Shape Matching
Wanho Choi
(wanochoi.com)
•
What is the
best
4 by 4
similarity transformation
matrix
between two point clouds which have same # of points?
Shape Matching Problem
P
= p
{ }
iQ
= q
{ }
ii
= 1,0,!,n
(
)
(
i
= 1,0,!,n
)
(translation, rotation, and isotropic scale)
argmin
M
Mp
i
− q
i
2
i
=1
n
∑
⎡
⎣⎢
⎤
⎦⎥
Shape Matching Problem
argmin
M
Mp
i
− q
i
2
i
=1
n
∑
⎡
⎣⎢
⎤
⎦⎥
A : 3 by 3 matrix rotation & scale
(
)
t : 3D translation
Translation is easy!
A : 3 by 3 matrix rotation & scale
(
)
t : 3D translation
c
p
=
1
n
i
=1
p
i
n
∑
c
q
=
1
n
i
=1
q
i
n
∑
t
= c
q
− c
p
Mp
i
= q
i
⇔ Ap
i
+ t = q
i
argmin
M
Mp
i
− q
i
2
i
=1
n
∑
⎡
⎣⎢
⎤
⎦⎥
Relative Positions
p
i
= p
i
− c
p
q
i
= q
i
− c
q
A : 3 by 3 matrix rotation & scale
(
)
t : 3D translation
Mp
i
= q
i
⇔ Ap
i
+ t = q
i
argmin
M
Mp
i
− q
i
2
i
=1
n
∑
⎡
⎣⎢
⎤
⎦⎥
Ap
i
= q
i
Minimization
argmin
A
Ap
i
− q
i
2
i
=1
n
∑
⎡
⎣⎢
⎤
⎦⎥
Minimization
∂
∂A
Ap
i− q
i 2 i=1 n∑
⎡
⎣⎢
⎤
⎦⎥
argmin
A
Ap
i
− q
i
2
i
=1
n
∑
⎡
⎣⎢
⎤
⎦⎥
Minimization
∂
∂A
Ap
i− q
i 2 i=1 n∑
⎡
⎣⎢
⎤
⎦⎥
= ∂
∂A
Ap
1− q
1 2⎡
⎣
⎤
⎦ +
∂A
∂
Ap
2− q
2 2⎡
⎣
⎤
⎦ +!+
∂A
∂
Ap
n− q
n 2⎡
⎣
⎤
⎦
argmin
A
Ap
i
− q
i
2
i
=1
n
∑
⎡
⎣⎢
⎤
⎦⎥
Minimization
∂
∂A
Ap
i− q
i 2 i=1 n∑
⎡
⎣⎢
⎤
⎦⎥
= ∂
∂A
Ap
1− q
1 2⎡
⎣
⎤
⎦ +
∂A
∂
Ap
2− q
2 2⎡
⎣
⎤
⎦ +!+
∂A
∂
Ap
n− q
n 2⎡
⎣
⎤
⎦
∵ ∂ ∂A Api − qi 2 ⎡ ⎣ ⎤⎦ = 2 Ap(
i − qi)
pi T ⎛ ⎝⎜ ⎞⎠⎟= 2 Ap
(
1− q
1)
p
1T+ 2 Ap
(
2− q
2)
p
2T+!+ 2 Ap
(
n− q
n)
p
nTargmin
A
Ap
i
− q
i
2
i
=1
n
∑
⎡
⎣⎢
⎤
⎦⎥
Minimization
∂
∂A
Ap
i− q
i 2 i=1 n∑
⎡
⎣⎢
⎤
⎦⎥
= ∂
∂A
Ap
1− q
1 2⎡
⎣
⎤
⎦ +
∂A
∂
Ap
2− q
2 2⎡
⎣
⎤
⎦ +!+
∂A
∂
Ap
n− q
n 2⎡
⎣
⎤
⎦
∵ ∂ ∂A Api − qi 2 ⎡ ⎣ ⎤⎦ = 2 Ap(
i − qi)
pi T ⎛ ⎝⎜ ⎞⎠⎟= 2 Ap
(
1− q
1)
