# Definition of an induced matrix norm.

Could someone explain the second equality in the definition of a induced matrix norm to me?

Let $\| \cdot \|$ be a norm for $\cdot \in \mathbb{R}^n$, then the induced matrixnorm for $A\in \mathbb{R}^{n\times n}$ is given by:

$$\|A\| = \sup_{x\not = 0} \frac{\|Ax\|}{\|x\|} \color{red}{\stackrel{?}{=}} \max_{\|x\|= 1} \|Ax\|$$

### Problem:

Why does $\color{red}{=}$ hold?

I know $\|Ax\|$ is continous for all $x\in \mathbb{R}^n$ and $\{x \in \mathbb{R}^n :\|x\| = 1 \}$ is compact which implies $\dfrac{\|Ax\|}{\|x\|}$ to reach a maximum according to Weierstrass.

But why is this maximum also the maximum for all $x$? (even those with norm bigger or less then 1?)

• Note that you can write any $x\neq 0$ uniquely as $\alpha x'$, where $\alpha=\|x\|$ and $\|x'\|=1$. Does that help? Commented Feb 22, 2015 at 23:31
• Because it is possible to write it as a normalized vector right? $$x = \|x\| \cdot \left( \frac{x}{\|x\|}\right)$$ And then $\sup_{x\not =0}\frac{\|A\alpha x'\|}{\|\alpha x'\|} = \sup_{\|x'\|=1} \|Ax'\|$, where the supremum can be replaced by maximum according to Weierstrass? (Awesome!) Commented Feb 23, 2015 at 0:11
• In the first one it is $\sup$ because the set $x\neq0$ is an open set. If you use the properties of norm and get $\sup_{x\neq0}\frac{\|Ax\|}{\|x\|}=\sup_{x\neq0}\left\|A\frac{x}{\|x\|}\right\|=sup_{\|y\|=1}\|Ay\|$, where we made the substitution $y=\frac{x}{\|x\|}$. Also note that the set $\|y\|=1$ is compact, i.e., $\sup$ value is actually $\max$. Commented Aug 6, 2018 at 22:24

I guess it is clear that we have $$\sup_{x\neq 0} \frac{\Vert Ax\Vert}{\Vert x\Vert} \geq \max_{\Vert x\Vert = 1} \Vert A x\Vert$$ because $A:=\{ x \in \mathbb{R}^n \mid \Vert x \Vert = 1 \} \subseteq \{ x\in\mathbb{R}^n \mid x \neq 0\}=:B$.

The other way around, assume that $y\in B$. Then we have $z:= \frac{y}{\Vert y\Vert} \in A$ and thus $$\frac{\Vert Ay\Vert}{\Vert y\Vert}= \left\Vert A\frac{y}{\Vert y\Vert}\right\Vert= \Vert Az\Vert \leq \max_{x\in A}\Vert Ax\Vert = \max_{\Vert x\Vert = 1} \Vert A x\Vert.$$ Because $y\in B$ was arbitrary this inequality holds for all $y\in B$. In particular, we get $$\sup_{y \neq 0} \frac{\Vert Ay\Vert}{\Vert y\Vert} = \sup_{y\in B} \frac{\Vert Ay\Vert}{\Vert y\Vert} \leq \max_{\Vert x\Vert = 1}\Vert A x\Vert\text{.}$$ Both inequalities together give the desired equality.