Inertia tensor and principal axes

Remarks

By the spectral theorem, there is an orthonormal basis of $\R^3$ made of eigenvectors of $I_O$, so every body has three mutually perpendicular principal axes at each point. For a unit vector $n$, the number $n^T I_O n$ is the moment of inertia of the body about the line through $O$ in the direction $n$. For a body turning about the fixed point $O$ with angular velocity $\omega$, the angular momentum about $O$ is $I_O \omega$ and the kinetic energy is $\tfrac12 \omega^T I_O \omega$. The matrix $I_O$ is positive semidefinite, and it is positive definite unless all the mass lies on one line through $O$. For point masses $m_i$ at positions $x^{(i)}$, the integral is replaced by the sum $\sum_i m_i \big(\abs{x^{(i)}}^2 \delta_{jk} - x^{(i)}_j x^{(i)}_k\big)$.

Example: a uniform box of mass $M$ and side lengths $a, b, c$, centred at $O$ with edges along the coordinate axes. The products $x_j x_k$ with $j \ne k$ integrate to $0$ by symmetry, and $\int_B \rho\, x_1^2 \, dV = M a^2 / 12$. So $I_O = \operatorname{diag}\big(M(b^2 + c^2)/12, \ M(a^2 + c^2)/12, \ M(a^2 + b^2)/12\big)$, and the coordinate axes are principal axes.

Non-example: for two point masses $m$ at $(1, 1, 0)$ and $(-1, -1, 0)$, the off-diagonal entry is $(I_O)_{12} = -2m \ne 0$. So the $x_1$-axis is not a principal axis; the principal axes are the lines spanned by $(1, 1, 0)$, $(1, -1, 0)$ and $(0, 0, 1)$.