How can I prove $\mathrm{max \, mag}(A)=\frac{1}{\mathrm{min \, mag}(A^{-1})}$, and $\mathrm{max \, mag}(A^{-1})=\frac{1}{\mathrm{min \, mag}(A)}$?

Question

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

I found this quite simple to prove using a proposition stating that $$\kappa(A)=\frac{\mathrm{max \, mag}(A)}{\mathrm{min \, mag}(A)}$$ 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?

Answer 1

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

$\rightarrow \mathrm{max \, mag}(A) = \|Ax\|/\|x\| = \|y\|/\|x\|$

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

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

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

Answer 2

This is basically the same thing @Pratik Deoghare has stated, but in a more convincing fashion.

Starting from

$$\mathrm{min \, mag}(A)=\min_{x \neq 0} \frac{||Ax||}{||x||}$$

Invert the expression inside the min. Be a bit careful here while taking a reciprocal, since just inverting the expression inside the minimum to make it a max is not right. Think about it.

$$\mathrm{min \, mag}(A) = \frac{1}{\max \limits_{x\neq 0}\frac{||x||}{||Ax||}}$$

Now let $Ax = y$, which gives $x=A^{-1}y$

$$\mathrm{min \, mag}(A) = \frac{1}{\max \limits_{x\neq 0}\frac{||A^{-1}y||}{||y||}}$$

Since $x\neq 0 \implies y\neq 0$

$$\mathrm{min \, mag}(A) = \frac{1}{\max \limits_{y\neq 0}\frac{||A^{-1}y||}{||y||}}$$

which is, by the definition of $\mathrm{max \, mag}$,

$$\mathrm{min \, mag}(A) = \frac{1}{\mathrm{max \, mag}(A^{-1})}$$

The other result can be obtained similarly.

Answer 3

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 $$\frac{\|Ax\|}{\|x\|}\mathrm{min \, mag}(A^{-1})\leq \frac{\|Ax\|}{\|x\|}\left \Vert A^{-1}\frac{Ax}{\|Ax\|}\right \Vert=1$$ so taking the supremum over these $x$ we get $\mathrm{max \, mag}(A)\cdot\mathrm{min \, mag}(A^{-1})\leq 1$. We have, if $A^{-1}x\neq 0$ $$\mathrm{max \, mag}(A)\frac{\|A^{-1}x\|}{\|x\|}\geq \|A\frac{A^{-1}x}{\|A^{-1}x\|}\|\frac{\|A^{-1}x\|}{\|x\|}=1$$ so $\mathrm{min \, mag}(A^{-1})\mathrm{max \, mag}(A)\geq 1$ and $\mathrm{min \, mag}(A^{-1})\mathrm{max \, mag}(A)=1$. Since the unit ball of $\mathbb R^n$ is compact, $\mathrm{min \, mag}(A^{-1})\neq 0$ and we get the result.