MATLAB: Zscore and solving by definition differ

MATLABnormalizationprobabilitystatistics

I am trying to recreate some steps of finding eigenvalues and eigenvectors from a data set. The one I used is Matlab's cities. The code is:
clear all
close all 
load cities
A=zeros(329,9);
L=zeros(9,9);
V=zeros(9,9);
%normalization
for i=1:1:9
    means(i)=sum(ratings(:,i))/size(ratings,1);
    stdev(i)=std(ratings(:,i));
end
for i=1:1:9
    for j=1:1:329
        A(j,i)=(ratings(j,i)-means(i))/stdev(i);
    end
end
%covariance matrix
C=cov(A);
%eigenvalues
[vectors,lambda]=eig(C);
%sorting
sort(lambda,'descend'); %variances
for k=1:1:9
    V(:,k)=vectors(:,10-k);
end
L=flip(flip(lambda,2),1);
Z=zscore(ratings);
[coefs,scores,variances,t2]=pca(Z);
Outputs: V =
    0.2064    0.2178   -0.6900    0.1373   -0.3691    0.3746   -0.0847    0.3623   -0.0014
    0.3565    0.2506   -0.2082    0.5118    0.2335   -0.1416   -0.2306   -0.6139   -0.0136
    0.4602   -0.2995   -0.0073    0.0147   -0.1032   -0.3738    0.0139    0.1857    0.7164
    0.2813    0.3553    0.1851   -0.5391   -0.5239    0.0809    0.0186   -0.4300    0.0586
    0.3512   -0.1796    0.1464   -0.3029    0.4043    0.4676   -0.5834    0.0936   -0.0036
    0.2753   -0.4834    0.2297    0.3354   -0.2088    0.5022    0.4262   -0.1887   -0.1108
    0.4631   -0.1948   -0.0265   -0.1011   -0.1051   -0.4619   -0.0215    0.2040   -0.6858
    0.3279    0.3845   -0.0509   -0.1898    0.5295    0.0899    0.6279    0.1506    0.0255
    0.1354    0.4713    0.6073    0.4218   -0.1596    0.0326   -0.1497    0.4048   -0.0004
coefs =
    0.2064   -0.2178    0.6900   -0.1373   -0.3691    0.3746   -0.0847   -0.3623   -0.0014
    0.3565   -0.2506    0.2082   -0.5118    0.2335   -0.1416   -0.2306    0.6139   -0.0136
    0.4602    0.2995    0.0073   -0.0147   -0.1032   -0.3738    0.0139   -0.1857    0.7164
    0.2813   -0.3553   -0.1851    0.5391   -0.5239    0.0809    0.0186    0.4300    0.0586
    0.3512    0.1796   -0.1464    0.3029    0.4043    0.4676   -0.5834   -0.0936   -0.0036
    0.2753    0.4834   -0.2297   -0.3354   -0.2088    0.5022    0.4262    0.1887   -0.1108
    0.4631    0.1948    0.0265    0.1011   -0.1051   -0.4619   -0.0215   -0.2040   -0.6858
    0.3279   -0.3845    0.0509    0.1898    0.5295    0.0899    0.6279   -0.1506    0.0255
    0.1354   -0.4713   -0.6073   -0.4218   -0.1596    0.0326   -0.1497   -0.4048   -0.0004
Mathematically, they should be the same. But, some of the eigenvectors are totally equal, where some of them (columns 2,3,4,8) are negative (multiplied by -1). What do you think is happening? I need to find the reason and source for this.

Best Answer

Source? Basic mathematics. An eigenvector is NOT unique.
If A is a square matrix, with eigenvalue lambda, and eigenvector X, then we learn that
A*X = lambda*X
But then we can multiply by ANY scalar.
k*A*X = k*lambda*X
Since scalar multiplication will commute here, we see that
A*(k*X) = lambda*(k*X)
Therefore k*X is ALSO an eigenvector. What did you find? k=-1. SURPRISE!
Some of your eigenvectors are the same, others are essentially the same, but with s sign flip. Still the same. Since the eigenvectors will be normalized to have unit norm, they can still have an arbitrary factor of -1 applied to any eigenvector.
Finally, in some cases, where an eigenvalue has multiplicity higher than 1, then the corresponding eigenvectors are even less unique. But that is not your problem here.
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