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Principal components analysis (Lecture 1: Sept. 10, 2014)

Introduction

Principal Component Analysis (PCA), first invented by Karl Pearson in 1901, is a statistical technique for data analysis. Its main purpose is to reduce the dimensionality of the data.

Suppose there is a set of data points in a d-dimensional space. The goal of PCA is to find a linear subspace with lower dimensionality p (p [math]\displaystyle{ \leq }[/math] d), such that maximum variance is retained in this lower-dimensional space. The linear subspace can be specified by d orthogonal vectors, such as [math]\displaystyle{ u_1 , u_2 , ... , u_d }[/math] which form a new coordinate system, called 'principle components'.In other words, PCA aims to reduce the dimensionality of the data, while preserving its information (or minimizing the loss of information). Information comes from variation. In other words, the more variation is captured in the lower dimension, the more information is preserved. For example, if all data points have the same value along one dimension, that dimension does not carry any information. So, to preserve information, the subspace need to contain components (or dimensions) along which, data has its most variability. However, finding a linear subspace with lower dimensionality which include all data points is not possible in practical problems and loss of information is inevitable. But, we try to reduce this loss and capture most of the features of data.

Figure below, demonstrates an example of PCA. Data is transformed from original 3D space to 2D coordinate system where each coordinate is a principal component.

• 

  Data in original Space

• 

  Extraction of the principal components

• 

  Reduction of dimensionality by keeping the first two principal components and removing the third one

Consider now the ability of two of the above examples' components (PC₁ and PC₂) to retain information from the original data. The data in the original space is projected onto each of these two components separately in the below figure. Notice that PC₁ is better able to capture the variation in the original data than PC₂ alone. If reducing the original d=3 dimensional data to p=1 dimension, PC₁ is therefore preferable to PC₂.

Comparison of two different components used for dimensionality reduction

PCA Plot example

For example<ref> https://onlinecourses.science.psu.edu/stat857/sites/onlinecourses.science.psu.edu.stat857/files/lesson05/PCA_plot.gif </ref>, in the top left corner of the image below, the point [math]\displaystyle{ x_1 }[/math] showed in two-dimensional and it's coordinates are [math]\displaystyle{ (x_1, 1) }[/math] and [math]\displaystyle{ (x_1, 2) }[/math]

All the red points in the plot represented by their projected values on the two original coordinators (Feature 1, Feature 2).

In PCA technique, it uses new coordinators (showed in green). As coordinate system changed, the points also are shown by their new pair of coordinate values for example [math]\displaystyle{ (z_1, 1) }[/math] and [math]\displaystyle{ (z_1, 2) }[/math]. These values are determined by original vector, and the relation between the rotated coordinates and the original ones.

In the original coordinates the two vectors [math]\displaystyle{ x_1 }[/math] and [math]\displaystyle{ x_2 }[/math] are not linearly uncorrelated. In other words, after applying PCA, if there are two principal component, and use those as the new coordinates, then it is guaranteed that the data along the two coordinates have correlation = 0

By rotating the coordinates we performed an orthonormal transform on the data. After the transform, the correlation is removed - which is the whole point of PCA.

As an example of an extreme case where all of the points lie on a straight line, which in the original coordinates system it is needed two coordinates to describe these points.

In short, after rotating the coordinates, we need only one coordinate and clearly the value on the other one is always zero. This example shows PCA's dimension reduction in an extreme case while in real world points may not fit exactly on a line, but only approximately.

PCA applications

As mentioned, PCA is a method to reduce data dimension if possible to principal components such that those PCs cover as much data variation as possible. This technique is useful in different type of applications which involve data with a huge dimension like data pre-processing, neuroscience, computer graphics, meteorology, oceanography, gene expression, economics, and finance among of all other applications.

Data usually is represented by lots of variables. In data preprocessing, PCA is a technique to select a subset of variables in order to figure our best model for data. In neuroscience, PCA used to identify the specific properties of a stimulus that increase a neuron’s probability of generating an action potential.

Figures below show some example of PCA in real world applications. Click to enlarge.

• 

  Finding two principal components of original data in 2D space. Components of data are highly correlated in original space. So they can be substituted by just one principal component, PC1, which has more than %99 of total variance of data.

• 

  PCA in Neuroscience. Data in different classes in 3 dimensional space is mapped to a 2 dimensional space using PCA. The mapped data is clearly linearly sparable.

Mathematical details

PCA is a transformation from original space to a linear subspace with a new coordinate system. Each coordinate of this subspace is called a Principle Component. First principal component is the coordinate of this system along which the data points has the maximum variation. That is, if we project the data points along this coordinate, maximum variance of data is obtained (compared to any other vector in original space). Second principal component is the coordinate in the direction of the second greatest variance of the data, and so on.

Lets denote the basis of original space by [math]\displaystyle{ \mathbf{v_1} }[/math], [math]\displaystyle{ \mathbf{v_2} }[/math], ... , [math]\displaystyle{ \mathbf{v_d} }[/math]. Our goal is to find the principal components (coordinate of the linear subspace), denoted by [math]\displaystyle{ \mathbf{u_1} }[/math], [math]\displaystyle{ \mathbf{u_2} }[/math], ... , [math]\displaystyle{ \mathbf{u_p} }[/math] in the hope that [math]\displaystyle{ p \leq d }[/math]. First, we would like to obtain the first principal component [math]\displaystyle{ \mathbf{u_1} }[/math] or the components in the direction of maximum variance. This component can be treated as a vector in the original space and so is written as a linear combination of the basis in original space.