p
1T+ 2 Ap
(
2− q
2)
p
2T+!+ 2 Ap
(
n− q
n)
p
nT= 2 A p
⎡⎣
( )
1p
1T− q
( )
1p
1T⎤⎦+ 2 A p
⎡⎣
(
2p
2T)
− q
( )
2p
2T⎤⎦+!+ 2 A p
⎡⎣
(
np
nT)
− q
( )
np
nT⎤⎦
argmin
A
Ap
i
− q
i
2
i
=1
n
∑
⎡
⎣⎢
⎤
⎦⎥
Minimization
∂
∂A
Ap
i− q
i 2 i=1 n∑
⎡
⎣⎢
⎤
⎦⎥
= ∂
∂A
Ap
1− q
1 2⎡
⎣
⎤
⎦ +
∂A
∂
Ap
2− q
2 2⎡
⎣
⎤
⎦ +!+
∂A
∂
Ap
n− q
n 2⎡
⎣
⎤
⎦
∵ ∂ ∂A Api − qi 2 ⎡ ⎣ ⎤⎦ = 2 Ap(
i − qi)
pi T ⎛ ⎝⎜ ⎞⎠⎟= 2 Ap
(
1− q
1)
p
1T+ 2 Ap
(
2− q
2)
p
2T+!+ 2 Ap
(
n− q
n)
p
nT= 2 A p
⎡⎣
( )
1p
1T− q
( )
1p
1T⎤⎦+ 2 A p
⎡⎣
(
2p
2T)
− q
( )
2p
2T⎤⎦+!+ 2 A p
⎡⎣
(
np
nT)
− q
( )
np
nT⎤⎦
= 2A
p
ip
iT i=1 n∑
⎛
⎝⎜
⎞
⎠⎟
− 2
q
ip
i T i=1 n∑
⎛
⎝⎜
⎞
⎠⎟
argmin
A
Ap
i
− q
i
2
i
=1
n
∑
⎡
⎣⎢
⎤
⎦⎥
Minimization
argmin
A
Ap
i
− q
i
2
i
=1
n
∑
⎡
⎣⎢
⎤
⎦⎥
∂
∂A
Ap
i− q
i 2 i=1 n∑
⎡
⎣⎢
⎤
⎦⎥
= ∂
∂A
Ap
1− q
1 2⎡
⎣
⎤
⎦ +
∂A
∂
Ap
2− q
2 2⎡
⎣
⎤
⎦ +!+
∂A
∂
Ap
n− q
n 2⎡
⎣
⎤
⎦
∵ ∂ ∂A Api − qi 2 ⎡ ⎣ ⎤⎦ = 2 Ap(
i − qi)
pi T ⎛ ⎝⎜ ⎞⎠⎟= 2 Ap
(
1− q
1)
p
1T+ 2 Ap
(
2− q
2)
p
2T+!+ 2 Ap
(
n− q
n)
p
nT= 2 A p
⎡⎣
( )
1p
1T− q
( )
1p
1T⎤⎦+ 2 A p
⎡⎣
(
2p
2T)
− q
( )
2p
2T⎤⎦+!+ 2 A p
⎡⎣
(
np
nT)
− q
( )
np
nT⎤⎦
= 2A
p
ip
iT i=1 n∑
⎛
⎝⎜
⎞
⎠⎟
− 2
q
ip
i T i=1 n∑
⎛
⎝⎜
⎞
⎠⎟
= 0
Minimization
∂
∂A
Ap
i− q
i 2 i=1 n∑
⎡
⎣⎢
⎤
⎦⎥
= ∂
∂A
Ap
1− q
1 2⎡
⎣
⎤
⎦ +
∂A
∂
Ap
2− q
2 2⎡
⎣
⎤
⎦ +!+
∂A
∂
Ap
n− q
n 2⎡
⎣
⎤
⎦
∵ ∂ ∂A Api − qi 2 ⎡ ⎣ ⎤⎦ = 2 Ap(
i − qi)
pi T ⎛ ⎝⎜ ⎞⎠⎟= 2 Ap
(
1− q
1)
p
1T+ 2 Ap
(
2− q
2)
p
2T+!+ 2 Ap
(
n− q
n)
p
nT= 2 A p
⎡⎣
( )
1p
1T− q
( )
1p
1T⎤⎦+ 2 A p
⎡⎣
(
2p
2T)
− q
( )
2p
2T⎤⎦+!+ 2 A p
⎡⎣
(
np
nT)
− q
( )
np
nT⎤⎦
= 2A
p
ip
iT i=1 n∑
⎛
⎝⎜
⎞
⎠⎟
− 2
q
ip
i T i=1 n∑
⎛
⎝⎜
⎞
⎠⎟
= 0
∴A =
q
ip
iT i=1 n∑
⎛
⎝⎜
⎞
⎠⎟
p
ip
i T i=1 n∑
⎛
⎝⎜
⎞
⎠⎟
−1argmin
A
Ap
i
− q
i
2
i
=1
n
∑
⎡
⎣⎢
⎤
⎦⎥
