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Prove: The Additive Inverse of any vector is unique

  1. Oct 6, 2015 #1

    RJLiberator

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    1. The problem statement, all variables and given/known data
    Let v, w, ∈ V. Prove that if v + w = 0, then w = -v.
    V is a complex vector space.


    2. Relevant equations
    Axioms of a vector space.

    3. The attempt at a solution

    So, this solution was pretty easy to come up with.
    My question is, have I proven that w = -v or have I simply proven that -v is unique? Let's see:

    Proof by contradiction:
    Suppose w and w' are additive inverses of v.
    w
    = w + 0 (By zero addition, axiom)
    w = w + (v + w') (by giving assumption)
    w = (w + v) + w' (by axiom 2 of vector spaces)
    w = w' (by giving assumption, w is additive inverse of v).

    Here, we have shown that the additive inverse of v is unique. But does this mean that w = -v as the question seems to state?

    Thanks.
     
  2. jcsd
  3. Oct 6, 2015 #2

    Krylov

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    Yes, because from ##\mathbf{v} + \mathbf{w} = \mathbf{0}## it follows that ##\mathbf{w}## is an additive inverse of ##\mathbf{v}##, hence (by uniqueness) the additive inverse of ##\mathbf{v}##. This additive inverse is denoted by ##-\mathbf{v}##, so indeed ##\mathbf{w} = -\mathbf{v}##.
     
  4. Oct 6, 2015 #3

    RJLiberator

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    Excellent, that was the clarification I was looking for here.

    Thank you kindly.
     
  5. Oct 6, 2015 #4

    Mark44

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    Your post is not as clear as it should be. From the thread title, you are supposed to prove that a given vector has a unique additive inverse. The problem statement is to prove that if v + w = 0, then w = -v. Those are not the same thing.

    In your proof by contradiction you start by assuming that w and w' are additive inverses. To get the contradiction you're looking for, you need to assume that ##w \ne w'##. In the last line of your proof, you have w = w' -- that is the contradiction you're looking for.
     
  6. Oct 6, 2015 #5

    RJLiberator

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    So here is my statement for proof by contradiction:

    Suppose w and w' are additive inverses of v, but w ≠ w'.

    That last addition makes it more clear what I am trying to show?

    I, too, saw this as a problem. The official question asked: "Let v, w, ∈ V. Prove that if v + w = 0, then w = -v. (In other words, the additive inverse of any vector is unique.)

    This is why I made this topic, I don't know if what I have shown by contradiction was what I needed to show. I know that what I've shown is that the additive inverse of a vector is unique, but have I truly shown that w=-v?
     
  7. Oct 6, 2015 #6

    Mark44

    Staff: Mentor

    Then it seems to me that the problem, as stated, is unclear. The first part (prove that w = -v) is showing that w and v are additive inverses. The second part is proving that the additive inverse is unique.

    In a completely unrelated problem, just to show the difference between existence and uniqueness, suppose the question is: Find a 2x2 matrix A such that A2 = 0. An obvious answer is
    $$A = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$$
    Clearly, a solution to the equation A2 = 0 exists, but is this solution unique? No, since
    $$A = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}$$
    is also a solution to A2 = 0, among many others.
     
  8. Oct 6, 2015 #7

    vela

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    I think your original proof is fine; it's just not a proof by contradiction.

    You showed that if w and w' are additive inverses of v, then w=w'. That essentially proves the uniqueness of the additive inverse. You could add a follow-up step by taking w' = -v, which you know exists because V is a vector space, so w = -v.
     
  9. Oct 6, 2015 #8

    RJLiberator

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    Thanks for the discussion on this.

    I will re-check my wording on the contradiction.
    And I have added w' = -v at the end as well.
     
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