[math]\displaystyle{ \mathbf{u_1}=w_1\mathbf{v_1}+w_2\mathbf{v_2}+...+w_d\mathbf{v_d} }[/math]

Vector [math]\displaystyle{ \mathbf{w} }[/math] contains the weight of each basis in this combination.

[math]\displaystyle{ \mathbf{w}=\begin{bmatrix} w_1\\w_2\\w_3\\...\\w_d \end{bmatrix} }[/math]

Suppose we have n data points in the original space. We represent each data points by [math]\displaystyle{ \mathbf{x_1} }[/math], [math]\displaystyle{ \mathbf{x_2} }[/math], ..., [math]\displaystyle{ \mathbf{x_n} }[/math]. Projection of each point [math]\displaystyle{ \mathbf{x_i} }[/math] on the [math]\displaystyle{ \mathbf{u_1} }[/math] is [math]\displaystyle{ \mathbf{w}^T\mathbf{x_i} }[/math].

Let [math]\displaystyle{ \mathbf{S} }[/math] be the sample covariance matrix of data points in original space. The variance of the projected data points, denoted by [math]\displaystyle{ \Phi }[/math] is

[math]\displaystyle{ \Phi = Var(\mathbf{w}^T \mathbf{x_i}) = \mathbf{w}^T \mathbf{S} \mathbf{w} }[/math]

we would like to maximize [math]\displaystyle{ \Phi }[/math] over set of all vectors [math]\displaystyle{ \mathbf{w} }[/math] in original space. But, this problem is not yet well-defined, because for any choice of [math]\displaystyle{ \mathbf{w} }[/math], we can increase [math]\displaystyle{ \Phi }[/math] by simply multiplying [math]\displaystyle{ \mathbf{w} }[/math] in a positive scalar greater than one. So, we add the following constraint to the problem to bound the length of vector [math]\displaystyle{ \mathbf{w} }[/math]:

[math]\displaystyle{ max \Phi = \mathbf{w}^T \mathbf{S} \mathbf{w} }[/math]

subject to : [math]\displaystyle{ \mathbf{w}^T \mathbf{w} =1 }[/math]

Using Lagrange Multiplier technique we have:

[math]\displaystyle{ L(\mathbf{w} , \lambda ) = \mathbf{w}^T \mathbf{S} \mathbf{w} - \lambda (\mathbf{w}^T \mathbf{w} - 1 ) }[/math]

By taking derivative of [math]\displaystyle{ L }[/math] w.r.t. primary variable [math]\displaystyle{ \mathbf{w} }[/math] we have:

[math]\displaystyle{ \frac{\partial L}{\partial \mathbf{w}}=( \mathbf{S}^T + \mathbf{S})\mathbf{w} -2\lambda\mathbf{w}= 2\mathbf{S}\mathbf{w} -2\lambda\mathbf{w}= 0 }[/math]

Note that [math]\displaystyle{ \mathbf{S} }[/math] is symmetric so [math]\displaystyle{ \mathbf{S}^T = \mathbf{S} }[/math].

From the above equation we have:

[math]\displaystyle{ \mathbf{S} \mathbf{w} = \lambda\mathbf{w} }[/math].

So [math]\displaystyle{ \mathbf{w} }[/math] is the eigenvector and [math]\displaystyle{ \lambda }[/math] is the eigenvalue of the matrix [math]\displaystyle{ \mathbf{S} }[/math] .(Taking derivative of [math]\displaystyle{ L }[/math] w.r.t. [math]\displaystyle{ \lambda }[/math] just regenerate the constraint of the optimization problem.)

By multiplying both sides of the above equation to [math]\displaystyle{ \mathbf{w}^T }[/math] and considering the constraint, we obtain:

[math]\displaystyle{ \Phi = \mathbf{w}^T \mathbf{S} \mathbf{w} = \mathbf{w}^T \lambda \mathbf{w} = \lambda \mathbf{w}^T \mathbf{w} = \lambda }[/math]

The interesting result is that objective function is equal to eigenvalue of the covariance matrix. So, to obtain the first principle component, which maximizes the objective function, we just need the eigenvector corresponding to the largest eigenvalue of [math]\displaystyle{ \mathbf{S} }[/math]. Subsequently, the second principal component is the eigenvector corresponding to the second largest eigenvalue, and so on.

Principal component extraction using singular value decomposition

Singular Value Decomposition (SVD), is a well-known way to decompose any kind of matrix [math]\displaystyle{ \mathbf{A} }[/math] (m by n) into three useful matrices.

[math]\displaystyle{ \mathbf{A} = \mathbf{U}\mathbf{\Sigma}\mathbf{V}^T }[/math].

where [math]\displaystyle{ \mathbf{U} }[/math] is m by m unitary matrix, [math]\displaystyle{ \mathbf{UU^T} = I_m }[/math], each column of [math]\displaystyle{ \mathbf{U} }[/math] is an eigenvector of [math]\displaystyle{ \mathbf{AA^T} }[/math]

[math]\displaystyle{ \mathbf{V} }[/math] is n by n unitary matrix, [math]\displaystyle{ \mathbf{VV^T} = I_n }[/math], each column of [math]\displaystyle{ \mathbf{V} }[/math] is an eigenvector of [math]\displaystyle{ \mathbf{A^TA} }[/math]

[math]\displaystyle{ \mathbf{\Sigma} }[/math] is m by n diagonal matrix, non-zero elements of this matrix are square roots of eigenvalues of [math]\displaystyle{ \mathbf{AA^T} }[/math].

Now, comparing the concepts of PCA and SVD, one may find out that we can perform PCA using SVD.

