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Math Help - Find the general relationship between coordinates

  1. #1
    Member
    Joined
    Jan 2011
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    Find the general relationship between coordinates

    Hi, this is the question from My book, i Will show
    My progress aswell

    Problem: A vector u has got the coordinates
    (1,2) in Two different basis (e1,e2) and (f1,f2).
    We know That f1-f2 = e1+e2. Find the
    General relationship between the coordinates
    In both of the basis


    My progress

    First of all.

    I use Two formulas

    (1) y=t(inverse)*x. T = change of basis Matrix
    (2) x=t*y

    So i try to set up two different change of basis
    Matrices one for the basis of e and with f
    To be exact i mean e1+2e2 and F1+2f2

    So i use these values to create Two different matrices

    T(f)= 1 1
    -1 2

    And t(e) = 1 1
    1 2


    I have written the basis values from their
    Rows to columns and placed them in the
    First columns

    I need the general relationship
    So i assume That i need the inverse
    Of the f:s basis and because formula
    With the y has got t inverse
    Y represents f basis and i assume
    That i need to invert the matrix
    Of t(f)


    Through gaussian elimination
    Or determinants i get inverse of t(f)

    1/3. * 2 -1
    1 1

    How can i proceed to get
    The general relationship between
    The coordinates in the Two basis?


    Can someone Solve this for me


    The right answer should be


    (y1,y2) = (x1, -4*x1 + 3*x2)
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  2. #2
    MHF Contributor

    Joined
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    Tejas
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    Re: Find the general relationship between coordinates

    i don't understand where you are getting those two matrices. suppose T is the matrix that turns the f-basis coordinates into e-basis coordinates. so T =

    [a b]
    [c d], and we know that:

    a+2b = 1
    c+2d = 2, since T([1,2]f) = [1,2]e.

    we also know that:

    T([1,-1]f) = [1,1]e (because f1-f2 = e1+e2).

    thus a+b = 1 and c-d = -1.

    from the two equations involving a and b, we have b = 0, so a = 1. from the two equations involving c and d, we have:

    3d = 1, so d = 1/3, so c = 4/3. that gives us T, but what we really WANT is T-1, which is found to be:

    T^{-1} = \begin{bmatrix}1&0\\-4&3 \end{bmatrix}.

    this, of course, means that:

    T-1(x1,x2) = (x1,-4x1+3x2).

    it is easy to check that (1,2) is invariant under T-1:

    T-1(1,2) = (1,(-4)(1) + (3)(2)) = (1,-4+6) = (1,2) and that:

    T-1(1,1) = (1,(-4)(1) + (3)(1)) = (1,-4+3) = (1,-1), as required.
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