Moment of inertia tensor and symmetry of the object What information does the moment of inertia tensor give on the structure of an item.
I was told that its eigenvectors give the principal axes of the object. 
Do you know more about this?
 A: Using the inertia tensor $I_{jk}$ of an object, you can construct an ellipsoid satisfying the equation $$1=x_{j}I_{jk}x_{k} .$$ As you said, the eigenvectors of $I_{jk}$ are the principal axis $\vec{r}_{j}$ of your object with the respective moment of inertia $\theta_{j}$ as eigenvalue. Thus, changing coordinates to the principal axis of the object diagonalises the inertia tensor and the equation for the ellipsoid is then $$1=\theta_{1}x_{1}^{2}+\theta_{2}x_{2}^{2}+\theta_{3}x_{3}^{2} .$$ The semi-axles of this ellipsoid are parallel to the principal axles of your object and their length is the inverse square root of the moment of inertia in that direction $a_{j}=\frac{1}{\sqrt{\theta_{j}}}$.
A: The inertia tensor is a bit more descriptive in the spherical tensor basis (so instead of having nine basis dyads made from combinations like $\hat x \hat y$, you have 9 basis tensors that transform like the nine $Y_l^m$ for $l \in \{0,1,2\}$).
Since $I_{ij}$ is antisymmetric, all $l=1$ spherical tensors are zero. The $l=0$ portion is:
$$ I^{(0,0)} = \frac 1 3 {\rm Tr}(I)\delta_{ij}$$
and that is the spherically symmetric part of the object.
Removing the spherically symmetric part leaves a "natural" (read: symmetric, trace-free) rank-2 tensor:
$$ S_{ij} = I_{ij} -I^{(0,0)}$$
The spherical comments are:
$$ S^{(2,0)} = \sqrt{\frac 3 2}S_{zz} $$
This tells you if your object is prolate or oblate.
$$ S^{(2,\pm 2)} = \frac 1 2 [S_{xx}-S_{yy}\pm 2iS{xy}]$$
You will find that $S^{(2,+2)} = (S^{(2,-2)})^*$, and that if you are in diagonal coordinates, they are real and equal. If the value is 0, then the object is cylindrically symmetric.
$$ S^{(2,\pm 1)} = \frac 1 2 [S_{zx}\pm iS_{zy}]$$
Here: $S^{(2,+1)} = -(S^{(2,-1)})^*$, and the term is zero in diagonal coordinates.
