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Vector space

  1. Jul 11, 2012 #1
    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?

    Thanks in advance!
     
  2. jcsd
  3. Jul 11, 2012 #2
    Does anyone have any thoughts?
     
  4. Jul 11, 2012 #3

    HallsofIvy

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    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.)
     
  5. Jul 11, 2012 #4
    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.
     
  6. Jul 11, 2012 #5

    Zondrina

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    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[itex]\ni[/itex]V ( Which is equivalent to saying V≠∅ )
    2. If u,w[itex]\ni[/itex]V then u+w[itex]\ni[/itex]V
    3. If u[itex]\ni[/itex]V and α[itex]\ni[/itex]F then αu[itex]\ni[/itex]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.
     
  7. Jul 11, 2012 #6
    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.:confused:
     
  8. Jul 11, 2012 #7
    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'.
     
  9. Jul 11, 2012 #8

    Zondrina

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

    V = {(x,y)[itex]\ni[/itex]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)[itex]\ni[/itex]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)[itex]\ni[/itex]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
  10. Jul 11, 2012 #9

    HallsofIvy

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    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.
     
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