Prove: The Additive Inverse of any vector is unique

1. Oct 6, 2015

RJLiberator

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. Oct 6, 2015

Krylov

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 $\mathbf{w} = -\mathbf{v}$.

3. Oct 6, 2015

RJLiberator

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

Thank you kindly.

4. Oct 6, 2015

Staff: Mentor

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.

5. Oct 6, 2015

RJLiberator

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?

6. Oct 6, 2015

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.

7. Oct 6, 2015

vela

Staff Emeritus
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.

8. Oct 6, 2015

RJLiberator

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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