Affine and Euclidean Geometry
Affine and Euclidean Geometry
Affine and Euclidean Geometry
Rueda
2. AFFINE TRANSFORMATIONS
Definition Let (A, V, φ) and (A0, V 0, φ0) be two real affine spaces. We will say
that a map
f : A −→ A0
is an affine transformation if there exists a linear transformation f¯: V −→ V 0
such that:
f¯(P Q) = f (P ) f (Q), ∀P, Q ∈ A.
This is equivalent to say that for every P ∈ A and every vector ū ∈ V we
have
f (P + u) = f (P ) + f¯(u).
We call f¯ the linear transformation associated to f .
AFFINE AND PROJECTIVE GEOMETRY, E. Rosado & S.L. Rueda
Proposition Let (A, V, φ) and (A0 , V 0 , φ0 ) be two affine subspaces and let f : A −→ A0 be
an affine transformation with associated linear transformation f¯: V −→ V 0 . The following
statements hold:
1. f is injective if and only if f¯ is injective.
2. f is surjective if and only if f¯ is surjective.
3. f is bijective if and only if f¯ is bijective.
Let (A, V, φ) and (A0, V 0, φ0) be two affine subspaces and let f : A −→ A0 be
an affine transformation with associated linear transformation f¯: V −→ V 0.
We consider affine coordinate systems R = {O; B}, B = {e1, . . . , en} and
R0 = {O0; B 0}, B 0 = {e01, . . . , e0m} of the spaces A, A0 respectively. Let us
assume that:
O0f (O) = b1e01 + · · · + bme0m,
f¯(e1) = a11e01 + · · · + am1e0m
...
¯
f (en) = a1ne01 + · · · + amne0m
Let P be the point with coordinates (x1, . . . , xn) with respect to R and let
(y1, . . . , ym) be the coordinates of f (P ) ∈ A0.
AFFINE AND PROJECTIVE GEOMETRY, E. Rosado & S.L. Rueda
Then:
1 1 0 ··· 0 1
y1 b1 a11 · · · a1n x1
= ..
.. ... . . . ... ..
. . .
ym bm am1 · · · amn xn
We will write
1 0 ··· 0
t
1 0 b1 a11 · · · a1n
0
Mf (R, R ) = =
... ... . . . ...
b Mf¯(B, B 0)
bm am1 · · · amn
where b are the coordinates of f (O) in the coordinate system R0 and Mf¯(B, B 0)
is the matrix associated to the linear transformation f¯ taking in V the basis
B and in V 0 the basis B 0.
AFFINE AND PROJECTIVE GEOMETRY, E. Rosado & S.L. Rueda
Example 1
Let (A, V, φ) be an affine space with affine coordinate system R = {O; B =
{e1, e2, e3}}, and let (A0, V 0, φ0) be an affine space with affine coordinate sys-
tem R0 = {O0; B 0 = {e01, e02}}. Is the transformation f : A −→ A0, f (x, y, z) =
(x − 2y + 5, x − z + 1) an affine transformation? Give its associated linear
transformation and obtain the matrix associated to f in the coordinate sys-
tems R, R0.
Solution.
To see if f is an affine transformation we have to check if there exists a
linear transformation f¯: V −→ V 0 such that f (P )f (Q) = f¯ P Q for every
pair of points P, Q ∈ A. We take P (x1, y1, z1) and Q(x2, y2, z2) then
P Q = Q − P = (x2 − x1, y2 − y1, z2 − z1)
and
AFFINE AND PROJECTIVE GEOMETRY, E. Rosado & S.L. Rueda
The coordinates of the origin O in the coordinate system R are the cooor-
dinates of the vector OO = (0, 0, 0) in the basis B where e1 = (1, 0, 0)B ,
e2 = (0, 1, 0)B and e3 = (0, 0, 1)B .
AFFINE AND PROJECTIVE GEOMETRY, E. Rosado & S.L. Rueda
Then:
f (O) = f (0, 0, 0) = (5, 1),
f¯(e1) = f¯(1, 0, 0) = (1, 1),
f¯(e2) = f¯(0, 1, 0) = (−2, 0),
f¯(e3) = f¯(0, 0, 1) = (0, −1).
So,
1 0 0 0
Mf (R, R0) = 5 1 −2 0 .
