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Geometric Transformations and Wallpaper
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Isometries of the Plane
Lance Drager
Department of Mathematics and Statistics
Texas Tech University
Lubbock, Texas
2010 Math Camp
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Outline
Lance Drager
Introduction
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Orthogonal
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Classifying
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Matrices
Exercises
Classifying
Isometries
Translations
First
Classification
Final
Classification
Exercises
1 Introduction
2 Orthogonal Matrices
Orthogonal Matrices and Dot Product
Classifying Orthogonal Matrices
Exercises
3 Classifying Isometries
Translations
First Classification
Final Classification
Exercises
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Isometries
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Translations
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Final
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• A transformation T of the plane is an isometry if it is
one-to-one and onto and
dist(T (p), T (q)) = dist(p, q),
for all points p and q.
• If S and T are isometries, so is ST , where
(ST )(p) = S(T (p)).
• If T is an isometry, so is T −1 .
• Our goal is to find all the isometries.
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Final
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• A matrix A is orthogonal if kAxk = kxk for all x ∈ R2 .
• An orthogonal matrix gives an isometry of the plane since
dist(Ap, Aq) = kAp − Aqk
= kA(p − q)k = kp − qk = dist(p, q)
• If A and B are orthogonal, so is AB.
• If A is orthogonal, it is invertible and A−1 is orthogonal.
• Rotation matrices are orthogonal.
y
q
kp − qk
p
x
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Dot Product 1
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Theorem
A is orthogonal if and only if Ax · Ay = x · y for all x, y ∈ R2 .
Thus, an orthogonal matrix preserves angles.
Proof.
(=⇒)
kAxk2 = Ax · Ax = x · x = kxk2
(⇐=) Recall
2x · y = kxk2 + ky k2 − kx − y k2 .
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Proof Continued.
So, we have
2Ax · Ay = kAxk2 + kAy k2 − kAx − Ay k2
= kAxk2 + kAy k2 − kA(x − y )k2 (distributive law)
Classifying
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= kxk2 + ky k2 − kx − y k2
Translations
First
Classification
Final
Classification
Exercises
(A is orthogonal)
= 2x · y ,
and dividing by 2 gives the result.
Theorem
A matrix is orthogonal if and only if its columns are orthogonal
unit vectors.
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Proof.
Orthogonal
Matrices
(=⇒)
Orthogonal
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kAei k = kei k = 1.
Ae1 · Ae2 = e1 · e2 = 0.
(⇐=) Let A = [u | v ] where u and v are orthogonal unit
vectors. Note Ax = x1 u + x2 v .
kAxk2 = Ax · Ax
= (x1 u + x2 v ) · (x1 u + x2 v )
= x12 u · u + 2x1 x2 u · v + x22 v · v
= x12 (1) + 2x1 x2 (0) + x22 (1)
= x12 + x22 = kxk2
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Classifying Orthogonal Matrices 1
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• If A is orthogonal, Col1 (A) = (cos(θ), sin(θ)) for some θ.
So the possibilities are
cos(θ) − sin(θ)
A=
= R(θ),
sin(θ)
cos(θ)
cos(θ)
sin(θ)
A=
= S(θ).
sin(θ) − cos(θ)
• In the first case A = R(θ) is a rotation.
• What about S(θ)?
• There are orthogonal unit vectors u and v so that
S(θ)u = u and S(θ)v = −v .
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Classifying Orthogonal Matrices 2
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• In fact, let u = (cos(θ/2), sin(θ/2)) and
v = (− sin(θ/2), cos(θ/2)).
•
S(θ)u =
cos(θ)
sin(θ)
sin(θ) − cos(θ)
cos(θ/2)
sin(θ/2)
cos(θ) cos(θ/2) + sin(θ) sin(θ/2)
=
sin(θ) cos(θ/2) − cos(θ) sin(θ/2)
cos(θ − θ/2)
cos(θ/2)
=
=
= u.
sin(θ − θ/2)
sin(θ/2)
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•
S(θ)v =
cos(θ)
sin(θ)
sin(θ) − cos(θ)
− sin(θ/2)
cos(θ/2)
− cos(θ) sin(θ/2) + sin(θ) cos(θ/2)
=
− sin(θ) sin(θ/2) − cos(θ) cos(θ/2)
sin(θ − θ/2)
=
− cos(θ − θ/2)
sin(θ/2)
− sin(θ/2)
=
=−
= −v .
− cos(θ/2)
cos(θ/2)
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Classifying Orthogonal Matrices 4
• S(θ)u = u and S(θ)v = −v . If x = tu + sv , then
S(θ)x = tu − sv .
y
x
sv
Classifying
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M
tu
v
u
−sv
x
S(θ)x
• S(θ) is a reflection with its mirror line at an angle of θ/2.
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Exercises
• S(θ)2 = I , so S(θ)−1 = S(θ).
• Let A be an orthogonal matrix. Then A is a rotation (or
the identity) if and only if det(A) = 1 and A is a reflection
if and only if det(A) = −1.
a c
• If A =
define the transpose of A by
b d
a b
AT =
. Show that A is orthogonal if and only if
c d
A−1 = AT .
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2
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Exercises Continued
• Use the addition laws for sine and cosine to verify the
important identities
R(θ)S(φ) = S(θ + φ),
S(φ)R(θ) = S(φ − θ),
S(θ)S(φ) = R(θ − φ)
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• For v ∈ R2 , define Tv : R2 → R2 by Tv (x) = x + v . We
say Tv is translation by v.
