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Proving a vector space.

  1. Oct 19, 2011 #1
    1. The problem statement, all variables and given/known data
    Let V be the set of ordered pairs (x, y) of real numbers with the operations of vector addition and scalar multiplication given by:
    (x, y) + (x', y') = (y + y', x + x')
    c(x, y) = (cx, cy)

    V is not a vector space. List one of the properties from the definition of vector space that fails to hold for V. Justify your answer.


    2. Relevant equations
    N/A


    3. The attempt at a solution

    I really do not have any idea on how to get started on this question. I've tried all 8 properties of a vector space and, at least to me, it holds for all 8 properties and should thus be a vector space, but the question explicitly says it is not. I have the answer to the question, but would like to try a different approach to proving it is not a vector space. Any help?
     
  2. jcsd
  3. Oct 19, 2011 #2

    lanedance

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    not the first one interchanges x & y, try and use that to find a counterexample... does a zero element exist such that v+0=0 for all v?
     
  4. Oct 19, 2011 #3

    HallsofIvy

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    lanedance means "v+ 0= v".

    What is (x, y)+ (0, 0)?

    You should see that (0, 0) is NOT the "0 vector" but then you have to show that no other vector can be:
    (x, y)+ (u, v)= (y+ v, x+ u)= (x, y) gives what?
     
  5. Oct 19, 2011 #4
    How exactly did you get (x, y) from (y + v, x + u)?

    As to your first question: (x, y) + (0, 0) = (y + 0, x + 0) = (y, x)

    But, (x, y) - (y, x) = (y - x, x - y) ≠ (0, 0) assuming, of course, that x ≠ y.

    Is this sufficient to prove that V is not a vector space?
     
  6. Oct 19, 2011 #5

    HallsofIvy

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    No, that only shows that (0, 0) is not the 0 vector. It doesn't mean that some other vector, a, say, does not have the property that v+ a= v for all v.

    That was why I wrote (x, y)+ (u, v)= (y+ v, x+ u)= (x, y).
    I did NOT get (x,y) from (y+ v, x+ u), I am saying that if (u, v) is the 0 vector then that must be true. Now, show that u and v cannot be independent of x and y.
     
  7. Oct 19, 2011 #6
    Ah, yes. I was thinking in terms of subspaces (where the standard zero vector exists in the vector space it is a subset of).

    Let's say (a, b) is part of the set V, then using the associativity property of addition:

    (a,b) + ((x, y) + (u, v)) = (y + v, x + u) + (a, b) = (x + u + b, y + v + a)

    (x, y) + ((a, b) + (u, v)) = (b + v, a + u) + (x, y) = (a + u + y, b + v + x)

    Since (a + u + y, b + v + x) ≠ (x + u + b, y + v + a), the associativity property of vector addition does not hold, and hence, V is not a vector space. Correct?
     
  8. Oct 19, 2011 #7

    lanedance

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    exactly, cheers
     
  9. Oct 19, 2011 #8

    lanedance

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    good enough for me
     
  10. Oct 19, 2011 #9

    lanedance

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    thw way I was hinting about the zero vector was as follows:

    say the additive identity element 0 exists, lets call it 0=(p,q)

    now take an arbitrary element (x,y), it must satisfy
    (x,y)+(p,q) = (x,y)

    but
    (x,y)+(p,q) = ((y+q), (x+p))

    for the "zero: property to be true we must have
    q=x-y
    p=y-x

    hence no universal 0=(p,q) element exists
     
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