Lets construct a matrix p by n with our n data points such that each column of this matrix represent one data point in p-dimensional space:

[math]\displaystyle{ \mathbf{X} = [\mathbf{x_1} \mathbf{x_2} .... \mathbf{x_n} ] }[/math]

and make another matrix [math]\displaystyle{ \mathbf{X}^* }[/math] simply by subtracting the mean of data points from [math]\displaystyle{ \mathbf{X} }[/math].

[math]\displaystyle{ \mathbf{X}^* = \mathbf{X} - \mu_X }[/math]

Then we will get a zero-mean version of our data points for which [math]\displaystyle{ \mathbf{X^*} \mathbf{X^{*^T}} }[/math] is the covariance matrix.

So, [math]\displaystyle{ \mathbf{\Sigma} }[/math] and [math]\displaystyle{ \mathbf{U} }[/math] give us the eigenvalues and corresponding eigenvectors of covariance matrix, respectively. We can then easily extract the desired principal components.

MATLAB example

In this example we use different pictures of a man's face with different facial expressions. But, these pictures are noisy. So the face is not easily recognizable. The data set consists of 1965 pictures each 20 by 28. So dimensionality is 20*28= 560.

Our goal is to remove noise using PCA and of course by means of SVD function in matlab. We know that noise is not the main feature that makes these pictures look different, so noise is among least variance components. We first extract the principal components and then remove those which correspond to the least values of eigenvalues of covariance matrix. Here, eigenvalues are sorted in a descending order from column to column of matrix S. For example, the first column of matrix U, which corresponds to the first element in S, is the first principal component.

We then reconstruct the picture using first d principal components. If d is too large, we can not completely remove the noise. If it is too small, we will loose some information from original data, for example we may loose the facial expression (smile, sadness and etc.). We can change d until we achieve a reasonable balance between noise reduction and information loss.

File:noisy-face.jpg

Noise reduced version of a picture in MATLAB example

 >> % loading the noisy data, file "noisy" stores our variable X which contains the pictures 
 >> load noisy;
 >>  
 >> 
 >> % show a sample image in column 1 of matrix X
 >> imagesc(reshape(X(:,1),20,28)')
 >> 
 >> % set the color of image to grayscale
 >> colormap gray
 >>
 >> % perform SVD, if X matrix if full rank, will obtain 560 PCs
 >> [U S V] = svd(X);
 >> 
 >> d= 10;
 >> 
 >> % reconstruct X using only the first d principal components and removing noise
 >> XX = U(:, 1:d)* S(1:d,1:d)*U(:,1:d)' ;
 >> 
 >> % show image in column 1 of XX which is a noise reduced version
 >> imagesc(reshape(XX(:,1),20,28)')

PCA continued, Lagrange multipliers, singular value decomposition (SVD) (Lecture 2: Sept. 15, 2014)

Principal component analysis (continued)

PCA is a method to reduce dimensions in data or to extract features from data.

Given a data point in vector form x the goal of PCA is to map x to y where x [math]\displaystyle{ \isin }[/math] [math]\displaystyle{ \real }[/math]^d and y [math]\displaystyle{ \isin }[/math] [math]\displaystyle{ \real }[/math]^p such that p is much less than d. For example we could map a two-dimensional data point onto any one-dimensional vector in the original 2D space or we could map a three-dimensional data point onto any 2D plane in the original 3D space.

In terms of data visualization, it is fairly clear why PCA is important; high dimensional data is sometimes impossible to visualize. Even a 3D space can be fairly difficult while a 2D space, which can be printed on a piece of paper, is much more easily visualized. If a higher dimensional dataset can be reduced to only 2 dimensions, it can be easily represented by plots and graphs.

In the case of many data points x_i in d-dimensional space:

X_dxn = [x₁ x₂ ... x_n]

There is any infinite amount of vectors in [math]\displaystyle{ \real }[/math]^p to which the points in X can be mapped. When mapping these data points to lower dimensions information will be lost. To preserve as much information as possible the points are mapped to a vector in [math]\displaystyle{ \real }[/math]^p which will preserve as much variation in the data as possible. In other words, the data is mapped to the vector in [math]\displaystyle{ \real }[/math]^p that will have maximum variance. In the images below, the data points mapped onto u in Mapping 1 have greater variance than in Mapping 2, and thus preserves more information in the data.

• 

  Two dimensional point mapped onto one dimensional vector

• 

  Mapping 1: Mapping of a two dimensional data set onto a one dimensional vector

• 

  Mapping 2: Mapping of a two dimensional data set onto a one dimensional vector

To project n data points onto u compute Y = u^TX.

The variance of Y = u^TX can be calculated easily as Y is a 1xn vector.

In higher dimensional space [math]\displaystyle{ \real }[/math]^q where q > 1 the concept is Covariance.

The variance of projected points in the direction of vector u is given by:

Var(u^TX) = u^TSu where S is the sample covariance matrix of X.

In finding the first principal component the objective is to find the direction u which will have the greatest variance of the points projected onto it. This can be expressed as the following optimization problem:

max_u u^TSu
s.t. u^Tu = 1

The restriction u^Tu = 1 is to bound the problem. If the length of u were not restricted, u^TSu would go to infinity as the magnitude of u goes to infinity. The selection of 1 as the bound is arbitrary.

Lagrange multipliers

The Lagrange multiplier is a method for solving optimization problems. After constraining the principal component analysis output vector to an arbitrary size the task of maximizing variance in lower dimensional spaces becomes an optimization problem.

