Solutions Geometrical Transformations
CS1BA1
Exercise on Rotation:
1. Compute the determinant of a rotation matrix.
Solution:
"
det
cos θ
− sin θ
sin θ
cos θ
#!
= cos2 θ + sin2 θ = 1
2. Compute the eigenvalues of a rotation matrix. For which value of θ do we have eigenvalues?
Solution:
the characteristic equation is
"
det
cos θ
− sin θ
sin θ
cos θ
"
#
−λ
1
0
0
1
#!
= λ2 − 2λ cos θ + 1 = 0
The characteristic equation has roots if ∆ = (2 cos θ)2 − 4 ≥ 0. This is equivalent to writting
cos2 θ ≥ 1 (or cos2 θ = 1 since | cos θ| ≤ 1). The only possibility for the characteristic equation
to have roots is that:
(
cos θ = 1 or θ = 0
cos θ = −1 or θ = π
So we have roots for the characteristic equation only in two cases: when θ = 0 and θ = π.
(a) Case θ = 0. In this case the rotation matrix becomes:
"
cos θ
− sin θ
sin θ
cos θ
#
"
=
1
0
0
1
#
The rotation matrix is simply the identity matrix I and the eigenvalue is λ = 1 (this is a
repeated eigenvalue). The eigenvectors are any vector in the plane since ∀x ∈ R2 , Ix =
x.
(b) Case θ = π. In this case the rotation matrix becomes:
"
cos θ
− sin θ
sin θ
cos θ
#
"
=
−1
0
0
−1
#
The rotation matrix is simply the matrix −I . The eigenvalue is λ = −1 (this is a repeated
eigenvalue). The eigenvectors are any vector in the plane since ∀x ∈ R2 , −Ix = −x.
If θ 6= 0, π then there is no eigenvalues and eigenvectors for the rotation matrix.
3. In the case when eigenvalues exist, compute the eigenvectors.
Solution:
If θ = 0 or π, the eigenvectors are any vectors v in the plane R2 :
1
v=x
!
0
+y
0
!
, with x ∈ R and y ∈ R
1
Exercise on Reection:
1. Compute the determinant of a reection matrix.
Solution:
"
det
cos θ
sin θ
sin θ
− cos θ
#!
= − cos2 θ − sin2 θ = −1
2
2. Draw the house when it has been transformed by a reection matrix with θ = π4 .
Solution:
see graph.
3. Compute the eigenvalues and the eigenvectors.
Solution:
the characteristic equation is
"
det
"
#
cos θ
sin θ
sin θ
− cos θ
−λ
1
0
0
1
#!
= λ2 − 1 = (λ − 1)(λ + 1) = 0
So we have two eigenvalues: λ = 1 and λ = −1.
(a) Case λ = 1. The augmented matrix is:
"
It is reduced by R2 →
R2
sin θ
−
cos θ − 1
sin θ
0
sin θ
− cos θ − 1
0
R1
cos θ−1
#
:
"
cos θ − 1
sin θ
0
0
0
0
#
so for an eigenvector v = (x, y), we have the following relation:
(cos θ − 1)x + sin θ · y = 0 or
so
v =x·
(1 − cos θ)
x
sin θ
y=
!
1
∀x ∈ R∗
(1−cos θ)
sin θ
θ)
You can draw the line y = (1−cos
sin θ x (using trigonometry, this equation is equivalent to
y = tan θ2 · x ) and check that this is the axis of symmetry for the reection.
(b) Case λ = 1. The augmented matrix is:
"
It is reduced by R2 →
R2
sin θ
−
cos θ + 1
sin θ
0
sin θ
− cos θ + 1
0
R1
cos θ−1
#
:
"
cos θ + 1
sin θ
0
0
0
0
#
so for an eigenvector v = (x, y), we have the following relation:
(cos θ + 1)x + sin θ · y = 0
so
v =x·
1
−(1+cos θ)
sin θ
or
y=
−(1 + cos θ)
x
sin θ
!
∀x ∈ R∗
4. Draw the line or the eigenspace corresponding to the eigenvalue λ = 1.
Solution:
see graph.
5. Check that you have a good reection by folding the graph on this eigen-line.
Solution:
see graph.
3
Exercise on Afne Transformations:
1. Consider a transformation combining a translation and a linear transformation such that
(1)
x0 = Ax + t
a xed point (or vector) is a vector x such that it remains at the same place
after afne transformation (i.e. we have x = Ax + t)
Fixed point:
(a) what is the xed point of equation (1)?
Solution: The xed point is the vector x such that
x
= Ax + t
or
(I − A)x = t
so if (I − A) is an invertible matrix then you have the xed point dened as:
x = (I − A)−1 t
(b) on what condition this xed point exists?
Solution: We need (I − A) to be invertible so we need det(I − A) 6= 0 (or det(A − I) 6= 0).
It means that A should not have 1 as an eigenvalue so that (I − A) is invertible.
(c) does a pure translation (i.e. if A = I ) have a xed point?
Solution: For A = I , there is no xed point.
2.
Combination of transformations.
(1) applied to a vector x.
Consider two successive transformations as in equation
(a) is the resulting transformation linear? afne ?
Solution: The rst transformation is x1 = A1 x + t1 , and then the second is applied to x1
such that x2 = A2 x1 + t2 . We want the output x2 w.r.t. the rst input x:
x2
= A2 (A1 x + t1 ) + t2
= A2 A1 x + A2 t1 + t2
| {z }
| {z }
t
A
So the resulting transformation of x in x2 is afne.
(b) Check that the combination of those two transformations can be expressed simply by a
multiplication of matrices using homogeneous coordinates.

and
we have then:

x2

x1



b1
tx1

 
 y1  =  c1

 
1
0
d1


 y 
ty1 


1
1

a2
b2
tx2

 
 y2  =  c2

 
0
1
d2




ty2 
  y1 
1
1
x2



0
0


x

a1
x1


b2
tx2
a1
b1
tx1

 
 y2  =  c2

 
1
0
d2


ty2 
  c1
1
0
d1




ty1 
 y 
1
1
0
0
x

a2
4
you should check that the relation in between
the same! as in the
! x and x2 is exactly
!
previous question, assuming A1 =
tx2
ty2
3.
a1
b1
c1
d1
, A2 =
a2
b2
c2
d2
, t1 =
tx1
ty1
and t2 =
!
.
The projection of the 3D space (world) to a 2D plane (an image) by
a camera can also be expressed via a transformation of the form:
Camera calibration.


x


 y =P


1

X



 Y 


 Z 


1
(2)
(a) what is the size of the matrix P ?
Solution: P is a matrix of order (3 × 4)
(b) How would you nd the coefcients of the matrix P ?
Solution: you nd corresponding points in between the 3D world and its projection in
the image plane. Once you have several corresponding pairs (xi , Xi ), you can solve the
unknown coefcients in P .
Figure 1: Graph for the exercise on reection.