Results 1 to 5 of 5

Math Help - Exercise: Matrix transformations

  1. #1
    Newbie
    Joined
    Jan 2011
    Posts
    7

    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.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    4
    Awards
    2
    What ideas have you had so far? Also, I assume you meant "reflection about L", right?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor FernandoRevilla's Avatar
    Joined
    Nov 2010
    From
    Madrid, Spain
    Posts
    2,162
    Thanks
    45
    A little help:

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

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


    Fernando Revilla
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,598
    Thanks
    1421
    To find the matrix corresponding to a linear transformation from R^2 to R^2, it is enough to see what it does to the "basis vectors", <1, 0> and <0, 1>.

    Notice that \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
    \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.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor FernandoRevilla's Avatar
    Joined
    Nov 2010
    From
    Madrid, Spain
    Posts
    2,162
    Thanks
    45
    Alternative 3 :

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

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

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

    As a consequence the simmetry matrix on H is

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

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

    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}<br />



    Fernando Revilla
    Last edited by FernandoRevilla; January 13th 2011 at 09:23 AM.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. An Exercise: Matrix Multiplication
    Posted in the Advanced Algebra Forum
    Replies: 3
    Last Post: January 8th 2011, 07:44 PM
  2. Matrix transformations
    Posted in the Advanced Algebra Forum
    Replies: 3
    Last Post: October 28th 2009, 05:00 PM
  3. Jacobian matrix exercise
    Posted in the Calculus Forum
    Replies: 3
    Last Post: August 2nd 2009, 12:46 AM
  4. Exercise in Matrix Properties
    Posted in the Advanced Statistics Forum
    Replies: 0
    Last Post: July 28th 2009, 03:56 PM
  5. A strange exercise with matrix norm!
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: March 22nd 2009, 10:41 AM

Search Tags


/mathhelpforum @mathhelpforum