−1
• Tu Tv = Tu+v and Tv
= T−v .
• Translations are isometries
dist(Tv (x), Tv (y )) = kTv (x) − Tv (y )k
= k(x + v ) − (y + v )k
= kx + v − y − v k
= kx − y k
= dist(x, y ).
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Three Distances Determine a Point
• Let p1 , p2 and p3 be noncolinear points. A point x is
uniquely determined by the three numbers r1 = dist(x, p1 ),
r2 = dist(x, p2 ) and r3 = dist(x, p3 ).
Classifying
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p1
p2
p3
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• If an isometry T fixes three noncolinear points p1 , p2 , p3
then T is the identity.
dist(x, pi ) = dist(T (x), T (pi )) = dist(T (x), pi ),
i = 1, 2, 3,
=⇒ x = T (x).
Theorem
Every isometry T can be written as T (x) = Ax + v where A is
an orthogonal matrix, i.e., T is multiplication by an orthogonal
matrix followed by a translation.
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Proof.
1
The points 0, e1 , e2 are noncolinear.
2
Let u = −T (0). Then Tu T (0) = 0.
3
dist(Tu T (e1 ), 0) = 1, so we can find a rotation matrix R
so that RTu T (e1 ) = e1 .
4
RTu T (e2 ) = ±e2 .
5
If RTu T (e2 ) = −e2 , let S = S(0) be reflection through
the x-axis, otherwise let S = I .
6
SRTu (0) = 0, SRTu T (e1 ) = e2 , and SRTu T (e2 ) = e2
7
SRTu T = I , so T = T−u R −1 S −1
8
Let A = R −1 S −1 and v = −u. Then T (x) = Ax + v
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• If T (x) = Ax + v write T = (A | v ).
• Calculate the product (composition) in this notation
(A | u)(B | v )x = (A | u)(Bx + v )
= A(Bx + v ) + u
Classifying
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= ABx + Av + u
Translations
First
Classification
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= (AB | u + Av )x.
• (A | u)(B | v ) = (AB | u + Av )
• The identity transformation is (I | 0). Translation by v is
(I | v ).
• (A | u)−1 = (A−1 | − A−1 u)
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• Consider (R | v ) where R = R(θ) 6= I is a rotation.
• Look for a fixed point p, i.e., (R | v )p = p
Rp + v = p
⇐⇒ p − Rp = v
⇐⇒ (I − R)p = v .
• If (I − R) is invertible, there is a unique solution p.
• (I − R) is invertible because
(I − R)x = 0 ⇐⇒ x = Rx ⇐⇒ x = 0. (Use the Big
Theorem.)
• There is a unique fixed point p, and we can write
(R | v ) = (R | p − Rp).
• (R | p − Rp)x = Rx + p − Rp = p + R(x − p).
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• y = (R | p − Rp)x = p + R(x − p) says that this isometry
is rotation though angle θ around the point p,which is
called the center of rotation or rotocenter.
y
y
R(x − p)
Translations
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Classification
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Classification
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θ
x−p
x
p
x
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• Consider (S | w ) where S = S(θ) is a reflection matrix.
• Let u and v be the vectors with Su = u and Sv = −v .
We can write w = αu + βv , so (S | w ) = (S | αu + βv ).
• First Case: α = 0. So we have (S | βv ).
• The point p = βv /2 is fixed.
(S | βv )p = S(βv /2) + βv = −βv /2 + βv = βv /2 = p.
• The fixed points are exactly x = p + tu.
(S | 2p)(p + tu) = Sp + tSu + 2p = −p + tu + 2p = p + tu.
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• The line through p parallel to u is parametrized by
x = p + tu, for t ∈ R.
y
tu
p
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Exercises
x
v
u
x
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• We can write a point x as x = p + tu + sv . Then
y = (S | 2p)x = p + tu − sv .
y
x
sv
tu + sv
M
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tu
p
−sv
tu − sv
v
y
u
x
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• (S | βv ) = (S | 2p) is reflection through the mirror line M
given by y = p + tu.
• Second Case: α 6= 0. We have (S | αu + βv ).
(S | αu + βv ) = (I | αu)(S | βv )
This is a reflection followed by a translation parallel to the
mirror line. This is called a glide refection or just a glide.
The mirror line of the reflection is called the glide line.
• A glide has no fixed points.
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Glide Reflection Picture
• A glide with a horizontal glide line, and the translation
vector shown in blue.
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Theorem
Every isometry of the plane falls into on of the following five
mutually exclusive classes.
1
The identity.
2
A translation (not the identity).
3
A rotation about some point (not the identity);
4
A reflection through some mirror line.
5
A glide along some glide line.
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Exercises
1
Consider the cases for the product T1 T2 of two isometries
T1 and T2 . Describe what happens geometrically (e.g.,
rotation angle, location of the mirror line, etc.). In what
cases do the isometries commute, i.e., when does
T1 T2 = T2 T1
2
Project: For a rotation (R | v ) give a geometric description
of how to find the rotocenter p = (I − R)−1 v .
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3
As part of the first exercise, given two rotations (with
possibly different rotocenters) show that the composition
is a rotation or a translation.
4
Project: Given two rotations with different rotocenters
give a geometric description of how to find the rotocenter
of the composition.