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in the answer of this post: I don't understand how How $\|Ax\|\leq ||A|| ||x||$?

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  • $\begingroup$ What is $x,X$? ${}{}$ $\endgroup$ – copper.hat Oct 18 '19 at 1:38
  • $\begingroup$ I edited it. $x$ is a vector. $\endgroup$ – user 42493 Oct 18 '19 at 1:39
  • $\begingroup$ How do you define $\|A\|$? $\endgroup$ – copper.hat Oct 18 '19 at 1:40
  • $\begingroup$ $$\lVert A\rVert=\sup\limits_{\lVert x\rVert =1}\{\lVert Ax\rVert :x\in K^n\}$$ $\endgroup$ – user 42493 Oct 18 '19 at 1:40
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    $\begingroup$ So, if $x \neq 0$, then $\|A {x \over \|x\|} \| \le \|A\|$. $\endgroup$ – copper.hat Oct 18 '19 at 1:43
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By definition, $\|Ax\| \leq \|A\|$ for all $x$ with $\|x\|=1$. In particular, if $v$ an arbitrary vector (distinct from zero) in $\textsf K^n$ then we know that $\|tv\|=1$ where $t=1/\|v\|$.

So, $\|A(tv)\| \leq \|A\|$ and observe that $\|A(tv)\|=\|t(Av)\|=|t|\|Av\|=t\|Av\|$. In conclusion, we have $$t\|Av\|\leq\|A\|$$ and then $$\|Av\|\leq\frac1t\|A\|=\|v\|\|A\|=\|A\|\|v\|$$

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