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Computer Graphics Recitation 5. 2 Motivation – Shape Matching What is the best transformation that...
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Computer Graphics
Recitation 5
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Motivation – Shape Matching
What is the best transformation that aligns the dinosaur with the dog?
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Motivation – Shape Matching
The above is not a good matching….
Regard the shapes as sets of points and try to “match” these setsusing a linear transformation
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Motivation – Shape Matching
Regard the shapes as sets of points and try to “match” these setsusing a linear transformation
To find the best transformation we need to know SVD…
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Finding underlying dimensionality
x
y
Given a set of points, find the best line thatapproximates it – reduce the dimension of the data set…
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Finding underlying dimensionality
x
y
Given a set of points, find the best line thatapproximates it – reduce the dimension of the data set…
x’y’
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Finding underlying dimensionality
x
y
When we learn PCA (Principal Component Analysis),we’ll know how to find these axes that minimize
the sum of distances2
x’y’
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PCA and SVD
PCA and SVD are important tools not only in graphics but also in statistics, computer vision and more.
To learn about them, we first need to get familiar with eigenvalue decomposition.
So, today we’ll start with linear algebra basics reminder.
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The plan today
Basic linear algebra review. Eigenvalues and eigenvectors.
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Vector space
Informal definition: V (a non-empty set of vectors)
v, w V v + w V (closed under addition)
v V, is scalar v V (closed under multiplication by scalar)
Formal definition includes axioms about associativity and distributivity of the + and operators.
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Subspace - example
Let l be a 2D line though the origin L = {p – O | p l} is a linear subspace of R2
x
y
O
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Subspace - example
Let be a plane through the origin in 3D V = {p – O | p } is a linear subspace of R3
y
z
x
O
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Linear independence
The vectors {v1, v2, …, vk} are a linearly independent set if:
1v1 + 2v2 + … + kvk = 0 i = 0 i
It means that none of the vectors can be obtained as a linear combination of the others.
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Linear independence - example
Parallel vectors are always dependent:
Orthogonal vectors are always independent.
vw
v = 2.4 w v + (2.4)w = 0
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Basis
{v1, v2, …, vn} are linearly independent
{v1, v2, …, vn} span the whole vector space V:
V = {1v1 + 2v2 + … + nvn | i is scalar}
Any vector in V is a unique linear combination of the basis.
The number of basis vectors is called the dimension of V.
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Basis - example
The standard basis of R3 – three unit orthogonal vectors x, y, z: (sometimes called i, j, k or e1, e2, e3)
y
z
x
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Matrix representation
Let {v1, v2, …, vn} be a basis of V Every vV has a unique representation
v = 1v1 + 2v2 + … + nvn
Denote v by the column-vector:
The basis vectors are therefore denoted:
n
v
1
1
0
0
0
1
0
0
0
1
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Linear operators
A : V W is called linear operator if: A(v + w) = A(v) + A(w) A(v) = A(v)
In particular, A(0) = 0 Linear operators we know:
Scaling Rotation, reflection Translation is not linear – moves the origin
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Linear operators - illustration
Rotation is a linear operator:
v
w
v+w
R(v+w)
v
w
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Linear operators - illustration
Rotation is a linear operator:
v
w
v+w
R(v+w)
v
wR(w)
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Linear operators - illustration
Rotation is a linear operator:
v
w
v+w
R(v+w)
v
wR(w)R(v)
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Linear operators - illustration
Rotation is a linear operator:
v
w
v+w
R(v+w)
v
wR(w)R(v)
R(v)+R(w)
R(v+w) = R(v) + R(w)
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Matrix representation of linear operators
Look at A(v1),…, A(vn) where {v1, v2, …, vn} is a basis.
For all other vectors: v = 1v1 + 2v2 + … + nvn
A(v) = 1A(v1) + 2A(v2) + … + nA(vn) So, knowing what A does to the basis is enough The matrix representing A is:
|||
)()()(
|||
21 nA vAvAvAM
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Matrix representation of linear operators
|
)(
|
0
0
1
|||
)()()(
|||
121 vAvAvAvA n
|
)(
|
0
1
0
|||
)()()(
|||
221 vAvAvAvA n
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Matrix operations
Addition, subtraction, scalar multiplication – simple…
Matrix by column vector:
iimi
iii
nmnm
n
ba
ba
b
b
aa
aa
1
1
1
111
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Matrix operations
Transposition: make the rows columns
(AB)T = BTAT
mnn
m
T
mnm
n
aa
aa
aa
aa
1
111
1
111
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Matrix operations
Inner product can in matrix form:
vwwvwv TT ,
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Matrix properties
Matrix A (nn) is non-singular if B, AB = BA = I B = A-1 is called the inverse of A A is non-singular det A 0
If A is non-singular then the equation Ax=b has one unique solution for each b.