1 1 0 −1
AFFINE AND PROJECTIVE GEOMETRY, E. Rosado & S.L. Rueda
Example 2
Let (A, V, φ) be an affine space with affine coordinate system R = {O; B =
{e1, e2}}, and let (A0, V 0, φ0) be an affine space with affine coordinate system
R0 = {O0; B 0 = {e01, e02, e03}}. Determine the affine transformation f : A −→
A0, such that
f (1, 2) = (1, 2, 3),
f¯(e1) = e01 + 4e02,
f¯(e2) = e01 − e02 + e03.
Find the matrix associated to f in the coordinate systems R, R0.
Solution.
We know the value of f at the point P (1, 2) and the linear transformation
associated to f , therefore we can determine f .
f¯(1, 0) = (1, 4, 0)B 0 ,
f¯(0, 1) = (1, −1, 1)B 0 ,
AFFINE AND PROJECTIVE GEOMETRY, E. Rosado & S.L. Rueda
As
1 0 0 1
1
−2 1 1 x + x − 2
1 2
x1 =
0 4 −1 4x1 − x2
x2
1 0 1 x2 + 1
we have:
f (x1, x2) = (x1 + x2 − 2, 4x1 − x2, x2 + 1) .
AFFINE AND PROJECTIVE GEOMETRY, E. Rosado & S.L. Rueda
Example 3
Let (R2, R2, φ) be an affine space with affine coordinate system R = {O; B =
{e1, e2}}. Determine the affine transformation f : R2 −→ R2 such that
f (P0) = f (1, 1) = (7, 5), f (P1) = f (1, 2) = (11, 4), f (P2) = f (2, 1) = (8, 8).
Solution.
To determine an affine transformation f : R2 −→ R2 we need three points
that form an affine coordinate system and their images.
First method
Since P0P1 = (0, 1) and P0P2 = (1, 0) , then we know that
f¯(e1) = f¯(1, 0) = f¯(P0P2) = f (P2) − f (P0) = (1, 3),
f¯(e2) = f¯(0, 1) = f¯(P0P1) = f (P1) − f (P0) = (4, −1).
Also OP0 = e1 + e2 so we have:
f (P0) = f (O) + f¯(OP0) = f (O) + f¯(e1 + e2)
= f (O) + f¯(e1) + f¯(e2)
AFFINE AND PROJECTIVE GEOMETRY, E. Rosado & S.L. Rueda
then
f (O) = f (P0) − f¯(e1) − f¯(e2) = (7, 5) − (1, 3) − (4, −1)
= (2, 3).
finally,
1 0 0
Mf (R, R) = 2 1 4
3 3 −1
and f (x1, x2) = (2 + x1 + 4x2, 3 + 3x1 − x2).
Second method
The set R0 = {P0(1, 1), P1(1, 2), P2(2, 1)} is an affine coordinate system since
P0P1 = (0, 1) and P0P2 = (1, 0) is a basis of R2. We have:
f (P0) = f (1, 1) = (7, 5),
f¯(P0P1) = f (P0)f (P1) = f (P1) − f (P0) = (11, 4) − (7, 5) = (4, −1),
f¯(P0P2) = f (P0)f (P2) = f (P2) − f (P0) = (8, 8) − (7, 5) = (1, 3).
AFFINE AND PROJECTIVE GEOMETRY, E. Rosado & S.L. Rueda
Therefore,
1 0 0
Mf (R, R0) = 7 4 1
5 −1 3
To obtain Mf (R, R)
Mf (R, R) = Mf (R0, R)M (R, R0) = Mf (R0, R)M (R0, R)−1
−1
1 0 0 1 0 0 1 0 0
= 7 4 1 1 0 1 = 2 1 4
5 −1 3 1 1 0 3 3 −1
Therefore f (x1, x2) = (2 + x1 + 4x2, 3 + 3x1 − x2).
AFFINE AND PROJECTIVE GEOMETRY, E. Rosado & S.L. Rueda
Example 4
Determine the affine transformation f : A3 −→ A3 which transforms the
points P0(0, 0, 0), P1(0, 1, 0), P2(1, 1, 1) and P3(1, 1, 4) in the points Q0(2, 0, 2),
Q1(2, −1, 1), Q2(2, 1, 3) and Q3(5, 7, 6) respectively.
Solution.
To determine an affine transformation f : A3 −→ A3 we will need four points
that form an affine coordinate system and their images.
The set R0 = {P0(0, 0, 0), P1(0, 1, 0), P2(1, 1, 1), P3(1, 1, 4)} is an affine coordi-
nate system as P0P1 = (0, 1, 0), P0P2 = (1, 1, 1) and P0P3 = (1, 1, 4) is a basis
of R3 because
dim(P0P1, P0P2, P0P3) = 3.
