#covariance-matrix

Question 1: Single choice

mediumAI-checkedDimensionality Reduction

A d2l exercise asks for the eigenvectors of $\mathbf{A} = \begin{bmatrix} 2 & 1 \\ 0 & 2 \end{bmatrix}$ . The eigenvalue $2$ is repeated, yet there is only one independent eigenvector, $(1, 0)^\top$ , so $\mathbf{A}$ has no eigendecomposition. Why can this never happen to the covariance matrix that PCA eigendecomposes?

Covariance matrices are symmetric, and real symmetric matrices always have an orthonormal eigenbasis.1
Covariance matrices always have distinct eigenvalues, so a repeated eigenvalue like this one cannot occur.2
Centering the data removes all off-diagonal covariances, so PCA only eigendecomposes a diagonal matrix.3
Covariance matrices are always invertible, and every invertible matrix has a full set of eigenvectors.4

Timed test