Lagrange multipler theorem

The method of Lagrange multipliers states that given the optimization problem:

   [math]\displaystyle{ max\ f(x,y) }[/math] 
      [math]\displaystyle{ s.t.\ g(x,y) = c  }[/math]

The Lagrangian is defined by:

[math]\displaystyle{ \ L(x,y,\lambda)= f(x,y) - \lambda(g(x,y)-c) }[/math]

Which implies, where [math]\displaystyle{ \bigtriangledown\ }[/math] is the gradient, that:

[math]\displaystyle{ \bigtriangledown\ f(x,y) = \lambda \bigtriangledown g(x,y) }[/math]

[math]\displaystyle{ \bigtriangledown\ f(x,y) - \lambda \bigtriangledown g(x,y) = 0 }[/math]

In practice,

[math]\displaystyle{ \bigtriangledown\ L }[/math] is calculated by computing the three partial derivatives of [math]\displaystyle{ \ L }[/math] with respect to [math]\displaystyle{ \ x, y, \lambda }[/math], setting them to 0, and solving the resulting system of equations.

That is to say, the solution is given by taking the solution of:

[math]\displaystyle{ \ \bigtriangledown L = \begin{bmatrix} \frac{\partial L}{\partial x}\\\frac{\partial L}{\partial y}\\\frac{\partial L}{\partial \lambda}\end{bmatrix}= 0 }[/math]

which maximize the objective function f(x,y).

Intuitive explanation

The Lagrange multiplier utilizes gradients to find maximum values subject to constraints. A gradient [math]\displaystyle{ \bigtriangledown }[/math] is a vector which collects all of the partial derivatives of a function. Intuitively, this can be thought of as a vector which contains the slopes of all of a graph's tangent lines, or the direction which each point on the graph points.

Finding a point where the gradient of f(x,y) and g(x,y) differ by a factor lambda: [math]\displaystyle{ \bigtriangledown f(x,y) = \lambda \bigtriangledown g(x,y) }[/math] defines a point where our restraint and objective functions are both pointing in the same direction. This is important because if our objective function is pointing in a higher direction than g, increasing our objective function will increase our output, meaning we are not at a maximum. If the restraint is pointing in a higher direction than f, then reducing our input value will yield a higher output. The factor lambda is important because it is adjusted to ensure that the constraint is satisfied and maximized at our current location.

Lagrange multiplier example

Consider the function: f(x,y) = x - y
s.t. g(x,y) = x² + y² = 1

We want to solve the problem :

max (x-y)
x2 + y2 = 1
  

Calculating the Lagrangian gives:
L(x,y,[math]\displaystyle{ \lambda }[/math]) = x - y - [math]\displaystyle{ \lambda }[/math]g(x,y) = x - y - [math]\displaystyle{ \lambda }[/math](x² + y² - 1)

Which gives three partial derivatives:
[math]\displaystyle{ \frac{\partial L}{\partial x} = 1 - 2\lambda x = 0 }[/math]

[math]\displaystyle{ \frac{\partial L}{\partial y} = -1 - 2\lambda y = 0 }[/math]

[math]\displaystyle{ \frac{\partial L}{\partial \lambda} = x^2 + y - 1 = 0 }[/math]

This gives two solutions:

Solution 1:

[math]\displaystyle{  x = \frac{\sqrt{2}}{2}, y = \frac{-\sqrt{2}}{2}, \lambda = \frac{1}{\sqrt{2}} }[/math]

Solution 2:

[math]\displaystyle{  x = \frac{-\sqrt{2}}{2}, y = \frac{\sqrt{2}}{2}, \lambda = \frac{1}{\sqrt{2}} }[/math]

Both solutions give extreme values of our objective function. Because this is an optimization problem we are interested in the solution which gives the maximum output, which is solution 1.

Singular value decomposition revisited

Any matrix X can be decomposed to the three matrices: [math]\displaystyle{ \ U \Sigma V^T }[/math] where:

[math]\displaystyle{ \ U }[/math] is an m x m matrix of the eigenvector XX^T

[math]\displaystyle{ \ \Sigma }[/math] is an m x n matrix containing singular values along the diagonal and zeroes elsewhere

[math]\displaystyle{ \ V^T }[/math] is an orthogonal transpose of a n x n matrix of the eigenvector X^TX

Connection of Singular Value Decomposition to PCA

If X is centered (has a mean of 0) then XX^T is the covariance matrix.

Using the SVD of X we get:

[math]\displaystyle{ \ XX^T = (U \Sigma V^T)(U \Sigma V^T)^T }[/math]

[math]\displaystyle{ \ XX^T = (U \Sigma V^T)(V \Sigma U^T) }[/math]

[math]\displaystyle{ \ XX^T = U \Sigma V^T V \Sigma U^T }[/math]

[math]\displaystyle{ \ XX^T = U \Sigma I \Sigma U^T }[/math] (By the definition of V)

[math]\displaystyle{ \ XX^T = U \Sigma^2 U^T }[/math]

It is important to remember that U and V are unitary matricies, i.e.: [math]\displaystyle{ U^TU=UU^T=I }[/math] and [math]\displaystyle{ V^TV=VV^T }[/math], each equality using the appropriate dimensional identity matrix.. It is also interesting that the nonsquare matrix formed by taking a subset of the eigenvectors contained in [math]\displaystyle{ U }[/math] or [math]\displaystyle{ V }[/math] has the property that [math]\displaystyle{ V^TV=I }[/math] (the appropriate lower dimensional identity matrix)

Matlab example of dimensionality reduction using PCA

The following example shows how principal component analysis can be used to identify between a handwritten 2 and a handwritten 3. In the example the original images are each 64 pixels and thus 64 dimensional. By reducing the dimensionality to 2D the 2's and 3's are visualized on a 2D plot and can it could be estimated as to which digit one was.

