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Math Help - Change of base

  1. #1
    Member Jones's Avatar
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    Change of base

    Hi,

    I've just started Linear Algebra
    I have one question i'd like some help with.

    The line l Oe_{1}e_{2 is described by the equation 2x_{1} + 3x_{2} = 0

    Find an equation for the line in OeŽ_{1}eŽ_{2}
    if ~~\begin{cases} eŽ_{1} = \frac{1}{13}(5e_{1} + 12e_{2}) \\<br />
eŽ_{2} =\frac{1}{13}(-12e_{1} + 5e_{2}) \end{cases}

    Any advice?
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  2. #2
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    suppose we knew how to transform e'-coordinates into e-coordinates, by using some matrix A. it stands to reason that A^-1 would translate e-coordinates to e'-coordinates.

    now the matrix that sends e'-coordinates to e-coordinates is very easy to write down. in e'-coordinates, e1' is (1,0). and for ANY 2x2 matrix A, A(1,0) is the first column of the matrix. similarly, in e'-coordinates, e2' is (0,1), and A(0,1) is the second column of the matrix A. so the first column of A is A(e1') = (5/13, 12/13) (in e-coordinates). and the second column of A is e2' in e-coordinates, or (-12/13,5/13) so A =

    [5/13 -12/13]
    [12/13 ..5/13]

    by a standard computation, A^-1 =


    [ 5/13 12/13]
    [-12/13 5/13]

    so in e' coordinates, e1 = (5/13,-12/13) = (5/13)e1' - (12/13)e2', while e2 = (12/13,5/13) = (12/13)e1' + (5/13)e2'.

    in e'-coordinates the points (x1,x2) that lie on the line now have coordinates ((5/13)x1+(12/13)x2, (5/13)x2 - (12/13)x1)), so the equation for our line becomes:

    2(5/13)x1 + 2(12/13)x2 + 3(5/13)x2 - 3(12/13)x1 = 0, or upon simplfying:

    -2x1 + 3x2 = 0
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  3. #3
    Member Jones's Avatar
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    Hm, are you sure about this?

    The key to this question says it's 46x_{1} - 9x_{2}
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  4. #4
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    2x_1+ 3x_2= 0 is the same as saying (x_1e_1+ x_2e_2)\cdot(2e_1+ 3e_2)= 0. That is, that the line is perpendicular to the vector 2e_1+ 3e_2

    Given that e_1'= \frac{5}{13}e_1+ \frac{12}{13}e_2 and e_2'= -\frac{12}{13}e_1+ \frac{5}{13}e_2, then the matrix having those coefficients as columns,
    \begin{pmatrix}\frac{5}{13} & -\frac{12}{13} \\ \frac{12}{13} & \frac{5}{13}\end{pmatrix}
    will change a vector in e_1' and e_2' components to the same vector in e_1 and e_2 components. What we want to do is change the vector 2e_1+ 3e_2 to e_1' and e_2' components. We can do that by using the inverse matrix which happens to be
    \begin{pmatrix}\frac{5}{13} & \frac{12}{13} \\ -\frac{12}{13} & \frac{5}{13}\end{pmatrix}

    So we are saying that our vector, in x_1'e_1'+ x_2'e_2' components must be perpendicular to
    \begin{pmatrix}\frac{5}{13} & \frac{12}{13} \\ -\frac{12}{13} & \frac{5}{13}\end{pmatrix}\begin{pmatrix}2 \\ 3\end{pmatrix}= \begin{pmatrix}\frac{46}{13} \\ \frac{-9}{13}\end{pmatrix}

    That is, \left(\frac{46}{13}e_1'- \frac{9}{13}e_2'\right)\cdot\left(x_1'e_1'+ x_2'e_1')= \frac{46}{13}x_1'- \frac{9}{13}x_2'= 0

    Now, multiply both sides of that equation by 13.
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  5. #5
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    Essentially the same thing but slightly different point of view:
    solve e_1'= \frac{5}{13}e_1+ \frac{12}{13}e_2, e_2'= -\frac{12}{13}e_1+ \frac{5}{13}e_2 for e_1 and e_2 in terms of e_1' and e_2'. (That is equivalent to finding the inverse matrix above.) Once you have done that, replace e_1 and e_2 in 2e_1+ 3e_2 by their expression in terms of e_1' and e_2' to find the vector our x_1'e_1'+ x_2'e_2' must be perpendicular to.
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  6. #6
    Member Jones's Avatar
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