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On the axioms of vector space

  1. Jun 3, 2013 #1
    Hello, I'd like to make a, probably stupid, question regarding the axioms that define a vetor space. Among them, there are the axioms:

    λ[itex]\cdot[/itex](μ[itex]\cdot[/itex]X) = (λμ)[itex]\cdot[/itex]Χ (1) and 1[itex]\cdot[/itex]Χ=Χ (2)

    for all λ,μ in the field and for all X in the vector space, where 1 is the identity of the multiplication on the field and "[itex]\cdot[/itex]" indicates the multiplication of a scalar with a vector. So, my question is: can axiom (2) derived as a special case of (1) and, therefore, is not really an axiom?

    I ask this because, setting λ = μ = 1 in (1), we get:

    1[itex]\cdot[/itex](1[itex]\cdot[/itex]X) = 1[itex]\cdot[/itex]Χ

    Since X is an abitrary vector, 1[itex]\cdot[/itex]Χ will also be an arbitrary vector, say Y. Then the above equation reads:

    1[itex]\cdot[/itex]Y = Y

    for all Y in the vector space. But this is "axiom" (2)! I understand that, since both axioms are used in the standard definition of vector space, they should indeed be indepentend, despite the above objection. So what is the flaw in my thought about this?
  2. jcsd
  3. Jun 3, 2013 #2


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    As far as showing it for all vectors in the vector space, is it obvious that if I give you an arbitrary vector Y that you can find some vector X such that 1⋅X = Y?

    In fact here's a simple counterexample: Let my "vector space" be R2 with regular addition and with scalar multiplication defined by

    [tex] \lambda(a,b) = (\lambda a, 0)[/tex]

    Let's check the other axioms. Since addition is untouched all the addition only axioms are still true. For scalar multiplication:
    [tex] (\lambda+\mu)(a,b) = \left( (\lambda+\mu)a,0 \right) = (\lambda a,0) + (\mu a, 0) = \lambda(a,b) + \mu(a,b) [/tex]
    Other distributivity:
    [tex] \lambda\left( (a,b) + (c,d) \right) = \lambda(a+c,b+d) = (\lambda a+\lambda c, 0) = (\lambda a,0) + (\lambda c,0) = \lambda(a,b) + \lambda(c,d) [/tex]
    Associativity of multiplication:
    [tex]\mu \left( \lambda(a,b) \right) = \mu (\lambda a, 0) = (\mu \lambda a, 0) = \left( \mu \lambda \right) (a,b) [/tex]

    So great! We have ourselves a good ol' fashion vector space
    Last edited: Jun 3, 2013
  4. Jun 3, 2013 #3

    D H

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    The flaw is that Y is not an arbitrary vector in the space. It is a member of the subset of the vector space that can be expressed as [itex]1 \cdot X[/itex], where X is some member of the vector space. What axiom (2) adds is that it says all members of of the vector space can be expressed in this form, and in particular, [itex]1 \cdot X = X[/itex].
  5. Jun 3, 2013 #4
    So, that would be true if the function:

    1[itex]\cdot[/itex] : V→V , 1[itex]\cdot[/itex]:X→1[itex]\cdot[/itex]X

    is surjective? But this is an extra axiom, more complicated than it's substitute (that is (2)). Better keep (2) as it is... :smile:

    Thank's for your replies!
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