# Prove that the additive inverse -v of an element v in a vector space is unique.

1. Jan 18, 2010

### mammarf

1. The problem statement, all variables and given/known data

Prove that the additive inverse -v of an element v in a vector space is unique.

2. Relevant equations

For each v in V, there is an element -v in V such that v + (-v) = 0.

3. The attempt at a solution

Assume that the additive inverse is not unique and there exists different y,z in V such that
A + y = 0
A + z = 0
which implies y = -A and z = -A => y=z which is a contradiction.
Hence, the additive inverse is unique.

Correct? sumthin missing?

2. Jan 18, 2010

### CompuChip

You say: this implies y = -A and z = -A.
What is -A? Is it the element you are trying to prove is unique? The one which you just called y (or z, which should be the same, but you're trying to prove that)?

I propose looking at the expression A + y + z. You may use all the other axioms of a vector space (x + y = y + x, x + 0 = x, 0 x = x 0 = 0, ...)

3. Jan 18, 2010

### vela

Staff Emeritus
Doesn't work because this could happen:

A+y = 0 -> y+A+y = y+0 -> y = y
A+z = 0 -> z+A+z = z+0 -> z = z

which doesn't say anything about y and z.

What you do have is A+y=0 and A+z=0, so A+y=A+z, right?

4. Jan 18, 2010

### mammarf

My bad. A and -A are elements in V and I am trying to prove that -A is unique. (basically replace A with v in the relevant equations.)

5. Jan 18, 2010

### mammarf

Yes, I concur.
"A+y=A+z => y=z which is a contradiction"
Is this correct?

(u prolly noticed that i'm really weak in this, hence y i'm looking everywhere for answers.)

6. Jan 19, 2010

### vela

Staff Emeritus
You should show how you get rid of the A's.

If you assume there's more than one additive inverse of A, then it's a contradiction, so the initial assumption was wrong: there must not be more than one additive inverse. If you don't assume y and z are distinct but are additive inverses of A, you've proved that they have to be equal, so there's only one additive inverse for each A. Either way works.

7. Jan 20, 2010

### CompuChip

Err, I was talking about A + y + z, not A + y + y or A + z + z.
If A + y = 0, then A + y + z = (A + y) + z = 0 + z = z.
If A + z = 0, then A + y + z = (A + z) + y = 0 + y = y.

And I hope you agree that A + y + z is equal to itself?

8. Jan 20, 2010

### vela

Staff Emeritus
I was replying to mammarf's original post. You and I replied at the same time.