Notes 18: Rotations in R2, R3
Lecture November 10, 2010
Problem 1. Let L be a line through the origin in R3 . Find explicitly the matrix for
rotation about L by a given angle θ.
We sketch the solution. We find a basis of R3 so that we can easliy write down the
matrix for this rotation using this basis. Then we use our ability to change coordinates to
find the matrix in the usual coordinates in R3 .
We begin by finding the matrix of some simple rotations in R3 . Let
 
 
 
1
0
0





E = {e1 = 0 , e2 = 1 , e3 = 0}.
0
0
1
Case 1. Assume that L is the z-axis and let Rz denote rotation by angle θ about L. Note
that Rz (e3 ) = e3 . Note that Rz (e1 ) and R(e2 ) lie in the xy-plane, that is, the plane spanned
by e1 , e2 . This means that the third coordinate of
) and
is 0.
Rz (e1 Rz (e2 ) cos(θ)
− sin(θ)
The first two coordinates of Rz (e1 ), Rz (e2 ) are
,
. Set c = cos(θ), s =
sin(θ)
cos(θ)
sin(θ). We find that the matrix of Rz is


c −s 0
Rz = s c 0 .
0 0 1
Case 2. We assume that L is the y-axis and let Ry rotate about L so that the x-axis moves
towards the z-axis. Note that Ry (e2 ) = e2 . Note also that Ry (e1 ) and Ry (e3 ) remain in
the e1 , e2 plane. This means that the second coordinate of Ry (e1 ), Ry (e3 ) is 0. In the e1 , e3
plane Ry behaves like rotation in R2 . We conclude that


c 0 −s
Ry = 0 1 0  .
s 0 c
Definition 1. We say that a basis F = {f1 , f2 , · · · , fn } of Rn is orthonormal provided the
length of fi , i = 1, · · · , n is 1 and fi is orthogonal to fj if i 6= j.
Case 3. The matrix of rotation by angle θ in R2 with respect to any orthonormal basis is
c −s
.
s c
Case 4. Let F = {f1 , f2 , f3 } be an orthonormal basis of R3 . Rotation by angle θ about the
line through the origin and f3 is


c −s 0
R = s c 0 .
0 0 1
1
We have done the first step towards solving our problem. We have found the matrix of
the rotation about a line L through the origin. We just use an orthonormal basis F with
the third basis element passing through the line L.
We have to find the change of basis matrix from the basis F to the basis E. Recall
that the change of basis matrix 1E←F has for columns the expresssion in E-coordinates of
the vectors fi . Hence our task is to find the E-coordinates of a basis that is orthonormal
and the vector f3 is in L.
Since rotations preserved distances and angles, a rotation matrix maps an orthonormal
basis to an orthonormal basis. If we can find a rotation or a composition of rotations that
maps e3 to a vector in L we will have reached our objective. The columns of such a matrix
will be the basis F expressed in E coordinates. Indeed it will be the change of basis matrix
1E←F .
We describe the line L. We identify origin of R3 with the center of the earth. The line
L passes through the origin and a point on the earth described by longitude and latitude.
Recall that latitude is measured from the equator (where the latitude is 0) up to the
north pole where the latitude is 90. Longitude is measured from Greenwich, England. The
longitude of Kansas City is about 90 west. In this set up, the y-axis passes through the
equator near the Ivory Coast and the x-axis passes through the equator near the Galopagos
Islands.
We now explain how to use the rotations Rx and Rz to move e3 so that it passes
through, for example, Brooklyn.
• We rotate about the x axis so that we move e3 so that it points through the correct
line of latitude. The latitude of Brooklyn is α1 = 40 . Since latitude is measured
from the equator, we replace α1 with π/2 − α1 . Second note that the rotation moves
the y-axis towards the z axis and we wish to rotate in the opposite direction. Thus
we replace π/2 − α1 with α Let Rx denote the rotation we have constructed by angle
alpha. Let f denote Rx (e3 ).
• We rotate about the z-axis bringing f so that it lands on Brooklyn. We need to
rotate by β = −74, where 74◦ W is the longitude of Brooklyn. Let Rz denote the
rotation about the z-axis by angle β.
• Let A = Rz ◦ Rx . The columns of A are the elements of the basis F . We have
constructed F so that it is orthonormal and A(e3 ) = f3 points to Brooklyn. The
change of basis matrix from F coordinates to E coordinates is 1E←F = A.
We are now in a position to write down the matrix which rotates about an axis through
the center of the earth and Brooklyn by angle theta. We take the matrix R which gives
the rotation in the F coordinates and apply the change of coordinates, that is
1E←F ◦ R ◦ 1F ←E = ARA−1 .
Complex Numbers and Rotations in R2
2
p
Given a1 + ia2 ∈ C, we can write a1 + ia2 = ρ(b1 + ib2 ) with ρ = a21 + a22 . From this
we see that b21 + b22 = 1, so that there is an θ so that cos(θ) = b1 , sin(θ) = b2 . Putting this
together we see that we can write
a1 + ia2 = ρ(cos(θ) + i sin(θ)).
This is called the polar representation of the complex number a1 + ia2 .
Consider the map
m:C→C
z 7→ (a1 + ia2 )z.
a1 −a2
This has the matrix representation M =
with respect to the basis {1, i} of C.
a2 a1
From our calculations above we can write this as
a1 −a2
b1 −b2
c −s
M=
=ρ
=ρ
a2 a1
b2 b1
s c
if we set s = sin(θ), c = cos(θ). We see that the map m is the composition of dilation
by ρ and rotation by θ. This answers the question: What is the geometry of complex
multiplication?
We give an application of this insight. Let z = cos(θ)+i sin(θ), w = cos(φ)+i sin(φ). let
mz , mw denote multiplication by z, w in the complex numbers. We have mz ◦ mw = mzw .
We have
zw = (cos(θ) cos(φ) − sin(θ) sin(φ)) + i(cos(θ) sin(φ) + cos(φ) sin(θ)).
On the other hand mz ◦ mw is the composition of rotation by φ with rotation by θ). This
is rotation by θ + φ which is given by multiplication by
cos(θ + φ) + i sin(θ + φ).
Comparing the two displayed equations gives us the addition laws for trig functions.
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