We have:
f (P0) = Q0 = (2, 0, 0),
f¯(P0P1) = f (P0)f (P1) = f (P1) − f (P0) = Q1 − Q0 = (0, −1, −1),
f¯(P0P2) = f (P0)f (P2) = f (P2) − f (P0) = Q2 − Q0 = (0, 1, 1),
f¯(P0P3) = f (P0)f (P3) = f (P3) − f (P0) = Q3 − Q0 = (3, 7, 4).
Therefore
1 0 0 0
2 0 0 3
0
Mf (R , R) =
0 −1 1 7
0 −1 1 4
AFFINE AND PROJECTIVE GEOMETRY, E. Rosado & S.L. Rueda
To obtain Mf (R, R)
Mf (R, R) = Mf (R0, R)M (R, R0) =
−1
1 0 0 0 1 0 0 0
2 0 0 3 0 0 1 1
=
0 −1 1 7 0 1 1 1
0 −1 1 4 0 0 1 4
1 0 0 0
2 −1 0 1
=
0 0 −1 2
0 1 −1 1
Proposition Let (A, V, φ) and (A0, V 0, φ0) two affine spaces and let f : A −→
A0 be an affine transformation with associated linear transformation f¯: V −→
V 0. The following statements hold:
1. If L ⊂ A is an affine subspace of A then
f (L) = {P 0 ∈ A0 | there exists P ∈ L such that f (P ) = P 0}
is an affine subspace of A0.
2. If L0 ⊂ A0 is an affine subspace of A0 then the set
L = {P ∈ A | f (P ) ∈ L0}
is an affine subspace of A, called the inverse image of L0 and denoted
f −1 (L0).
Example
Obtain the fixed points of the affine transformation
f (x, y) = (−2y + 1, x + 3y − 1).
Solution.
The matrix associated to f is
1 0 0
Mf (R, R) = 1 0 −2
−1 1 3
AFFINE AND PROJECTIVE GEOMETRY, E. Rosado & S.L. Rueda
The fixed points of f are the solutions P (x, y) of the following matrix equa-
tion:
0 = (A − I) P + b
this is
0 −1 −2 x 1
= + ⇐⇒ x + 2y − 1 = 0
0 1 2 y −1
Therefore, the points of the line x + 2y − 1 = 0 are the fixed points of f .
AFFINE AND PROJECTIVE GEOMETRY, E. Rosado & S.L. Rueda
transformation f¯
2. P f (P ) ∈ hui
AFFINE AND PROJECTIVE GEOMETRY, E. Rosado & S.L. Rueda
Example
Obtain the invariant subspaces of the transformation f of the former exam-
ple.
Solution.
To search for the invariant subspaces of f first we compute the eigenvalues
of f¯. The characteristic polynomial A is
−λ −2
det(A − λI) = = λ2 − 3λ + 2 = (λ − 1) (λ − 2)
1 3−λ
and, therefore, the eigenvalues of A are λ = 1, 2.
AFFINE AND PROJECTIVE GEOMETRY, E. Rosado & S.L. Rueda
Therefore, the lines with associated vector space V (2) = h(1, −1)i are inva-
riant lines of f because
f¯(1, −1) = 2(1, −1)
P f (P ) ∈ V (2)
If x + 2y − 1 = 0 (this is the line of fixed points of f ) then P f (P ) = 0 ∈ V (1).
Example
Let (A3, V, φ) be an affine space and R = {O; {e1, e2, e3}} a coordinate sys-
tem of A3. Determine the affine transformation f : A3 −→ A3 such that the
plane π ≡ x + 2y − z = 1 is a plane of fixed points of f and the vector e1 is
an eigenvector of f¯ associated to the eigenvalue 3.
Solution.
The plane π is a plane of fixed points, any point of the plane is a fixed
point of f . For example, the point P (1, 0, 0) ∈ π is a fixed point of f ; this is,
f (P ) = P . Also, we know that the vectors of the vector subspace associated
to π, this is, vectors from the plane π ≡ x + 2y − z = 0, are eigenvectors
associated to the eigenvalue 1.
For example, for
u = (1, 0, 1) ∈ π,v = (0, 1, 2) ∈ π
AFFINE AND PROJECTIVE GEOMETRY, E. Rosado & S.L. Rueda
2.4.1 Translations
Given a vector v ∈ V , we define the translation of vector v as the transfor-
mation Tv : A −→ A such that f (P ) = P + v.
