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Math Help - Subsets of R^3

  1. #1
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    Subsets of R^3

    I have a question about the following 2 subset of R^3

    S = ((2,2,2),(2,3,1)),
    T = ((x,y,2x-y):x,y which are elements of real numbers).

    Show S is contained within T and write down a vector in R^3 that does not belong to T.

    Can anyone help?
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  2. #2
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    if (2,2,2) is contained in T, then we have:

    2 = x
    2 = y
    2 = 2x - y = 2(2) - 2 = 4 - 2 = 2, which is true. see if you can find an x and y so that (x,y,2x-y) = (2,3,1).

    to find a vector NOT in T, pick some x, and some y, and calculate 2x-y. pick anything BUT that (value for 2x-y) for the 3rd coordinate.
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  3. #3
    MHF Contributor FernandoRevilla's Avatar
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    Any vector of S has de form



    verify



    Try the second question.


    Edited: Sorry, I didn`t see Deveno's post.
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  4. #4
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    I need to show that S is a basis for T.

    Any ideas anyone?
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  5. #5
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    Quote Originally Posted by Arron View Post
    I need to show that S is a basis for T.
    Arron, do you ever plan to show any of your own effort?
    I have not seen any in the questions you have posted.
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  6. #6
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    I get your point.

    How about the following,

    First we check that S is linearly independant.

    Suppose that

    \alpha (2,2,2)+ \beta (2,3,1)=0
    that is
    (2 \alpha +2 \beta ,2 \alpha +3 \beta ,2 \alpha + \beta )=(0,0,0)
    Equating corresponding coordinates, we obtain the simultaneous equations

    2 \alpha +2\beta =0
    2 \alpha +3\beta =0
    2 \alpha +\beta =0

    Subtracting the third from the first gives \beta =0 and substituting \beta =0 into the third gives \alpha =0, so the set is linearly independent.

    Alternatively the set {(2,2,2),(2,3,1)} is linearly indepentent, as the second vector is not a multiple of the first.

    We then check whether each vector in T can be written as a linear combination of the vectors in the set. Choosing (x,y,2x-y) as the vector in T, we seek numbers \alpha ,\beta such that
    (x,y,2x-y)= \alpha (2,2,2)+\beta (2,3,1)

    Equating corresponding coordinates, we obtain the simultaneous equations

    2 \alpha +2\beta =x
    2 \alpha +3\beta =y
    2\alpha +\beta =2x-y

    Subtracting the third equation form the first gives \beta =-x+y, and substituting this into the third gives \alpha =1/2(3x-2y).So

    (x,y,2x-y)=1/2(3x-2y)(2,2,2)+(-x+y)(2,3,1)
    and the set of vecotrs S spans T.
    Since the two conditions hold, the set S is a basis for T.
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  7. #7
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    Hi!

    Could anyone just confirm that my analysis is correct on the last post.

    Thanks
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  8. #8
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    your analysis looks fine to me.
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  9. #9
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    Thanks.

    T= {(x,y,2x-y):x,y E R}

    I found the dimension of T to be 2.
    However I am not sure how to describe it geometrically.

    Can anyone help?
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  10. #10
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    the plane containing (2,2,2) and (2,3,1)?
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  11. #11
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    And it contains (0, 0, 0). A plane is determined by three points.
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  12. #12
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    i was thinking of (2,2,2) and (2,3,1) being vectors, rather than points, and thus (implicitly) including the point (0,0,0). a better choice of words would have been: the plane through the origin...
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