# Zero vector and -v question

1. Oct 1, 2012

### v1ru5

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

If we give X = R2 the non-standard operations

(x, y,) ⊕ (x' , y') = (x + x' - 1, y + y' +2) (vector addition)
and
k~(x, y) = (kx - k + 1, ky + 2k -2) (multiplication by scalars)

then X is a real vector space.

- What is the zero vector of X?
- If v = (x, y) is in X then what is -v?

2. Relevant equations

3. The attempt at a solution

So attempted to answer both these questions. For the first one if you make x and y = 0 then you are left with (-1, 2) so that is what I thought the zero vector of X was. For the second answer I did (-1)x(x,y) so I got -v = (-x+2, -y-4). Just wondering if someone could help me out here and confirm if these are correct.

Thanks

2. Oct 1, 2012

### LCKurtz

To get the zero vector (a,b) you need to have (x,y) ⊕ (a,b) = (x,y). Solve that for a and b.

3. Oct 1, 2012

### v1ru5

So therefore (a,b) = (0,0)? Also, is my answer for the second question correct?

4. Oct 1, 2012

### LCKurtz

No. Show your work how you got (a,b). No point in worrying about your second question until you get the first one correct because you will need -v ⊕ v = 0 and you don't have 0 figured out yet.

5. Oct 1, 2012

### v1ru5

so then would it be (x,y,) ⊕ (a,b) = (x+a-1, y+b+2) then if (a,b) = (0,0) then = (x-1, y+2). Is that how it should be? Thanks for your help!

6. Oct 1, 2012

### LCKurtz

I don't know where you are getting (0,0) from. You are trying to find the additive identity -- when you ⊕ it to a vector you get that vector back. The red tells you exactly what to do.

7. Oct 1, 2012

### v1ru5

Sorry lol im really confused. If (x,y) ⊕ (a,b) = (x,y) then (x+a, y+b) = (x,y).Is that how it should be?

8. Oct 1, 2012

### LCKurtz

No. That isn't how you do ⊕. Do it correctly. Then figure out the a and b that make it work.

On another note, one thing that may be confusing you is the fact the problem as calling what you seek the "zero vector", which you are thinking of as (0,0). But this is a different vector space, and ⊕ is not + and the "zero vector" would better be called the "additive identity". And it isn't (0,0). You can see that by taking (x,y)⊕(0,0) and seeing you don't get (x,y).

9. Oct 1, 2012

### v1ru5

Sorry I am really trying to understand this but I can't :(

The ⊕ symbol is basically implying vector addition. So if you want me to do (x,y) ⊕ (a,b) = (x,y) it would basically be (x+a, y+b) = (x,y). Is that right?

Sorry for not understanding but I am new to all this linear algebra stuff. Everything else is pretty easy.

10. Oct 1, 2012

### LCKurtz

NO!! (x,y) ⊕ (a,b) is not (x+a, y+b). Your first post tells how to do ⊕. Look at your formula above. It is not standard vector addition. Use that formula.

11. Oct 1, 2012

### v1ru5

Oh okay, so like this:

(x, y,) ⊕ (a , b) = (x + a - 1, y + b +2)

then a = 1 - x and b = -2 - y?

12. Oct 1, 2012

### LCKurtz

No. Do exactly what I suggested in post #2, which I have quoted above.

13. Oct 1, 2012

### v1ru5

Okay so this is what you quoted:

(x, y,) ⊕ (x' , y') = (x + x' - 1, y + y' +2)

To find the zero vector (a, b), I need to do this (x, y) ⊕ (a, b). So basically I am replacing (x', y') with (a, b) which would give me

(x, y) ⊕ (a, b) = (x + a - 1, y + b - + 2)

Like that?

14. Oct 1, 2012

### LCKurtz

Man, you are making this way too difficult. What part of (x,y) ⊕ (a,b) = (x,y) don't you understand? It is an equation with two sides and an = sign. And you know how to calculate the left side but you keep ignoring the right side.

15. Oct 1, 2012

### v1ru5

So then (x + a - 1, y + b +2) = (x, y)?

16. Oct 2, 2012

### LCKurtz

and so...?

17. Oct 2, 2012

### v1ru5

(x + a - 1, y + b +2) = (x, y)

x + a - 1 = x
a - 1 = 0

y + b + 2 = y
b + 2 = 0

so (a-1, b+2) is the zero vector?

18. Oct 2, 2012

### LCKurtz

Read my post #2 again. What does it say to do with the equation?

19. Oct 2, 2012

### v1ru5

Oh so a = 1 and b = -2 so (1, -2) is the zero vector?

20. Oct 2, 2012

### LCKurtz

Now, you don't have to ask me that. I want you to check it yourself. If it is the additive identity then if you ⊕ it with any vector (c,d) you should get (c,d) back. That's what an additive identity does. Does (1,-2) do that?