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Determine if the space is a subspace testing both closure axioms

  1. Jun 9, 2009 #1
    1. The problem statement, all variables and given/known data
    determine if the space is a subspace testing both closure axioms.

    in R^2 the set of vectors (a,b) where ab=0


    2. Relevant equations



    3. The attempt at a solution

    i just used the sum and product which are the closure axioms.

    But at the end how do you tell if the resulting vector is a subspace?

    (a,b) + (c,d) = (a+c, b+d)

    (a+c)(b+d)=0 then ab+cb + ad+dc=0 ab+cb = -ad-dc

    then ???

    x=constant

    x(a,b) = (xa,xb) then abx^2 =0


    then?????
     
  2. jcsd
  3. Jun 9, 2009 #2

    CompuChip

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    Re: Subspace

    How about you look at (1, 1) and (1, -1) ?
     
  4. Jun 9, 2009 #3

    Cyosis

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    Re: Subspace

    If (a,b) and (c,d) are in the set. For it to be a subspace (a+c,b+d) needs to be in the set. To be in the set (a+c).(b+d)=0 needs to be satisfied. Proof that the equality holds. Then do the same for scalar multiplication. If both equalities hold you can conclude it is a subspace.
     
    Last edited: Jun 9, 2009
  5. Jun 9, 2009 #4
    Re: Subspace

    so if i set
    a=1 , b=1 , c=1 , d=-1


    and add (a,b) +(c,d)
    i will have 1+1-1-1 = 0

    Is that the way to prove it?


    is isn't the roduct beetween a and b has to be equal to 0. I understand by that that one of the 2 terms has to be 0 right?
     
  6. Jun 9, 2009 #5
    Re: Subspace

    then if i set (0,12) and (45,0) both are in the set but when i add them

    (45,12) they are not in the space anymore right?

    45*12 is not equal to 0 right?
     
  7. Jun 9, 2009 #6

    Cyosis

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    Re: Subspace

    What you say in post 4 is not correct. The sum of both may be zero but separately they aren't even in the subspace, since (1,1), 1*1=1!=0.

    What you say in post 5 is correct, because 0*12=45*0=0.

    That said is this really the problem? As you've stated the problem it could be read that a and b are vectors in R^2 and they are a subspace when a.b=0 (inner product). I think CompuChip interpreted your question like this.
     
    Last edited: Jun 9, 2009
  8. Jun 9, 2009 #7

    Pengwuino

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    Re: Subspace

    Yes, that is an example of how it works. I'm not sure why one would use (1,1), (1,-1). They aren't in the subspace.

    As Cyosis began, you can use two arbitrary vectors, X=(a,b) and Y=(c,d) where ab=0 and cd=0. However, investigating closure under addition, X+Y = (a+c,b+d). Let's let e=a+c, f=b+d meaning X + Y = (e,f). For closure to be true, ef=0. Plug back in what e and f are and you get (a+c)(b+d). Does this equal 0? That is, is it in the subspace for arbitrary a,b,c,d?
     
  9. Jun 9, 2009 #8
    Re: Subspace

    so under addition does not work.... by it does work under product....

    in the end is not a subspace.....

    it's so good learning the material like that.....thanks to all of you....
     
  10. Jun 11, 2009 #9

    CompuChip

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    Re: Subspace

    Ah, I must apologize.
    I misread the condition of "ab equal to 0" as "ab not equal to 0".
     
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