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Showing that "zero vector space" is a vector space

  1. Oct 9, 2016 #1
    1. The problem statement, all variables and given/known data
    Let ## \mathbb{V} = \{0 \}## consist of a single vector ##0## and define ##0 + 0 = 0## and ##c0 = 0## for each scalar in ##\mathbb{F}##. Prove that ##\mathbb{V}## is a vector space.

    2. Relevant equations


    3. The attempt at a solution

    Proving that the first six axioms of a vector space are true is trivial, I am just on the distributive axioms.

    So if I want to show that ##a(x + y) = ax + ay## is true, where a is a scalar and x and y are vectors, is it sufficient to make the argument that ##a(0 + 0) = a0 = 0 = a0 + a0##? Does this show that a distributes over the vector ##0##?
     
  2. jcsd
  3. Oct 9, 2016 #2

    Simon Bridge

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    I guess ... since the space only contains one vector you can also show that the last two are equivalent to the previous ones.
    ie. u=v, then a(u+v)=a(u+u)=2au = scalar multiplication axiom already tested.
    ie u=v, then RHS: a(u+v)=2au and LHS=au+av = 2au so RHS=LHS
     
  4. Oct 9, 2016 #3

    PeroK

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    Yes, that's exactly how to do it.
     
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