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Vector spaces + proving of properties

  1. Dec 30, 2007 #1
    [SOLVED] vector spaces + proving of properties

    Im aware in vector spaces that there are 3 properties associated with it
    Note v is an element in a vector space, 0 is the additive identity in the vector space and c is a field element

    1) 0.v = 0

    2) c.0 = 0

    3) (-c).v = c.(-.v) = -(c.v)

    atm Im trying to prove 2), however I can only properly prove 2) if I use 1). As mathematical properties are meant to be unique, is my current solution considered as bad practice ?
    Last edited: Dec 30, 2007
  2. jcsd
  3. Dec 30, 2007 #2
    did you already prove 1? if so it could go something like this:

    [tex]c \cdot \vec{0}=c \cdot (0 \cdot \vec{v})[/tex] by 1
    [tex] (c \cdot 0)\cdot \vec{v}= 0 \cdot \vec{v}=0[/tex] by 1 again

    i'm not sure if this is rigorous though
  4. Dec 30, 2007 #3


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    1) look at [itex](c+ 0)\vec{v}= c\vec{v}+ 0\vec{v}[/itex]

    2) Look at [itex]c(\vec{v}+ \vec{0})= c\vec{v}+ c\vec{0}[/itex]
    3) Look at [itex](c- c)\vec{v}= c\vec{v}- c\vec{0}[/itex] showing that [itex]-c\vec{0}[/itex] is the inverse of [itex]c\vec{v}[/itex]: i.e. (-c)\vec{v}= -(c\vec{v}). Look at c(\vec{v}- \vec{v})= c\vec{v}+ c(-\vec{v})[/itex] to show that c(-\vec{v})= -(c\vec{v}).
    Last edited by a moderator: Dec 31, 2007
  5. Dec 30, 2007 #4


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    0v + 0v = (0+0)v = 0v, implies that 0v = 0.
  6. Dec 30, 2007 #5
    mine was no good?

    mathwonk why does that imply that? ah nm because you can just subtract one of the 0v from the left
  7. Dec 31, 2007 #6
    thanks guys for your insights. I will look into the given suggestions and get back to you

    btw Happy 2008 to everyone !! :)
  8. Dec 31, 2007 #7
    My question arose as a result of using 1). The part which concerned me was the [itex]0\vec{v}[/itex] because that is associated with the 1st property of vector spaces.

    However, my textbook gives me the proof for [itex]0\vec{v}[/itex] = 0 (where 0 is the additive identity of the vector space) :D, so I guess it should be okay to use 1), provided you can prove [itex]0\vec{v}[/itex] = 0.

    Any thoughts or comments ??
    Last edited: Jan 1, 2008
  9. Jan 2, 2008 #8
    Anyone have anything they like to add in regards to my previous post ??

    If not, I will mark this as solved
  10. Jan 3, 2008 #9
    Why is it that you want to avoid using property 1 when proving property 2? I don't see any reason to care about that.
  11. Jan 3, 2008 #10
    I thought properties of anything in mathematics are meant to be independent of each other. This was why I initally wanted to avoid the use of property 1 when proving property 2.
  12. Jan 3, 2008 #11
    how bout me, someone tell me if my proof is rigorous?
  13. Jan 3, 2008 #12
    In light of our discussion, I think your approach would be ok, provided you include the proof of property 1. This would make it more understandable and convincing on how you
    reached your answer.

    Also for this line of working

    [tex] (c \cdot 0)\cdot \vec{v}= 0 \cdot \vec{v}=0[/tex] by 1 again

    do you mean

    [tex] (c \cdot 0)\cdot \vec{v}= 0 \cdot \vec{v}=vec{0}[/tex]

    (where vec0 is the additive identity in the vector space)
    instead ?
    Last edited: Jan 3, 2008
  14. Jan 4, 2008 #13


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    The idea is fine, but I'd specify a specific vector v (for instance the 0 vector).

    Axioms are meant to be independent of each other -- after all, having redundant axioms serves no purpose -- but once you have a set of axioms for some structure and you start establishing some of its properties, then they definitely don't have to be independent.
  15. Jan 4, 2008 #14
    why would you specify the vector v?
  16. Jan 4, 2008 #15


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    Because you haven't qualified what v is at all. I suppose you could also say "where v is a fixed vector in our vector space".
  17. Jan 4, 2008 #16
    Axiom = Unverified (unproven) property of a structure ?
  18. Jan 4, 2008 #17


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    Axiom = generator of a theory. The theory of vector spaces contains a lot of statements; but they all have the property they are logical consequences of the vector space axioms. They are 'abstract'; they do not refer to any particular structure, mathematical or otherwise.

    They are useful becuase if you have some mathematical structure, and you:

    (1) Interpret the vector space language in that structure (for example, this inclides specifying what vector addition means in your structure)

    (2) Prove that the vector space axioms are valid in this interpretation

    Then it follows that the entire theory of vector spaces is valid in the given interpretation.
  19. Jan 5, 2008 #18

    According to wolfram mathworld

    An axiom is a proposition regarded as self-evidently true without proof

    So, based on your understanding and the mathworld definition, an axiom is basically the proposition that needs a theory to validate it ?
  20. Jan 5, 2008 #19
    no an axiom is a granted truth from which theorems are derived and proven.
  21. Jan 6, 2008 #20
    I just want to verify that I understand what an axiom is

    An axiom is basically a math-based statement which is assumed to be true. (i.e. 3 < 4,
    2+9=11, 4.0 = 0 etc....).

    A theorem is similar to an axiom, except it uses a proof process to validate it (is this correct as well ??). Since properties of anything in mathematics require a proof process to validate them, are properties the same as a theorem ?

    thanks everyone so far for your contributions, Im enjoying what Im learning from this discussion.
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