A Simple Proof On
∂A
∂
⎡⎣
Ap
− q
2⎤⎦ = 2 Ap−q
(
)
p
TA Simple Proof On
∂A
∂
⎡⎣
Ap
− q
2⎤⎦ = 2 Ap−q
(
)
p
T Let A= a b c d ⎡ ⎣ ⎢ ⎤ ⎦ ⎥, p =⎡ xy ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥, and q= uv ⎡ ⎣ ⎢ ⎤ ⎦ ⎥.A Simple Proof On
∂A
∂
⎡⎣
Ap
− q
2⎤⎦ = 2 Ap−q
(
)
p
T Let A= a b c d ⎡ ⎣ ⎢ ⎤ ⎦ ⎥, p =⎡ xy ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥, and q= uv ⎡ ⎣ ⎢ ⎤ ⎦ ⎥.For simplicity, let's just look at the 2D case.
When θ∈!, ∂θ ∂A ≡ ∂θ ∂a ∂θ ∂b ∂θ ∂c ∂θ ∂d ⎡ ⎣ ⎢ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ ⎥ .
A Simple Proof On
∂A
∂
⎡⎣
Ap
− q
2⎤⎦ = 2 Ap−q
(
)
p
T Let A= a b c d ⎡ ⎣ ⎢ ⎤ ⎦ ⎥, p =⎡ xy ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥, and q= uv ⎡ ⎣ ⎢ ⎤ ⎦ ⎥.For simplicity, let's just look at the 2D case.
When θ∈!, ∂θ ∂A ≡ ∂θ ∂a ∂θ ∂b ∂θ ∂c ∂θ ∂d ⎡ ⎣ ⎢ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ ⎥ . ∂ ∂A Ap− q 2 ⎡⎣ ⎤⎦
A Simple Proof On
∂A
∂
⎡⎣
Ap
− q
2⎤⎦ = 2 Ap−q
(
)
p
T Let A= a b c d ⎡ ⎣ ⎢ ⎤ ⎦ ⎥, p =⎡ xy ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥, and q= uv ⎡ ⎣ ⎢ ⎤ ⎦ ⎥.For simplicity, let's just look at the 2D case.
When θ∈!, ∂θ ∂A ≡ ∂θ ∂a ∂θ ∂b ∂θ ∂c ∂θ ∂d ⎡ ⎣ ⎢ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ ⎥ . ∂ ∂A Ap− q 2 ⎡⎣ ⎤⎦ = ∂A∂ ⎡⎣⎢ a bc d ⎤ ⎦ ⎥⎡ xy ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥− uv ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ 2 ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥
A Simple Proof On
∂A
∂
⎡⎣
Ap
− q
2⎤⎦ = 2 Ap−q
(
)
p
T Let A= a b c d ⎡ ⎣ ⎢ ⎤ ⎦ ⎥, p =⎡ xy ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥, and q= uv ⎡ ⎣ ⎢ ⎤ ⎦ ⎥.For simplicity, let's just look at the 2D case.
