Norms in R^n .... and Aut(R^n) .... .... D&K Corollary 1.8.12 .... ....

[Math Amateur]

I am reading "Multidimensional Real Analysis I: Differentiation" by J. J. Duistermaat and J. A. C. Kolk ...

I am focused on Chapter 1: Continuity ... ...

I need help with an aspect of Corollary 1.8.12 ... ...

Duistermaat and Kolk's Corollary 1.8.12 and the preceding definition of \(\displaystyle \text{Aut} \mathbb{R}^n\) read as follows:View attachment 7747
I can see how D&K arrive at the result:

\(\displaystyle c_1 \mid \mid x \mid \mid \le \mid \mid Ax \mid \mid \le c_2 \mid \mid x \mid \mid\)... BUT ... how, exactly, did D&K derive the result ...\(\displaystyle c_2^{ -1 } \mid \mid x \mid \mid \le \mid \mid A^{ -1 } x \mid \mid \le c_1^{ -1 } \mid \mid x \mid \mid
\)
Hope that someone can help ...

Peter=========================================================================================

***NOTE***

It may help MHB members reading the above post to have access to the results preceding Corollary 1.8.12 ... so I am providing the same ... as follows:https://www.physicsforums.com/attachments/7745
View attachment 7746Hope that helps ...

Peter

 

[GJA]

Hi, Peter.

The trick is to write $x=AA^{-1}x$. Now set $y=A^{-1}x$ and see if you can use the bounds you know to be true on $Ay.$

 

[Math Amateur]

Hi, Peter.

The trick is to write $x=AA^{-1}x$. Now set $y=A^{-1}x$ and see if you can use the bounds you know to be true on $Ay.$

Thanks ...

Just reflecting on this ...

Peter

 

[Math Amateur]

Thanks ...

Just reflecting on this ...

Peter

Sorry GJA ... despite your help, I am not making progress ...

In particular having trouble determining the bounds on Ay ...

Can you please help further ...

Peter

 

[GJA]

We have $x=AA^{-1}x$. Setting $y=A^{-1}x$ and using the known inequality we obtain
$$\|x\|=\|AA^{-1}x\|=\|Ay\|\leq c_{2}\|y\|=c_{2}\|A^{-1}x\|.$$
Dividing through by $c_{2}$ we get
$$c_{2}^{-1}\|x\|\leq \|A^{-1}x\|,$$
as desired. A similar argument will give the second inequality.

 

[Math Amateur]

We have $x=AA^{-1}x$. Setting $y=A^{-1}x$ and using the known inequality we obtain
$$\|x\|=\|AA^{-1}x\|=\|Ay\|\leq c_{2}\|y\|=c_{2}\|A^{-1}x\|.$$
Dividing through by $c_{2}$ we get
$$c_{2}^{-1}\|x\|\leq \|A^{-1}x\|,$$
as desired. A similar argument will give the second inequality.

Thanks for your guidance and assistance GJA ... I appreciate your help ...

With your help I am beginning to understand analysis in \(\displaystyle \mathbb{R}^n\) ...

Thanks again ...

Peter

 

[GJA]

Any time, Peter!