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Vectorial Subspace

  1. Jan 22, 2010 #1
    There's not really a problem statement here.
    I just want to know :
    If I have a vector starting on the origin (like a position vector), then it will always correspond to a vectorial subspace, right?

    For example:

    [tex] (b, 2a + b ) : a, b \in R [/tex]

    is a vectorial subspace

    but is

    [tex] (b, 2a + 1 ) : a, b \in R [/tex]

    a subspace too?
    And if not, why is that?
     
  2. jcsd
  3. Jan 22, 2010 #2

    Mark44

    Staff: Mentor

    For a subset of a vector space to actually be a subspace of that vector space, the subset has to satisfy three conditions:
    The set has to have the 0 vector.
    If u and v are in the subset, then u + v is also in the subset.
    If u is in the subset, and c is any scalar, then cu is in the subset.

    The set of vectors (b, 2a + 1) is not a subspace of R2, because at least one of the three conditions is not met.
     
  4. Jan 22, 2010 #3

    Dick

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    Uh, but (b,2a+1) spans all of R^2.
     
  5. Jan 22, 2010 #4

    Mark44

    Staff: Mentor

    Right, but maybe we aren't talking about the same thing. I'm thinking in terms of the set {(a, 2b + 1) | a, b are real}. This set isn't closed under addition, so isn't a subspace of R2.
     
  6. Jan 22, 2010 #5

    Dick

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    (a1,2b1+1)+(a2,2b2+1)=(c,2d+1) where c=a1+a2 and d=b1+b2+1/2. The question looks like a different kind of question (i.e. is (a,2b,1) a subspace?). But it's not. 2b+1 is ANY real number, just like b and independent of a.
     
  7. Jan 22, 2010 #6

    Mark44

    Staff: Mentor

    Dick, I wasn't confusing it with (a, 2b, 1). I think I got thrown by the lack of dependence of a and b.
     
  8. Jan 22, 2010 #7

    Dick

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    That's true. (a,2a+1) would also be a whole different story.
     
  9. Jan 22, 2010 #8

    Mark44

    Staff: Mentor

    That's exactly where I was coming from. My eyes must have glazed over...
     
  10. Jan 22, 2010 #9
    you got me kinda lost here.

    so it is indeed a subspace since it spans all R^2?
     
  11. Jan 22, 2010 #10

    Dick

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    R^2 is a subspace of R^2, isn't it? Check the conditions to be a subspace Mark44 was referring to.
     
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