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Thread: Exercise: Matrix transformations

  1. #1
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    Exercise: Matrix transformations

    Let L denote the line through the origin in R2 with slope m. Show that reflection in L has matrix

    Exercise: Matrix transformations-matrix.png

    Please help me do this exercise. Thanks so much.
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  2. #2
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    What ideas have you had so far? Also, I assume you meant "reflection about L", right?
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  3. #3
    MHF Contributor FernandoRevilla's Avatar
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    A little help:

    Por every $\displaystyle (x,y)\in \mathbb{R}^2$ find the symmetric point $\displaystyle (x',y')$ of $\displaystyle (x.y)$ with respect to $\displaystyle Y=mX$ . You'll obtain an expression of the form:

    $\displaystyle \begin{bmatrix}{x'}\\{y'}\end{bmatrix}=A\begin{bma trix}{x}\\{y}\end{bmatrix}$


    Fernando Revilla
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  4. #4
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    To find the matrix corresponding to a linear transformation from $\displaystyle R^2$ to $\displaystyle R^2$, it is enough to see what it does to the "basis vectors", <1, 0> and <0, 1>.

    Notice that $\displaystyle \begin{bmatrix}a_{11} & a_{12} \\ a_{21} & a_{22}\end{bmatrix}\begin{bmatrix}1 \\ 0\end{bmatrix}= \begin{bmatrix}a_{11} \\ a_{21}\end{bmatrix}$
    and than
    $\displaystyle \begin{bmatrix}a_{11} & a_{12} \\ a_{21} & a_{22}\end{bmatrix}\begin{bmatrix}0 \\ 1\end{bmatrix}= \begin{bmatrix}a_{12} \\ a_{22}\end{bmatrix}$
    so what the linear transformation does to <1, 0> and <0, 1> gives you the two columns of the matrix.

    Now, the line, through the origin with slope m, y= mx, contains the points (0, 0) and (1, m) and so is in the direction of the vector <1, m>. To find the "reflection" of <1, 0> in that line, find its projection onto the vector <1, m>, then subtract that from <1, 0> to find its component perpendicular to the line. Subtract twice that vector from <1, 0> (subtracting once puts you on the line, subtracting again, on the other side) to find the reflection in the line.
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  5. #5
    MHF Contributor FernandoRevilla's Avatar
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    Alternative 3 :

    Some courses of Linear Algebra cover the following and well known theorem:

    If $\displaystyle H:a_1x_1+\ldots a_nx_n=0$ is a hyperplane of $\displaystyle \mathbb{R}^n$ , then the projection matrix on $\displaystyle H$ with the usual inner product is:

    $\displaystyle P=I-\dfrac{uu^t}{u^tu},\quad u=(a_1,\ldots,a_n)^t$

    As a consequence the simmetry matrix on $\displaystyle H$ is

    $\displaystyle S=2P-I=I-2\dfrac{uu^t}{u^tu}$

    In our case, $\displaystyle u=(m,-1)^t$ so,

    $\displaystyle S=\begin{bmatrix}{1}&{0}\\{0}&{1}\end{bmatrix}-\dfrac{2}{m^2+1}\begin{bmatrix}{m}\\{-1}\end{bmatrix}\begin{bmatrix}{m}&{-1}\end{bmatrix}=\dfrac{1}{m^2+1}\begin{bmatrix}{1-m^2}&{2m}\\{2m}&{m^2-1}\end{bmatrix}
    $



    Fernando Revilla
    Last edited by FernandoRevilla; Jan 13th 2011 at 09:23 AM.
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