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I found a proof in an online course note which purports to show that two definitions of matrix norms are equivalent, however, I have some doubts regarding the proof, I would like a second pair of eyes

Claim: $\|A\| = \max\limits_{ x \neq 0} \dfrac{\|Ax\|}{\|x\|} = \max\limits_{\|x\| = 1} \|Ax\|$


Proof:

$(\text{first show} \max\limits_{ x \neq 0} \dfrac{\|Ax\|}{\|x\|} \leq \max\limits_{\|x\| = 1} \|Ax\|)$

Suppose $x \neq 0$, then $\max\limits_{ x \neq 0} \dfrac{\|Ax\|}{\|x\|} = \max\limits_{ x \neq 0} \|A\dfrac{x}{\|x\|}\|$. Let $z = \dfrac{x}{\|x\|}$, $\|z\| = 1$. Therefore, $\max\limits_{ x \neq 0} \dfrac{\|Ax\|}{\|x\|} \leq \max\limits_{ \|z\| =1} \|Az\| = \max\limits_{ \|x\| =1} \|Ax\|$

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$(\text{next show:}\max\limits_{ x \neq 0} \dfrac{\|Ax\|}{\|x\|} \geq \max\limits_{\|x\| = 1} \|Ax\|)$

Suppose that $\|x\| = 1$, therefore $\max\limits_{\|x\| = 1} \|Ax\| = \max\limits_{\|x\| = 1} \dfrac{\|Ax\|}{\|x\|} \leq \max\limits_{x \neq 0} \dfrac{\|Ax\|}{\|x\|}$, where the inequality follows from maximization over a superset of $\{x\in \mathbb{R}^n:\|x\|=1\}$.

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Thus, $\max\limits_{ x \neq 0} \dfrac{\|Ax\|}{\|x\|} = \max\limits_{\|x\| = 1} \|Ax\|$.

It seems that there is no justification for $\max\limits_{ x \neq 0} \dfrac{\|Ax\|}{\|x\|} \leq \max\limits_{ \|z\| =1} \|Az\|$. It doesn't make sense, because we are maximizing over a much larger set than the unit circle. Can someone see if there is indeed a problem with the proof and if there is some opportunity to fix it?

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  • $\begingroup$ I think you may have a misunderstanding. Usually, "equivalent" for norms $N$ and $M$ means that there are constants $C_1,C_2$ such that $C_1M(A) \leq N(A) \leq C_2M(A)$ for all $A$. $\endgroup$ – AlexanderJ93 Jan 18 '18 at 22:00
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    $\begingroup$ you dont need the inequalities, actually $\frac{\|Ax\|}{\|x\|}=\|A(x/\|x\|)\|$ from the definition of norm. $\endgroup$ – Masacroso Jan 18 '18 at 22:00
  • $\begingroup$ Why don't use the bijection between unit vectors and nonzero vectors? $$x \leftrightarrow x / ||x||$$ $\endgroup$ – GNUSupporter 8964民主女神 地下教會 Jan 18 '18 at 22:01
  • $\begingroup$ @GNUSupporter That's not a bijection. $\endgroup$ – user296602 Jan 18 '18 at 22:21
  • $\begingroup$ @user296602 Thx for pointing this out. You're right. It's better to read Masacroso's comment. $\endgroup$ – GNUSupporter 8964民主女神 地下教會 Jan 18 '18 at 22:37
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Note that, if $x\neq0$, then$$\frac{\|Ax\|}{\|x\|}=\left\|A\left(\frac x{\|x\|}\right)\right\|$$and that $\left\|\frac x{\|x\|}\right\|=1$.

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For $x\ne 0$, then $\dfrac{\|Ax\|}{\|x\|}=\|x\|^{-1}\|Ax\|=\|\|x\|^{-1}Ax\|=\left\|A\left(\dfrac{x}{\|x\|}\right)\right\|\leq\max_{\|z\|=1}\|Az\|$.

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  • $\begingroup$ How do you justify the last inequality? $\endgroup$ – Shamisen Expert Jan 18 '18 at 23:08
  • $\begingroup$ So $\left\|\dfrac{x}{\|x\|}\right\|=1$ then put $z=\dfrac{x}{\|x\|}$, so the left side is just $\|Az\|$ and hence it must be $\|Az\|\leq\max_{\|z\|=1}\|Az\|$. $\endgroup$ – user284331 Jan 18 '18 at 23:09

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