Please read this introduction first before looking through the solutions. Here’s a quick index to all the problems in this section.
1. Show that no collineation of type V can be periodic.
Multiplying the standard form of the matrix of a collineation of type V nn times with itself, we get (kn10nakn−10kn1nbkn−100kn1) As a≠0 and b≠0, this can never be a scalar multiple of the identity matrix and hence a collineation of type V can not be periodic.
2. Under what conditions, if any, can a collineation of type IV be periodic?
Multiplying the standard form of the matrix of a collineation of type IV n times with itself, we get (kn100na21kn−1kn10tna32kn−1kn1) As a21≠0 and a32≠0, this can never be a scalar multiple of the identity matrix and hence a collineation of type IV can not be periodic under any conditions.
3. Show that the composition of two collineations of type V with the same axis is always a collineation of type V.
Let x3=0 be the common axis and the two centers be (a,b,0) and (c,d,0). Let the repeated characteristic roots of the collineation of type V be k1 and k2 respectively. Then the matrix of transformation of points for the composition of the two collineations of type V will be (k10a0k1b00k1)(k20c0k2d00k2)=(k1k20k1c+k2a0k1k2k1d+k2b00k1k2)
This is a collineation of type V with center (k1c+k2a,k1d+k2b,0), axis x3=0 and repeated characteristic root k1k2.
4. Show that the composition of two collineations of type V with distinct axes will also be a collineation of type V if and only if the center of each transformation is the point of intersection of the two axes.
This is a dual of the result in exercise #3 and as such needs no extra proof but we’ll do it anyway for fun and clarity.
Let (0,0,1) be the common center and the two axes be [a,b,0] and [c,d,0]. Let the repeated characteristic roots of the collineation of type V be k1 and k2 respectively. Then the matrix of transformation of lines for the composition of the two collineations of type V will be (k10a0k1b00k1)(k20c0k2d00k2)=(k1k20k1c+k2a0k1k2k1d+k2b00k1k2) This is a collineation of type V with axes [k1c+k2a,k1d+k2b,0], center (0,0,1) and repeated characteristic root k1k2.
5. Find the equations of the most general collineation of type IV having the fixed elements
(a) F1:(0,1,0),f1:x3=0
(b) F1:(1,0,1),f1:x2=0
(c) F1:(1,0,0),f1:x2−x3=0
(a) Under this transformation the point (0,1,0) will get transformed to (0,k1,0). This implies that a12=a32=0 and a22=k1. Further, as x3=0 is the invariant line, any point with the last coordinate zero will be transformed to another point with the last coordinate zero. This implies a31=0. This means the matrix of transformation will look like A=(a110a13a21k1a2300a33) For this matrix to have a single repeated characteristic root k1, a11=a33=k1 and for the characteristic matrix A−k1I to have rank two a21≠0 and a13≠0. Hence the equations of the most general collineation of type IV with the given fixed elements is x′1=k1x1+a13x3 x′2=a21x1+k1x2+a23x3 x′3=k1x3 For this to be a collineation, i.e to keep the matrix nonsingular, k1≠0.
(b) Under this transformation the point (1,0,1) will get transformed to (k1,0,k1). This implies k1=a11+a13⟹a11=k1−a13 0=a21+a23⟹a21=−a23 k1=a31+a33⟹a33=k1−a31 As x2=0 is the invariant line, any point with the second coordinate zero will be transformed to another point with the second coordinate zero. This implies a21=a23=0. This means the matrix of transformation will look like A=(a11a12a130a220a31a32a33) The characteristic polynomial of this matrix is (a22−k)((a11−k)(a33−k)−a13a31) For k1 to be the single repeated root a22=k1. Substituting the values of a11 and a33 we derived earlier in the second component of the equation above, we get (k1−a13−k)(k1−a31−k)−a13a31 As k1 is a repeated root of this quadratic polynomial, the discriminant must be zero and a13+a31=0 This implies a13=−a31 a11=k1+a31 So the equations of transformation will be x′1=(k1+a31)x1+a12x2−a31x3 x′2=k1x2 x′3=a31x1+a32x2+(k1−a31)x3 For this to be a nonsingular transformation k1≠0 and for the matrix $A
- k_1Itohaverank2,a_{31} \ne 0anda_{12} \ne a_{32}$.
