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Showing Uniqueness of Elements of a Vector Space

  1. Jan 18, 2012 #1
    1. The problem statement, all variables and given/known data

    http://img267.imageshack.us/img267/8924/screenshot20120118at121.png [Broken]

    3. The attempt at a solutionWe have that X = A + B. To show that X is unique, let two such sums be denoted by X1 X2 such that X1X2. We write,

    X1 = A + B
    X2 = A + B

    The equations imply,

    X1 - A - B = 0
    X2 - A - B = 0

    Which imply,

    X1 - A - B = X2 - A - B. If we add vectors to both sides,

    X1 - A - B + A + B = X2 - A - B + A + B

    X1 + 0 + 0 = X2 + 0 + 0
    X1 = X2, which contradicts our assertion that X1X2. This shows that such an X is unique.
     
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Jan 18, 2012 #2

    lanedance

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    don't you mean A+X=B?
     
  4. Jan 18, 2012 #3

    lanedance

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    and though on the right track 3 points,
    - i think you need to justify the minus operation by invoking the existence of an additive inverse
    - eqauting to zero is an unecessary step, i would start from B=B in this case
    - you need to show both existence and uniqueness
     
  5. Jan 18, 2012 #4
    Yeah, you're totally right. I just typed X =A + B by accident. As for proving the existence of such an X, how do I begin that? I think proving uniqueness should be, with some adjustment, what I have in #1.
     
  6. Jan 18, 2012 #5

    lanedance

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    just find a vector that exists that satisfies the requirement
    A+X=B
    A+X+(-A)=B+(-A)
    X+A+(-A)=B+(-A)
    X+0=B+(-A)
    X=B+(-A)

    So somewhat obviously B+(-A) does the job
     
  7. Jan 18, 2012 #6

    lanedance

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    its a little mechanical, but each one of the above operations is justified by a vector space axiom
     
  8. Jan 18, 2012 #7
    Okay, that's pretty clear. Thanks!

    Another question I have is why the set of polynomials of degree = 2 is not a vector space. It isn't obvious to me which axiom is not being satisfied here. Could you recommend an axiom to examine more closely?
     
  9. Jan 18, 2012 #8

    lanedance

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    does it contain the zero vector?
     
  10. Jan 18, 2012 #9
    So I understood that degree = 2 indicated polynomials of the form [itex]a_0 + a_1 x +a_2 x^2[/itex] if [itex]a_0[/itex],[itex]a_1[/itex],[itex]a_2 = 0[/itex], then certainly the entire polynomial goes to zero.

    I doubt, somehow, that this is the conclusion I am supposed to arrive at. Does degree = 2 indicate rather polynomials of the form [itex]a_2 x^2[/itex]? But even if this is so, I still think I could let the coefficient be zero to get a zero vector.

    What does "degree = 2" actually mean?
     
  11. Jan 19, 2012 #10

    lanedance

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    I would interpret degree=2 as [itex]a_2\neq0 [/itex]..

    And degree<=2 as putting no constraints on a_0,a_1,a_2... but thats just my interpretation

    Its also probably biased by that i'm pretty sure all polynomials with degree<=2 is a vector space
     
  12. Jan 19, 2012 #11

    HallsofIvy

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    Also the sum of the two second degree polynomials, [itex]1+ x- x^2[/itex] and [itex]2- 2x+ x^2[/itex], is [itex]3- x[/itex] which is not a second degree polynomial.

    (The set of all polynomials of degree less than or equal to is a vector space.)
     
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