# Symmetry axis and products of inertia

So if we have an object that has a symmetry axis let us say $z$-axis is a symmetry axis does this mean that the product of inertia $I_{zx} = I_{zy} = 0$? And if that is true why is it true mathematically speaking?

• Do you mean moments of inertia? No, they will not be zero, not even for a symmetric object. Look at the definition $I=\int \rho (r) r^2dV$. Because physical densities are always positive the integral can only be zero if $\rho (r)=0$ $\forall r>0$, which is physically impossible. – CuriousOne Jun 30 '15 at 8:52

Indeed, since $z$ is a rotational symmetry axis, it defines an eigenspace of the inertial operator $I$. Since $I$ is a symmetric linear operator, it admits an orthonormal basis of eigenvectors, one such vector is ${\bf e}_z$. This unit vector can be completed into a basis of eigenvectors just adding some pair of unit orthogonal vectors. However, in view of the axial symmetry around $z$, each pair of mutually orthogonal vectors which are also orthogonal to ${\bf e}_z$ must define such a basis.
We conclude that a basis of eigenvectors of $I$ is ${\bf e}_x, {\bf e}_y, {\bf e}_z$.
Writing $I$ with respect to that basis, it gives rise to a diagonal matrix. Consequently $I_{ij}=0$ if $i\neq j$. This happens, in particular, for $i=z$ and $j=x,y$.