 >> % load file containing 200 handwritten 2's and 200 handwritten 3's
 >> load 2_3;
 >>  
 >> who?
 >> % displays that X is 64x400 matrix containing the handwritten digits
 >> 
 >> imagesc(reshape(X(:,1),8,8)')
 >> colormap gray
 >> % displays first image, a handwritten 2
 >>
 >> M = mean(X,2)*ones(1,400);
 >> Xb = X-M
 >> % M is matrix of mean of rows repeated 400 times
 >> % Xb is data centred about mean
 >> 
 >> % Now use SVD to find eigenvectors, eigenvalues of Xb
 >> [U S V] = svd(Xb);
 >> 
 >> % 2D projection on points
 >> Y = U(:,1:2)'*X;
 >> 
 >> % 2D projections of 2's
 >> plot(Y(1,1:200),Y(2,1:200),'.')
 >> 
 >> % 2D projections of 3's
 >> plot(Y(1,201:400),Y(2,201:400),'.') 

PCA Algorithm, Dual PCA, and Kernel PCA (Lecture 3: Sept. 17, 2014)

PCA Algorithm (Algorithm 1)

Recover basis

To find the basis for PCA we first center the data by removing the mean of the sample, then calculate the covariance:

[math]\displaystyle{ XX^T = \sum_{i=1}^n x_i x_i^T }[/math]

Let [math]\displaystyle{ U }[/math] be the eigenvectors of [math]\displaystyle{ XX^T }[/math] corresponding to the top [math]\displaystyle{ \,d }[/math] eigenvalues.

Encode training data

To encode our data using PCA we let

[math]\displaystyle{ \,Y = U^TX }[/math]

where [math]\displaystyle{ Y }[/math] is an encoded matrix of the original data

Reconstruct training data

To project our data back to the higher dimension

[math]\displaystyle{ \hat{X} = UY = UU^T X }[/math]

Encode test example

[math]\displaystyle{ \,y = U^T x }[/math]

where [math]\displaystyle{ \,y }[/math] is an encoding of [math]\displaystyle{ \,x }[/math]

Reconstruct test example

[math]\displaystyle{ \hat{x} = Uy = UU^T x }[/math]

Dual principal component analysis

Motivation

In some cases the dimension of our data can be much larger than the number of data points [math]\displaystyle{ ( d \gt \gt n ) }[/math]. For example, suppose we are interested in weekly closing rates of the Dow Jones Industrial Average, S&P 500 Index, and the NASDAQ Composite Index over the past forty years. In this example we only have three data points [math]\displaystyle{ ( n = 3 ) }[/math], but our dimension is over two thousand! [math]\displaystyle{ ( d \approx 2080 ) }[/math] Consequently, the matrix [math]\displaystyle{ XX^T }[/math] used to find the basis [math]\displaystyle{ U }[/math] is [math]\displaystyle{ 2080\times2080 }[/math], which is computationally heavy to calculate. As another example is studying genes for patients since the number of genes studied is likely much larger than the number of patients. When the data's dimensionality is high, a less compute-intensive way of recovering the basis is desirable.

Recall from Lecture 2 how we used singular value decomposition for PCA:

[math]\displaystyle{ \,X = U \Sigma V^T }[/math]

Multiply both sides by V to get:

[math]\displaystyle{ \,X V = U \Sigma V^T V }[/math]

Since [math]\displaystyle{ V^T V = I }[/math] (the identity matrix) the SVD can be rearranged as

[math]\displaystyle{ \,X V = U \Sigma }[/math]

Then, multiply both sides by [math]\displaystyle{ \Sigma }[/math]^-1 to get:

[math]\displaystyle{ \,U = X V \Sigma }[/math]^-1

And so [math]\displaystyle{ U }[/math] can be calculated in terms of [math]\displaystyle{ V }[/math] and [math]\displaystyle{ \Sigma }[/math]. [math]\displaystyle{ V }[/math] is the eigenvectors of [math]\displaystyle{ X^T X_{n \times n} }[/math], so it is much easier to find than [math]\displaystyle{ U }[/math] because [math]\displaystyle{ n\lt \lt d }[/math].

Algorithm

We can replace all instances of [math]\displaystyle{ \,U }[/math] in Algorithm 1 with the dual form, [math]\displaystyle{ \,U = X V \Sigma^{-1} }[/math], to obtain the algorithm for Dual PCA.

Recover Basis:
For the basis of Dual PCA, we calculate

[math]\displaystyle{ XX^T }[/math]

then let [math]\displaystyle{ V }[/math] be the eigenvectors of [math]\displaystyle{ XX^T }[/math] with respect to the top [math]\displaystyle{ d }[/math] eigenvalues. Finally we let [math]\displaystyle{ \Sigma }[/math] be the diagonal matrix of square roots of the top [math]\displaystyle{ d }[/math] eigenvalues.

