Re: tangential riddle

[John Denker <jsd@MONMOUTH.COM>]

At 11:41 AM 11/8/99 -0600, Carl E. Mungan wrote:

> Aren't you really asking whether the principal axes must be orthogonal?

That's almost what I was asking.

> I would therefore say the answer is yes: start with some arbitrary 3x3
> inertia tensor I. Then it is always possible to find a rotation matrix from
> xyz to x'y'z' which will diagonalize I. But by construction x'y'z' is a
> mutually orthogonal coordinate system.

That proves there must exist at least *one* set of eigenvectors that are
mutually orthogonal.  That's fine.

But I was fishing for a stronger result.  Can you prove that *every* set of
eigenvectors *must* have the property of being mutually
orthogonal?  Answer:  almost but not quite.

The strongest statement that can be made is that every eigenvector is
orthogonal to any other eigenvector that has a different
eigenvalue.  That's a remarkably strict result.  It's useful, too.

(You may wonder what happens if the eigenvalues are not all
different.  Suppose you have an eigenvalue repeated R times.  Then there
will be a R-dimensional subspace in which every vector is an
eigenvector.  You can form innumerable orthogonal and/or nonorthogonal
bases in that subspace.)