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Finding the stable line of a Glide reflection

  1. May 8, 2012 #1
    1. For the following matrix, determine whether it is a translation, rotation, mirror reflection, or a glide reflection and find its fixed points and stable lines.


    2. Relevant equations
    Given the matrix D=
    .8, .6, 2
    .6, -.8, 1
    0, 0, 1


    3. The attempt at a solution
    First I got determinant of the matrix which is -1, so the it must be an indirect isometry. Next I found that it has no fixed points by setting D*(u, v, 1) = (u, v, 1).
    i.e. u = .8u + .6v +2, and v = .6u - .8v + 1
    I solved for u and got u = 3v + 10, then plugging the new u into the 2nd equation I got that v = v +7. Hence there are no fixed points, and therefore it is a glide reflection.

    Next to find the stable lines I found the inverse matrix which was,
    .8, .6, -2.2
    .6, -.8, -.4
    0, 0, 1

    Then I found the eigenvalues which are 1 and -1. From there I plugged them into the equation D-1 - Iλ. For λ=1 the matrix was

    -.2, .6, -2.2
    .6, -1.8, -.4
    0, 0, 0
    and for λ=-1 the matrix was

    1.8, .6, -2.2
    .6, .2, -.4
    0, 0, 2

    I then multiplied the matrices by [a, b, c] which gave me

    -.2a + .6b = 0 ====> b = 1/3a
    .6a - 1.8b = 0 ====> b = 1/3a
    -2.2a -.4b = 0 ====> b = -5.5a
    for λ=1 and

    1.8a + .6b = 0 ====> b = -3a
    .6a + .2b = 0 ====> b = -3a
    -2.2a -.4b +2c = 0 ===> b = -5.5a -5c
    for λ=-1

    This is where I become confused. I know that if they b equaled 1/3a for all 3 equations then there would be a family of parallel lines [a, 1/3a, c] but that is not the case and therefore I am stumped.

    I know that glide reflections have 1 stable line but I'm not sure how to go about finding it from these equations.
     
  2. jcsd
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