Please read this introduction first before looking through the solutions. Here’s a quick index to all the problems in this section.
1. Prove Theorem 2
Composing two general euclidean transformations, we get a transformation of the form below. (a11b11−a12b12ka12b11k+a11b12a12b23k+a11b13+a13−(a11b12k+a12b11)a11b11−a12b12ka11b23k−a12b13k+a2300a33b33)
This is clearly a similarity transformation.
Next, let’s consider the fraction relevant to euclidean transformations (a11b11−a12b12k)2+(a11b12k+a12b11)2a233b233
Expanding this we get (a211+a212)(b211+b212)a233b233 =(a211+a212)a233(b211+b212)b233
As the value of each of these terms is 1, the value of the whole fraction will also be 1. Thus, two euclidean transformations are closed under the operation of composition.
The inverse of a general euclidean transformation has the form 1a33k(a211+a212)(a11a33k−a12a33a12a23−a11a13ka12a33ka11a33−(a12a13k+a11a23)00k(a211+a212))
Let’s consider the fraction relevant to euclidean transformations k2a233(a211+a212)k2(a211+a212)2 =a233(a211+a212)
The value of this is clearly 1. Hence the inverse is also an euclidean transformation.
Finally, it is obvious, from the rules of matrix multiplication, that the composition of euclidean transformations is associative and that the identity is included among the euclidean transformations.
Hence the set of all euclidean transformations is a group under the operation of composition.
2. Are there euclidean transformations of all six types? Give an example of each possible type.
Type I
a11≠0, a12≠0 and a211+a212=a233.
(a11a12a13a12−a11a2300a33)
Type II
a23≠0
(a110a130−a11a2300−a11)
Type III
(a11000a11000−a11)
Type IV
As a euclidean transformation is also a similarity and there is no similarity transformation of type IV, there can be no euclidean transformation of type IV.
Type V
a13≠0,a23≠0
(a110a130a11a2300a11)
Type VI
(100010001)
3. Prove Theorem 1 using Theorem 1, Sec. 5.11, and the result of Exercise 9, Sec. 5.11.
The entire field of trigonometry attests to the fact that the angle made by two lines is a function of the lengths of sides of a right triangle that contains the angle. Hence, as euclidean transformations preserve distances, they also preserve angles, which further imply that euclidean transformations are similarities.
This means the matrix of transformation for a euclidean transformation by Theorem 1, Sec. 5.11 takes the form (a11a12a13−ka12ka11a2300a33)
where aij real, k=±1.
By the result of Exercise 9, Sec. 5.11, we know that the length of a general segment is transformed by a factor of $\sqrt{a_{11}^2
- a_{12}^2}underasimilaritytransformwitha_{33} = 1$. Combining this with the fact that this factor is 1 in a euclidean transformation, we have
(a11a33)2+(a12a33)2=1 ⟹a211+a212=a233
All these taken together form Theorem 1.
4. Show that every euclidean transformation induces an involution on the ideal line which is the identity, or has I:(1,i,0) and J:(1,−i,0) as its fixed points, or has I and J as a pair of mates. Determine the geometric nature of the euclidean transformations with each of these properties.
I might be wrong here but I think this question is incorrect for two reasons.
- Every euclidean transformation does not induce an involution on the ideal line.
- The identity is not an involution; it is a transformation of period 1.
I think the question should instead be as follows.
Show that every euclidean transformation that induces an involution on the ideal line has I:(1,i,0) and J:(1,−i,0) as its fixed points, or has I and J as a pair of mates. Determine the geometric nature of the euclidean transformations with each of these properties.
We know that a euclidean transformation is affine and hence leaves the ideal line invariant. Furthermore, as the last coordinate of every point on the ideal line is 0, the projectivity induced on the ideal line is given by
(a11a12−ka12ka11)
with (1,0,0) and (0,1,0) as the base points.
From Theorem 5, Sec. 4.8, we know that this projectivity can be an involution only if a11+ka11=0. For this to be true, either a11=0 or k=−1.
If k=−1, the transformation is an indirect similarity and from Exercise 4, Sec. 5.11 we know that an indirect similarity interchanges the circular points at infinity. This will be a rotation about the origin followed by a translation and a reflection.
If a11=0 and k=1, the transformation is a direct similarity and again from Exercise 4, Sec. 5.11 we know that a direct similarity leaves the circular points at infinity invariant. This will be a rotation of 90∘ about the origin followed by a translation.
5. Show that the composition of two reflections in parallel lines is a translation in the direction perpendicular to the two lines.
Without loss of generality, let the two lines be x=0 and x=d. Then the composition of matrices of reflections in these lines will be
(1000−10001)(1000−12d001)=(10001−2d001)
This is clearly a translation in the direction parallel to the y axis which is perpendicular to the direction of the lines chosen as the axis of reflection.
6. Show that the composition of two reflections in lines which intersect in a finite point is a rotation about the point of intersection of the lines.
Without loss of generality, we can take the point of intersection of the lines to be the origin. This makes both transformations Householder reflections1 implying that a13=a23=b13=b23=0 and k=−1 for both transformations. Taking a11=cos(θ),a12=sin(θ),b11=cos(ω),b12=sin(ω), the composition of the two reflections will be
(cos(θ)sin(θ)0sin(θ)−cos(θ)0001)(cos(ω)sin(ω)0sin(ω)−cos(ω)0001) =(sin(θ)sin(ω)+cos(θ)cos(ω)cos(θ)sin(ω)−sin(θ)cos(ω)0sin(θ)cos(ω)−cos(θ)sin(ω)sin(θ)sin(ω)+cos(θ)cos(ω)0001) =(cos(ω−θ)sin(ω−θ)0−sin(ω−θ)cos(ω−θ)0001)
which is a rotation of ω−θ around the origin as asserted.
7. What is the composition of two reflections in distinct points?
Reflection about a point is equivalent to a rotation of 180∘ about the point2. Using the form of the matrix for a rotation of 180∘ about an arbitrary point that we derived in Exercise 10, Sec. 5.11, we can write the composition of two reflections in distinct points as follows. (−10a130−1a23001)(−10b130−1b23001) =(10a13−b1301a23−b23001)
which is clearly a translation in the direction determined by the centers of reflection/rotation.
8. Under what conditions will a euclidean transformation be of period 2?
As euclidean transformations are a subgroup of similarity transformations and as the type of similarities of period 2 are also euclidean, those are the only type of euclidean transformations of period 2.
As per Exercise 10, Sec. 5.11, these are rotations of an angle of 180∘ about an arbitrary point and reflection about an arbitrary axis.
9. Under what conditions will a euclidean transformation be of period 3?
A euclidean transformation is the composition of rotations, translations and reflections. As translations and reflections can’t have periods of 3, the euclidean transformation having a period of 3 must be a rotation. Rotations that have a period of 3 are rotations of n3603∘=n120∘.
10. Under what conditions will two euclidean transformations commute?
For two general euclidean transformations, with matrices A and B, to commute the following must hold.
k=1
a12b23+a11b13+a13=b12a23+b11a13+b13
a11b23−a12b13+a23=b11a23−b12a13+b23
For examples of commuting euclidean transformations, please see Commuting Isometries.
References
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Wikipedia. Householder Transformation. https://en.wikipedia.org/wiki/Householder_transformation. ↩︎
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Wikipedia. Point Reflections, Examples. https://en.wikipedia.org/wiki/Point_reflection#Examples. ↩︎