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A principal body frame is an orthonormal basis ${\mathbf{e}}_a(t)$ fixed in the rotating body and aligned with its principal axes. If $R(t)$ is the rotation matrix whose columns are these vectors, differentiating $R^{T}R=I$ shows that $\dot{R}{R}^T$ is skew-symmetric. Every skew-symmetric linear map in three dimensions is cross product with a unique vector $\mathbf{\omega }$, so

For any vector $\mathbf{A}=A_a{\mathbf{e}}_a$,Applying this to the conserved angular momentumgives the Euler equations for a torque-free rigid body:

Multiplying these equations respectively by ${\omega }_i$ and summing proves conservation of kinetic energy; multiplying by $I_i{\omega }_i$ and summing proves conservation of squared angular momentum. Thus the torque-free rigid-body invariants areSolving these two linear equations for the first two squared components givesUsing $I_3{\dot{\omega }}_3=(I_1-I_2){\omega }_1{\omega }_2$ yields the quartic equation for one component of torque-free angular velocity:

Multiply the three Euler equations for a torque-free rigid body by ${\omega }_1,{\omega }_2,{\omega }_3$ and add. The cubic terms cancel, giving conservation of the rotational kinetic energyInstead multiply the equations by $I_1{\omega }_1,I_2{\omega }_2,I_3{\omega }_3$ and add. The cubic terms again cancel, givingthe squared magnitude of the conserved angular momentum. For an asymmetric body these two quadratic forms are not proportional, so the torque-free rigid-body invariants are independent.

The rotational kinetic energy and squared angular momentum are the torque-free rigid-body invariantsandDifferentiating the energy and inserting the Euler equations givesSimilarly,