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Vector space and fields question

  1. Feb 17, 2015 #1
    1. The problem statement, all variables and given/known data
    Let V be a vector space over the field F. The constant, a, is in F and vectors x, y in V.

    (a) Show that a(x - y) = ax - ay in V .

    (b) If ax = 0_V show that a = 0_F or x = 0_V .

    2. Relevant equations
    axiom 1: pv in V, if v in V and p in F.
    axiom 2: v + v' in V if v, v' in V

    3. The attempt at a solution

    (a) x - y = x + (-y)


    ====> (x+ (-y)) in V (axiom 2)
    ====> a(x + (-y)) in V (axiom 1)


    (b)...?

    how would I start part (b)?
     
    Last edited: Feb 17, 2015
  2. jcsd
  3. Feb 17, 2015 #2

    Orodruin

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    How about starting by assuming that both a and x are different from their resoective zeros and trying to find a contradiction?
     
  4. Feb 17, 2015 #3
    if they are both non-zero then they cannot bultiply together to give a zero vector i.e. either one (or both) have to be their zero's.

    I feel as though im missing something, my solution is too simple.
     
  5. Feb 17, 2015 #4

    Orodruin

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    Yes, but why can they not multiply to zero if they are both non-zero? This is the entire question.
     
  6. Feb 17, 2015 #5
    just looked up all the axioms from a Field and a vector space, from what I can see:

    a Field doesn't have the multiplicative axiom: 0a = 0. On the other hand, a vector space does have this multiplicative axiom: 0v = 0.

    (b) If ax = 0_V show that a = 0_F or x = 0_V .

    ax =/= 0_F because that would assume that an element in the Field contains this specific multiplicative axiom.

    ergo, ax = 0_V iff a = 0_F or x = 0_V
     
  7. Feb 17, 2015 #6

    Orodruin

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    Note that, while not an axiom, it is trivial to show that this is true from the distributive property of multiplication.

    I am not really sure what you are trying to argue here.
     
  8. Feb 17, 2015 #7
    a in F. v in V. Let's refer to it as an axiom for the sake of simplicity.

    since a is in the field, you cannot have 0a = 0, because this multiplicative axiom is not in any Field, given the axioms in a Field.

    Therefore, ax =/= 0_F, because if x=0_V, then you have a0 = 0_F, and this is a contradiction (given the axioms in a Field).

    Therefore, ax = 0_V, because a vector space can contain the axiom 0v = 0, and it holds when either a = 0_F or v = 0_V.






    I could be wrong, or perhaps i didnt explain it properly.
     
  9. Feb 17, 2015 #8

    Orodruin

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    As I mentioned, the relation 0a = 0 is true for all fields. You have that:
    (b + 0)a = (b)a = ba
    But at the same time, the distributive property gives
    (b + 0)a = ba + 0a
    Adding the additive inverse of ba to both these relations gives
    ba + (-ba) = 0 = ba + 0a + (-ba) = 0a
    And therefore 0a = 0.
     
  10. Feb 17, 2015 #9
    i misinterpreted your reply.

    If that's the case, could we use the inverse?

    assume a and x are non-zero's:

    ax = 0_V

    ax*a^-1 = 0_V * a^-1
    x = 0_V

    contradiction. x must equal 0_V.

    on the other hand, there is no multiplicative inverse axiom within a vector subspace, therefore if

    ax = 0_V

    you CANNOT do:

    ax*x^-1 = 0_V x^-1

    i.e. you cannot eliminate the x on the LHS, ergo:

    ax = 0_V (=/= 0_F), iff a = 0_F or x = 0_V
     
  11. Feb 17, 2015 #10

    Orodruin

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    This is sufficient. You do not need to worry about the existence or non-existence of a multiplicative inverse for V. You assumed the anti-hypothesis of what you wanted to prove, i.e., that both were non-zero, and arrived at a contradiction, thus proving that the anti-thesis is false, meaning that the hypothesis is true.
     
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