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Linear Transformation check

  1. Nov 17, 2011 #1
    1. The problem statement, all variables and given/known data

    L:R3->R3 is a dialiation by a factor of 3 of points in the plane W given by the equation z = 0 and a contraction along the line L = span({(1,0,0)}) by a factor of 3.
    Find [L], but I'm mainly concerned with finding L(e31)


    2. Relevant equations

    z = 0
    L(1,0,0) = (3,0,0)
    L(0,1,0) = (0,3,0)
    L(span(1,0,1)) = (1/3,0,1/3)


    3. The attempt at a solution

    L(e1) = a(1,0,0) + b(0,1,0) + c(1,0,1)
    e1 = aL(1,0,0) + bL(0,1,0) + cL(1,0,1)
    e1 = a(3,0,0) + b(0,3,0) + c(1/3,0,1/3)
    e1 = [(3,0,0),(0,3,0),(1/3,0,1/3)][a,b,c]
    [(1/3,0,0),(0,1/3,0),(-1/3,0,3)]e1 = [a,b,c]
    [(1/3,0,0),(0,1/3,0),(-1/3,0,3)][1,0,0] = [a,b,c]
    [a,b,c] = [1/3,0,0]

    Can anyone check my work?
     
  2. jcsd
  3. Nov 17, 2011 #2

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Would you please recheck the problem? The line, L, is in the plane, W. You can't "dilate by a factor of 3" and "contract by a factor of 3" the same points!

    Aha! L is spanned by (1, 0, 1), not (1, 0, 0). But you don't mean to say "L(span(1,0,1))= (1/3,0,1/3)" What is true is that L(1, 0, 1)= (1/3, 0, 1/3) and that L(a, 0, a)= (a/3, 0, a/3) for any a.

    Looks like a good start- but you don't want just "a b c", you are looking for a 3 by 3 matrix. In a slightly better notation (using LaTeX)
    [tex][L]= \begin{bmatrix}a & b & c \\ d & e & f \\ g & h & i\end{bmatrix}[/tex]
    So
    [tex][L]e_1= \begin{bmatrix}a & b & c \\ d & e & f \\ g & h & i\end{bmatrix}\begin{bmatrix}1 \\ 0 \\ 0 \end{bmatrix}= \begin{bmatrix}a \\ d\\ g\end{bmatrix}= \begin{bmatrix}3 \\ 0 \\ 0\end{bmatrix}[/tex]

    Similarly
    [tex][L]e_2= \begin{bmatrix}a & b & c \\ d & e & f \\ g & h & i\end{bmatrix}\begin{bmatrix}0 \\ 1 \\ 0\end{bmatrix}= \begin{bmatrix} b \\ e \\ i\end{bmatrix}= \begin{bmatrix}0 \\ 3\\ 0\end{bmatrix}[/tex]

    Finally, we have
    [tex][L]e_2= \begin{bmatrix}a & b & c \\ d & e & f \\ g & h & i\end{bmatrix}\begin{bmatrix}1 \\ 0 \\ 1\end{bmatrix}= \begin{bmatrix} a+ c \\ d+ f \\ g+ h\end{bmatrix}= \begin{bmatrix}1/3 \\ 0\\ 1/3\end{bmatrix}[/tex]
     
    Last edited: Nov 17, 2011
  4. Nov 18, 2011 #3
    That's exactly what the problem said. I believe it means that the line is contracted by a factor of 3 only and the points are dilated by 3.

    What I had thought was that I was supposed to pick 2 points from W and then use the span as the third point. And then like you did set [L]en = xnvn; where v is the vectors and x is the dilate/contract.

    I know I'm looking for the whole matrix, I was just concerned with L(e1) for the moment. What I had thought was that I would just apply the same steps for L(e2) and L(e3).

    I'm still a little confused when you calculated [L]e2. For the last one, I'm not sure if you meant [L]e3 or [L]e2?

    So would this make the matrix [L] = [tex]\begin{bmatrix}3 & 0 & 1/3 \\ 0 & 3 & 0 \\ 0 & 0 & 1/3\end{bmatrix}[/tex] ?
     
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