Proof: Principal Axes Orthogonal

Take two principal axes $\vec{\omega}^{(a)}$ and $\vec{\omega}^{(b)}$.

$$I_{ik}\omega_k^{(a)}=I^{(a)}\omega_i^{(a)}$$

$$I_{ik}\omega_k^{(b)}=I^{(b)}\omega_i^{(b)}$$

Dot each of the two equations into the other $\vec{\omega}$.

$$I_{ik}\omega_k^{(a)}\omega_i^{(b)}=I^{(a)}\omega_i^{(a)}\omega_i^{(b)}$$ first eq.

$$I_{ik}\omega_k^{(b)}\omega_i^{(a)}=I^{(b)}\omega_i^{(a)}\omega_i^{(b)}$$ second eq.

$$I_{ik}\omega_i^{(a)}\omega_k^{(b)}=I^{(b)}\omega_i^{(a)}\omega_i^{(b)}$$ commute omegas

$$I_{ki}\omega_k^{(a)}\omega_i^{(b)}=I^{(b)}\omega_i^{(a)}\omega_i^{(b)}$$ rename indices lhs

$$I_{ik}\omega_k^{(a)}\omega_i^{(b)}=I^{(b)}\omega_i^{(a)}\omega_i^{(b)}$$ inerta tensor is symmetric

In penultimate step we have renamed the indices $i$ and $k$ and in the last step we have used the fact that the tensor is symmetric. Now the left hand sides of the two equations are the same and we can set the right hand sides to be equal.

$$I^{(a)}\omega_i^{(a)}\omega_i^{(b)}=I^{(b)}\omega_i^{(a)}\omega_i^{(b)}$$

The dot product $\omega_i^{(a)}\omega_i^{(b)}$ must be zero if the moments are not equal. If they are equal, we can just choose orthogonal axes.