Eigenvectors of covariance matrix and inertia tensor

Question

The moment of inertia tensor from physics looks very similar to the covariance matrix, used for PCA. How are their eigenvectors and eigenvalues related?

Answer 1

First, let's set up the moment of inertia tensor for $N$ points, where point $n$ has mass $m_n$ and coordinates $(x_m^{(1)}, x_m^{(2)}, ...)$ and define $C_{ij} = \sum_{n=1}^N m_n x_n^{(i)} x_n^{(j)} $. Then the moment of inertia tensor is

$\mathbf{J} = tr(C)\mathbf{I}-\mathbf{C}$

The symbol $\mathbf{I}$ denotes the identity matrix, and $\mathbf{J}$ is the moment of inertia tensor. If $\mathbf{v}$ is an eigenvector of $\mathbf{C}$ with eigenvalue $\lambda$ then $\mathbf{C} \lambda = \lambda \mathbf{v}$. Also, we have $tr(C)\mathbf{v} = tr(C)\mathbf{I} \mathbf{v}$. Subtracting these two equations gives:

$(tr(C)-\lambda)\mathbf{v} = ( tr(C)\mathbf{I}-\mathbf{C})\mathbf{v}$

so $tr(C)-\lambda$ is an eigenvalue of $\mathbf{J}$ and $\mathbf{v}$ is an eigenvector of $\mathbf{J}$. And finally, the weighted covariance matrix is $\mathbf{K}=\mathbf{C}/M$ where $M$ is the total mass and the covariance weights correspond to the masses.