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

[mammarf]

Homework Statement

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

Homework Equations

Additive Inverse in V

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

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?

 

[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, ...)

 

[vela]

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?

 

[mammarf]

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, ...)

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.)

 

[mammarf]

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?

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.)

 

[vela]

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.

 

[CompuChip]

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?

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?

 

[vela]

Err, I was talking about A + y + z, not A + y + y or A + z + z.

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