(c) Under this transformation the point (1,0,0) will get transformed to (k1,0,0). This implies a11=k1 and a21=a31=0. Further, as x2−x3=0 is the invariant line, any point that has the same last two coordinates must be transformed to another point which has identical second and third coordinates. This implies that a22+a23=a32+a33. The characteristic polynomial of this matrix is (a11−k1)((a22−k)(a33−k)−a32a23) For k1 to be the single repeated root a22+a33=2k1 and a23=−a32. Expressing values in terms of a22 we get a33=2k1−a22 a23=k1−a22 a32=−k1+a22 Hence, the equations of transformation will be x′1=k1x1+a12x2+a13x3 x′2=a22x2+(k1−a22)x3 x′3=−(k1−a22)x2+(2k1−a22)x3 For this to be a nonsingular transformation k1≠0 and for the matrix $A
- k_1Itohaverank2,a_{12} \ne -a_{13}anda_{22} \ne k$.
6. Find the equations of the most general collineation of type V having the fixed elements:
(a) F1:(s,0,t),f1:x2=0
(b) F1:(1,0,1),f1:x1−x3=0
(c) F1:(1,1,1),f1:x1−x2=0
(a) Under the transformation, every point (p1,0,p3) will get transformed to (k1p1,0,k1p1). This implies that a11=a33=k1 and a21=a23=a13=a31=0. Furthermore as k1 is a repeated triple root of the matrix of transformation, a22=k1. Hence the equations of transformation will be x′1=k1x1+a12x2 x′2=k1x2 x′3=a32x2+k1x3 Applying the collinearity condition of each point and its image with F1, we get a12t−a32s=0. This means that a12=αs and a32=αt for some α≠0. So the final equations of transformation can also be written as x′1=k1x1+αsx2 x′2=k1x2 x′3=αtx2+k1x3
(b) Under the transformation, every point (p1,p2,p1) will get transformed to (k1p1,k1p2,k1p1). This implies that a11+a13=k1 a21+a23=0 a31+a33=k1 a22=k1 a12=a32=0 The matrix of transformation look like A=(k1−a130a13−a23k1a23k1−a330a33) For k1 to be a repeated root of this matrix, we must have a33−a13=k1. So expressing the values in terms of a13 we have a33=k1+a13 a31=−a13 So, the equations of transformation will be x′1=(k1−a13)x1+a13x3 x′2=−a23x1+k1x2+a23x3 x′3=−a13x1+(k1+a13)x3 Applying the collinearity condition of each point and its image with F1, we get a23=0. Hence, the final equations of transformation can also be written as x′1=(k1−a13)x1+a13x3 x′2=k1x2 x′3=−a13x1+(k1+a13)x3
(c) Under the transformation, every point (p1,p1,p3) will get transformed to (k1p1,k1p1,k1p3). This implies that a13=a23=0 a11+a12=k1 a21+a22=k1 a31+a32=0 a33=k1 The matrix of transformation look like A=(k1−a12a120k1−a22a220−a32a32k1) For k1 to be a repeated root of this matrix, we must have a22−a12=k1. So expressing the values in terms of a12 we have a22=k1+a12 a21=−a12 So, the equations of transformation will be x′1=(k1−a12)x1+a12x2 x′2=−a12x1+(k1+a12)x2 x′3=−a32x1+a32x2+k1x3 Applying the collinearity condition of each point and its image with F1, we get a32=a12. Hence, the final equations of transformation can also be written as x′1=(k1−a12)x1+a12x2 x′2=−a12x1+(k1+a12)x2 x′3=−a12x1+a12x2+k1x3
7. Under what conditions, if any, will two collineations of type V commute with each other?
From the solutions to exercise #3 and its dual, #4, we can clearly see that two collineations of type V that either share the same axis or the same center commute.
Let’s assume that two commuting collineations, TA and TB, have different axes.
Consider the transformation of a general point, P, on the axis of TA, different from the point of intersection of the axes of the two collineations.
Under the transformation TBTA, P will first be transformed by TA into itself, as it is invariant under TA, and then transformed by TB into another point not equal to P, say P′, as P is not invariant under TB. So the image of P under TBTA will be P′.
Under the transformation TATB, P will be transformed by TB to P′ just as before, and P′ will be invariant under TA only if it lies on the axis of TA. Now as the line defined by a point and its image must pass through the center of the transformation, PP′ must pass through the center of TB. In other words, for a point on the axis of TA to have the same image under both TATB and TBTA, the axis of TA must pass through the center of TB.
Applying the same logic for the transformation of a point on the axis of TB we will find that for a point on the axis of TB to have the same image under both TATB and TBTA, the axis of TB must pass through the center of TA. Both conditions taken together imply that the two collineations can not commute if they do not share the same axis.
