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Subspace of R2 involving absolute value

  1. Feb 10, 2013 #1
    1. The problem statement, all variables and given/known data

    {(x1,x2)T| |x2|=|x2|}

    So my first thought is we would have to check for both cases (x1,x1) and (-x1,-x1)

    a=(x1,x1)T b=(v1,v1)T


    βa=(βx1,βx1)T for the case where a<0 βa(-βx1,-βx1)T

    thus it is closed under scalar multiplication.

    a+b=(x1+v1,x1+v1) for the case where a and b<0 [-(x1+v1,x1+v1)]T
    thus closed under addition.

    I would think this would be a subset, however the book says this is false. Where did I go wrong?


    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Feb 10, 2013 #2

    Dick

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    You are basically ignoring the absolute value. Are (1,1) and (1,-1) in your subspace? How about the sum?
     
  4. Feb 10, 2013 #3
    No because a+b would give you (2,0) and x1 and x2 have to be equal?
     
  5. Feb 10, 2013 #4

    Dick

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    No, because |2| is not equal |0|. You don't have to have x1=x2 to be in the subspace. x1=(-x2) also works. But (2,0) doesn't fit either one.
     
  6. Feb 10, 2013 #5

    SammyS

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    Did you really mean |x2| = |x2| ?
     
  7. Feb 10, 2013 #6
    Oops that was a typo it was suppose to be |x1|=|x2|.

    Also Dick, I'm a bit confused on why you chose (1,1) (1,-1)
     
  8. Feb 10, 2013 #7

    Dick

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    I was just trying to find two vectors in your set which didn't sum to be something in the set. Choosing one that satisfies x1=x2 and another that satisfies x1=(-x2) will do it. Nothing special about the 1's.
     
  9. Feb 10, 2013 #8
    But they must be equal? for instance we could use (2,2) and (2,-2) or (-2,2) and (2,2)
     
  10. Feb 10, 2013 #9

    Dick

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    (1,1) and (3,-3) work fine too. Their sum isn't in the set either.
     
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