Linear Algebra Review
Why do we need Linear Algebra?
• We will associate coordinates to
– 3D points in the scene
– 2D points in the CCD array
– 2D points in the image
• Coordinates will be used to
– Perform geometrical transformations
– Associate 3D with 2D points
• Images are matrices of numbers
– We will find properties of these numbers
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Octavia I. Camps
2
Matrices
Anm
 a11 a12
a
 21 a22
 a31 a32


 
an1 an 2
Example:
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 a1m 
 a2 m 
 a3m 

  
 anm 
Sum:
Cnm  Anm  Bnm
cij  aij  bij
A and B must have the same
dimensions
2 5 6 2 8 7
3 1  1 5  4 6

 
 

Octavia I. Camps
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Matrices
Product:
Cn p  Anm Bm p
m
cij   aik bkj
A and B must have
compatible dimensions
Ann Bnn  Bnn Ann
k 1
Examples:
2 5 6 2 17 29
3 1.1 5  19 11


 

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6 2 2 5 18 32
1 5.3 1  17 10 


 

Octavia I. Camps
4
Matrices
Transpose:
Cmn  A nm
cij  a ji
( A  B)  A  B
T
If
AT  A
T
T
( AB)  B A
T
T
T
A is symmetric
Examples:
T
T
6 2 
6 1
1 5  2 5




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T
6 2 
1 5  6 1 3
2 5 8




3 8
Octavia I. Camps
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Matrices
Determinant:
A must be square
 a11 a12  a11 a12
det 

 a11a22  a21a12

a21 a22  a21 a22
 a11 a12
det a21 a22
a31 a32
a13 
a22

a23   a11
a32

a33 
Example:
2 5
det 
 2  15  13

3 1
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a23
a33
 a12
a21 a23
a31 a33
Octavia I. Camps
 a13
a21 a22
a31 a32
6
Matrices
Inverse:
Ann A
A must be square
1
nn
A
1
nn
1
Ann  I
 a11 a12 
1

a

a
a11a22  a21a12
21
22


 a22  a12 
 a

a
21
11


1
Example:
6 2 
1  5  2
1 5  28  1 6 




1
6 2 6 2 1  5  2 6 2 1 28 0  1 0
1 5 .1 5  28  1 6 .1 5  28  0 28  0 1

 





 

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Octavia I. Camps
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2D Vector
x2
v  ( x1 , x2 )
Magnitude:
|| v || x1  x2
If || v || 1 ,
v
2
2
P
v

x1
Is a UNIT vector
 x1
v
x2 
 Is a unit vector
 
,
|| v ||  || v || || v || 
Orientation:
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 x2 
  tan  
 x1 
Octavia I. Camps
1
8
Vector Addition
v  w  ( x1 , x2 )  ( y1 , y2 )  ( x1  y1 , x2  y2 )
v
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V+w
w
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Vector Subtraction
v  w  ( x1 , x2 )  ( y1 , y2 )  ( x1  y1 , x2  y2 )
v
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V-w
w
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Scalar Product
av  a( x1 , x2 )  (ax1 , ax2 )
av
v
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Inner (dot) Product
v

w
v.w  ( x1 , x2 ).( y1 , y2 )  x1 y1  x2 . y2
The inner product is a SCALAR!
v.w  ( x1 , x2 ).( y1 , y2 ) || v ||  || w || cos
v.w  0  v  w
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Octavia I. Camps
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Orthonormal Basis
P
x2
j
v

i
x1
v  ( x1 , x2 )
i  (1,0)
|| i || 1
j  (0,1)
|| j || 1
ij 0
v  x1.i  x2 .j
v.i  ( x1.i  x2 .j).i  x1.1  x2 .0  x1
v.j  ( x1.i  x2 .j).j  x1.0  x2 .1  x2
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Vector (cross) Product
u w
u  v w

v
Magnitude:
Orientation:
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The cross product is a VECTOR!
|| u ||  || v.w |||| v ||| w || sin 
u  v  u  v  (v  w)  v  0
u  w  u  w  (v  w)  w  0
Octavia I. Camps
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Vector Product Computation
k  (0,0,1)
j  (0,1,0)
|| k || 1
|| j || 1
i  (1,0,0)
|| i || 1
i  j  0, i  k  0, j  k  0
u  v  w  ( x1 , x2 , x3 )  ( y1 , y2 , y3 )
i
j k
u w

