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Field axioms, subspaces

  1. Feb 28, 2015 #1
    1. The problem statement, all variables and given/known data
    Let [itex]F_{2} = {0, 1}[/itex] denote a field with 2 elements.

    Let V be a vector space over [itex]F_{2}[/itex]. Show that every non-empty set W of V which is closed under addition is a subspace of V.




    3. The attempt at a solution

    subspace axioms: 0 elements, closed under scalar multiplication, closed under vector addition.

    We can skip the latter axiom as it's given in the question.

    proof of 0 element:

    0, 1 ∈ F_2
    x ∈ W

    [itex]0x = 0_{W}[/itex]

    therefore there exists a zero element.

    proof of scalar multiplication:

    0, 1 ∈ F_2
    x ∈ W

    [itex]1x = x[/itex]

    this is true due to the scalar multiplication identity.

    ----------------------------------

    I believe this could be wrong, I feel as though I am missing something. Thanks.
     
    Last edited: Feb 28, 2015
  2. jcsd
  3. Feb 28, 2015 #2

    Dick

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    The words describing what you are doing could be clearer, but basically all the steps are there. All you need to do is prove that W contains the zero vector (which you did) and is closed under scalar multiplication. You showed that if x is in W then 1x is in W. For completeness you might want to mention why 0x is in W. It really is a pretty easy problem.

    Actually thinking about it again, you also need to show that there are additive inverses. 'Closed under addition' doesn't imply that.
     
  4. Feb 28, 2015 #3
    I think I've found a more robust proof for scalar multiplication:

    0, 1 in F_2
    x in W

    consider all possible combinations:

    0x, 0x + 1x, 1x + 1x, 0x + 0x, 1x + 1x + 1x

    [itex]0x = 0_{W}[/itex]
    [itex]0x + 1x = 0_{W} + x = x[/itex]
    [itex]1x + 1x = (1 + 1)x = 0x = 0_{W}[/itex]
    [itex]0x + 0x = 0_{W}[/itex]
    [itex]1x + 1x + 1x = (1 + 1)x + 1x = 0x + 1x = 0_{W} + x = x[/itex]



    all of them belong to W. But if you compare the second and last line you can see they are the same, and so we're going around in circles, I've shown all possible outcomes between the (very small) Field and the subspace W.

    I don't think I need to show that it contains the zero element since it's non-empty and so by definition, it contains the zero element.
     
  5. Feb 28, 2015 #4

    pasmith

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    [itex]W[/itex] is a subset of [itex]V[/itex]. If [itex]W[/itex] is to be a subspace of [itex]V[/itex] then you need [itex]0_V \in W[/itex].

    You have to show why the condition that [itex]W[/itex] is closed under vector addition requires that [itex]0_V \in W[/itex], so that if [itex]w \in W[/itex] then both [itex]1w = w \in W[/itex] and [itex]0w = 0_V \in W[/itex].

    Let [itex]w \in W[/itex]. Since [itex]W[/itex] is closed under addition, we have [itex]w + w \in W[/itex]. What do you know about [itex]w + w = (1 + 1)w[/itex]?
     
  6. Feb 28, 2015 #5

    Dick

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    You've added a bunch of unnecessary stuff and removed some important stuff. It's not adding robustness, it's adding confusion. Could you say what you want to prove (and why!) before the actual proof. "proof for scalar multiplication" doesn't do it. WHAT about scalar multiplication?
     
  7. Feb 28, 2015 #6
    show that W is a subspace of the vector space V (and V is a vector space over the Field F = {0, 1}).

    1.x = x

    since 1 is in the field and x is in W (and W is a subspace of V) then we can use the scalar multiplication identity from V. Ergo, scalar multiplication holds for this subspace.

    Proof of zero element:

    1x + 1x = (1+1)x (distributivity law from the vector space, V, onto subspace, W).#

    (1+1)x = 0x = 0_w

    since 1x + 1x ∈ W, since elements in a subspace added together are also in the subspace, then 0_w must also be in the subspace. Therefore 0 element exists.



    What do you think of my proof now? im trying to connect all the dots. Thanks.
     
    Last edited: Feb 28, 2015
  8. Feb 28, 2015 #7

    Dick

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    Getting better! The first part could be clearer. What you need to prove there is that the set W is closed under scalar multiplication. I.e. if x is in W then 0x and 1x are also in W. You might want to prove that the zero vector is in W first. I'd also suggest you comment on why additive inverses of elements of W are also in W.
     
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