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Vector space

  1. Feb 18, 2006 #1
    decide whether this is a vector space or not
    a(x,y,z) = (2ax,2ay,2az)

    all the addition axoims hold easily
    for the scalar multiplications axioms
    for some real scaral a
    [tex] a(x,y,z) = (ax,ay,az) \in 2(ax,ay,az) [/tex]
    [tex] a(x_{1}+x_{2},y_{1}+y_{2},z_{1}+z_{2}) = a(x_{1},y_{1},a(z_{1}) + a(x_{2},y_{2},a(z_{2}) \in (2ax,2ay,2az) [/tex] this doesnt seem to hold because it would only be possible if the wo vecotrs being added were not distinct.

    [tex] (a+b)(x,y,z) = (ax+bx,ay+by,az+bz) \in (2az,2ay,2az) [/tex]

    [tex] a(bx,by,bz) = (ab)(x,y,z) [/tex] this would seem t ohold if b was 2... not not anytrhing else? Not sure here?

    multiplcation by 1 gives us what we want. so the last axiom holds
    i can post hte axioms if u want

    my text says it is not a vector space because of the failure of scalara multiplication where i have stated the doubts mysefl. Are those hte valid reasons for that??
     
    Last edited: Feb 18, 2006
  2. jcsd
  3. Feb 18, 2006 #2

    matt grime

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    what the hell are x,y,z, or a?
     
  4. Feb 18, 2006 #3

    HallsofIvy

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    a(x,y,z)= (2ax,2ay,2az) doesn't make sense. Are you asked to show that the set of all (x,y,z) with ordinary addition but scalar multiplication defined by a(x,y,z)= (2a,2ay,2az) is a vector space?
     
  5. Feb 19, 2006 #4
    that is what i am asked to prove
     
  6. Feb 19, 2006 #5

    matt grime

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    Multiplication by must send v to v, ie 1.v=v for all v in the space.
     
  7. Feb 19, 2006 #6

    HallsofIvy

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    I don't believe for a moment that your problem says exactly
    "decide whether this is a vector space or not a(x,y,z) = (2ax,2ay,2az)"
    There isn't even a set that could be a vector space there!
     
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