# Off-diagonal elements of the inertia tensor and its eigenvalues [duplicate]

Given a certain rigid object, its inertia tensor $$\tilde I$$ is given by $$\begin{pmatrix} I_{xx} & I_{xy} & I_{xz}\\ I_{yx}& I_{yy}& I_{yz}\\ I_{zx}& I_{zy}& I_{zz} \end{pmatrix}$$

I understand that the entries in the main diagonal of the inertia tensor are the moments of inertia around the $$x$$, $$y$$ and $$z$$ axis. However, what is the physical meaning of the off-diagonal entries in the matrix. Additionaly, and more importantly, what is the physical meaning of the eigenvectors of the inertia tensor?

• Do you know the concept of eigenvalues and eigenvectors of a matrix? – Semoi Dec 6 '19 at 19:47
• @Semoi Yeah. I know, that the eigenvectors are only going to be multiplied by the associated eigenvalue, which is going to be the moment of inertia if the objects rotation vector was parallel to the eigenvector, but I can't seem to understand the physical meaning of the eigenvector. I believe it has to do something with the off-diagonal elements being zero if we change the reference frame to coincide with the 3 orthogonal eigenvectors, but I don't understand why, physically, the off-diagonal elements become zero, and what's the physical meaning of the eigenvectors. – Expain Dec 6 '19 at 20:10

These off-diagonal terms are called the products of inertia. If we have an angular velocity vector $$\boldsymbol{\omega}$$, then the product of inertia $$I_{ij}$$ is the proportionality constant that measures how much the $$j$$th component of $$\boldsymbol{\omega}$$ contributes to the $$i$$th component of the angular momentum. This is a simple generalization of normal moment of inertia $$(I_{ii})$$, which measures how much the $$i$$th component of angular velocity affects the $$i$$th component of the angular momentum.