Impose Uniqueness on Diagonalization of Inertia Tensor?

[robg]

Given an inertia tensor of a rigid body I, one can always find a rotation that diagonalizes I as I = R^T I₀ R (let's say none of the value of the inertia in I₀ equal each other, though). R is not unique, however, as one can always rotate 180 degrees about a principal axis, or rearrange the entries of I₀ via rotations. I'm curious if there a set of conditions that one can impose on R to make it unique, however. One can eliminate the ordering issue by insisting that the entries of I₀ are in sorted order. What about the 180 degree rotation issue, is there an additional condition that one can impose to eliminate this ambiguity?

 

[vanhees71]

You can always diagonalize a symmetric matrix with a orthogonal transformation. The direction of the principal axes is unique if no two eigenvalues are the same. So the only freedom you have after ordering the eigenvalues in the diagonal for $R$ is the direction of the axes. As an additional constraint you can only impose the condition that the transformation matrix is not only orthogonal but even a rotation, i.e., an SO(3) matrix (with determinant +1). Then still you have the freedom to switch any two of the chosen eigenvectors.

So I don't think that you can make the transformation matrix unique, but why should you want this anyway? It's good enough to have one body-fixed basis where the inertia tensor is diagonal.

 

[Andy Resnick]

<snip>I'm curious if there a set of conditions that one can impose on R to make it unique, however. <snip>

Interesting question. I wonder if you can impose conditions based on the symmetry (or lack) of the material itself: Cosserat media, materials that internally generate forces and moments ('active' media), or has some other chiral property that does not have inversion symmetry. Not sure, tho.