Encode Training Data:
Using Dual PCA to encode the data, let

[math]\displaystyle{ \,Y = U^TX = \Sigma V^T }[/math]

where [math]\displaystyle{ Y }[/math] is an encoded matrix of the original data

Reconstruct Training Data:
Project the data back to the higher dimension by

[math]\displaystyle{ \hat{X} = UY = U \Sigma V^T = X V \Sigma^{-1} \Sigma V^T = X V V^T }[/math]

Encode Test Example:

[math]\displaystyle{ \,y = U^T x = \Sigma^{-1} V^T X^T x = \Sigma^{-1} V^T X^T x }[/math]

where [math]\displaystyle{ \,y }[/math] is an encoding of [math]\displaystyle{ \,x }[/math]

Reconstruct Test Example:

[math]\displaystyle{ \hat{x} = Uy = UU^T x = X V \Sigma^{-2} V^T X^T x = X V \Sigma^{-2} V^T X^T x }[/math]

Note that the steps of Reconstructing training data and Reconstructioning test example still depend on [math]\displaystyle{ d }[/math], and therefore still will be impractical in the case that the original dimensionality of the data ([math]\displaystyle{ d }[/math]) is very large

Kernel principle component analysis

Kernel methods

Kernel methods are a class of algorithms used for analyzing patterns and measuring similarity. They are used for finding relations within a data set. Kernel methods map data into a higher-dimensional space to make it easier to find these relations. While PCA is designed for linear variabilities, a lot of higher-dimensional data are nonlinear. This is where Kernel PCA can be useful to help model the data variability when it comes to a nonlinear manifold. Generally speaking, Kernels is a measure of "similarity" between data points using different methods.

Some common examples of kernels include:

1. Linear Kernel:

[math]\displaystyle{ k_{ij} = \lt x_{i},x_{j}\gt }[/math]

2. Polynomial Kernel:

[math]\displaystyle{ k_{ij} = (1+\lt x_{i},x_{j}\gt )^P }[/math]

3. Gaussian Kernel:

[math]\displaystyle{ k_{ij} = exp(||x_{i}-x_{j}||^2/2\sigma^2) }[/math]

Other kernel examples can be found here:

Source:http://mlpy.sourceforge.net/docs/3.2/dim_red.html

Kernel PCA

With the Kernel PCA, we take the original or observed space and map it to the feature space and then map the feature space to the embedded space of a lower dimension.

[math]\displaystyle{ X_{observed space} \rightarrow \Eta_{feature space} \rightarrow Y_{embedded space} }[/math]

In the Kernel PCA algorithm, we can use the same argument as PCA and again use the Standard Value Decomposition (SVD) where:

[math]\displaystyle{ \Phi(X) = U \Sigma V^T }[/math]

where [math]\displaystyle{ U }[/math] contains the eigenvectors of [math]\displaystyle{ \Phi(X)\Phi(X)^T }[/math]

The algorithm is similar to the Dual PCA algorithm except that the training and the test data cannot be reconstructed since [math]\displaystyle{ \Phi(X) }[/math] is unknown.

Algorithm

Recover Basis:
For the basis of Kernel PCA, we calculate

[math]\displaystyle{ K = \Phi(X)^T \Phi(X) }[/math]

using the kernel [math]\displaystyle{ K }[/math] and let [math]\displaystyle{ V }[/math] be the eigenvectors of [math]\displaystyle{ \Phi(X)^T \Phi(X) }[/math] with respect to the top [math]\displaystyle{ p }[/math] eigenvalues. Finally we let [math]\displaystyle{ \Sigma }[/math] be the diagonal matrix of square roots of the top [math]\displaystyle{ p }[/math] eigenvalues.

Encode Training Data:
Using Kernel PCA to encode the data, let

[math]\displaystyle{ \,Y = U^T \Phi(X) = \Sigma V^T }[/math]

Recall [math]\displaystyle{ U = \Phi(X) V \Sigma^{-1} }[/math]:

[math]\displaystyle{ \,Y = U^T \Phi(X) = \Sigma^{-1} V^T \Phi(X)^T \Phi(X) = \Sigma^{-1} V^T K(X,X) }[/math]

where [math]\displaystyle{ Y }[/math] is an encoded matrix of the image of the original data in the function [math]\displaystyle{ \Phi }[/math]. [math]\displaystyle{ Y }[/math] is computable without knowledge of [math]\displaystyle{ \Phi(X) }[/math] via the kernel function.

Reconstruct Training Data:

[math]\displaystyle{ \hat{\Phi(X)} = UY = UU^T\Phi(X)=\Phi(X)V\Sigma^{-1}\Sigma V^T=\Phi(X) V V^T }[/math]

[math]\displaystyle{ \hat{X} = \Phi^{-1}(\hat{\Phi(X)}) }[/math]

However, [math]\displaystyle{ \Phi(X) }[/math] is unknown so we cannot reconstruct the data.

Encode Test Example:

[math]\displaystyle{ \,y = U^T \Phi(x) = \Sigma^{-1} V^T \Phi(X)^T \Phi(x) = \Sigma^{-1} V^T K(X,x) }[/math]

This is possible because we know that [math]\displaystyle{ \Phi(X)^T \Phi(X) = K(X,x) }[/math].
Recall
[math]\displaystyle{ U = \Phi(X) V \Sigma^{-1} }[/math]:
[math]\displaystyle{ \Phi(X) }[/math] is the transformation of the training data [math]\displaystyle{ X_{d \times n} }[/math] while [math]\displaystyle{ \Phi(x) }[/math] is the transformation of the test example [math]\displaystyle{ x_{d \times 1} }[/math]

Reconstruct Test Example:

[math]\displaystyle{ \hat{\Phi(x)} = Uy = UU^T \Phi(x) = \Phi(X) V \Sigma^{-2} V^T \Phi(X)^T \Phi(x) = \Phi(X) V \Sigma^{-2} V^T K(X,x) }[/math]

Once again, since [math]\displaystyle{ \Phi(X) }[/math] is unknown, we cannot reconstruct the data.

Centering for kernel PCA and multidimensional scaling (Lecture 4: Sept. 22, 2014)

Centering for kernel PCA

In the derivation of kernel PCA we assumed that [math]\displaystyle{ \sum_{i=1}^n\Phi(x_i)=0 }[/math] which is improbable for a random data set and an arbitrary mapping [math]\displaystyle{ \Phi }[/math]. We must be able to ensure this condition though we don't know [math]\displaystyle{ \Phi }[/math].