When θ∈!, ∂θ ∂A ≡ ∂θ ∂a ∂θ ∂b ∂θ ∂c ∂θ ∂d ⎡ ⎣ ⎢ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ ⎥ . ∂ ∂A Ap− q 2 ⎡⎣ ⎤⎦ = ∂A∂ ⎡⎣⎢ a bc d ⎤ ⎦ ⎥⎡ xy ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥− uv ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ 2 ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ = ∂∂A ax+ by cx+ dy ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥− uv ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ 2 ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥
A Simple Proof On
∂A
∂
⎡⎣
Ap
− q
2⎤⎦ = 2 Ap−q
(
)
p
T Let A= a b c d ⎡ ⎣ ⎢ ⎤ ⎦ ⎥, p =⎡ xy ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥, and q= uv ⎡ ⎣ ⎢ ⎤ ⎦ ⎥.For simplicity, let's just look at the 2D case.
When θ∈!, ∂θ ∂A ≡ ∂θ ∂a ∂θ ∂b ∂θ ∂c ∂θ ∂d ⎡ ⎣ ⎢ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ ⎥ . ∂ ∂A Ap− q 2 ⎡⎣ ⎤⎦ = ∂A∂ ⎡⎣⎢ a bc d ⎤ ⎦ ⎥⎡ xy ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥− uv ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ 2 ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ = ∂∂A ax+ by cx+ dy ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥− uv ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ 2 ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ = ∂A∂ ax+ by − u cx+ dy − v ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ 2 ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥
A Simple Proof On
∂A
∂
⎡⎣
Ap
− q
2⎤⎦ = 2 Ap−q
(
)
p
T Let A= a b c d ⎡ ⎣ ⎢ ⎤ ⎦ ⎥, p =⎡ xy ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥, and q= uv ⎡ ⎣ ⎢ ⎤ ⎦ ⎥.For simplicity, let's just look at the 2D case.
When θ∈!, ∂θ ∂A ≡ ∂θ ∂a ∂θ ∂b ∂θ ∂c ∂θ ∂d ⎡ ⎣ ⎢ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ ⎥ . ∂ ∂A Ap− q 2 ⎡⎣ ⎤⎦ = ∂A∂ ⎡⎣⎢ a bc d ⎤ ⎦ ⎥⎡ xy ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥− uv ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ 2 ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ = ∂∂A ax+ by cx+ dy ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥− uv ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ 2 ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ = ∂A∂ ax+ by − u cx+ dy − v ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ 2 ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ = ∂ ∂A
(
ax+ by − u)
2 + cx + dy − v(
)
2 ⎡⎣ ⎤⎦A Simple Proof On
∂A
∂
⎡⎣
Ap
− q
2⎤⎦ = 2 Ap−q
(
)
p
T Let A= a b c d ⎡ ⎣ ⎢ ⎤ ⎦ ⎥, p =⎡ xy ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥, and q= uv ⎡ ⎣ ⎢ ⎤ ⎦ ⎥.For simplicity, let's just look at the 2D case.
When θ∈!, ∂θ ∂A ≡ ∂θ ∂a ∂θ ∂b ∂θ ∂c ∂θ ∂d ⎡ ⎣ ⎢ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ ⎥ . ∂ ∂A Ap− q 2 ⎡⎣ ⎤⎦ = ∂A∂ ⎡⎣⎢ a bc d ⎤ ⎦ ⎥⎡ xy ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥− uv ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ 2 ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ = ∂∂A ax+ by cx+ dy ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥− uv ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ 2 ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ = ∂A∂ ax+ by − u cx+ dy − v ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ 2 ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ = ∂ ∂A
(
ax+ by − u)
2 + cx + dy − v(
)
2 ⎡⎣ ⎤⎦ = ∂ ∂a(ax+ by − u) 2 ∂ ∂b(ax+ by − u) 2 ∂ ∂c(cx+ dy − v) 2 ∂ ∂d(cx+ dy − v) 2 ⎡ ⎣ ⎢ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ ⎥A Simple Proof On
∂A
∂
⎡⎣
Ap
− q
2⎤⎦ = 2 Ap−q
(
)
p
T Let A= a b c d ⎡ ⎣ ⎢ ⎤ ⎦ ⎥, p =⎡ xy ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥, and q= uv ⎡ ⎣ ⎢ ⎤ ⎦ ⎥.For simplicity, let's just look at the 2D case.
