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Squaring a vector and subspace axioms

  1. May 25, 2009 #1
    If a problem I'm doing asks to find

    V2 where V is a vector

    is it simply the dot product of the vector, or the cross product?

    The question: Which of the following sets of vectors v = {v1,...,vn} in Rn are subspaces of Rn (n>=3)

    iii) All v such that V2=V12

    He proved it by saying it's not closed under addition (axiom of a subspace)

    By (1,1,0) + (-1,1,0) = (0,2,0)

    And that concludes his proof, but I'm not seeing what he's proved there at all.

    Personally I would have done let a = (a1,...,an) and b = (b1,...,bn)

    then a+b = (a1+b1,....,an+bn)

    and so (a+b)2=(a1+b1,....,an+bn)2=(a1+b1,....,an+bn).(a1+b1,....,an+bn)

    = a12+2a1b1+b12+.. (dot product)

    Which is a scalar field not a vector field, so it's not closed under addition.

    Am I wrong?
     
    Last edited: May 25, 2009
  2. jcsd
  3. May 25, 2009 #2

    Dick

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    They aren't asking you to square the vector. The v's are the components of the vector (v1,v2,v3). The question is take the subset of these vectors such that v2=v1^2. Like (2,4,1) and the first two vectors in your example.
     
  4. May 25, 2009 #3
    I was assuming v was a matrix of vectors, and each v1 etc was a vector. hence '...following sets of vectors v = {v1,...,vn}'

    I still don't understand the rest, especially his proof
     
  5. May 25, 2009 #4

    Dick

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    It's not supposed to be a matrix. If V is the subset of R^3 such that v2=v1^2, then (1,1,0) (since 1^1=1) and (-1,1,0) (since (-1)^2=1) are in V. Their sum (0,2,0) is not in V since 0^2 is not equal to 2.
     
  6. May 25, 2009 #5
    ok thanks alot I get it now :)
     
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