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Why different result with PCA and SVD in Matlab?

Date: 2022-11-30 12:20:34

I have implemented my PCA function in Matlab in the following way: 

 

function e = myPCA(X) [D, N] = size(X); m = mean(X, 2); X = X - repmat(m, 1, N); [e, ~, ~] = svd(X,'econ'); end

When I use now the built in Matlab function [e, ~, ~] = pca(X'); I'm getting the same absolute values as with my function but for some columns of U the sign is flipped.

Which ones is the right way to go and why is there this difference in sign? 

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Expert Answer:

: 

It is true that the matrix you denote by e has columns which are the basis in which the covariance matrix is diagonal, as should be in PCA. However, an orthogonal basis stays an orthogonal basis if you flip any basis vector. For example, in 2D if the basis e you find via Matlab's pca is the axes x and y, then -x and -y is also a basis. Or stated slightly different, since for pca you find the eigenvectors of the covariance matrix, and since if v is an eigenvector then -v is also an eigenvector (with the same eigenvalue), we see that the principal components are defined up to a sign. Since svd and pca are implemented differently, you don't have a guaranty to get the same signs. 

 

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