- #1
kungfu420
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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.
Given the matrix D=
.8, .6, 2
.6, -.8, 1
0, 0, 1
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.
Homework Equations
Given the matrix D=
.8, .6, 2
.6, -.8, 1
0, 0, 1
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.