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

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Browder's Proposition 8.6 defines the distance function \(\rho(S, T)\) as \(\| S - T \|\), prompting questions about the definition of \(-T\). It is clarified that \(-T\) is indeed \((-1)T\) and that \(\| -T \| = \| T \|\). An example using the transformation \(T: \mathbb{R}^2 \to \mathbb{R}^2\) demonstrates that both \(\| T(x,y) \|\) and \(\| -T(x,y) \|\) yield the same result. The discussion confirms the correctness of the example provided. Understanding these concepts is essential for grasping the implications of the distance function in linear algebra.
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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:
View attachment 9392
In the above proposition, Browder defines the distance function \rho (S, T) as follows:

$$\rho (S, T) = \| S - T \| $$... but just some basic questions ...
How do we define $$-T$$?Is $$-T = ( -1) T$$?Is $$\| -T \| = \| T \| $$?

A simple example that shows the way things work as I see it follows:Consider $$T: \mathbb{R}^2 \to \mathbb{R}^2 $$Let $$T(x,y) = ( x - y, 2y )$$... then ...$$- T (x,y) = (-1) T(x,y) = ( -x + y, -2y)$$ and then it follows that ...$$\| T(x,y) \| = \| ( x - y, 2y ) \| = \sqrt{ (x - y)^2 + (2y)^2 }$$and ...$$\| -T(x,y) \| = \| ( -x + y, -2y) \| = \sqrt{ (-x + y)^2 + (-2y)^2 } = \| T(x,y) \| $$
Is the above example correct?Peter
 

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Yes, it looks like you have everything correct.
 
LCKurtz said:
Yes, it looks like you have everything correct.
Thanks LCKurtz

Peter
 

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