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

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.

• $||A||_\infty = \sup_{x\neq 0} \frac{||Ax ||_{\infty}}{||x||_\infty}$ from definition, so I don't really understand. – Jakobian Jun 23 at 19:25
• @Jakobian I gave our definition in the yellow box. So we did not define it that way, or am I misunderstanding something? – Cornman Jun 23 at 19:26
• You should write it more clearly then. I didn't understand it's the definition. The other comment is how you usually define norm for operators. – Jakobian Jun 23 at 19:27
• @Jakobian I wrote now, that this is our definition. But how do you define $\|Ax\|_\infty$ when your definition of $\|A\|_\infty$ involves that norm? This seems confusing? – Cornman Jun 23 at 19:30
• @user1551 Yes, i meant it how you wrote it. My bad. Sorry. – Cornman Jun 23 at 19:50

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

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

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 $$\sup_{x\neq \vec0} \frac{||Ax||}{||x||} = \sup_{x\neq \vec0,\ ||x||\leq 1} ||Ax||$$.

• How do you get the last equality? What if $|x_i|>1$ – Cornman Jun 23 at 19:51
• Because $\sup_{x\neq \vec0} \frac{||Ax||}{||x||} = \sup_{x\neq \vec0} ||A\frac{x}{||x||}|| = \sup_{x\neq \vec0,\ ||x|| = 1} ||Ax || = \sup_{x\neq \vec0, \ ||x||\leq 1} ||Ax||$. – Jakobian Jun 23 at 19:55
• Ask if one of the equalities is still unclear. – Jakobian Jun 23 at 19:57
• It is not fully clear to me at the moment. I will think about it. Why it is not needed to observe the case $|x_i|>1$. – Cornman Jun 23 at 19:58
• Because of the last equality in the answer. We need to only care about supremum of vectors with norm $\leq 1$. – Jakobian Jun 23 at 19:59