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

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The discussion centers on Proposition 8.6 from Andrew Browder's "Mathematical Analysis: An Introduction," specifically regarding the distance function defined as \(\rho(S, T) = \| S - T \|\). Participants clarify the definition of \(-T\) as \(-T = (-1)T\) and confirm that \(\| -T \| = \| T \|\). An example involving the transformation \(T: \mathbb{R}^2 \to \mathbb{R}^2\) is presented, demonstrating that the norms of \(T\) and \(-T\) are indeed equal, validating the participant's understanding of the concepts discussed.

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  • Understanding of linear transformations in \(\mathbb{R}^2\)
  • Familiarity with norm definitions and properties
  • Basic knowledge of mathematical analysis concepts
  • Ability to interpret propositions in mathematical literature
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  • Study the properties of norms in vector spaces
  • Explore linear transformations and their geometric interpretations
  • Review additional propositions in Browder's "Mathematical Analysis: An Introduction"
  • Learn about differentiable maps and their applications in analysis
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Students of mathematical analysis, particularly those studying linear algebra and differentiable maps, as well as educators seeking to clarify concepts related to distance functions and transformations in \(\mathbb{R}^2\).

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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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