From - Tue Jul 15 21:35:51 1997
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From: dc@cage.rug.ac.be (Denis Constales)
Newsgroups: sci.math
Subject: Re: Matrix puzzle
Date: Tue, 15 Jul 1997 11:22:38 +0200
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Not so long ago, in a message that has now expired on my news server by the
Old Salt antics (grr!), Hauke Redmann (fc3a501@GEO.math.uni-hamburg.de)
asked for the determination of all real orthogonal 3x3 matrices whose
diagonal elememts are 0.1, 0.2 and 0.3. I posted a Brute Force solution,
but asking myself where the result came from, I now realise that the
quaternion formulation leads to a fast and elegant solution.

Remember that any 3D rotation can be written x -> q^(-1) x q, where x is
represented by a purely imaginary quaternion x = x1 i + x2 j + x3 k, and q
is a quaternion of modulus 1, q = q0+q1 i + q2 j + q3 k with
q0^2+q1^2+q2^2+q3^2=1.

Then the 1,1 element of the corresponding rotation matrix turns out to be
q0^2 + q1^2 - q2^2 - q3^2 (with cyclic permutation on 1,2,3 for the other
diagonal elements). The representation is not unique: replacing q by -q
gives the same rotation.

This leads to a system of equations, a, b, c standing for the consecutive
diagonal elements

  q0^2 + q1^2 - q2^2 - q3^2 = a
  q0^2 - q1^2 + q2^2 - q3^2 = b
  q0^2 - q1^2 - q2^2 + q3^2 = c

adding the modulus 1 restriction:

  q0^2 + q1^2 + q2^2 + q3^2 = 1

we have a non-singular 4x4 *linear* system in q0^2,...,q3^2 which is easily
solved. Then there are four signs to be chosen arbitrarily (one for each
q<i>), but since q and -q are the same rotation, only three choices are
essential; they produce the eight rotation solutions posted before.

For the solutions of determinant -1, use the represemtation x -> - q^(-1) x
q and obtain the eight other solutions.
--
Dr. Denis Constales - dcons@world.std.com - http://cage.rug.ac.be/~dc/