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Math Help - Showing Transformation is True

  1. #1
    Member alexgeek's Avatar
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    Showing Transformation is True

    I'm confused by this question:
    A transformation T in the plane consists of a reflection in the line x + y = 0 followed by a translation in which the point (x,y) is transformed to the point (x+h,y+k).

    (a). Show that the matrix representing T is

    \begin{bmatrix} 0 & -1 & h \\ -1 & 0 & k \\ 0 & 0 & 1 \end{bmatrix}
    The question only mentions x and y so I assume this is supposed to be an R^2 transformation but the matrix is a 3x3.. don't see how the multiplication would work.

    I tried transforming the unit square but didn't do much to help me understand.

    Thanks
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  2. #2
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    this is in R^3

    T transforms (x,y,z) to (-y+zh,-x+zk,z) so that the z coordinate is unchanged

    the first two rows reflect in the line y=-x and then add zh to the x and zk to the y coordinate respectively, the third row leaves the z coordinate unaltered.

    if (x,y) is transformed to (x+h,y+k) then we are in the plane z=1.

    So i guess
    \begin{bmatrix} 0 & -1 & -h \\ -1 & 0 & -k \\ 0 & 0 & 1 \end{bmatrix}

    is the same transformation but in the plane z=-1
    Last edited by Krahl; September 22nd 2010 at 07:21 PM.
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  3. #3
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    This is an example of "projective coordinates". Essentially, a point (x, y) is represented by (x, y, 1) with the provision that the triple (x, y, a) represents the same point as (x/a, y/a, 1).

    Working in two dimensions, we can represent a rotation about the origin, through angle \theta as a matrix multiplication:
    \begin{bmatrix} cos(\theta) & -sin(\theta) \\ sin(\theta) & cos(<br />
theta)\end{bmatrix}\begin{bmatrix}x \\ y \end{bmatrix}= \begin{bmatrix}x cos(\theta)- y sin(\theta) \\ x sin(\theta)+ y cos(\theta)\end{bmatrix}
    but translations have to be represented as a vector addition:
    \begin{bmatrix}x \\ y\end{bmatrix}+ \begin{bmatrix}h \\ k\end{bmatrix} = \begin{bmatrix} x+ h \\ y+ k\end{bmatrix}

    Using projective coordinates we can combine both into a single matrix:
    \begin{bmatrix} cos(\theta) & -sin(\theta) & h \\ sin(\theta) & cos(\theta) & k \\ 0 & 0 & 1\end{bmatrix}\begin{bmatrix}x \\ y \\ 1\end{bmatrix}= \begin{bmatrix}x cos(\theta)- y sin(\theta)+ h \\ x sin(\theta)+ y cos(\theta)+ k \\ 1\end{bmatrix}

    "Projective Coordinates" are derived from "Projective Geometry". If, in three dimensional Euclidean geometry, you define two points to be "equivalent" if and only if they lie on the same straight line through the origin and define "points" in your new geometry to be the equivalence classes, then you have the "Projective Plane".
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  4. #4
    Member alexgeek's Avatar
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    Thanks for the reply, looking over it now. Think I need to read over this a few thousand more times.
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