Download 4. Topic

4. Topic
• Cramer’s Rule
• Speed of Calculating Determinants
• Projective Geometry
Cramer’s Rule
x1  2 x2  6
3x1  x2  8
~
1
 2  6
x1    x2     
 3
 1 8
Compare the sizes of these shaded boxes.
6 2
6 2
8 1

x1  1 2
x1  3 1
 x1
1 2
3 1
x1 
8 1
10

2
1 2
5
3 1
Cramer’s Rule:
Bi
xi 
A
If |A|  0, then A x = b has a unique solution give by
where Bi is obtained by replacing column i of A by b.
Example:
1 0 4  x   2 
 2 1 1   y    1 

   
 1 0 1   z   1 

   
→
1
2
4
2
1
1
1 1 1
y
1 0 4
2 1 1
1 0
1
18

6
3
Exercises 4.Topic.1.
1. The first picture in this Topic (the one that doesn’t use determinants)
shows a unique solution case. Produce a similar picture for the case of
infintely many solutions, and the case of no solutions.
Speed of Calculating Determinants
Computational complexity: P & NP problems.
Speed of calculation inversely  Number of arithematic operations required.
Permutation expansion: Sum of n! terms each involving n multiplications.
Stirling’s formula: ln n!  n ln n  n = ln (n/e) n → n!  nn.
Row reduction: 2 nested loops of n  n2 operations.
Projective Geometry
Perspective: Map (project) the 3-D scene to a 2-D image.
Parallel lines seem to meet at the vanishing point.
Central projection: from a single point to the canvas.
Non-orthogonal,
non-isometric.
The study of the effects of central projections is projective geometry.
3 types of central projection:
Movie projector:
( Mapper P
pushes S out to I )
Painter:
(Mapper P pulls
S back to I )
Pinhole:
(Mapper P inbetween S & I )
Railroad tracks that appear to converge to a point.
Push out
Pull back
In between.
Green line has no image.
Vanishing point not image
of point on S.
Projective geometry:
Points on the same line passing through 0 belong to the same class.
For any nonzero vR3, the associated point v in the projective plane RP2 is the set
 kv
k  R and k  0 
Each kv is a homogeneous coordinate vector for v.
Dome model: v is chosen to be on the upper hemi-sphere.
Deficient of the dome model:
Images of points below equator appear behind mapper P.
Remedy: Identify antipodal points on sphere.
(This device is used in the study of SO(3) & SU(2) groups.)
The intersect of a plane through 0 with the sphere is a great
circle, which defines a line in RP2 in the following manner.
A plane through 0 in R3 is the set
P a , b, c 





 x
 y  ax  by  cz  0
 
z
 





P(a,b,c) is associated with a line L in RP2 :
L   kL k  R , k  0 
where L   a
b c
Usually, we just refer to this as line L = ( a, b, c ).
A point v & a line L are incident (v is on L) iff L v = 0.
Proof: The normal of P(a,b,c) is the column vector LT .
By definition, any point on P(a,b,c) satisfies L v = 0.
is a row vector
(1-form).
Duality principle of projective geometry:
Interchanging ‘point’ & ‘line’ in any true statement results in another true one.
E.g., 2 distinct points in RP2 determine a unique line.

2 distinct lines in RP2 determine a unique point.
In contrast, 2 distinct lines in R3 may or may not intersect.
Projective geometry is simpler, more uniform, than Euclidean geometry.
The projective plane can be viewed as an extension of the Euclidean plane.
Points on equator are extra since they do
not project onto the Euclidean plane.
Equator = ideal line = line at infinity
Parallel lines in R3 intersect at equator of RP2.
Linear algebra is a natural tool for analytic projective geometry.
E.g., 3 points t, u, v are collinear (incident in a single line) iff
t1
u1
v1
t2
u2
v2  0
t3
u3
v3
By duality, 3 lines incident on a point iff the representative row vectors are L.D.
Equation of a point v is the equation satisfied by any line incident on it, i.e.,
v1 L1 + v2 L2 + v3 L3 = 0
The shaded triangles are in perspective from O
because their corresponding vertices are collinear.

Desargue’s
Theorem
Lemma:
If W, X, Y , Z are four points in the projective plane (no three of which are collinear)
then there are homogeneous coordinate vectors w, x, y, and z for the projective points,
and a basis B for R3, satisfying the following:
1
w   0 
0
 B
 0
x   1 
 0
 B
 0
y   0 
1
 B
 1
z   1
 1
 B
Proof:
Since W, X, Y are not on the same projective line,
any homogeneous coordinate vectors w0 , x0 , y0 do not lie on the same plane
through 0 in R3 and so form a spanning set for R3.
 z0 ,  a, b, c s.t.
z0  a w0  b x0  c y 0
Proof is complete by setting B =  w, x, y , where w = a w0 , x = b x0 , y = c y0 .
Desargue’s Theorem
Let
1
t1   0 
 0
 B
0
u1   1 
0
 B
Since T2 is incident on the projective line OT1,
 1
O   1
 1
 B
 t2 
 1
1
 
t 2  a  1  b  0    1 
1
 1
 0
 B
 B   B
Similarly,
1
u2   u2 
1
 B
 0
v1   0 
1
 B
1
v 2   1 
v 
 2 B
The projective line T1U1 is the image of a plane through 0 in R3 and satisfies
1 0 x
0 1
y 0
0 0
z
→
z=0
For T2U2 , we have

t2
1
x
1
u2
y 0
1
1
z
T1U1
→
 t2  1 
T2U 2   1  u2 
 0 


1  u2  x  1  t2  y  t2u2  1 z  0
Similarly
These are on same projection line since their sum = 0.
QED
T1V1
U 1V1
 1  t2 
T2V2   0 
 v  1
 2 
 0 
U 2V2   u2  1
1  v 
2

Every projective theorem has a translation to a Euclidean version, although
the Euclidean result is often messier to state and prove.
Euclidean pictures can be thought of as figures from projective geometry for
a model of very large radius.
(Projective plane is ‘locally Euclidean’.)
The projective plane is not orientable:
Exercises 4.Topic.3.
1. What is the equation of this point?
2.
(a)
1
0
 
0
 
Find the line incident on these points in the projective plane.
 1  4
 2 ,  5
   
 3  6
   
(b) Find the point incident on both of these projective lines.
1
2 3 ,  4 5 6