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The components in Euler equations for a torque-free rigid body are evaluated in the body-fixed principal-axis frame, with $L_i=I_i{\omega }_i$. In an inertial space frame the angular momentum vector is constant; its components in the rotating body frame need not be.

Assume $\Omega \ne 0$ and distinct positive principal moments of inertia. Write $\omega =(\Omega +{\delta }_1,{\delta }_2,{\delta }_3)$. To first order,soThe coefficient is positive when $I_1$ lies between the other moments, giving an exponentially growing mode and instability. When $I_1$ is an extreme moment, it is negative and the transverse perturbations oscillate.

To establish actual nonlinear Lyapunov stability in the latter case, use the conserved kinetic energy $E=\frac{1}{2}\sum I_j{\omega }_j^2$ and squared angular momentum $L^2=\sum I_j^2{\omega }_j^2$. Their combination isFor an extreme $I_1$ its two coefficients have the same sign. Conservation controls both transverse components by their initially small values, while conservation of $E$ controls ${\omega }_1^2$ near ${\Omega }^2$. Continuity preserves its initial sign for sufficiently small perturbations. Thus nonzero steady rotation is stable exactly about the largest or smallest principal moment, the intermediate axis theorem. It is not asymptotically stable, since energy is conserved. If $\Omega =0$, energy makes the zero angular-velocity state stable for every ordering; the nonzero-rotation assumption is essential to the stated equivalence.

Set ${\omega }_1=u$, ${\omega }_2=\Omega +v$, ${\omega }_3=w$ and retain terms linear in the perturbations in Euler equations for a torque-free rigid body. Using $I_3=I_1+I_2$ givesConsequently $\ddot{u}={\beta }^2{\Omega }^{2}u$, where $\beta =\sqrt{(a^2-b^2)/(a^2+b^2)}>0$. For $\Omega >0$ the growing solution isThere is also a decaying mode with the opposite exponent. For negative $\Omega$, select the exponent $+\beta |\Omega |$ as the growing one. This is the intermediate axis theorem: steady rotation about the intermediate principal inertia axis is unstable whenever $\Omega \ne 0$.

First consider rotation predominantly about the third principal axis, writing ${\omega }_3=\Omega +O({\omega }_1^2+{\omega }_2^2)$. To first order the third Euler equation gives $\dot{\Omega }=0$, while the transverse components satisfyThereforeand the same equation holds for ${\omega }_2$.

Order the principal moments as $I_1<I_2<I_3$. The coefficient above is negative, so perturbations about axis three oscillate and remain bounded. Cyclically applying the same linear stability of principal-axis rotation calculation about axis one givesso that rotation is also stable. About axis two, however,which has an exponentially growing solution. Thus the intermediate axis theorem gives

Setting ${\omega }_1={\omega }_3=0$ makes all three Euler equations stationary. The invariants then giveso the two rotations arewhere $L=\sqrt{L^2}$.

Linearize about $(0,\Omega ,0)$, where $\Omega$ is either of these values. Writing the transverse perturbations as $(\xi ,\zeta )=(\delta {\omega }_1,\delta {\omega }_3)$ givesConsequentlyThe transverse eigenvalues are thereforeone of which is positive. Both rotations are linearly unstable, which is the intermediate axis theorem.

For three distinct positive principal moments of inertia, the denominator of $p^2$ is positive. If $I_3$ is the largest moment, the two numerator factors have opposite signs, so $p^2<0$. If it is the smallest, they again have opposite signs. If $I_3$ is intermediate, the two factors have the same sign, so $p^2>0$. Thus the intermediate axis theorem givesFor the stable cases the body-frame wobble frequency isThe perturbation does not decay in the torque-free conservative model. For the intermediate moment, a generic small impulse excites the growing branch and eventually leaves the linear regime.

The principal moments of inertia about a specified point $O$ are the eigenvalues $I_1,I_2,I_3$ of the inertia tensor ${\mathbf{I}}_O$. Their associated unit vectors that are eigenvectors specify principal axes through $O$. The spectral theorem for real symmetric matrices guarantees an orthonormal basis of these eigenvectors, even for an irregular rigid body. Choose their signs to make a right-handed basis and collect them as the columns of an orthogonal matrix $Q$:Thus diagonalization removes the off-diagonal coupling in the rotational kinetic energy. The eigenvalues belong to the mass distribution and reference point, independently of which orthonormal basis initially describes them. ^[ref]

In this principal body frame, write angular velocity as $\mathbf{\omega }={\sum }_i{\omega }_i{\mathbf{e}}_i$. ThenUse these as total angular momentum and kinetic energy only under the reference-point conditions stated under inertia tensor. Along a single principal axis with $I_i>0$, angular momentum is parallel to angular velocity. For a mixture of directions with different principal moments of inertia, it generally is not. A principal axis need not be a geometric symmetry axis: it is selected by the complete mass distribution. A principal axis also need not give stable free rotation; for three distinct principal moments of inertia, the intermediate axis theorem supplies the stability distinction.

For any unit vector $\mathbf{n}={\sum }_i{n}_i{\mathbf{e}}_i$,The axial moment of inertia is a weighted average of the principal moments of inertia. At fixed angular speed, the least and greatest rotational kinetic energies occur in the corresponding eigenspaces. At fixed nonzero angular momentum magnitude $L$, instead, $K=\frac{1}{2}{\sum }_i{L}_i^2/I_i$, so the largest principal moment of inertia gives the smallest rotational kinetic energy. These comparisons hold the reference point fixed.

There is an additional physical restriction beyond positive semidefiniteness. In principal axes coordinates,with two analogous cyclic inequalities. Therefore three arbitrary positive numbers do not always represent a possible inertia tensor. For a distribution in the $xy$ plane through $O$, $I_3=I_1+I_2$, as in the planar inertia tensor of paired point masses. A uniform solid ellipsoid of mass $M$ and semiaxes $a,b,c$ supplies a continuous three-dimensional example: using its mass second-moment tensor, its central principal moments of inertia are $M(b^2+c^2)/5$, $M(a^2+c^2)/5$, $M(a^2+b^2)/5$. Setting $a=b=c=R$ recovers the moment of inertia of a uniform solid sphere.