_{12} \\ a_{21} & a_{22}\end{bmatrix}\cdot \begin{bmatrix} x_{1} \\ x_{2}\end{bmatrix}\right\| }{\left\|\begin{bmatrix} x_{1} \\ x_{2}\end{bmatrix}\right\|}
= \max_{x\not=0} \frac{|a_{11}x_1+a_{12}x_2|+|a_{21}x_1+a_{22}x_2|}{|x_1|+|x_2|}
$$
That's what I have gotten so far, but I don't really see how this is related to the max of the column sum. Can anyone help me explain this?
Best Answer
Let's denote the columns of $A$ by $A_1,\, \dotsc,\, A_n$. Then for every $x \in \mathbb{R}^n$, we have
$$\begin{align} \lVert Ax \rVert_1 &= \left\lVert\sum_{\nu=1}^n x_\nu\cdot A_\nu \right\rVert_1\\ &\leqslant \sum_{\nu=1}^n \lVert x_\nu\cdot A_\nu\rVert_1\\ &= \sum_{\nu=1}^n \lvert x_\nu\rvert\cdot\lVert A_\nu\rVert_1\\ &\leqslant \max \left\{\lVert A_\nu\rVert_1 : 1 \leqslant \nu \leqslant n\right\} \left(\sum_{\nu=1}^n \lvert x_\nu\rvert\right)\\ &= \max \left\{\lVert A_\nu\rVert_1 : 1 \leqslant \nu \leqslant n\right\}\cdot \lVert x\rVert_1. \end{align}$$
That shows that
$$\lVert A\rVert_1 \leqslant \max \left\{\lVert A_\nu\rVert_1 : 1 \leqslant \nu \leqslant n\right\},$$
and choosing $x = e_m$, where $m$ is the index where the absolute column sum has its maximum shows the converse inequality, hence equality.