A is non-singular the rows of A are linearly independent set of vectors.
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Vector product
v w is a new vector: ||v w|| = ||v|| ||w|| |sin(v,w)| v w v, v w w v, w, v w is a right-hand system
In matrix form in 3D:
),,(
ˆˆˆ
det
),,(,),,(
122113312332
321
321
321321
wvwvvwvwwvwv
www
vvv
kji
wv
wwwwvvvv TT
v
w
vw
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Orthogonal matrices
Matrix A (nn) is orthogonal if A-1 = AT
Follows: AAT = ATA = I The rows of A are orthonormal vectors!
Proof:
I = ATA = v1
v2
vn
= viTvj = ij
v1 v2 vn
<vi, vi> = 1 ||vi|| = 1; <vi, vj> = 0
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Orthogonal operators
A is orthogonal matrix A represents a linear operator that preserves inner product (i.e., preserves lengths and angles):
Therefore, ||Av|| = ||v|| and (Av,Aw) = (v,w)
.,
)()(,
wvwvwIv
AwAvAwAvAwAvTT
TTT
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Orthogonal operators - example
Rotation by around the z-axis in R3 :
In fact, any orthogonal 33 matrix represents a rotation around some axis and/or a reflection detA = +1 rotation only detA = 1 with reflection
100
0cossin
0sincos
R
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Eigenvectors and eigenvalues
Let A be a square nn matrix v is eigenvector of A if:
Av = v ( is a scalar) v 0
The scalar is called eigenvalue
Av = v A(v) = (v) v is also eigenvector
Av = v, Aw = w A(v+w) = (v+w) Therefore, eigenvectors of the same form a linear
subspace.
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Eigenvectors and eigenvalues
An eigenvector spans an axis that is invariant to A – it remains the same under the transformation.
Example – the axis of rotation:
O
Eigenvector of therotation transformation
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Finding eigenvalues
For which is there a non-zero solution to Ax = x ? Ax = x Ax – x = 0 Ax – Ix = 0 (A- I)x = 0 So, non trivial solution exists det(A – I) = 0
A() = det(A – I) is a polynomial of degree n. It is called the characteristic polynomial of A. The roots of A are the eigenvalues of A. Therefore, there are always at least complex
eigenvalues. If n is odd, there is at least one real eigenvalue.
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Example of computing A
411
303
201
A
)32)(3(
)3(2)3()1()3(2)3)4()(1(
411
33
201
det)(
2
2
A
Cannot be factorized over ROver C: )21)(21( ii
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Computing eigenvectors
Solve the equation (A – I)x = 0 We’ll get a subspace of solutions.
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Spectra and diagonalization
The set of all the eigenvalues of A is called
the spectrum of A. A is diagonalizable if A has n independent
eigenvectors. Then:
nnn vAv
vAv
vAv
222
111
A v2v1 vn= v2v1 vn
1
2
n
AV = VD
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Spectra and diagonalization
Therefore, A = VDV1, where D is diagonal A represents a scaling along the eigenvector
axes!
nnn vAv
vAv
vAv
222
111
A = v2v1 vn
1
2
n
v2v1 vn
1
A = VDV1
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Spectra and diagonalization
A is called normal if AAT = ATA. Important example: symmetric matrices A = AT. It can be proved that normal nn matrices have
exactly n linearly independent eigenvectors (over C).
If A is symmetric, all eigenvalues of A are real, and it has n independent real orthonormal eigenvectors.
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Spectra and diagonalization
A Drotationreflection
orthonormalbasis
otherorthonormal
basis
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Spectra and diagonalization
A
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Spectra and diagonalization
A
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Spectra and diagonalization
A
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Spectra and diagonalization
A
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Why SVD…
Diagonalizable matrix is essentially a scaling Most matrices are not diagonalizable – they do
other things along with scaling (such as rotation)
So, to understand how general matrices behave, only eigenvalues are not enough
SVD tells us how general linear transformations behave, and other things…
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See you next time