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

  • $\begingroup$ You still need $x\ne0$ in the second definition, as $\frac{x}{\|x\|}$ is undefined at $0$. $\endgroup$ – user1551 Dec 19 '16 at 9:36
  • $\begingroup$ 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. $\endgroup$ – 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

$a=||A(\frac{z}{||z||})|| \in M_2.$

  • $\begingroup$ 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 $ ? $\endgroup$ – Jake B. Dec 19 '16 at 9:54
  • $\begingroup$ 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 $a \in M_1$ $\endgroup$ – Fred Dec 19 '16 at 9:58

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