Proof Vector Spaces: Unique Vector Satisfying "u + 0 = u

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SUMMARY

The discussion focuses on proving the uniqueness of the zero vector in vector spaces, specifically addressing the equation "u + 0 = u". Participants clarify that while the existence of a zero vector is an axiom in vector spaces, its uniqueness can be established through contradiction. The key argument presented is that if u + 0_1 = u + 0_2, then it follows that 0_1 must equal 0_2, confirming the zero vector's uniqueness. The conversation highlights the relationship between vector spaces and group theory, noting that uniqueness of the identity element is an axiom in groups but can be derived in vector spaces.

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soul
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Hi eveyone,

I was trying to prove that for the vector spaces, there is a unique vector that satisfy "u + 0 = u" and I used contradiction technique. The last point that I reached is u + 0_1 = u + 0_2. However, I don't know whether I can say 0_1 = 0_2 after this statement or there are some other operations that I must do (like this statement needs a proof as well?).

Thank you.
 
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? Why do you need to prove that? Isn't that an axiom that every vector space has to satisfy? Namely that every vector space has a unique zero vector? What axioms do you start off with?

You can "prove" this by noting that along with your last step, -u also exists in the same vector space.
 
Sorry, I guess I wrote my question wrong. I was trying to prove there is a unique identitiy element in summation and what I did is to select two different vectors and at the end of it, to show they are the same. I used the axiom in the question.
 
Uniqueness of the identity is an axiom in groups but can be proved in a vector space.

In any vector space, there exist additive inverses and addition is commutative. Add the additive inverse of u to both sides of your equation.
 
HallsofIvy said:
Uniqueness of the identity is an axiom in groups but can be proved in a vector space.

In any vector space, there exist additive inverses and addition is commutative. Add the additive inverse of u to both sides of your equation.

Is this true? Almost every group theory book I have looked at proves the uniqueness of the identity as a theorem.
 
HallsofIvy said:
Uniqueness of the identity is an axiom in groups but can be proved in a vector space.

In any vector space, there exist additive inverses and addition is commutative. Add the additive inverse of u to both sides of your equation.

The axioms say there must exist a zero vector. It does not say it is unique or must be unique. You prove that it is unique if there exists such a vector.
 
It is impossible to reply. You do not said definition of the vector space, and do not said about preceding procedure.
 

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