How to prove that the negative of any $\vec{v}\in V$ is unique?

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In a group $G$, the inverse is unique. Let $G$ be the group of vectors of your vector space under addition. Show this.

Now in a group $G$, if $b,c$ both invert an element $a$, i.e. $ab = ac$ = 1, then cancel $a$ to show that $b = c$. This was done in multiplicative notation but it also applies to an additive notationed group.

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Mirrana
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Mirrana

Updated on August 01, 2022

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  • Mirrana
    Mirrana over 1 year

    I can't help but come up with a solution that feels wrong. I am using a fact to prove something, when I am asked to prove that fact...

    $\vec{v}$ is a negative of $\vec{w}$ if $\vec{v}+\vec{w}=\vec{0}$.

    Assume $\vec{u}_{1}$ and $\vec{u}_{2}$ are distinct vectors such that they both are the negative of $\vec{w}$. Therefore, $\vec{u}_{1}+\vec{w}=\vec{0}$ and $\vec{u}_{2}+\vec{w}=\vec{0}$. Therefore, $\vec{u}_{1}+\vec{w}=\vec{u}_{2}+\vec{w}$, and $\vec{u}_{1}=\vec{u}_{2}$.

    Is this right? It certainly feels like I'm not doing it right, since I'm "subtracting" $\vec{w}$ from both sides of the equation to arrive at the conclusion that the two constructed vectors are equal.

    • Daniel Fischer
      Daniel Fischer about 10 years
      Right, that's a problem. A small modification however makes it work: $\vec{u}_1 = \vec{u}_1 + \vec{0} = \vec{u}_1 + (\vec{w} + \vec{u}_2) = (\vec{u}_1 + \vec{w}) + \vec{u}_2 =\dotsc$ I guess you know how it continues.
    • EuYu
      EuYu about 10 years
      I don't really see anything wrong with "subtracting $\mathbf{w}$". It is established that we have an additive inverse after all. To be absolutely explicit, we can continue as follows $$\mathbf{u}_1 + \mathbf{w} = \mathbf{u}_2 + \mathbf{w}\iff \mathbf{u}_1 + (\mathbf{w} + \mathbf{u}_1) = \mathbf{u}_2 + (\mathbf{w} + \mathbf{u}_1) \iff \mathbf{u}_1 + \mathbf{0} = \mathbf{u}_2 + \mathbf{0} \iff \mathbf{u}_1 = \mathbf{u}_2$$ Not as pretty as Daniel Fischer's approach, but it gets the job done.
    • Vedran Šego
      Vedran Šego about 10 years
      @EuYu +1, although I do think that some more writing (like what you did) is needed, beyond just "we subtract" (as the OP did). This is simply because "subtracting" does not exist as an operation, but it is actually "adding an inverse", and we're currently proving an important property of inverses, so it is not so clear that we can "subtract". I believe your addition is sufficient to make the OP's proof complete (although, I do prefer Daniel's elegant approach).
  • Mirrana
    Mirrana about 10 years
    I think this is a bit advanced of where I am in class currently. We never talked anything about groups... We only covered the basics of vector spaces and subspaces so far.
  • DiagramChasingBlues
    DiagramChasingBlues about 10 years
    Look up the definition of group and prove that your vector space is an abelian group under $+$. It would be good math practice.