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Quick Subspace Question

  1. Apr 21, 2012 #1
    1. The problem statement, all variables and given/known data
    Hey, I'm trying to figure out whether A = [x+1,0] is a subspace... I know it's probably simple but what i'm confused is that....


    2. Relevant equations



    3. The attempt at a solution
    u+v must be an element of A. let u = [x1+1,0] and v=[x2+1,0]. Adding them together gives you [x1+x2+2,0], where (x1+x2) = x. Therefore it wouldn't be a subspace because you now have the form [x+2,0] which is not the same?

    But if I look at it, [x1+x2+2,0] should be a subspace because you can get the same value with [x+1, 0] (let x1, x2 = -1 and let x = -1, both get 0 vector, for example)
    Thanks for any help!
     
  2. jcsd
  3. Apr 21, 2012 #2

    Dick

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    I think your second answer is closer. You want to show [x+1,0] is a subspace of R^2 if I understand what you are asking. What things do you have show to prove A is a subspace? And yes, [x1+x2+2,0]=[(x1-1+x2)+1,0]. So it is closed under addition if I've understood you correctly.
     
  4. Apr 21, 2012 #3
    Well I know the three conditions. However, what I'm not getting is that you should have your u+v in the form [x+1,0], where x is made up of x1 + x2, or am I getting the definition wrong? In this case i have (x1 + x2 + 1)...
     
  5. Apr 21, 2012 #4

    Dick

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    [x1+x2+2,0]=[(x1+1+x2)+1,0] is in the form [x+1,0] where x=x1+1+x2 as you said, isn't it? Sorry, I had a typo in the previous post.
     
  6. Apr 21, 2012 #5
    Oh yeah! So you can rearrange it anyway so that you get the correct form? I thought that the initial variable had to be represented by only variables (if that makes sense)...
     
  7. Apr 21, 2012 #6

    Dick

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    Yeah, you are probably thinking too hard. The set of vectors [x+1,0] is really the same thing as the set of vectors [x',0]. Where x'=x+1. Where both x and x' can be any real number.
     
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