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.