3D Geometry forComputer Graphics

Lesson 2: PCA & SVD

2

Last week - eigendecomposition

A

We want to learn how the matrix A works:

3

Last week - eigendecomposition

A

If we look at arbitrary vectors, it doesn’t tell us much.

4

Spectra and diagonalization

A

If A is symmetric, the eigenvectors are orthogonal (and there’s always an eigenbasis).

A = UΛUT

⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛

nλ

λλ

2

1

Λ= Aui = λi ui

5

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 enoughSVD tells us how general linear transformations behave, and other things…

6

The plan for today

First we’ll see some applications of PCA –Principal Component Analysis that uses spectral decomposition.Then look at the theory.SVD

Basic intuitionFormal definitionApplications

7

PCA finds an orthogonal basis that best represents given data set.

The sum of distances2 from the x’ axis is minimized.

PCA – the general idea

x

y

x’

y’

8

PCA – the general idea

PCA finds an orthogonal basis that best represents given data set.

PCA finds a best approximating plane (again, in terms of Σdistances2)

3D point set instandard basis

x y

z

9

Application: finding tight bounding box

An axis-aligned bounding box: agrees with the axes

xminX maxX

y

maxY

minY

10

Application: finding tight bounding box

Oriented bounding box: we find better axes!

x’y’

11

Application: finding tight bounding box

This is not the optimal bounding box

x y

z

12

Application: finding tight bounding box

Oriented bounding box: we find better axes!

13

Usage of bounding boxes (bounding volumes)

Serve as very simple “approximation” of the objectFast collision detection, visibility queriesWhenever we need to know the dimensions (size) of the object

The models consist of thousands of polygonsTo quickly test that they don’t intersect, the bounding boxes are testedSometimes a hierarchy of BB’s is used

The tighter the BB – the less “false alarms” we have

Sample

14

Scanned meshes

15

Point clouds

Scanners give us raw point cloud dataHow to compute normals to shade the surface?

normal

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Point clouds

Local PCA, take the third vector

17

Notations

Denote our data points by x1, x2, …, xn ∈ Rd

11 11 2

22 21 2

1 2

, , ,

n

n

dd dn

xx xxx x

xx x

⎛ ⎞⎛ ⎞ ⎛ ⎞⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟⎜ ⎟ ⎜ ⎟= = = ⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠ ⎝ ⎠

1 2 nx x x…

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The origin is zero-order approximation of our data set (a point)It will be the center of mass:

It can be shown that:

The origin of the new axes

1

1

n

ini=

= ∑m x

2

1

argminn

ii=

= ∑x

m x - x

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Scatter matrix

Denote yi = xi – m, i = 1, 2, …, n

where Y is d×n matrix with yk as columns (k = 1, 2, …, n)

TS YY=

1 1 1 1 21 2 1 1 12 2 2 1 21 2 2 2 2

1 21 2

dn

dn

d d d dn n n n

y y y y y yy y y y y y

S

y y y y y y

⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟= ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

Y YT

20

Variance of projected points

In a way, S measures variance (= scatterness) of the data in different directions.Let’s look at a line L through the center of mass m, and project our points xi onto it. The variance of the projected points x’i is:

Original set Small variance Large variance

21

1

var( ) || ||n

ini

L=

′= −∑ x m

L L L L

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Variance of projected points

Given a direction v, ||v|| = 1, the projection of xionto L = m + vt is:

|| || , / || || , Ti i i i′ − = < − > = < > =x m v x m v v y v y

vm

xi

x’iL

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Variance of projected points

So,

2 2 21 1 1

1 1

21 1 1 1 1

var( ) || || ( ) || ||

|| || , ,

n n

i in n ni i

T T Tn n n n n

L Y

Y Y Y YY S S= =

′= = = =

= = < > = = = < >

∑ ∑ T T

T T T

x -m v y v

v v v v v v v v v

( ) ( )

2 21 1 1 11 2

2 2 2 22 1 2 1 2 21 2

1 1

1 2

( ) || ||

i nn n

T d di n

i id d d di n

y y y yy y y y

v v v v v v Y

y y y y= =

⎛ ⎞⎛ ⎞ ⎛ ⎞⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟⎜ ⎟ ⎜ ⎟= = =⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠

∑ ∑ Tiv y v

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Directions of maximal variance

So, we have: var(L) = <Sv, v>Theorem: Let f : {v ∈ Rd | ||v|| = 1} → R,

f (v) = <Sv, v> (and S is a symmetric matrix).

Then, the extrema of f are attained at the eigenvectors of S.

So, eigenvectors of S are directions of maximal/minimal variance!

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Summary so far

We take the centered data vectors y1, y2, …, yn ∈ Rd

Construct the scatter matrixS measures the variance of the data pointsEigenvectors of S are directions of maximal variance.

TS YY=

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Scatter matrix - eigendecomposition

S is symmetric⇒ S has eigendecomposition: S = VΛVT

S = v2v1 vd

λ1λ2

λd

v2

v1

vd

The eigenvectors formorthogonal basis

26

Principal components

Eigenvectors that correspond to big eigenvalues are the directions in which the data has strong components (= large variance).If the eigenvalues are more or less the same –there is no preferable direction.

