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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.