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Homework Help: Proofs using axioms

  1. Sep 9, 2012 #1
    Hey all,

    So I'm just starting a course in linear algebra, but I don't have much experience with proofs. This problem has been giving me some difficulty.

    So we have a scalar "a" and vector x. V is a linear space, and x is contained in V. I have to show that if ax=0, where 0 is the zero element of V, then either a=0 (scalar) or x=0 (vector).

    However, I must do the proof using the following axioms. I will briefly summarize them.

    1. for every v1 and v2 there is a unique element in V equal to the sum of v1 and v2
    2. same deal as above but with multiplication of a scalar and a vector
    3. v1+v2=v2+v1
    4. (v1+v2)+v3= v1 + (v2+v3)
    5. existence of the element 0 (v+0=v)
    6. v + (-1)v = 0
    7. for scalars a and b, a(bv)=(ab)v
    8. a(v1+v2)=av1 + av2
    9. same deal as 8 but with one vector being multiplied by (a+b)
    10. 1v=v

    I was messing around with dividing things, but I don't think I can relate it to the axioms
  2. jcsd
  3. Sep 9, 2012 #2


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    Hey TrapMuzik and welcome to the forums.

    Do you assume the vector space axioms hold or do you have to prove them? (It looks like you assume them since you are talking about a general space).

    The first part is just using the first axiom directly. The axioms are given here:


    A lot of what you are saying follows directly from the definitions. For example the one involving three terms is just a special case of the one with two terms.
  4. Sep 9, 2012 #3
    I should add that there are 3 proofs that I believe build on each other and may be of use for this problem.

    The first is a proof that 0x= the zero vector

    Let z=0x
    z+z= 0x + 0x = (0+0)x (axiom 8) = 0x = z
    so we have z+z=z, and by axiom 5 we get that z=0

    The second proof is for a0=0
    It's the same method as used for the above proof.

    The third proof is to show that (-a)x=-(ax)=a(-x)
    let z = (-a)x
    z + ax = (-a)x + ax = (-a+a)x = 0x = 0 (using axiom 9)
    From this we get z= -(ax).
    In the same way, if we add a(-x) to ax and use axiom 8/what we proved in the second proof, we get a(-x)=-(ax)

    Okay so I think we have to somehow build off of these proofs for the next one...
    I'm sure there is a simple proof that is evading me. Any tips/ideas?
  5. Sep 9, 2012 #4
    Hi chiro,

    I may have to go about showing that this space meets all of the axioms, now that you mention it. Thanks for your help!
  6. Sep 9, 2012 #5


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    The zero vector is part of the axioms for a general vector space so you can use that for free.

    If you have to prove the axioms for a general vector space, then that's a different matter altogether. Do you have to prove the rest of the axioms for the vector space given the initial ones as opposed to proving a set of specific results where you can assume all the base axioms?
  7. Sep 9, 2012 #6
    For this question we are assuming the base axioms. There are later problems where we have to define addition, check closure, etc.
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