Note: the eigenvalues are always non-negative.

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Principal components

There’s no preferable directionS looks like this:

Any vector is an eigenvector

TV Vλ

λ⎛ ⎞⎜ ⎟⎝ ⎠

There is a clear preferable directionS looks like this:

μ is close to zero, much smaller than λ.

TVV ⎟⎟⎠

⎞⎜⎜⎝

⎛μ

λ

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How to use what we got

For finding oriented bounding box – we simply compute the bounding box with respect to the axes defined by the eigenvectors. The origin is at the mean point m.

v2v1

v3

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For approximation

x

yv1

v2

x

y

This line segment approximates the original data set

The projected data set approximates the original data set

x

y

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For approximation

In general dimension d, the eigenvalues are sorted in descending order:

λ1 ≥ λ2 ≥ … ≥ λd

The eigenvectors are sorted accordingly.To get an approximation of dimension d’ < d, we take the d’ first eigenvectors and look at the subspace they span (d’ = 1 is a line, d’ = 2 is a plane…)

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For approximation

To get an approximating set, we project the original data points onto the chosen subspace:xi = m + α1v1 + α2v2 +…+ αd’vd’ +…+αdvd

Projection:

xi’ = m + α1v1 + α2v2 +…+ αd’vd’ +0⋅vd’+1+…+ 0⋅ vd

SVD

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Geometric analysis of linear transformations

We want to know what a linear transformation A does Need some simple and “comprehendible” representation of the matrix of A. Let’s look what A does to some vectors

Since A(αv) = αA(v), it’s enough to look at vectors v of unit length

A

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The geometry of linear transformations

A linear (non-singular) transform A always takes hyper-spheres to hyper-ellipses.

A

A

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The geometry of linear transformations

Thus, one good way to understand what A does is to find which vectors are mapped to the “main axes” of the ellipsoid.

A

A

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Geometric analysis of linear transformations

If we are lucky: A = V Λ VT, V orthogonal (true if A is symmetric)The eigenvectors of A are the axes of the ellipse

A

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Symmetric matrix: eigen decomposition

In this case A is just a scaling matrix. The eigendecomposition of A tells us which orthogonal axes it scales, and by how much:

[ ] [ ]1

21 2 1 2

Tn n

n

A

λλ

λ

⎡ ⎤⎢ ⎥⎢ ⎥=⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

v v v v v v… …

A

11

λ2

λ1

iiiA vv λ=

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General linear transformations: SVD

In general A will also contain rotations, not just scales:

A

[ ] [ ]1

21 2 1 2

Tn n

n

A

σσ

σ

⎡ ⎤⎢ ⎥⎢ ⎥=⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

u u u v v v… …

1 1 σ2σ1

TA U V= ∑

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General linear transformations: SVD

A

[ ] [ ]1

21 2 1 2n n

n

A

σσ

σ

⎡ ⎤⎢ ⎥⎢ ⎥=⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

v v v u u u… …

1 1 σ2σ1

AV U= ∑

, 0i iA σ σ= ≥i iv u

orthonormal orthonormal

40

SVD more formally

SVD exists for any matrixFormal definition:

For square matrices A ∈ Rn×n, there exist orthogonal matrices U, V ∈ Rn×n and a diagonal matrix Σ, such that all the diagonal values σi of Σ are non-negative and

TA U V= Σ

=

A U Σ TV

41

SVD more formally

The diagonal values of Σ (σ1, …, σn) are called the singular values. It is accustomed to sort them: σ1 ≥ σ2≥ … ≥ σn

The columns of U (u1, …, un) are called the left singular vectors. They are the axes of the ellipsoid.The columns of V (v1, …, vn) are called the right singular vectors. They are the preimages of the axes of the ellipsoid.

TA U V= Σ

=

A U Σ TV

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Reduced SVD

For rectangular matrices, we have two forms of SVD. The reduced SVD looks like this:

The columns of U are orthonormalCheaper form for computation and storage

A U Σ TV

=

43

Full SVD

We can complete U to a full orthogonal matrix and pad Σ by zeros accordingly

A U Σ TV

=

44

Some history

SVD was discovered by the following people:

E. Beltrami(1835 − 1900)

M. Jordan(1838 − 1922)

J. Sylvester(1814 − 1897)

H. Weyl(1885-1955)

E. Schmidt(1876-1959)

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SVD is the “working horse” of linear algebra

There are numerical algorithms to compute SVD. Once you have it, you have many things:

Matrix inverse → can solve square linear systemsNumerical rank of a matrixCan solve least-squares systemsPCAMany more…

46

Matrix inverse and solving linear systems

Matrix inverse:

So, to solve

( ) ( )1

1 11 1 1

1

1n

T

T T

T

A U V

A U V V U

V Uσ

σ

− −− − −

= ∑

= ∑ = ∑ =

⎡ ⎤⎢ ⎥

= ⎢ ⎥⎢ ⎥⎣ ⎦

1 T

AV U−

=

= ∑

x bx b

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Matrix rank

The rank of A is the number of non-zero singular values

A U Σ TV

=m

n

σ1σ2

σn

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Numerical rank

If there are very small singular values, then A is close to being singular. We can set a threshold t, so that numeric_rank(A) = #{σi| σi > t}

If rank(A) < n then A is singular. It maps the entire space Rn onto some subspace, like a plane (so Ais some sort of projection).

