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Math Help - Basis problem , finding coordinates

  1. #1
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    Basis problem , finding coordinates

    Hi , the problem is written like this:
    In a basis (e1,e2,e3) in r^3
    F1=(1,0,1) , F2 = (0,1,1) f3 = (1,1,0)
    Show That (F1,f2,f3) is also a basis in r^3 and give the relationship Between
    the coordinates in both of the basis. Determine the coordinates for e1 + 2*f3



    My progress:

    Well first i use two fornulas

    Y= t^-1 * x (1)

    This is the formula to find coordinates of f basis
    Where y is the basis f:s coordinates , and where t^-1
    is the inverse of the change of basis matrix, and x is. The coordinates which belong
    To the basis e

    X= t*y. (2) T is the change of basis matrix


    So

    First of all i put up t

    T is. 1 0 1
    0 1 1
    1 1 0

    I have put the rows as columns


    I find t through gauss elimination and get

    T inverse. Is

    1/2. *. 1 -1 1
    -1 1 1
    1 1 -1


    The NeXT step is finding the coordinates

    So i begin with formula. (1)

    X = e1. And because. E2 and E3 is 0

    Therefore the coordinates for x Will be
    (1,0,0)

    I use x And multiply it with t inverse
    And get. (1/2 , -1/2 , 1/2). This is wrong according to the answer
    It should be. (1/2, -1/2, 5/2) this in (e)

    NeXT i go on using formula (2)

    I multiply t with 2f3

    And get. (2,2,0)

    I add (2,2,0) + (1,0,0) = (3,2,0) this is correct



    What am i doing wrong? Is there better
    Way to Solve this problen , pls aid me

    Anywa
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  2. #2
    MHF Contributor

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    Tejas
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    Re: Basis problem , finding coordinates

    ok, we have two sets:

    B = \{e_1,e_2,e_3\} = \{(1,0,0),(0,1,0),(0,0,1)\} and

    B' = \{f_1,f_2,f_3\} = \{(1,0,1),(0,1,1),(1,1,0)\}.

    the first task is to show that B' is indeed a basis:

    since we know \mathrm{dim}_{\mathbb{R}}(\mathbb{R}^3) = 3, it suffices to show that B' is linearly independent.

    so suppose c_1f_1 + c_2f_2 + c_3f_3 = (0,0,0), that is:

    c_1(1,0,1) + c_2(0,1,1) + c_3(1,1,0) = (c_1,0,c_1) + (0,c_2,c_2) + (c_3,c_3,0)

     = (c_1+c_3,c_2+c_3,c_1+c_2) = (0,0,0). then:

    c_1+c_3 = 0
    c_2+c_3 = 0
    c_1+c_2 = 0

    hence c_3 = -c_1 = -c_2, so c_1 = c_2 thus c_1+c_2 = 2c_1 = 0 \implies c_1 = 0 and thus c_2 = c_3 = 0.

    so B' is indeed linearly independent and thus a basis for \mathbb{R}^3.

    now if T is the matrix that sends [f_i]_{B'} \to [f_i]_B (so that T([x]_{B'}) = [x]_B),

    then clearly the matrix that turns B-coordinates into B'-coordinates is T^{-1}.

    you indeed have found T correctly, and so now we must find T-1.

    we COULD use row-reduction to do this, but an easier way is to express the e_i as linear combinations of the f_i.

    now (1,0,0) = (1/2)(1,0,1) + (-1/2)(0,1,1) + (1/2)(1,1,0)

    (0,1,0) = (-1/2)(1,0,1) + (1/2)(0,1,1) + (1/2)(1,1,0) and

    (0,0,1) = (1/2)(1,0,1) + (1/2)(0,1,1) + (-1/2)(1,1,0), so:

    T^{-1} = \begin{bmatrix}\frac{1}{2}&-\frac{1}{2}&\frac{1}{2}\\ -\frac{1}{2}&\frac{1}{2}&\frac{1}{2}\\ \frac{1}{2}&\frac{1}{2}&-\frac{1}{2} \end{bmatrix}

    this agrees with what you found.

    now B-coordinates are just "the usual cooordinates" that is: [(a,b,c)]_B = (a,b,c).

    so the B-coordinates of e_1 + 2f_3 are: (1,0,0) + 2(1,1,0) = (1,0,0) + (2,2,0) = (3,2,0).

    however, the B'-coordinates of e_1 + 2f_3 will be:

    T^{-1} \begin{bmatrix}3\\2\\0 \end{bmatrix}= \begin{bmatrix}\frac{3}{2}-\frac{2}{2}\\-\frac{3}{2}+\frac{1}{2}\\\frac{3}{2}+\frac{2}{2} \end{bmatrix} = \begin{bmatrix}\frac{1}{2}\\ -\frac{1}{2}\\ \frac{5}{2}\end{bmatrix},

    so the answer you were given is correct (a sign error perhaps?)
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  3. #3
    Member
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    Jan 2011
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    Re: Basis problem , finding coordinates

    Thank you man, the way you Solve basis and coordinates with, in which book can i read more about it, which book would u recommend, im using swedish book
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