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It is well known that rigid body inertia tensors are 3 by 3 positive semidefinite matrices, which is the same as saying that their eigenvalues are all non-negative.

A little less known is the fact that those eigenvalues also satisfy the triangle inequalities, which means that the sum of any two eigenvalues is always greater or equal than the remaining one. This arises from the definition of the tensor itself which involves integrals of an always non-negative distribution of mass. Per https://physics.stackexchange.com/a/48273/116038 :

In other words, if a semi-positive definite symmetric real $3×3$ matrix with non-negative eigenvalues [...] does not satisfy the triangle inequality (1), it doesn't represent a physically possible distribution of mass.

When estimating inertia tensors using, for example, regression techniques, one can obtain matrices that do not represent physically possible rigid bodies (if we do not constrain the regression solution to physically consistent values).

Using a non-positive semidefinite $3×3$ symmetric matrix in places where an inertia tensor is expected -- e.g., in some rigid-body formulations, in simulation, in control schemes, etc. -- would entail major problems. For example, a non-positive semidefinite matrix used as inertia tensor would give rise to negative kinetic energies, $ E_k = \frac{1}{2} \omega^\top I \omega $.

Now, my question is: What problems (in formulations, simulation, control, etc.), if any, can arise if we use as inertia tensor some matrix that, while satisfying the positive semidefinite condition, do not satisfy the triangle inequality conditions on the eigenvalues?


Update: I've checked that "inertia tensors" that are positive semidefinite but do not satisfy the triangle inequalities still verify the energy conservation law with respect to kinetic energy and work.

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It is useful to look at inertia tensors in the spherical basis--in which the basis-tensors are eigentensors of rotations about the z-axis. So instead of considering 3 x 3 = 9 $T_{ij}$ for i, j $\in (x, y, z)$, on has 1 + 3 + 5 = 9 basis states as follows:

$T^{(0)} = \frac{1}{3}T_{ii}\ {\bf I}$

is spherically symmetric and proportional to the identity. As an inertial tensor, it corresponds to a sphere.

The antisymmetric part of a tensor:

$T^{(1)} = A_{ij} \equiv \frac{1}{2}(T_{ij}-T_{ji})$

transforms like a vector (much like a cross product), and has NO PLACE in the inertia tensors. There are 3 components.

Finally, the trace-free symmetric part has 5 components:

$T^{(2)} = S_{ij} \equiv \frac{1}{2}(T_{ij}+T_{ji})) - T^{(0)}$

that transform like spherical harmonics: $Y_{l=2}^m\ \ (\theta, \phi)$ for $m \in\ (-2, -1, 0, 1, 2) $.

The $|m|=1$ components:

$T^{(2, m=\pm 1)}\ \ \ \ = S_{xz} \pm i S_{yz} $

can be diagonalized away by a suitable coordinate rotation leaving:

$l=2, m=0$:

$T^{(2, 0)} \ \ = \sqrt{\frac{1}{2}} S_{zz} $

This part corresponds to the oblate/prolate-ness of the object; so for an ellipsoid, that would be with the eccentricity with the axis of symmetry aligned with the z-axis.

$l=2, |m|=2$:

$T^{(2, m=\pm 2)} \ \ \ \ \ = \frac{1}{2}(S_{xx} - S_{yy} \pm 2iS_{xy})$

For inertia tensors, the m=2 and m=-2 are the same, so that the Cartesian components are real. This part of the tensors corresponds to any cylindrical asymmetry--so it is zero for an ellipsoid of revolution.

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  • $\begingroup$ Thanks for this other way to see inertia tensors, but I could not get any further insights about my question. $\endgroup$ – Cristóvão D. Sousa Aug 29 '17 at 18:29

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