u  x1 x2 x3
v
y1 y2 y3
 ( x2 y3  x3 y2 )i  ( x3 y1  x1 y3 ) j  ( x1 y2  x2 y1 )k
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2D Geometrical
Transformations
2D Translation
P’
t
P
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2D Translation Equation
ty
P
P’
t
P  ( x, y )
t  (t x , t y )
y
x
tx
P'  ( x  t x , y  t y )  Pt
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2D Translation using Matrices
ty
P
P’
t
y
x
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P  ( x, y )
t  (t x , t y )
t
tx
 x
 x  t x  1 0 t x   
P'  

  y


 y  t y  0 1 t y   1 
 
Octavia I. Camps
P
19
Homogeneous Coordinates
• Multiply the coordinates by a non-zero
scalar and add an extra coordinate equal to
that scalar. For example,
( x, y )  ( x  z , y  z , z ) z  0
( x, y, z )  ( x  w, y  w, z  w, w) w  0
• NOTE: If the scalar is 1, there is no need
for the multiplication!
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Back to Cartesian Coordinates:
• Divide by the last coordinate and eliminate it. For
example,
( x, y , z ) z  0  ( x / z , y / z )
( x, y, z, w) w  0  ( x / w, y / w, z / w)
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2D Translation using
Homogeneous Coordinates
ty
P
P’
t
y
x
tx
P  ( x, y )  ( x, y,1)
t  (t x , t y )  (t x , t y ,1)
P
 x  t x  1 0 t x   x 
P'   y  t y   0 1 t y   y 
 1  0 0 1   1 
P'  T  P
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t
Octavia I. Camps
T
22
Scaling
P’
P
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Scaling Equation
P’
Sy.y
P  ( x, y )  ( x, y,1)
P'  ( s x x, s y y )  ( s x x, s y y,1)
P
y
x
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Sx.x
 sx x  sx



P'   s y y    0
 1   0
0
sy
P'  S  P
S
Octavia I. Camps
0
0  x 



0  y 
1  1 
24
Scaling & Translating
S
P’’=T.P
P’=S.P
T
P
P’’=T.P’=T.(S.P)=(T.S).P
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Scaling & Translating
P’’=T.P’=T.(S.P)=(T.S).P
1 0 t x   s x 0



P ' '  T  S  P  0 1 t y   0 s y
0 0 1   0 0
sx 0 t x   x   sx x  t x 






  0 s y t y   y   s y y  t y 
 0 0 1   1   1 
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Octavia I. Camps
0  x 



0  y  
1  1 
26
Translating & Scaling
 Scaling & Translating
P’’=S.P’=S.(T.P)=(S.T).P
 s x 0 0 1 0 t x  x 






P' '  S  T  P   0 s y 0 0 1 t y  y  
 0 0 1 0 0 1  1 
sx 0 sxt x   x   sx x  sxt x 






  0 s y s yt y   y   s y y  s yt y 
 0 0

1   1  
1
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Rotation
P
P’
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Rotation Equations
Counter-clockwise rotation by an angle 
Y’
P’

P
y
X’
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 x' cos
 y '   sin 
  
x
 sin    x 



cos   y 
P'  R.P
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Degrees of Freedom
 x' cos
 y '   sin 
  
R is 2x2
 sin    x 



cos   y 
4 elements
BUT! There is only 1 degree of freedom: 
The 4 elements must satisfy the following constraints:
R  RT  RT  R  I
det( R )  1
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Scaling, Translating & Rotating
Order matters!
P’ = S.P
P’’=T.P’=(T.S).P
P’’’=R.P”=R.(T.S).P=(R.T.S).P
R.T.S  R.S.T  T.S.R …
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3D Rotation of Points
Rotation around the coordinate axes, counter-clockwise:
Y’
P’

P
y
X’
z
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0
1
Rx ( )  0 cos 
0 sin 
x

 sin  
cos  
0 sin  
1
0 
0 cos  
 cos 
R y (  )   0
 sin 
cos   sin 
Rz ( )   sin  cos 
 0
0
Octavia I. Camps
0
0
0
1
32
3D Rotation (axis & angle)
n  n1 n2
n3  , angle 
T
 n12

R  I cos  I(1  cos ) n1n2
n n
 1 3
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n1n2
2
n2
n2 n3
n1n3 
 0

n2 n3   sin   n3
2
 n2
n3 
Octavia I. Camps
 n3
0
n1
33
n2 
 n1 
0 
3D Translation of Points
Translate by a vector t=(tx,ty,tx)T:
P’
t
Y’
z’
z
x’
P
x
1
0
T 
0

0
0 0 tx 

1 0 ty 
0 1 tz 

0 0 1
y
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