# Principal Component Analysis Singluar value

While studying PCA, I saw this question but not able to solve. Given a data matrix X that is taller than it is wide, prove that every right singular vector of X with singular value s is an eigenvector of the covariance matrix, cov(X), with eigenvalue s2.

This is a straightforward application of definitions. Write out the right-SVD of $$X$$ as: $$X=U\Sigma V^T$$. The covariance matrix is then $$XX^T = U(\Sigma\Sigma^T)U^T$$. Notice that this is of the form $$USU^T$$, where $$S$$ is a square matrix. Moreover, since $$\Sigma$$ is tall (say, $$m\times n$$ for $$m>n$$), then $$\Sigma$$ breaks into a diagonal $$n\times n$$ matrix and a $$(m-n)\times n$$ matrix of zeros. So show that this implies $$\Sigma\Sigma^T$$ is diagonal. Now you can just read off the eigenvectors and eigenvalues of $$XX^T$$.