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Prove that 0v = 0 for an arbitaryvector v belong to V

  1. Jan 16, 2010 #1
    1. The problem statement, all variables and given/known data

    Suppose V is a vector space over F.
    Prove that 0v(vector) = 0(vector) for an arbitrary vector v belong to V


    2. Relevant equations

    Vector additive identity

    3. The attempt at a solution

    Let v = (v1,...,vn) belong to V, then 0v = 0(v1,...,vn) = (0,...,0).
    Now (v1, ... ,v2) + (0,...,0) = (v1,...,vn), therefore (0,..,0) is the zero vector in V.

    Is this ok?
     
    Last edited: Jan 16, 2010
  2. jcsd
  3. Jan 16, 2010 #2
    There are probably many ways to do this. However, you are not close. A vector space does not necessarily have to be represented by euclidean vectors of the form (a,b). A vector space could even be the set of all differentiable functions. Anyway you have to use the full and general definition of a vector space: http://en.wikipedia.org/wiki/Vector_space


    Use these facts:

    1+(-1)=0 (1 being the identity)
    v+(-v)=0

    Also, your last step should be:
    (1-1)v=0
    0v=0

    Good luck!
     
  4. Jan 16, 2010 #3
    Okay so you're saying that my representation of V is not general and only apply to euclidean vectors right?

    Prove that 0v = 0(vector)

    Given
    1+(-1)=0,
    then 1+(-1)v=0(vector)
    (1-1)v = 0(vector)
    0v = 0(vector)
     
  5. Jan 17, 2010 #4

    HallsofIvy

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    And why is that last line true? Again, you are assuming what you want to prove!

    What you need is, rather, 0v= (1+ (-1))v= 1v+ (-1)v= what?
     
  6. Jan 17, 2010 #5
    Oooops! For some reason, I was proving something completely different, completely forgeting the OP.((-1)v=-v, I think) :redface:

    The last line is SUPPOSED to be (0+0)v=0v in which the theorem follows immediately.
     
  7. Jan 17, 2010 #6

    D H

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    How does the second line follow from the first?
     
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