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Show that dim U <= n

  1. Nov 17, 2009 #1
    1. The problem statement, all variables and given/known data

    Let U be a vector subspace of C2n such that
    sum(xi*yi) = 0 for 1 <= i <= 2n for any x, y ∈ U. Show that dim U <= n. Give an example of such a subspace U with dim U = n

    2. The attempt at a solution

    I tried just writing out the summation and was thinking along the lines of linear independence but I don't think that applies here (maybe it does, I'm not sure). Could I think of a linear map contained in U that maps two vectors x and y to be the sum = 0? I think I'm confusing myself here.
     
  2. jcsd
  3. Nov 17, 2009 #2

    lanedance

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    i'm not sure i understand the question correctly... so is that effectively the compex innner product of 2 vectors in the subsapce is always zero?

    [tex] <\texbf{x},\texbf{y}> = \sum_i x_i^* y_i [/tex]

    but if that were the case, as U is a vector space, if x is in U, then so is c.x, but
    [tex] <\texbf{x},\texbf{cx}> = c||x||^2 [/tex]
    contradiction?
     
  4. Nov 17, 2009 #3

    Hurkyl

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    Maybe by * he meant multiplication rather than complex conjugation? We'll have to wait for him to clarify, I guess.
     
  5. Nov 17, 2009 #4

    Mark44

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    Glad you guys (lanedance and Hurkyl) jumped in on this one. I was thinking along the lines that lanedance described, except I was thinking of this product of a vector with itself.
    [tex]\sum_{i = 1}^{2n} x_i*x_i~=~0[/tex]
    which suggests that all the x_i's are 0.
     
  6. Nov 17, 2009 #5
    it's multiplication not conjugate (sorry about the mix-up everybody!)
     
  7. Nov 18, 2009 #6

    Mark44

    Staff: Mentor

    brru25, You're sure you have given us the exact problem description, right?
     
  8. Nov 18, 2009 #7
    positive, word-for-word.....see why I'm confused? :-)
     
  9. Nov 18, 2009 #8

    lanedance

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    ok, think I'm getting it now, sounds like what Hurkyl was thinking...

    I haven't worked it, but would start with an example in the 2D case in [itex] \mathbb{C}^2[/itex], so n = 1

    so say you have a vector (a,b) which is in U, it satisfies the rule with itself
    [tex] \sum_i x_i^2 = a^2 + b^2 = 0[/tex]

    so, first can you find a vector that satisfies above... and 2nd can you show given a vector in U, there can be no other linearly independent vectors in U?
     
    Last edited: Nov 18, 2009
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