In a book I'm reading, it says:If beta is orthogonal to the set A,

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Discussion Overview

The discussion revolves around the concept of orthogonality in the context of linear algebra, specifically addressing the relationship between a vector beta and a set A, as well as the closure of the linear span of A. Participants explore the implications of beta being orthogonal to both the set A and its closure.

Discussion Character

  • Technical explanation
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • Some participants note that if beta is orthogonal to the set A, it follows that beta is orthogonal to the closure of the linear span of A, but they seek clarification on the necessity of mentioning closure.
  • One participant explains that while it is true beta is orthogonal to the span of A, in many applications, it is important for beta to be orthogonal to the closure of span(A).
  • Another participant provides a general argument stating that if a vector x is orthogonal to a set A, then x is also orthogonal to the closure of A, using a sequence argument to illustrate this point.
  • There is a question raised about whether the inner product with x needs to be shown as a continuous function, indicating a potential area of exploration or clarification.

Areas of Agreement / Disagreement

Participants generally agree on the relationship between orthogonality and closure, but there is some uncertainty regarding the necessity of emphasizing closure in the original statement. The discussion remains somewhat unresolved as participants seek further clarification.

Contextual Notes

There are assumptions regarding the continuity of the inner product that have not been explicitly addressed, and the discussion does not resolve whether the emphasis on closure is universally necessary in all contexts.

yifli
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In a book I'm reading, it says:

If beta is orthogonal to the set A, then beta is orthogonal to the closure of the linear span of A

It's easy to see beta is orthogonal to the linear span of A, but I don't understand why it has to mention closure here?
 
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Hi yifli! :smile:

What is your question exactly?

Do you wonder why they wrote this here? Well, because it's true. Of course it's also true that x is orthogonal to span(A), so they could have written this too. However, in most applications, you need x orthogonal to the closure of span(A). This is why they've written this here.

Do you ask why it is true? Well, in general, if x is orthogonal to a set A, then x is orthogonal to the closure of A. Indeed, pick a in the closure of A, then there exists a sequence an that converges to a. Since an is in A, we know that <x,an>=0. And thus [itex]<x,a>=\lim_{n\rightarrow +\infty}{<x,a_n>}=0[/itex]. Thus x is orthogonal to a.
 


micromass said:
Hi yifli! :smile:

What is your question exactly?

Do you wonder why they wrote this here? Well, because it's true. Of course it's also true that x is orthogonal to span(A), so they could have written this too. However, in most applications, you need x orthogonal to the closure of span(A). This is why they've written this here.

Do you ask why it is true? Well, in general, if x is orthogonal to a set A, then x is orthogonal to the closure of A. Indeed, pick a in the closure of A, then there exists a sequence an that converges to a. Since an is in A, we know that <x,an>=0. And thus [itex]<x,a>=\lim_{n\rightarrow +\infty}{<x,a_n>}=0[/itex]. Thus x is orthogonal to a.

micromass, thank you so much for your answer!
 


micromass said:
Hi yifli! :smile:

What is your question exactly?

Do you wonder why they wrote this here? Well, because it's true. Of course it's also true that x is orthogonal to span(A), so they could have written this too. However, in most applications, you need x orthogonal to the closure of span(A). This is why they've written this here.

Do you ask why it is true? Well, in general, if x is orthogonal to a set A, then x is orthogonal to the closure of A. Indeed, pick a in the closure of A, then there exists a sequence an that converges to a. Since an is in A, we know that <x,an>=0. And thus [itex]<x,a>=\lim_{n\rightarrow +\infty}{<x,a_n>}=0[/itex]. Thus x is orthogonal to a.

Do you need to show that the inner product with x is a continuous function?
 


lavinia said:
Do you need to show that the inner product with x is a continuous function?

Of course! :smile: But that's not that hard, I believe...
 

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