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Linear Algebra question regarding Matrices of Linear Transformations

  1. Mar 18, 2012 #1
    1. The problem statement, all variables and given/known data

    Find the matrix representations [T][itex]\alpha[/itex] and [T]β of the following linear transformation T on ℝ3 with respect to the standard basis:

    [itex]\alpha[/itex] = {e1, e2, e3}

    and β={e3, e2, e1}

    T(x,y,z)=(2x-3y+4z, 5x-y+2z, 4x+7y)

    Also, find the matrix representation of [T][itex]^{\alpha}_{\beta}[/itex]

    2. Relevant equations

    None

    3. The attempt at a solution

    T(e1) = (2, 5, 4)

    T(e2) = (-3, -1, 7)

    T(e3) = (4, 2, 0)

    So, I got [T][itex]\alpha[/itex] = (T(e1), T(e2), T(e3))

    but for [T]β, I got [T]β=(T(e3), T(e2), T(e1))
    However, the answers in the back of the book tell me that although my order for [T]β is correct, the vectors themselves are inverted.
    ie: T(e1) = (4, 5, 2)

    Why is this? And I'm not sure how to start the second half of the question...
     
  2. jcsd
  3. Mar 18, 2012 #2

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    The columns of the matrix of a linear transformation in a given ordered basis are the coefficents of the expression of the transformation of each basis vector, in turn, written as a linear combination of that ordered basis.

    The reason I emphasised "ordered" is that in the second basis, you are given [itex]\beta[/itex] as same vectors as in [itex]\alpha[/itex], just in a different order.
    [itex]T(\beta_1)= T(e_3)= (4, 2, 0)= 0e_3+ 2e_2+ 4e_1[/itex] so the first column is
    [tex]\begin{bmatrix}0 \\ 2 \\ 4\end{bmatrix}[/tex]
     
  4. Mar 19, 2012 #3
    Thank you, that cleared up some of my problems!
     
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