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Operator Norm and Distance Function ... Browder, Proposition 8.6 ...

Peter

Well-known member
MHB Site Helper
Jun 22, 2012
2,891
I am reading Andrew Browder's book: "Mathematical Analysis: An Introduction" ... ...

I am currently reading Chapter 8: Differentiable Maps and am specifically focused on Section 8.1 Linear Algebra ...

I need some help in fully understanding the concepts in Proposition 8.6 ...


Proposition 8.6 reads as follows:



Browder - Proposition 8.6 ... ....png



In the above proposition, Browder defines the distance function \rho (S, T) as follows:

\(\displaystyle \rho (S, T) = \| S - T \| \)


... but just some basic questions ...



How do we define \(\displaystyle -T\)?


Is \(\displaystyle -T = ( -1) T\)?


Is \(\displaystyle \| -T \| = \| T \| \)?




A simple example that shows the way things work as I see it follows:


Consider \(\displaystyle T: \mathbb{R}^2 \to \mathbb{R}^2 \)


Let \(\displaystyle T(x,y) = ( x - y, 2y )\)


... then ...


\(\displaystyle - T (x,y) = (-1) T(x,y) = ( -x + y, -2y)\)


and then it follows that ...


\(\displaystyle \| T(x,y) \| = \| ( x - y, 2y ) \| = \sqrt{ (x - y)^2 + (2y)^2 }\)


and ...


\(\displaystyle \| -T(x,y) \| = \| ( -x + y, -2y) \| = \sqrt{ (-x + y)^2 + (-2y)^2 } = \| T(x,y) \| \)



Is the above example correct?


Peter
 

LCKurtz

New member
Aug 10, 2018
10
Yes, it looks like you have everything correct.
 

Peter

Well-known member
MHB Site Helper
Jun 22, 2012
2,891