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Linear algebra: Transformations

  1. Aug 3, 2008 #1
    1. The problem statement, all variables and given/known data
    A linear transformation L : R2 -> R3 is defined by:

    [tex]L({\bf{x}}) = \left( {x_2 ,x_1 + x_2 ,x_1 - x_2 } \right)^T[/tex]

    I wish to find the matrix representation of L with respect to the orderes bases [u1, u2] and [b1, b2, b3], where

    u1 = (1,2)
    u2 = (3,1)

    and


    b1 = (1,0,0)
    b2 = (1,1,0)
    b3 = (1,1,1).

    3. The attempt at a solution
    Ok, I what I want to do is to find the matrix representation of L with respect to U and the standard basis E (I call this matrix A), and then find the matrix representation of L with respect to E and B (I call this matrix X). Then I will multiply these two matrices:

    [tex]\[
    A = \left( {\begin{array}{*{20}c}
    2 \hfill & 1 \hfill \\
    3 \hfill & 4 \hfill \\
    { - 1} \hfill & 2 \hfill \\
    \end{array}} \right)
    \]
    [/tex]

    and

    [tex]\[
    X = \left( {\begin{array}{*{20}c}
    { - 1} \hfill & 0 \hfill \\
    0 \hfill & 2 \hfill \\
    1 \hfill & { - 1} \hfill \\
    \end{array}} \right)
    \]
    [/tex].

    I believe that the matrix I am being asked for is X*A. But this wont work because of the dimensions. What am I missing here?

    Sincerely,
    Niles.
     
  2. jcsd
  3. Aug 3, 2008 #2
    If we define [itex]_u(x)[/itex] to be the coordinates of the vector x in the basis [itex]u = \{u_{1},u_{2}\}[/itex] (with x in [itex]\matbb{R}^{2}[/itex]) and [itex]_{b}(L(x))[/itex] the coordinates of the vector L(x) in the basis [itex]b = \{b_{1},b_{2},b_{3}\}[/itex], then we have:

    [tex]_b(L(x))=_{b}(L)_{u}\thinspace_{u}(x)[/tex]

    where [itex]_{b}(L)_{u}[/itex] is the matrix in which the j-th column is given by [itex]_b(L({u}_{j}))[/itex] (the coordinates of [itex]L({u}_{j})[/itex] in the basis [itex]b = \{b_{1},b_{2},b_{3}\}[/itex])

    The matrix representation of L is the matrix [itex]_{b}(L)_{u}[/itex]

    The first column would be calculated this way:

    [tex]L({u}_{1}) = (2,3,-1)^{T} = -b_{1} + 4b_{2} - b_{3}[/tex]

    (As you can check). And

    [tex]_{b}(L({u}_{1})) = (-1,4,-1)^{T}[/tex]

    gives you the first column.
     
  4. Aug 4, 2008 #3
    The way I was thinking of doing this is first finding L in the basis of (1,0), (0,1)-> (1,0,0),(0,1,0),(0,0,1) since this is easily done and then putting an transform matrix on each side of it. To the right of L you would want a matrix that goes from [tex]{u_1,u_2}[/tex] to (1,0),(0,1) and on the left side a matrix that transforms from (1,0,0),(0,1,0),(0,0,1) to [tex]{b_1,b_2,b_3}[/tex]. The L matrix should be 3 x 2. The right side transform matrix should be 2 x 2 and the left side transform matrix should be 3 x 3. Just my 2 cents as to how I think of approaching a problem like this.
     
  5. Aug 4, 2008 #4

    Defennder

    User Avatar
    Homework Helper

    You can approach the problem this way:

    Let [itex]x_1,x_2[/itex] be the coefficients of basis vectors u1, u2 and [itex]c_1,c_2,c_3[/itex] be the coefficients of the basis vectors b1,b2,b3. So by your notation, we have [tex]A(x_1\vec{u_1} + x_2\vec{u_2}) = c_1\vec{b_1} + c_2\vec{b_2} + c_3\vec{b_3}[/tex].

    So the matrix multiplication on the left can be seen as

    [tex]\left( \begin{array}{ccc} A\vec{u_1}&A\vec{u_2} \end{array} \right) \left(\begin{array}{c}x_1\\x_2 \end{array}\left) = (\vec{b_1} \ \vec{b_2} \ \vec{b_3}) \left( \begin{array}{c}c_1\\c_2\\c_3 \end{array} \right)[/tex]

    where Au1, Au2 and bn are column vectors in the matrix. Now, you should be able to find c1,c2,c3.
     
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