Vector space

1. Jul 11, 2012

greendays

Hi everyone,

I am having difficulty solving the following question:

Prove that the set V=R^2 with addition defined by (x,y)+(x'+y')=(x+x'+1, y+y') and scaler multiplication by k(x,y)=(kx+k-1, ky) is a vector space, find-(x+y) and the zero vector in this vector space.

the following was my attempt:

-(x,y)=(-1)(x,y)=((-1)x+(-1)-1), (-1)y)=(-x-1, -y) (Is this correct?)

Also, how to find the zero vector in this vector space? I do not know where to start for this one.

Any thoughts?

2. Jul 11, 2012

greendays

Does anyone have any thoughts?

3. Jul 11, 2012

HallsofIvy

Staff Emeritus
Do not "bump" a thread after just a few hours. People are not sitting around waiting for you to post a question!

Do you understand the basic definitions? In order to show that something is or is not a vector space you must show that all of the "axioms" for a vector space are true. Those are things like "associativity" of addition, the existence of a "zero" vector, the existence of a "negative" for each vector, etc. Do you know those axioms?

One of the axioms is the multiplying a vector, v, by the scalar -1 gives the "negative" (additive inverse) of v. The scalar multiplication defined here is k(x, y)= )=(kx+k-1, ky) so with k= -1, that would be (-x- 1- 1, -1y) which is NOT (-x-1, -y) so what you did there is not correct.

What is the additive identity? Denoting the additive identity, for now, as (a, b), we must have (x, y)+ (a, b)= (x, y) for all (x, y). Using this definition of addition, that is (x, y)+ (a, b)= (x+ a+ 1, y+ b) and for that to be equal to (x, y) we must have x+a+ 1= x, y+ b= y. What are a and b?

What is the additive inverse for a vector, (x, y)? (With this definition of "addition" of course.)

4. Jul 11, 2012

greendays

Clearly, that was a typo with my first post, "((-1)x+(-1)-1), (-1)y)=(-x-1, -y)", so the result should be (-x-2, -y), not (-x-1, -y)
so I guess I was on the right track.

When you mention " Denoting the additive identity, for now, as (a, b), we must have (x, y)+ (a, b)= (x, y) for all (x, y). Using this definition of addition, that is (x, y)+ (a, b)= (x+ a+ 1, y+ b) and for that to be equal to (x, y) we must have x+a+ 1= x, y+ b= y. What are a and b?" are you trying to find the zero vector in this vector space? I do not quite understand the rationale of doing this step, can you explain it a little more?

PS:I do understand the BASIC definitions, but as a beginner for this course, to understand a concept is one thing, and to apply it into a questions is another. That is why I came here to seek help. Sometimes, some questions may be a piece of cake for you, but could be very difficult for another, that is why you are a mentor, and other people need your help. However, as a mentor, you should also be careful with your choice of language, and/or your tune of voice, otherwise, many students will be easily discouraged by your post, which I suppose is not the intension of this forum. Hope you understand.

5. Jul 11, 2012

Zondrina

If you're trying to prove it's a vector space, you have to show all of the axioms hold individually ( All 10 of them ).

Something easier you can do though is employ the subspace test which has three easy conditions for you to prove ( Which is equivalent to showing that the 10 properties hold ).

For example, suppose you want to prove V is a vector space ( or more appropriately a subspace ) over a field F ( Like R^2 or R^N ). Then if :

1. 0$\ni$V ( Which is equivalent to saying V≠∅ )
2. If u,w$\ni$V then u+w$\ni$V
3. If u$\ni$V and α$\ni$F then αu$\ni$V

Sadly though if you're trying to prove it's a vector space on its own then you have to run through ALL of the axioms.

Hope that helps.

6. Jul 11, 2012

greendays

Hi, Zondrina,

Thanks for your input.
The original question was "Prove that the set V=R^2 with addition defined by (x,y)+(x'+y')=(x+x'+1, y+y') and scaler multiplication by k(x,y)=(kx+k-1, ky) is a vector space, find-(x+y) and the zero vector in this vector space."
I guess the question was to find -(x+y) and the zero vector (zero with an arrow on it).
But I do not know how to find the zero vector here.

7. Jul 11, 2012

Muphrid

You have a definition of addition. You should be able to solve for x', y' such that (x,y) + (x', y') = (x,y). Remember that x and y are constants for the purposes of this equation, not variables. You're not solving for them, just x' and y'.

8. Jul 11, 2012

Zondrina

So just to get things straight here you're working with :

V = {(x,y)$\ni$R2 | (x,y)+(x',y')=(x+x'+1, y+y'), α(x,y)=(αx+α-1, αy)}

and you are having trouble with certain axioms. By the looks of it you're asking about the additive identity and additive ineverse yes?

Additive identity : You seek a (a,b)$\ni$V such that (x,y)+(a,b) = (x,y). By the looks of it your identity in this vector space would be (-1,0), but i leave you to verify that.

Additive inverse: You seek a (a,b)$\ni$V such that (x,y)+(a,b) = (-1,0). Use the same logic i used to find the additive identity to find the additive inverse. Whatever (a,b) you find should add on to your (x,y) and spit out your zero vector ( aka your additive identity ).

Hope this helps.

Last edited: Jul 11, 2012
9. Jul 11, 2012

HallsofIvy

Staff Emeritus
The zero vector (or "zero" anything in a mathematical structure), e, is the additive identity. It has the property that for any vector, v, v+e= e+ v= v.

Here, your "vectors" consist of pairs of numbers so if you call the "0 vector", (a, b), you must have, for any vector (x, y), (a, b)+ (x, y)= (a+ x+1, b+ y)= (x, y) so we must have a+ x+ 1= x and b+ y= y. Solve those for a and b.