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Using $\mathrm{maxmag}(A)=\max_{x\neq 0}\frac{\|Ax\|}{\|x\|}$, and $\mathrm{minmag}(A)=\min_{x\neq 0}\frac{\|Ax\|}{\|x\|}$

I found this quite simple to prove using a proposition stating that $$\kappa(A)=\frac{\mathrm{maxmag}(A)}{\mathrm{minmag}(A)}$$ for all nonsingular A.

However, I think that propostion follows from the conclusion I am trying to prove. I don't think I can use it because I can't prove it without the using relationship I am trying to use it to prove, if that makes sense. How can I use only the definitions of maxmag and minmag to prove this?

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2 Answers 2

Let $y=Ax$ and $x = A^{-1} y$

$\rightarrow \mathrm{maxmag}(A) = \|Ax\|/\|x\| = \|y\|/\|x\|$

If $\|y\|/\|x\|$ is the maximum thing then $\|x\|/\|y\|$ is minimum.

$\rightarrow \|x\|/\|y\| = \|A^{-1}y\|/\|y\| = \mathrm{minmag}(A^{-1}) $

$$\mathrm{minmag}(A) \ \mathrm{minmag}(A^{-1}) = \frac{\|y\|/\|x\|}{\|x\|/\|y\|} =1 $$

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We just prove the first inequality, then we will apply the second on to $A^{-1}$. Let $x\in\mathbb R^n$. If $Ax\neq 0$ then $$\frac{||Ax||}{||x||}\operatorname{minmag}(A^{-1})\leq \frac{||Ax||}{||x||}\lVert A^{-1}\frac{Ax}{||Ax||}\rVert=1$$ so taking the supremum over these $x$ we get $\operatorname{maxmag}(A)\cdot\operatorname{minmag}(A^{-1})\leq 1$. We have, if $A^{-1}x\neq 0$ $$\operatorname{maxmag}(A)\frac{||A^{-1}x||}{||x||}\geq ||A\frac{A^{-1}x}{||A^{-1}x||}||\frac{||A^{-1}x||}{||x||}=1$$ so $\operatorname{minmag}(A^{-1})\operatorname{maxmag}(A)\geq 1$ and $\operatorname{minmag}(A^{-1})\operatorname{maxmag}(A)=1$. Since the unit ball of $\mathbb R^n$ is compact, $\operatorname{minmag}(A^{-1})\neq 0$ and we get the result.

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