1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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)


    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:

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


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

    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?

  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:


    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


    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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook