Properties of the components

Topic: Principal component analysis · more questions on this topic
‹ A two-variable analysis The effect of correlation ›

Using S=\Gamma\Lambda\Gamma^\top, show that the covariance matrix of the full vector of components \mathbf{Y}=\Gamma^\top(\mathbf{X}-\bar{\mathbf{x}}) is \Lambda, and that the generalised variance |S| equals \prod_j\lambda_j.

Worked answer (click to show)

Express the scores as an orthogonal change of coordinates. The covariance then becomes diagonal because the chosen directions diagonalise the same matrix used to define the components.

\operatorname{Cov}(\mathbf{Y})=\Gamma^\top S\Gamma=\Gamma^\top\Gamma\Lambda\Gamma^\top\Gamma=\Lambda,

by the linear transformation rule for covariances with coefficient matrix \Gamma^\top, and by orthogonality of \Gamma. The components are uncorrelated, with variances \lambda_j. For the determinant,

|S|=|\Gamma|\,|\Lambda|\,|\Gamma^\top|=|\Gamma|^2\prod_j\lambda_j=\prod_j\lambda_j,

since |\Gamma|^2=|\Gamma^\top\Gamma|=1.