Subspace Question: Why Not Closed Under Addition?

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Homework Help Overview

The discussion revolves around the concept of subspaces in linear algebra, specifically addressing the property of closure under addition. The original poster seeks clarification on why a certain set is not closed under addition, despite being closed under scalar multiplication.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants explore examples of vectors that satisfy the condition of the subspace and question whether their sum remains within the set. There is a focus on understanding the implications of closure under addition.

Discussion Status

Participants are actively engaging with the problem, questioning assumptions and clarifying definitions. Some have provided examples to illustrate their points, while others are seeking further explanations about the closure property.

Contextual Notes

There is a noted confusion regarding the definitions involved, particularly the conditions that define the subspace. Participants are also addressing potential mistakes in the original problem statement.

Nope
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Homework Statement



[PLAIN]http://img683.imageshack.us/img683/4530/unledkw.jpg
can someone please explain why it is not closed under addition?
My textbook did not explain very well, but I understand this can be zero vector and it is closed under scalar multiplication.
thanks!

Homework Equations


The Attempt at a Solution

 
Last edited by a moderator:
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[x,y]=[1,-1] satisfies x^2=y^2. So does [1,1]. Does their sum?
 
Last edited:
oh ,
(1+1, 1-1)
(2,0) not in (x,y)
am i correct?
or can you explain to me in word?
 
Last edited:
Nope said:
can you show me?, because I don't get the closed under addition part at all..

Do you agree [1,-1] and [1,1] are in your subspace? What's the sum [1,-1]+[1,1]? It's vector addition, right?
 
yes, (2,0)
 
Nope said:
yes, (2,0)

Is [2,0] in your set where x^2+y^2=1?
 
Nope said:
oh ,
(1+1, 1-1)
(2,0) not in (x,y)
am i correct?
or can you explain to me in word?

Right! (2,0) is not in (x,y) such that x^2=y^2. So your set is NOT closed under addition.
 
not in the set, but why is x^2+y^2 equal to one?
 
Nope said:
not in the set, but why is x^2+y^2 equal to one?

It isn't. I meant to write x^2=y^2 or x^2-y^2=0. My mistake.
 
  • #10
is there another way to prove it? like using x1 or y2
 
  • #11
Nope said:
is there another way to prove it? like using x1 or y2

The most direct way to prove a set is NOT closed under addition is to find two elements in the set whose sum is NOT in the set. I'm not sure why you would want another way.
 
  • #12
I got it, thanks!
 

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