2.4.2 Projections
2.4.3 Homotethy
Remark
A homotethy of ratio r has only one fixed point C called center of the ho-
mothety. The image of any other point P is obtained as follows:
f (P ) = C + rCP .
AFFINE AND PROJECTIVE GEOMETRY, E. Rosado & S.L. Rueda
Example 1
Study whether the affine transformation f (x, y, z) = (1 + 23 x, −1 + 23 y, 2 + 23 z)
has a fixed point or an invariant subspace.
Solution.
The matrix associated to f is
1 0 0 0
1 32 0 0
Mf (R) =
−1 0 23 0
2 0 0 23
and the matrix associated to the linear transformation f¯ is
2
Mf (B) = IR3 .
3
Therefore, f is a homothety of ratio r = 32 . The center of the homothety is:
1
C=P+ P f (P )
1 − 23
for every P ∈ A. We take P (0, 0, 0) then f (P ) = f (0, 0, 0) = (1, −1, 2) and
P f (P ) = (1, −1, 2), therefore
3
C= (1, −1, 2) = (3, −3, 6).
3−2
Example 2
Study whether the affine transformation f (x, y, z) = (x + 1, y + 2, z + 3) has
fixed points or invariant subspaces.
AFFINE AND PROJECTIVE GEOMETRY, E. Rosado & S.L. Rueda
Solution.
The matrix associated to f is
1 0 0 0
1 1 0 0
Mf (R) =
2 0 1 0
3 0 0 1
and the matrix associated to the linear transformation f¯ is the identity. The-
refore, f is a translation of vector v = Of (O) = (1, 2, 3) − (0, 0, 0) = (1, 2, 3).
The translations do not have fixed points.
The invariant subspaces of f are:
- The lines whose direction is the direction of the translation vector; this is,
lines r ≡ P + hvi.
- The planes whose direction contains the translation vector; this is, planes
of the form π ≡ P + hv, wi.
AFFINE AND PROJECTIVE GEOMETRY, E. Rosado & S.L. Rueda
Example 3
Study whether the affine transformation f (x, y, z) = (−2 + 2x − y, −4 + 2x −
y, z) has fixed points or invariant subspaces.
Solution.
The matrix associated to f is
1 0 0 0
−2 2 −1 0
Mf (R) =
−4 2 −1 0
0 0 0 1
equivalently
0=x−y−2
0 = 2x − 2y − 4
0=0
Example 4
Obtain the analytic expression of the affine tranformation f : A3 −→ A3 that
sends the origin to the point (3, 1, 1) and whose plane π of cartesian equa-
tion x1 + 2x2 − x3 + 1 = 0, is a plane of fixed points.
Solution.
As the plane π is a plane of fixed points, the vector plane associated with
π is a plane of eigenvectors associated with the eigenvalue λ = 1 of the
associated linear transformation f¯.
As π ≡ P + hu1, u2i with P (0, 0, 1), u1 = (1, 0, 1), u2 = (0, 1, 2) then P ∈ π (this
is, the coordinates of P are a solution of the equation of π) and the vectors
u1,u2 ∈ π (their coordinates are a solution of the associated homogeneous
equation: x1 + 2x2 − x3 = 0).
AFFINE AND PROJECTIVE GEOMETRY, E. Rosado & S.L. Rueda
Therefore, we have:
f (0, 0, 0) = (3, 1, 1)
f (0, 0, 1) = (0, 0, 1)
f¯(u1) = u1 =⇒ f¯(1, 0, 1) = (1, 0, 1)
f¯(u2) = u2 =⇒ f¯(0, 1, 2) = (0, 1, 2)
From the first two conditions we obtain
f¯(OP ) = f (P ) − f (O) = (0, 0, 1) − (3, 1, 1)
= (−3, −1, 0).
AFFINE AND PROJECTIVE GEOMETRY, E. Rosado & S.L. Rueda
Since
1 0 0 0
0 0 1 0
0
M (R , R) = ,
0 0 0 1
1 1 1 2
AFFINE AND PROJECTIVE GEOMETRY, E. Rosado & S.L. Rueda
we have
Mf (R, R) = Mf (R0, R) · M (R0, R)−1
−1
1 0 0 0 1 0 0 0
3 −3 1 0 0 0 1 0
1 −1 0 1 0 0 0 1
1 0 1 2 1 1 1 2
1 0 0 0
6 4 6 −3
= .
2 1 3 −1
1 1 2 0
So the analytic expression of f is:
f (x1, x2, x3) = (6 + 4x1 + 6x2 − 3x3, 2 + x1 + 3x2 − x3, 1 + x1 + 2x2).