When θ∈!, ∂θ ∂A ≡ ∂θ ∂a ∂θ ∂b ∂θ ∂c ∂θ ∂d ⎡ ⎣ ⎢ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ ⎥ . ∂ ∂A Ap− q 2 ⎡⎣ ⎤⎦ = ∂A∂ ⎡⎣⎢ a bc d ⎤ ⎦ ⎥⎡ xy ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥− uv ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ 2 ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ = ∂∂A ax+ by cx+ dy ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥− uv ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ 2 ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ = ∂A∂ ax+ by − u cx+ dy − v ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ 2 ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ = ∂ ∂A
(
ax+ by − u)
2 + cx + dy − v(
)
2 ⎡⎣ ⎤⎦ = ∂ ∂a(ax+ by − u) 2 ∂ ∂b(ax+ by − u) 2 ∂ ∂c(cx+ dy − v) 2 ∂ ∂d(cx+ dy − v) 2 ⎡ ⎣ ⎢ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ ⎥ = 2(ax+ by − u)x 2(ax + by − u)y2(cx+ dy − v)x 2(cx + dy − v)y ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥
A Simple Proof On
∂A
∂
⎡⎣
Ap
− q
2⎤⎦ = 2 Ap−q
(
)
p
T Let A= a b c d ⎡ ⎣ ⎢ ⎤ ⎦ ⎥, p =⎡ xy ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥, and q= uv ⎡ ⎣ ⎢ ⎤ ⎦ ⎥.For simplicity, let's just look at the 2D case.
When θ∈!, ∂θ ∂A ≡ ∂θ ∂a ∂θ ∂b ∂θ ∂c ∂θ ∂d ⎡ ⎣ ⎢ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ ⎥ . ∂ ∂A Ap− q 2 ⎡⎣ ⎤⎦ = ∂A∂ ⎡⎣⎢ a bc d ⎤ ⎦ ⎥⎡ xy ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥− uv ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ 2 ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ = ∂∂A ax+ by cx+ dy ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥− uv ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ 2 ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ = ∂A∂ ax+ by − u cx+ dy − v ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ 2 ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ = ∂ ∂A
(
ax+ by − u)
2 + cx + dy − v(
)
2 ⎡⎣ ⎤⎦ = ∂ ∂a(ax+ by − u) 2 ∂ ∂b(ax+ by − u) 2 ∂ ∂c(cx+ dy − v) 2 ∂ ∂d(cx+ dy − v) 2 ⎡ ⎣ ⎢ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ ⎥ = 2(ax+ by − u)x 2(ax + by − u)y2(cx+ dy − v)x 2(cx + dy − v)y ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ = 2
(ax+ by − u)x (ax + by − u)y (cx+ dy − v)x (cx + dy − v)y ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥
A Simple Proof On
∂A
∂
⎡⎣
Ap
− q
2⎤⎦ = 2 Ap−q
(
)
p
T Let A= a b c d ⎡ ⎣ ⎢ ⎤ ⎦ ⎥, p =⎡ xy ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥, and q= uv ⎡ ⎣ ⎢ ⎤ ⎦ ⎥.For simplicity, let's just look at the 2D case.
When θ∈!, ∂θ ∂A ≡ ∂θ ∂a ∂θ ∂b ∂θ ∂c ∂θ ∂d ⎡ ⎣ ⎢ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ ⎥ . ∂ ∂A Ap− q 2 ⎡⎣ ⎤⎦ = ∂A∂ ⎡⎣⎢ a bc d ⎤ ⎦ ⎥⎡ xy ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥− uv ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ 2 ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ = ∂∂A ax+ by cx+ dy ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥− uv ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ 2 ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ = ∂A∂ ax+ by − u cx+ dy − v ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ 2 ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ = ∂ ∂A
(
ax+ by − u)
2 + cx + dy − v(
)
2 ⎡⎣ ⎤⎦ = ∂ ∂a(ax+ by − u) 2 ∂ ∂b(ax+ by − u) 2 ∂ ∂c(cx+ dy − v) 2 ∂ ∂d(cx+ dy − v) 2 ⎡ ⎣ ⎢ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ ⎥ = 2(ax+ by − u)x 2(ax + by − u)y2(cx+ dy − v)x 2(cx + dy − v)y ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ = 2
(ax+ by − u)x (ax + by − u)y (cx+ dy − v)x (cx + dy − v)y ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ = 2 ax+ by − u cx+ dy − v ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥⎡⎣ x y ⎤⎦
A Simple Proof On
∂A
∂
⎡⎣
Ap
− q
2⎤⎦ = 2 Ap−q
(
)
p
T Let A= a b c d ⎡ ⎣ ⎢ ⎤ ⎦ ⎥, p =⎡ xy ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥, and q= uv ⎡ ⎣ ⎢ ⎤ ⎦ ⎥.For simplicity, let's just look at the 2D case.
