Principal component analysis

Features extraction and representation / Image processing

Farah Al-Tufaili

The reader is probably familiar with the common saying that goes something along the lines of ‘Why use a hundred words when ten will do?’ This idea of expressing information in it’s most succinct and compact form embodies very accurately the central idea behind Principal Component Analysis (PCA) a very important and powerful statistical technique.

Purpose: to reduce dimensionality of a vector image while maintaining

information as much as possible.

PCA- by Farah Al-Tufaili 2

Some of the more significant aspects of digital imaging in which it has found useful application include the classification of photographic film, remote sensing and data compression, and automated facial recognition and facial synthesis, an Application of PCA.

It is also commonly used as a low-level image processing tool for certain tasks, such as determination of the orientation of basic shapes.


PCA steps: transform an 𝑁×𝑑 matrix 𝑋 into an 𝑁×𝑚 matrix 𝑌:

Centralized the data (subtract the mean).

Calculate the 𝑑×𝑑 covariance matrix: C=1

𝑁−1𝑋𝑇𝑋

𝐶𝑖,𝑗=1

𝑁−1 𝑞=1𝑁 𝑋𝑞,𝑖 . 𝑋𝑞,𝑗

𝐶𝑖,𝑖 (diagonal) is the variance of variable i.

𝐶𝑖,𝑗 (off-diagonal) is the covariance between variables i and j.

Calculate the eigenvectors of the covariance matrix (orthonormal).


Let

Covariance matrix

K

k

T

xx

T

kk

T

xxxK

E1

1}))({( mmxxmxmxC

T

nxxx ]...[ 21x


A 2-D facial image can be represented as 1-D vector by concatenating each row (or column) into a long thin vector. Let’s suppose we have M vectors of size N (= rows of image × columns of image) representing a set of sampled images. pj’s represent the pixel values.

xi = [p1 ...pN]T,i = 1,...,M (1)

The images are mean centered by subtracting the mean image from each image vector. Let m represent the mean image.

(2)

And let wi be defined as mean centered image

wi = xi − m (3)

Our goal is to find a set of ei’s which have the largest possible projection onto each of the wi’s. We wish to find a set of M orthonormal vectors ei for which the quantity

(4)

is maximized with the orthonormality constraint


This example illustrates the primary and essential requirement for PCA to be useful: there must be some degree of correlation between the data vectors. In fact, the more strongly correlated the original data, the more effective PCA is likely to be.

To provide some concrete but simple data to work with, consider that we have just M=2 measurements on the height h and the weight w of a sample group of N=12 people as given in Table 9.2. Figure 9.7 plots the mean-subtracted data in the 2-D feature space (w, h) and it is evident that the data shows considerable variance over both


variables and that they are correlated. The sample covariance matrix for these

two variables is calculated as:

Where x=(h w)T and x=(h w)T is the simple mean.


Weight (X) Height (Y) X-mean Y-mean

65 170 -4.416666667 -3.25

75 176 5.583333333 2.75

53 154 -16.41666667 -19.25

54 167 -15.41666667 -6.25

61 171 -8.416666667 -2.25

88 184 18.58333333 10.75

70 182 0.583333333 8.75

78 190 8.583333333 16.75

52 166 -17.41666667 -7.25

95 168 25.58333333 -5.25

70 176 0.583333333 2.75

72 175 2.583333333 1.75

Mean: 69.41666667 173.25 Variance: 182.9924 90.5682

Applying the equation:

K

k

T

xx

T

kk

T

xxxK

E1

1}))({( mmxxmxmxC


clc

x=[65 75 53 54 61 88 70 78 52 95 70 72]

y=[170 176 154 167 171 184 182 190 166 168 176 175]

x_mean=mean(x) , y_mean=mean(y)

x1=x-x_mean; y1=y-y_mean;

var_x=sum(x1.^2)/(length(x)-1)

var_y=sum(y1.^2)/(length(y)-1)

cv=(x1).*(y1);

cov=sum(cv)/(length(x)-1)

Output:

x_mean =

69.4167

y_mean =

173.2500

var_x =

182.9924

var_y =

90.5682

cov =

73.4318PCA- by Farah Al-Tufaili 11

• The basic aim of PCA is to effect a rotation of the coordinate system and

thereby express the data in terms of a new set of variables or equivalently

axes which are uncorrelated.

• How are these found? The first principal axis is chosen to satisfy the following

criterion:

• The axis passing through the data points which maximizes the sum of the

squared lengths of the perpendicular projection of the data points onto that

axis is the principal axis.


Figure 9.7 Distribution of weight and height values within a 2-D feature space. The mean

value of each variable has been subtracted from the data


Thus, the principal axis is oriented so as to maximize the overall variance of the data with respect to it (a variance-maximizing transform). We note in passing that an alternative but entirely equivalent criterion for a principal axis is that it minimizes the sum of the squared errors (differences) between the actual data points and their perpendicular projections onto the said straight line. This concept is illustrated in Figure 9.8.

Let us suppose that the first principal axis has been found. To calculate the second principal axis we proceed as follows:

Calculate the projections of the data points onto the first principal axis.

Subtract these projected values from the original data. The modified set of data points is termed the residual data.

The second principal axis is then calculated to satisfy an identical criterion to the first (i.e. the variance of the residual data is maximized along this axis).


