TITLE: 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 [8 upvotes]: 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?

REPLY [2 votes]: 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.