Let [math]\displaystyle{ \mu=\frac{1}{n}\sum_{j=1}^n\Phi(x_j) }[/math]

Define the centered transformation (not computable)

[math]\displaystyle{ \tilde{\Phi}:x\mapsto(\Phi(x)-\mu) }[/math]

and the centered kernel (computable)

[math]\displaystyle{ \tilde{K}:(x,y)\mapsto \tilde{\Phi}(x)^T \tilde{\Phi}(y) }[/math]

Then

[math]\displaystyle{ \tilde{K}(x,y) = (\Phi(x)-\mu)^T (\Phi(y)-\mu)=\Phi(x)^T\Phi(y)-\mu^T\Phi(x)-\mu^T\Phi(y)+\mu^T\mu }[/math]

[math]\displaystyle{ =K(x,y)-\frac{1}{n}\sum_{j=1}^n\Phi(x_j)^T\Phi(x)-\frac{1}{n}\sum_{j=1}^n\Phi(x_j)^T\Phi(y)+\frac{1}{n^2}\sum_{j=1}^n\sum_{i=1}^n\Phi(x_j)^T\Phi(x_i) }[/math]
[math]\displaystyle{ =K(x,y)-\frac{1}{n}\sum_{j=1}^n K(x_j,x)-\frac{1}{n}\sum_{j=1}^n K(x_j,y)+\frac{1}{n^2}\sum_{j=1}^n\sum_{i=1}^n K(x_j,x_i) }[/math]
[math]\displaystyle{ =K(x,y)-\frac{1}{n}\bold{1}^TK(X,x)-\frac{1}{n}\bold{1}^TK(X,y)+\frac{1}{n^2}\bold{1}^TK(X,X)\bold{1} }[/math]
where [math]\displaystyle{ \bold{1}=([1,1,...,1]_{1\times n})^T }[/math]

Here [math]\displaystyle{ X_{d\times n} }[/math] and [math]\displaystyle{ \{x_j\}_{j=1}^n }[/math] denote the data set matrix and elements respectively and [math]\displaystyle{ x,y }[/math] represent test elements at which the centered kernel is evaluated.

Multidimensional scaling

Multidimensional Scaling (MDS) is a dimension reduction method that attempts to preserve the pair-wise distance between data points. For example, Suppose we want to map a set of points X, where each point [math]\displaystyle{ x \in \mathbb{R}^d }[/math], to a set of points Y, where each point [math]\displaystyle{ y \in \mathbb{R}^p }[/math]. We want to preserve the pair-wise distances such that [math]\displaystyle{ d(y_i, y_j) }[/math] is relatively the same as [math]\displaystyle{ d(y_i, y_j) }[/math]

MDS example

We can map points in 2 dimensions to points in 1 dimension by projecting the points on to a straight line:

Multidimensional scaling from 2D to 1D

Note that the pair-wise distances are relatively preserved. However, the triangle inequality is not preserved:

[math]\displaystyle{ d(x_1,x_2)+d(x_2,x_3) \gt d(x_1, x_3) }[/math], but [math]\displaystyle{ d(y_1,y_2)+d(y_2,y_3) = d(y_1, y_3) }[/math]

Another example that needs the concept of preserving the distances between data points in lower dimension space (MDS) is dealing with "distances between cities", as shown in here

Distance matrix

Given a [math]\displaystyle{ d x n }[/math] matrix, [math]\displaystyle{ X }[/math], the distance matrix is defined by:

[math]\displaystyle{ D^X_{n \times n} = \begin{bmatrix} d^{(X)}_{11} & d^{(X)}_{12} & ... & d^{(X)}_{1n}\\d^{(X)}_{21} & d^{(X)}_{22} & ... & d^{(X)}_{2n}\\...\\d^{(X)}_{n1} & d^{(X)}_{n2} & ... & d^{(X)}_{nn} \end{bmatrix} }[/math]

Where [math]\displaystyle{ d^{(X)}_{ij} }[/math] is defined as the euclidean distance between points [math]\displaystyle{ x_i }[/math] and [math]\displaystyle{ x_j }[/math] in [math]\displaystyle{ \mathbb{R}^d }[/math]

Important properties:

The matrix is symmetric ([math]\displaystyle{ d^{(X)}_{ij} = d^{(X)}_{ji} \; \forall \; 1 \leq i,j \leq n }[/math])

The matrix is always non-negative ([math]\displaystyle{ d^{(X)}_{ij} \ge 0 \; \forall \; 1 \leq i,j \leq n }[/math])

The diagonal of the matrix is always 0 ([math]\displaystyle{ d^{(X)}_{ii}= 0 \; \forall \; 1 \leq i \leq n }[/math])

If [math]\displaystyle{ x_i }[/math]'s don't overlap, then all elements that are not on the diagonal are positive

Triangle inequality: [math]\displaystyle{ d_{ab} + d_{bc} \gt = d_{ac} }[/math] whenever [math]\displaystyle{ a, b, c }[/math] are three different indices

For any mapping to matrix [math]\displaystyle{ Y_{p \times n} }[/math], we can calculate the corresponding distance matrix

[math]\displaystyle{ D^Y_{n \times n} = \begin{bmatrix} d^{(Y)}_{11} & d^{(Y)}_{12} & ... & d^{(Y)}_{1n}\\d^{(Y)}_{21} & d^{(Y)}_{22} & ... & d^{(Y)}_{2n}\\...\\d^{(Y)}_{n1} & d^{(Y)}_{n2} & ... & d^{(Y)}_{nn} \end{bmatrix} }[/math]