When θ∈!, ∂θ ∂A ≡ ∂θ ∂a ∂θ ∂b ∂θ ∂c ∂θ ∂d ⎡ ⎣ ⎢ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ ⎥ . ∂ ∂A Ap− q 2 ⎡⎣ ⎤⎦ = ∂A∂ ⎡⎣⎢ a bc d ⎤ ⎦ ⎥⎡ xy ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥− uv ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ 2 ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ = ∂∂A ax+ by cx+ dy ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥− uv ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ 2 ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ = ∂A∂ ax+ by − u cx+ dy − v ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ 2 ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ = ∂ ∂A
(
ax+ by − u)
2 + cx + dy − v(
)
2 ⎡⎣ ⎤⎦ = ∂ ∂a(ax+ by − u) 2 ∂ ∂b(ax+ by − u) 2 ∂ ∂c(cx+ dy − v) 2 ∂ ∂d(cx+ dy − v) 2 ⎡ ⎣ ⎢ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ ⎥ = 2(ax+ by − u)x 2(ax + by − u)y2(cx+ dy − v)x 2(cx + dy − v)y ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ = 2
(ax+ by − u)x (ax + by − u)y (cx+ dy − v)x (cx + dy − v)y ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ = 2 ax+ by − u cx+ dy − v ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥⎡⎣ x y ⎤⎦ = 2 a b c d ⎡ ⎣ ⎢ ⎤ ⎦ ⎥⎡ xy ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥− uv ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ x y⎡⎣ ⎤⎦
A Simple Proof On
∂A
∂
⎡⎣
Ap
− q
2⎤⎦ = 2 Ap−q
(
)
p
T Let A= a b c d ⎡ ⎣ ⎢ ⎤ ⎦ ⎥, p =⎡ xy ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥, and q= uv ⎡ ⎣ ⎢ ⎤ ⎦ ⎥.For simplicity, let's just look at the 2D case.
When θ∈!, ∂θ ∂A ≡ ∂θ ∂a ∂θ ∂b ∂θ ∂c ∂θ ∂d ⎡ ⎣ ⎢ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ ⎥ . ∂ ∂A Ap− q 2 ⎡⎣ ⎤⎦ = ∂A∂ ⎡⎣⎢ a bc d ⎤ ⎦ ⎥⎡ xy ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥− uv ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ 2 ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ = ∂∂A ax+ by cx+ dy ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥− uv ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ 2 ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ = ∂A∂ ax+ by − u cx+ dy − v ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ 2 ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ = ∂ ∂A
(
ax+ by − u)
2 + cx + dy − v(
)
2 ⎡⎣ ⎤⎦ = ∂ ∂a(ax+ by − u) 2 ∂ ∂b(ax+ by − u) 2 ∂ ∂c(cx+ dy − v) 2 ∂ ∂d(cx+ dy − v) 2 ⎡ ⎣ ⎢ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ ⎥ = 2(ax+ by − u)x 2(ax + by − u)y2(cx+ dy − v)x 2(cx + dy − v)y ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ = 2
(ax+ by − u)x (ax + by − u)y (cx+ dy − v)x (cx + dy − v)y ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ = 2 ax+ by − u cx+ dy − v ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥⎡⎣ x y ⎤⎦ = 2 a b c d ⎡ ⎣ ⎢ ⎤ ⎦ ⎥⎡ xy ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥− uv ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ x y⎡⎣ ⎤⎦ = 2 Ap − q