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Thread: Classifications of motions.

  1. #1
    Super Member Showcase_22's Avatar
    Sep 2006
    The raggedy edge.

    Classifications of motions.


    $\displaystyle A=\begin{pmatrix} \frac{3}{5} & \frac{4}{5} \\ \frac{4}{5} & -\frac{3}{5} \end{pmatrix}$

    and let

    $\displaystyle b=\begin{pmatrix} 10 \\ 0 \end{pmatrix}$

    Fit the map g defined by $\displaystyle x \rightarrow Ax+b$ into the classification of motions of $\displaystyle \mathbb{E}^2$.
    I've vome up with two possible ways to do this question which may work (if I wasn't so stuck!) so i'll list them both to see if either one of them is the way to go:

    This idea involves representing the matrix as elementary matrices and figure out what each of the separate matrices does (i'll deal with matrix b at the end, since it's just a translation 10 units right along the x axis). Here's A in this form:

    $\displaystyle \frac{1}{5} \begin{pmatrix} 3 & 4 \\4 & -3 \end{pmatrix}=\begin{pmatrix} 1 & -\frac{4}{3} \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 \\0 & -\frac{3}{5} \end{pmatrix} \begin{pmatrix} 1 & 0 \\-\frac{4}{5} & 1 \end{pmatrix} \begin{pmatrix} \frac{5}{3} & 0 \\ 0 & 1 \end{pmatrix}$

    I can see that the matrix on the furthest right takes $\displaystyle \begin{pmatrix} 1 \\ 0 \end{pmatrix}$ and maps it to $\displaystyle \begin{pmatrix} -\frac{5}{3} \\ 0 \end{pmatrix}$, so it's a "squishing" in the x direction (since the vertical component is left unchanged).

    My big problem is that "squishing" isn't a geometric motion, only the following are:
    • Translations
    • rotations
    • twists
    • reflections
    • glides
    • rotary reflections

    I also don't think that "squishing" can be put into any of the above criteria either since none of them involve scale factors.

    My second idea involves finding the eigenvalues and eigenvectors. These come out to be:

    $\displaystyle \lambda= \pm 1$ with $\displaystyle v_1=\begin{pmatrix} 1 \\ -2 \end{pmatrix}$ and $\displaystyle v_2=\begin{pmatrix} 2 \\ 1 \end{pmatrix}$

    So this means that if I take a point along the line $\displaystyle y=-2x$ or $\displaystyle y=\frac{x}{2}$, then it'll be moved along this line.

    My inital reaction was that it's a rotation (my coordinate axes have been rotated by arctan 1/2 anticlockwise). However, this doesn't match the rotation matrix when I try and find an angle of rotation.

    So which one of these ideas is the best idea to go down? I'm pretty much stuck on what i'm meant to do next!
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  2. #2
    MHF Contributor

    Apr 2005
    The matrix $\displaystyle \begin{pmatrix}\frac{3}{5} & \frac{4}{5} \\ \frac{4}{5} & \frac{3}{5}\end{pmatrix}$ has determinant -1 and so is a reflection. Its eigenvalues are, as you say, 1 and -1 and corresponding eigenvectors are <2, 1> and <1, -2> respectively. That tells you that it is a reflection about the line x= 2y, the line corresponding to eigenvalue 1.
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