31

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?

Rodrigo de Azevedo
  • 23,223
Cancan
  • 2,827

1 Answers1

51

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.

Daniel Fischer
  • 211,575
  • 1
    This first step does not seem to be right... $\lVert Ax \rVert_1 = \lVert \sum x_v \cdot A_v \rVert_1$ -- the right side of this equality is the 1-norm of a scalar, but the left side is the 1-norm of a vector. Did I miss something? – Robert Feb 22 '18 at 16:10
  • 3
    @Robert The "$\cdot$" in $\sum x_{\nu}\cdot A_{\nu}$ denotes scalar multiplication, not an inner product. $x_{\nu}$ is a real number. – Daniel Fischer Mar 03 '18 at 15:19