49

Back to PCA

We wanted to find principal components

x

yx’

y’

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PCA

Move the center of mass to the originpi’ = pi − m

x

y

x’y’

51

PCA

Constructed the matrix X of the data points.

The principal axes are eigenvectors of S = XXT

1 2

| | |

| | |nX

⎡ ⎤⎢ ⎥′ ′ ′= ⎢ ⎥⎢ ⎥⎣ ⎦

p p p

1T T

d

XX S U Uλ

λ

⎡ ⎤⎢ ⎥= = ⎢ ⎥⎢ ⎥⎣ ⎦

52

PCA

We can compute the principal components by SVD of X:

Thus, the left singular vectors of X are the principal components! We sort them by the size of the singular values of X.

2

( )T

T

T T T T

T TT

X U VXX U V U V

VU U UV U

= Σ

= Σ Σ =

= =Σ Σ Σ

53

Shape matching

We have two objects in correspondenceWant to find the rigid transformation that aligns them

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Shape matching

When the objects are aligned, the lengths of the connecting lines are small.

55

Shape matching – formalization

Align two point sets

Find a translation vector t and rotation matrix Rso that:

{ } { }1 1, , and , , .n nP Q= =p p q q… …

2

1( ) is minimized

n

i ii

R=

− +∑ p tq

56

Shape matching – solution

Turns out we can solve the translation and rotation separately.Theorem: if (R, t) is the optimal transformation, then the points {pi} and {Rqi + t} have the same centers of mass.

1

1 n

iin =

= ∑p p1

1 n

iin =

= ∑q q

( )1

1

1

1

n

ii

n

ii

R

R

n

R Rn

=

=

= +

⇓

⎛ ⎞= + = +⎜ ⎟⎝ ⎠

= −

∑

∑

p q

p q q t

t p

t

t

q

57

Finding the rotation R

To find the optimal R, we bring the centroids of both point sets to the origin:

We want to find R that minimizes

i i i i′ ′= − = −p p p q q q

2

1

n

i ii

R=

′ ′−∑ p q

58

Finding the rotation R

( ) ( )

( )

2

1 1

1

n nT

i i i i i ii i

nT T T T TT

i i i i i i i ii

R R R

R R R R

= =

=

′ ′ ′ ′ ′ ′− = − − =

′ ′ ′ ′ ′ ′ ′ ′= − − +

∑ ∑

∑

p q p q p q

p p p q q p q qI

These terms do not depend on R,so we can ignore them in the minimization

59

Finding the rotation R

( ) ( )

( )

( ) ( )

1 1

1 1

min max .

max 2 max

n nT T T T

i i i ii i i i iii i

TT T Ti ii i i i i

n nT

T T

T T

Ti i i i

i i

R R R R

R R R

R R

= =

= =

′ ′ ′ ′ ′ ′ ′ ′− − = +

′ ′ ′ ′ ′ ′= =

′ ′ ′ ′⇒ =

∑ ∑

∑ ∑

p q q p p q q p

q p q p p q

p q p q

this is a scalar

60

Finding the rotation R

( )1 1 1

1

n n nT T T

i i i i i ii i i

nT

i ii

Trace Trace

H

R R R= = =

=

⎛ ⎞ ⎛ ⎞′ ′ ′ ′ ′ ′= =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

′ ′=

∑ ∑ ∑

∑

p q q p q p

q p

3 ×1 1 ×3 = 3 ×3

=

1( )

n

iii

Trace A A=

= ∑

61

So, we want to find R that maximizes

Theorem: if M is symmetric positive definite (all eigenvalues of M are positive) and B is any orthogonal matrix then

So, let’s find R so that RH is symmetric positive definite. Then we know for sure that Trace(RH) is maximal.

Finding the rotation R

( )Trace RH

( ) ( )Trace M Trace BM≥

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This is easy! Compute SVD of H:

Define R:Check RH:

TH U V

Finding the rotation R

= ΣTR VU=

( )( )T T TVU U VR V VH = Σ = Σ

This is a symmetric matrix,Its eigenvalues are σi ≥ 0So RH is positive!

63

Summary of rigid alignment:

Translate the input points to the centroids:

Compute the “covariance matrix”

Compute the SVD of H:

The optimal rotation is

The translation vector is

i ii i′′ = =− −qp p qp q

1

nT

i ii

H=

′ ′= ∑ q p

TH U V= Σ

TR VU=

R= −t p q

64

Complexity

Numerical SVD is an expensive operationWe always need to pay attention to the dimensions of the matrix we’re applying SVD to.

65

Small somewhat related example

The End