# Thread: Prove that the additive inverse of a vector is unique

1. ## Prove that the additive inverse of a vector is unique

Suppose $\displaystyle V$ is a vector space over a field $\displaystyle F$. Prove that the additive inverse of a vector $\displaystyle v\in V$ is unique.

A hint provided for the question is as follows:
Suppose $\displaystyle x,y\in V$ are additive inverses of $\displaystyle v$. Prove that $\displaystyle x=y$ using the axioms of a vector space.

My work so far (uses the hint):
Suppose $\displaystyle \exists v\in V$ such that $\displaystyle x, y\in V$ are inverses of $\displaystyle v$.
Then $\displaystyle x=x+(v+y)=x+0=(x+v)+y=0+y=y$.
I'm pretty sure that my current answer is lacking substance, so I could use a hand in making it more solid.

The main issue I'm having is some of the wording. I feel as though there's something I could be missing in the question. The part on "axioms of a vector space" is one I don't entirely understand.

2. Originally Posted by Runty
Suppose $\displaystyle V$ is a vector space over a field $\displaystyle F$. Prove that the additive inverse of a vector $\displaystyle v\in V$ is unique.

A hint provided for the question is as follows:
Suppose $\displaystyle x,y\in V$ are additive inverses of $\displaystyle v$. Prove that $\displaystyle x=y$ using the axioms of a vector space.

My work so far (uses the hint):
Suppose $\displaystyle \exists v\in V$ such that $\displaystyle x, y\in V$ are inverses of $\displaystyle v$.
Then $\displaystyle x=x+(v+y)=x+0=(x+v)+y=0+y=y$.
I'm pretty sure that my current answer is lacking substance, so I could use a hand in making it more solid.

The main issue I'm having is some of the wording. I feel as though there's something I could be missing in the question. The part on "axioms of a vector space" is one I don't entirely understand.
Looks fine to me except I would switch your second and third expressions to get

$\displaystyle x=x+0=x+(v+y)=(x+v)+y=0+y=y$

You used associativity of addition, which is an axiom.

(Note: Since we never used commutativity, this proof also holds for non-abelian groups.)

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