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Subspace question

  1. Sep 26, 2011 #1
    1. The problem statement, all variables and given/known data

    [PLAIN]http://img683.imageshack.us/img683/4530/unledkw.jpg [Broken]
    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!


    2. Relevant equations



    3. The attempt at a solution
     
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Sep 26, 2011 #2

    Dick

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    [x,y]=[1,-1] satisfies x^2=y^2. So does [1,1]. Does their sum?
     
    Last edited: Sep 26, 2011
  4. Sep 26, 2011 #3
    oh ,
    (1+1, 1-1)
    (2,0) not in (x,y)
    am i correct?
    or can you explain to me in word?
     
    Last edited: Sep 26, 2011
  5. Sep 26, 2011 #4

    Dick

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    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?
     
  6. Sep 26, 2011 #5
    yes, (2,0)
     
  7. Sep 26, 2011 #6

    Dick

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    Is [2,0] in your set where x^2+y^2=1?
     
  8. Sep 26, 2011 #7

    Dick

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    Right! (2,0) is not in (x,y) such that x^2=y^2. So your set is NOT closed under addition.
     
  9. Sep 26, 2011 #8
    not in the set, but why is x^2+y^2 equal to one?
     
  10. Sep 26, 2011 #9

    Dick

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    It isn't. I meant to write x^2=y^2 or x^2-y^2=0. My mistake.
     
  11. Sep 26, 2011 #10
    is there another way to prove it? like using x1 or y2
     
  12. Sep 26, 2011 #11

    Dick

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    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.
     
  13. Sep 26, 2011 #12
    I got it, thanks!
     
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