The proof for the shared center case is the dual of the shared axis case.
Therefore, two collineations of type V commute with each other if and only if they share the same axis or the same center.
8. Show that in general the composition of two collineations of type V is a collineation of type I. Under what conditions, if any, will the composition be a collineation of type II? Of type III? Of type IV?
Consider two collineations of type V, TA and TB. In general they will not share the same axis or the same center.
Let’s call the point of intersection of the two axes as F1. This point will be invariant under the transformation defined by the composition of the two collineations as it lies on both their axes.
Let’s denote the line defined by the centers of the two collineations as f1. This line will be invariant under the transformation defined by the composition of the two collineations as it passes through both their centers.
By Theorem 1 Sec 5.3, the mapping induced of f1 onto itself by the composite collineation is a projectivity. As derived in Sec 4.8, this projectivity between cobasal ranges has a characteristic polynomial that, in general, has two distinct roots corresponding to the fixed points F2 and F3. This means that the composite collineation has a total of 3 fixed points, F1, F2 and F3 which is characteristic of a collineation of type I.
Hence, in general, a composition of two collineations of type V is a collineation of type I.
Note that the line f1 will be point-by-point invariant under the composite transformation of two collineations of type V only when they share the same axis. For a proof, consider the center of one of the collineations. Its image under the other collineation will be invariant only if the two share the same axis. Such a composite collineation will be of type V as we found out in #3.
When the projectivity induced on f1 has only one fixed point, then we have a total of two fixed points including F1 and two fixed lines. The line joining F1 to the other fixed point, f2, is not point-by-point invariant under the composite transformation. If it were point-by-point invariant then there must be another line f′2 that it will be mapped to by one collineation with the second collineation mapping f′2 back to f2 to maintain its invariance under the composition. This would mean that the lines between the points on f2 to their images on f′2 were invariant under both the collineations of type V. However this is not possible as the invariant lines can’t belong to two different pencils (centers). Hence f2 can’t be point-by-point invariant. Thus, this is a collineation of type II.
Using the same logic as above, we can deduce that no line other than a shared axis can be point-by-point invariant under a composite transformation of two collineations of type V. Hence, two collineations of type V can never compose to form a collineation of type III.
When the axis of one of the collineations of type V passes through the center of the other, the point of intersection is the center of the second collineation. No other point on the line will be invariant under the composite transformation. For a proof, take any point other than the point of intersection of the two axes. This point is invariant under the first transformation as it lies on its axis. The second collineation will then map this to some other point on the line as the point does not lie on the axis of the second collineation of type V. Hence this composite collineation with one fixed point and one fixed line is a collineation of type IV.
9. What is the locus of a point with the property that it and its image under a collineation of type IV are collinear with a given point O? What is the locus of images of such points?
The answer to this question had me scratching my head. Axis of transformation and center for a collineation of type IV? Then I realized it must’ve been another typo in the book (it has quite a few). The question is actually asking about a collineation of type V.
Consider a general point P:(p1,p2,p3) and its image P′:(p′1,p′2,p′3).
Using the equations of transformation of a general collineation of type V, we get
p′1=k1p1+ap3,p1=k1p′1−ap′3k21 p′2=k1p2+bp3,p2=k1p′2−bp′3k21 p′3=k1p3,p3=p′3k1
As per Sec 4.2, Theorem 1 if P and its image are to be collinear with a given point O:(o1,o2,o3), we get
|p1p2p3k1p1+ap3k1p2+bp3k1p3o1o2o3|=0
Expanding this we get the singular conic p3(bo3p1−a3p2+(ao2−bo1)p3)=0
This is the combination of two lines p3=0 which is the axis of the collineation and [bo3,−ao3,ao2−bo1] which is [o1,o2,o3]×[a,b,0] i.e. the line determined by the given point and the center of transformation.
Similarly, expressing in terms of image coordinates, we get |k1p′1−ap′3k21k1p′2−bp′3k21p′3k1p′1p′2p′3o1o2o3|=0
Expanding this we get the same singular conic as before.
10. Find the fixed points of the transformation X′=AX if:
(a) A=(3−116−232−12)
(b) A=(−2−4−1351−1−20)
(c) A=(32−483−884−9)
(d) A=(−2−1353−4−4−15)
I’m using maxima for this. You can use your favorite linear algebra software to find the eigenvectors.
(a) Type V, fixed line is [2,−1,1].
(b) Type IV, fixed point is (1,−1,1).
(c) Type V, fixed line is again [−2,−1,1].
(d) Type IV, fixed point is again (1,−1,1).