$\|A\|_\infty = \sup_{x\neq 0} \frac{\|Ax\|_\infty}{\|x\|_\infty}=\|A\operatorname{1}\|_\infty$

Question

Let $A$ be a $n\times m$ matrix, where every entry is either positiv (all entries are positiv) or negativ (all entries are negativ). Then holds:

$\|A\|_\infty = \sup_{x\neq 0} \frac{\|Ax\|_\infty}{\|x\|_\infty}=\|A\operatorname{1}\|_\infty$

where $1=(1,1,\dotso, 1)^T\in\mathbb{R}^m$

[It should be $x\in\mathbb{R}^m$ too, this is not mentioned...]

The equality $\|A\|_\infty=\|A\cdot 1\|_\infty$ is easy to see.

First of all without loss of generality, we can assume that every entry in $A$ is positiv.

Definition: It is $\|A\|_\infty=\displaystyle{\max_{i=1,\dotso, n}}\{\sum_{k=1}^m |a_{ik}|\}$

and

$\|A1\|_\infty = \|\begin{pmatrix}\sum_{k=1}^m a_{1k}\\\vdots\\\sum_{k=1}^m a_{nk} \end{pmatrix}\|_\infty=\displaystyle{\max_{i=1,\dotso n}}\{\sum_{k=1}^m a_{ik}\}$

Now, I want to see the other equality. Maybe, it is best to show something like

$\|A1\|_\infty\leq \sup_{x\neq 0} \frac{\|Ax\|_\infty}{\|x\|_\infty}\leq \|A\|_\infty$

Where the first inequality, is trivial.

So $\|A1\|_\infty\leq \sup_{x\neq 0} \frac{\|Ax\|_\infty}{\|x\|_\infty}$ holds for sure, because the RHS can not be smaller than $\|A1\|_\infty$, because we take the supremum over every $x\neq 0$, so espacially $x=1$ (here I mean equality of elements of $\mathbb{R}^m$). In that case, we have equality.

We are left to show

$\sup_{x\neq 0} \frac{\|Ax\|_\infty}{\|x\|_\infty}\leq \|A\|_\infty$

Do you have a hint how to show this? Writing out the definition of $\|\dot\|_\infty$ for the fraction did not help me for now.

Thanks in advance.

Answer 1

With

$$\left|\left| A \right|\right|_{\infty} = \max_{i=1,\ldots,n}\sum^{n}_{j=1}\left|A_{ij}\right|$$

we have

$$ \left|\left(A\textbf{x}\right)_i\right| = \left|\sum^{n}_{j=1}A_{ij}x_j\right|\leq \sum^{n}_{j=1} \left|A_{ij}\right|\left|x_j\right| \leq \left|\left| \textbf{x}\right|\right|_{\infty}\sum^{n}_{j=1}\left|A_{ij}\right| $$

and therefore

$$ \frac{\left|\left| A\textbf{x}\right|\right|_{\infty}}{\left|\left|\textbf{x}\right|\right|_{\infty}} = \frac{\max_{i=1}^{n}\left|\right(A\textbf{x}\left)_i \right|}{\left|\left| \textbf{x}\right|\right|_{\infty}}\leq\frac{\max^{n}_{i=1} \left|\left|\textbf{x}\right|\right|_{\infty}\sum^{n}_{j=1}\left|A_{ij}\right|}{\left|\left|\textbf{x}\right|\right|_{\infty}} = \max_{i=1,\ldots,n}\sum^{n}_{j=1}\left|A_{ij}\right| = \left|\left| A \right|\right|_{\infty} $$

hence

$$ \sup_{\textbf{x}\neq \textbf{0}}\frac{\left|\left| A\textbf{x}\right|\right|_{\infty}}{\left|\left|\textbf{x}\right|\right|_{\infty}} \leq \left|\left| A \right|\right|_{\infty} $$

Answer 2

I will get rid of $\infty$ subscript entirely for simplicity of notation. Take $x = (x_1, ..., x_n)\neq \vec0, \ |x_i|\leq 1$ $$||Ax|| = \sup_{1\leq i\leq n} |\sum_{k=1}^m a_{ik} x_k| \leq \sup_{1\leq i\leq n} \sum_{k=1}^m \left(a_{ik}|x_k|\right) \leq \sup_{1\leq i\leq n} \sum_{k=1}^m a_{ik} = ||A||$$ And we're done because $\sup_{x\neq \vec0} \frac{||Ax||}{||x||} = \sup_{x\neq \vec0,\ ||x||\leq 1} ||Ax||$.