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Problem With Change of Basis

  1. Mar 11, 2012 #1
    1. The problem statement, all variables and given/known data
    Skjermbilde_2012_03_11_kl_8_22_13_PM.png

    3. The attempt at a solution
    So first I thought to myself that the proper way of doing this problem was to construct each of the standard basis vectors as a linear combination of the basis given us. I have,

    [itex]T(1,0,0) = \frac{1}{2} T(1,0,1) + \frac{1}{2} T(1,0,-1) = (\frac{3}{2},0,1)[/itex]
    [itex]T(0,1,0) = (0,-1,0)[/itex]
    [itex]T(0,0,1) = \frac{1}{2}T(1,0,1) - \frac{1}{2} T(1,0,-1) = (\frac{3}{2},0,-1)[/itex]

    Thus I have the matrix with columns that are the images of the standard basis vectors.

    However, it seems conceivable that I should also be able to do this problem via change of basis. I have, taking the bases in matrix form,

    [itex]\left| \begin{array}{ccc}
    1&0&1 \\
    0&1&0 \\
    1&0&-1 \end{array} \right|[/itex]

    I have the transformation relative to the basis:

    [itex]\left| \begin{array}{ccc}
    3&0&0 \\
    0&-1&0 \\
    0&0&2 \end{array} \right|[/itex]

    and the inverse of the basis matrix,

    [itex]\left| \begin{array}{ccc}
    1/2&0&1/2 \\
    0&1&0 \\
    1/2&0&-1/2 \end{array} \right|[/itex]

    If I compose the matrices to find the matrix composition, I have

    [itex]\left| \begin{array}{ccc}
    1/2&0&1/2 \\
    0&1&0 \\
    1/2&0&-1/2 \end{array} \right|[/itex][itex]\left| \begin{array}{ccc}
    3&0&0 \\
    0&-1&0 \\
    0&0&2 \end{array} \right|[/itex][itex]\left| \begin{array}{ccc}
    1&0&1 \\
    0&1&0 \\
    1&0&-1 \end{array} \right|[/itex] = [itex]\left| \begin{array}{ccc}
    5/2&0&1/2 \\
    0&-1&0 \\
    1/2&0&5/2 \end{array} \right|[/itex]

    But clearly, my results for the two matrices are not identical, so what have I done wrong?
     
  2. jcsd
  3. Mar 11, 2012 #2

    vela

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    Your results for T(1,0,0) and the rest of the basis vectors are still expressed in terms of the given basis. In other words, the coordinate vector (3/2, 0, 1) corresponds to 3/2(1,0,1) + 1(1, 0,-1) = (5/2, 0, 1/2) with respect to the canonical basis.
     
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