EigenfaceCore.m
function [m, A, Eigenfaces] = EigenfaceCore(T)
% Use Principle Component Analysis (PCA) to determine the most?
% discriminating features between images of faces.
%
% Description: This function gets a 2D matrix, containing all training image vectors
% and returns 3 outputs which are extracted from training database.
%
% Argument: ? ? ?T ? ? ? ? ? ? ? ? ? ? ?- A 2D matrix, containing all 1D image vectors.
% ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? Suppose all P images in the training database?
% ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? have the same size of MxN. So the length of 1D?
% ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? column vectors is M*N and 'T' will be a MNxP 2D matrix.
%?
% Returns: ? ? ? m ? ? ? ? ? ? ? ? ? ? ?- (M*Nx1) Mean of the training database
% ? ? ? ? ? ? ? ?Eigenfaces ? ? ? ? ? ? - (M*Nx(P-1)) Eigen vectors of the covariance matrix of the training database
% ? ? ? ? ? ? ? ?A ? ? ? ? ? ? ? ? ? ? ?- (M*NxP) Matrix of centered image vectors
%
% See also: EIG
% Original version by Amir Hossein Omidvarnia, October 2007
% ? ? ? ? ? ? ? ? ? ? Email: aomidvar@ece.ut.ac.ir ? ? ? ? ? ? ? ? ?
%%%%%%%%%%%%%%%%%%%%%%%% Calculating the mean image?
m = mean(T,2); % Computing the average face image m = (1/P)*sum(Tj's) ? ?(j = 1 : P)
Train_Number = size(T,2);
%%%%%%%%%%%%%%%%%%%%%%%% Calculating the deviation of each image from mean image
A = [];
for i = 1 : Train_Number
temp = double(T(:,i)) - m; % Computing the difference image for each image in the training set Ai = Ti - m
A = [A temp]; % Merging all centered images
end
%%%%%%%%%%%%%%%%%%%%%%%% Snapshot method of Eigenface methos
% We know from linear algebra theory that for a PxQ matrix, the maximum
% number of non-zero eigenvalues that the matrix can have is min(P-1,Q-1).
% Since the number of training images (P) is usually less than the number
% of pixels (M*N), the most non-zero eigenvalues that can be found are equal
% to P-1. So we can calculate eigenvalues of A'*A (a PxP matrix) instead of
% A*A' (a M*NxM*N matrix). It is clear that the dimensions of A*A' is much
% larger that A'*A. So the dimensionality will decrease.
L = A'*A; % L is the surrogate of covariance matrix C=A*A'.
[V, D] = eig(L); % Diagonal elements of D are the eigenvalues for both L=A'*A and C=A*A'.
%%%%%%%%%%%%%%%%%%%%%%%% Sorting and eliminating eigenvalues
% All eigenvalues of matrix L are sorted and those who are less than a
% specified threshold, are eliminated. So the number of non-zero
% eigenvectors may be less than (P-1).
L_eig_vec = [];
for i = 1 : size(V,2)
if( D(i,i)>1 )
L_eig_vec = [L_eig_vec V(:,i)];
end
end
%%%%%%%%%%%%%%%%%%%%%%%% Calculating the eigenvectors of covariance matrix 'C'
% Eigenvectors of covariance matrix C (or so-called "Eigenfaces")
% can be recovered from L's eiegnvectors.
Eigenfaces = A * L_eig_vec; % A: centered image vectors