Where [math]\displaystyle{ d^{(Y)}_{ij} }[/math] is defined as the euclidean distance between points [math]\displaystyle{ y_i }[/math] and [math]\displaystyle{ y_j }[/math] in [math]\displaystyle{ \mathbb{R}^p }[/math]

Finding the optimal Y

MDS minimizes the metric

[math]\displaystyle{ \text{min}_Y \sum_{i=1}^n{\sum_{j=1}^n{(d_{ij}^{(X)}-d_{ij}^{(Y)})^2}} }[/math]

Where [math]\displaystyle{ d_{ij}^{(X)} = ||x_i - x_j|| }[/math], and [math]\displaystyle{ d_{ij}^{(Y)} = ||y_i - y_j|| }[/math]

The distance matrix [math]\displaystyle{ D^{(X)} }[/math] can be converted to a kernel matrix:

[math]\displaystyle{ K = -\frac{1}{2}HD^{(X)}H }[/math]

Where [math]\displaystyle{ H = I - \frac{1}{n}ee^T }[/math] and [math]\displaystyle{ e=\begin{bmatrix} 1 \\ 1 \\ ... \\ 1 \end{bmatrix} }[/math]

Here, K is a kernel matrix implies that K is positive semi-definite, so [math]\displaystyle{ x^TKx \geq 0 }[/math] [math]\displaystyle{ \forall x }[/math]

In fact, there is a one-to-one correspondence between symmetric matrices and kernel matrices

Theorem

If [math]\displaystyle{ D^{(X)} }[/math] is a distance matrix and if [math]\displaystyle{ K=-\frac{1}{2}HD^{(X)}H }[/math]
Then [math]\displaystyle{ D^{(X)} }[/math] is Euclidean iff [math]\displaystyle{ K }[/math] is Positive Semi-Definite
Where [math]\displaystyle{ H=I-\frac{1}{n}\bold{1}^T\bold{1} }[/math] and [math]\displaystyle{ \bold{1}=([1,1,...,1]_{1\times n})^T }[/math]

Relation between MDS and PCA

The goal of MDS is to find the [math]\displaystyle{ Y }[/math] at which the following minimum is obtained: [math]\displaystyle{ \text{min}_Y \sum_{i=1}^n{\sum_{j=1}^n{(d_{ij}^{(X)}-d_{ij}^{(Y)})^2}} }[/math]
K is positive semi-definite so it can be written [math]\displaystyle{ K=X^TX }[/math] . So

[math]\displaystyle{ \text{min}_Y \sum_{i=1}^n{\sum_{j=1}^n{(d_{ij}^{(X)}-d_{ij}^{(Y)})^2}} =\text{min}_Y(\sum_{i=1}^n\sum_{j=1}^n(x_i^Tx_j-y_i^Ty_j)^2) }[/math]

and

[math]\displaystyle{ \text{min}_Y(\sum_{i=1}^n\sum_{j=1}^n(x_i^Tx_j-y_i^Ty_j)^2)=\text{min}_Y(\text{Tr}((X^TX-Y^TY)^2)) }[/math]

From SVD we may make the decompositions [math]\displaystyle{ X^TX=V\Lambda V^T }[/math] and [math]\displaystyle{ Y^TY=Q\hat{\Lambda}Q^T }[/math]

[math]\displaystyle{ \text{min}_Y(\text{Tr}((X^TX-Y^TY)^2))=\text{min}_Y(\text{Tr}((V\Lambda V^T-Q\hat{\Lambda}Q^T)^2))=\text{min}_Y(\text{Tr}((\Lambda-V^TQ\hat{\Lambda}Q^TV)^2)) }[/math]

Let [math]\displaystyle{ G=Q^TV }[/math]

[math]\displaystyle{ \text{min}_{G,\hat{\Lambda}}(\text{Tr}((X^TX-Y^TY)^2))=\text{min}_{G,\hat{\Lambda}}(\text{Tr}((\Lambda-G^T\hat{\Lambda}G)^2))=\text{min}_{G,\hat{\Lambda}}(\text{Tr}(\Lambda^2+G^T\hat{\Lambda}GG^T\hat{\Lambda}G-2\Lambda G^T\hat{\Lambda}G)) }[/math]

[math]\displaystyle{ =\text{min}_{G,\hat{\Lambda}}(\text{Tr}(\Lambda^2+G^T\hat{\Lambda}^2G-2\Lambda G^T\hat{\Lambda}G)) }[/math]
It can be shown that for fixed [math]\displaystyle{ \hat{\Lambda} }[/math] that the minimum over [math]\displaystyle{ G }[/math] occurs at [math]\displaystyle{ G=I }[/math], which implies that

[math]\displaystyle{ \text{min}_{G,\hat{\Lambda}}(\text{Tr}((X^TX-Y^TY)^2))=\text{min}_{\hat{\Lambda}}(\text{Tr}(\Lambda^2-2\Lambda\hat{\Lambda}+\hat{\Lambda}^2)=\text{min}_{\hat{\Lambda}}(\text{Tr}((\Lambda-\hat{\Lambda})^2) }[/math]

Since [math]\displaystyle{ \Lambda, \hat{\Lambda} }[/math] are diagonal matrices containing the eigenvalues of [math]\displaystyle{ X^TX,\text{ and } Y^TY }[/math] respectively, and all said eigenvalues are positive, then the min is clearly obtained when the [math]\displaystyle{ p }[/math] nonzero diagonal elements in [math]\displaystyle{ Y^TY }[/math] are the [math]\displaystyle{ p }[/math] largest eigenvalues in [math]\displaystyle{ X^TX }[/math]