Note also that this is distinct from fitting a straight line by regression. In regression, x is an

independent variable and y the dependent variable and we seek to minimize the sum of the

squared errors between the actual y values and their predicted values.PCA- by Farah Al-Tufaili 15


In fact, this alignment is precisely the mechanism that decorrelatcs the data. Furthermore, as the

eigenvalues appear along the main diagonal of Cy ,λ i is the variance of component yi along

eigenvector ei . The two eigenvectors are perpendicular. The y-axes sometimes are called the

eigen axes for obvious reasons.PCA- by Farah Al-Tufaili 17

let the covariance matrix is : 7 33 −1

First Step: compute eigenvalues. To do this, we find the values of which satisfy the characteristic equation of the matrix A, namely those values of for which

det(A − I) = 0,

1. λ I= λ1 00 1

= λ 00 λ

2. A- λI= 7 33 −1

-λ 00 λ

= 7 − λ 33 −1 − λ

3. det7 33 −1

= (7- λ)(-1- λ)-(3)(3)= -7-7λ+λ+ λ2-9= λ2-6 λ-16

(λ-8)(λ+2)=0 λ1=8 λ2=-2

EigenValues

Let this

equation equal

to zero and

solve it


Second step: find corresponding eigenvectors.

1. For each eigenvalue λ, we have (A − λ I)V = 0, where x is the eigenvector associated with eigenvalue λ.

2. Find x by Gaussian elimination. That is, convert the augmented matrix (A − λ I ⋮ 0)to row echelon form, and solve the resulting linear system by back substitution.

We find the eigenvectors associated with each of the eigenvalues


−1𝑣1+ 3𝑣2 = 0 ………..(1

3𝑣1- 9𝑣2 = 0 ………..(2

2𝑣1- 6𝑣2 = 0 ………..(3

Lets 𝑠 =(2)2+(−6)2=6.3246

v1 = 2/6.3246= 0.3162

v2 = -6/6.3246= -0.9487


When λ1=8

A- λI=7 − 8 33 −1 − 8

= −1 33 −9

(A − λ I)V = 0 −1 33 −9

⋮𝑣1𝑣2

=0

−𝟏 × 𝐯𝟏 + 𝟑 × 𝐯𝟐 = 𝟎

𝟑 × 𝐯𝟏 + −𝟗 × 𝐯𝟐 = 𝟎


9𝑣1+ 3𝑣2 = 0 ………..(1

3𝑣1+1𝑣2 = 0 ………..(2

6𝑣1+2𝑣2 = 0 ………..(3

Lets 𝑠 =(2)2+(6)2=6.3246

v1 = -6/6.3246= -0.9487

v2 = -2/6.3246= -0.3162

When λ1=-2

A- λI=7 + 2 33 −1 + 2

= 9 33 1

(A − λ I)V = 0 9 33 1

⋮𝑣1𝑣2

=0

𝟗 × 𝐯𝟏 + 𝟑 × 𝐯𝟐 = 𝟎

𝟑 × 𝐯𝟏 + 𝟏 × 𝐯𝟐 = 𝟎

check our solution in Matlab:>> x=[7 3;3 -1]

x =

7 3

3 -1

>>[a b]=eig(x)

a =

0.3162 -0.9487

-0.9487 -0.3162

b =

-2 0

0 8


EigenValues

EigenVictor

There is a subtle difference between the two alternative derivations we have offered. In the first case we considered our variables to define the feature space and derived a set of principal axes. The dimensionality of our feature space was determined by the number of variables and the diagonalizing matrix of eigenvectors R to produce a new set of axes in that 2-D space. Our second formulation leads to essentially the same procedure (i.e. to diagonalize the covariance matrix), but it nonetheless admits an alternative viewpoint. In this instance, we view the observations on each of the variables to define our (higher dimensional) feature vectors. The role of the diagonalizing matrix of eigenvectors R here is thus to multiply and transform these higher dimensional vectors to produce a new orthogonal (principal) set. Thus, the dimensionality of the space here is determined by the number of observations made on each of the variables.


W=[65 75 53 54 61 88 70 78 52 95 70 72]' %1. Form data vector on weight

H=[170 176 154 167 171 184 182 190 166 168 176 175]' %Form data vector on height

XM=[mean(W).*ones(length(W),1) mean(H).*ones(length(H),1)] %matrix with mean values replicated

X=[W H]-XM %Form mean-subtracted data matrix

Cx=cov(X) %calculate covariance on data

[R,LAMBDA,Q]=svd(Cx) %Get eigenvalues LAMBDA and eigenvectors R

V=X*R %Calculate principal components

subplot(1,2,1), plot(X(:,1),X(:,2),'ko'); grid on; %2. display data on original axes

subplot(1,2,2), plot(V(:,1),V(:,2),'ro'); grid on; %display PCs as data in rotated space

XR=XM+V*R' %3. Reconstruct data in terms of PCs

XR-[W H] %Confirm reconstruction (diff = 0)

V'*V./(length(W)-1) %4. Confirm covariance terms

%MATLAB FUNCTIONS cov, mean, svd,PCA- by Farah Al-Tufaili 24

The weight–height data referred to the original axes x1 and x2 and to the principal axes v1 and v2PCA- by Farah Al-Tufaili 25

• It is self-evident that the basic

placement, size and shape of human

facial features are similar. Therefore,

we can expect that an ensemble of

suitably scaled and registered images

of the human face will exhibit fairly

strong correlation. Each image

consisted of 21 054 grey-scale pixel

values.

Figure 9.13 A sample of human faces scaled and

registered such that each can be described by the

same number of pixels (21 054).


Figure 9.15 The first six facial principal components from a sample of 290 faces.

The first principal is the average face. Note the strong male appearance coded by

principal component no. 3


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Principal component analysis