# Matrix norm, equivalent definitions [duplicate]

I find two main definitions for the norm of a matrix or linear operator.

First definition: $\lVert A\rVert := sup \{|A(\frac{x}{|x|})|: x\neq 0 \}$

Second definition: $\lVert A\rVert := sup \{|A(\frac{x}{|x|})|: |x| \leq 1 , x\neq 0 \}$

Why are those two equivalent? I am sure it has something to do with linearity, but I can't see the answer.

I know, since the unit ball is compact, the superemum is attained for some x on or inside the unit ball. Why must the maximum lie on the surface unit ball?

Help would be appreciated.

Thank you

• You still need $x\ne0$ in the second definition, as $\frac{x}{\|x\|}$ is undefined at $0$. – user1551 Dec 19 '16 at 9:36
• In both your definitions, $A(\frac{x}{\lVert x \rVert })$ is a vector. How do you find a supremum of a set of vectors? I suppose you need to take $\lVert A(\frac{x}{\lVert x \rVert })\rVert$ there. – tchao Dec 19 '16 at 9:43

Let $M_1=\{||A(\frac{x}{||x||})||: x\neq 0 \}$ and $M_2=\{||A(\frac{x}{||x||})||: 0 < ||x|| \le 1\}$
It is clear that $M_2 \subseteq M_1$. It remains to show that $M_1 \subseteq M_2$. To this end let $a \in M_1$ then $a=||A(\frac{x}{||x||})||$ for some $x \ne 0$. Let $z=\frac{x}{||x||}$, then $||z||=1$ hence $\frac{z}{||z||}=z=\frac{x}{||x||}$ and
$a=||A(\frac{z}{||z||})|| \in M_2.$
• Thank you for your answer. Unfortunately you explained the direction, that was clear to me before. Can you maybe also show why $M2 \subset M1$ ? – Jake B. Dec 19 '16 at 9:54
• If $a \in M_2$ then $a=||A(\frac{x}{||x||})||$ for some $x$ with $0 < ||x|| \le 1\}$. Thus $a=||A(\frac{x}{||x||})||$ for some $x \ne 0$ . This gives $a \in M_1$ – Fred Dec 19 '16 at 9:58