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Simple Complex Vector Space Proof Clarification

  1. Oct 6, 2015 #1

    RJLiberator

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    Gold Member

    1. The problem statement, all variables and given/known data
    Let v ∈ V and c ∈ ℂ, with c ≠ 0. Prove that if cv = 0, then v = 0.


    2. Relevant equations
    Vector space axioms.

    3. The attempt at a solution

    Simple proof overall, but I have one major clarification question.

    v = 1v
    = (c^(-1)c)v
    = c^(-1) (cv)
    = c^(-1) 0
    v
    = 0

    My question is, in a complex vector space, is it safe to assume that c^(-1) exists in this proof?
    If it does, I feel very confident about this proof.
    If it doesn't then I need to do something else.

    I don't see anywhere in the vector space axioms that states the multiplicative inverse exists.
    But, it makes sense to me that an inverse does exist for any possible scalar choice here.
     
  2. jcsd
  3. Oct 6, 2015 #2

    Krylov

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    Science Advisor
    Education Advisor

    Yes indeed, because ##\mathbb{C}## is a field. More generally, one always speaks of a vector space over a field. (Real and complex numbers are the common choices, but there are others.) So the existence of ##\mathbb{c}^{-1}## for ##\mathbb{c} \neq 0## really follows from the field axioms, not from the vector space axioms.
     
  4. Oct 6, 2015 #3

    RJLiberator

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    Gold Member

    Beautiful! I was searching the vector space axioms and had them in my head. The field axioms clearly state this.

    thank you, once again, Krylov.
     
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