Simple Complex Vector Space Proof Clarification

  • #1
RJLiberator
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Homework Statement


Let v ∈ V and c ∈ ℂ, with c ≠ 0. Prove that if cv = 0, then v = 0.


Homework Equations


Vector space axioms.

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.
 

Answers and Replies

  • #2
S.G. Janssens
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But, it makes sense to me that an inverse does exist for any possible scalar choice here.

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.
 
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  • #3
RJLiberator
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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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