Hello Fred, Interesting comment. How do you prove these results? I can’t see how you can use a spectral theorem For example, I don’t believe that all matrices in SO(n, 1) (The Lorentz group) can be diagonalized. What is a projective transformation incorporating translation? Best, — Jean
On Nov 3, 2020, at 2:38 PM, Fred Lunnon <fred.lunnon@gmail.com> wrote:
Worthwhile perhaps to clarify explicitly that a generalisation of Dan's observation yields the decomposition into axial coplanes and "angles" of an isometry in a general quadratic geometry --- for instance, projective n-space with homogeneous coordinates incorporating translations; Minkowski space (mathematician's conformal), boosts having imaginary "angles"; contact geometry (physicist's conformal, "oriented spheres").
WFL
On 11/2/20, Jean Gallier <jean@seas.upenn.edu <mailto:jean@seas.upenn.edu>> wrote:
Hello everybody,
(Dear moderator, I have been a member of this list for many years, but recently I don’t seem to be allowed to post messages).
The reason why a matrix Q in SO(n) (an orthogonal matrix of determinant +1) is similar (via an orthogonal transformation) to a block diagonal matrix consisting of 2D rotation matrices of angle theta, 0 < theta < pi, or the scalars, -1 in even number, or +1, is a corollary of the spectral theorem for (real) normal matrices (see below; I already posted this message a while ago).
If n is odd, then Q must have the eigenvalue +1.
Proof. First, it is well known that the eigenvalues of Q in SO(n) (in fact, Q in O(n)) have modulus 1 (if Qu = lambda u, u \not= 0, because Q is an isometry, <u, u> = <Qu, Qu> = | lambda |^2<u, u>, and since <u, u> \not= 0, | lambda | = 1). Since the characteristic polynomial of Q has real coefficients, its complex roots come in conjugate pairs. But det(Q) is the product of the eigenvalues, so the conjugate pairs contribute +1. There must be an even number of eigenvalues -1 since det(Q) = 1. But a real poly of degree n has n root over C, and the conjugate pairs and the even number of -1 contribute +1 to the product of the eigenvalues, so if n is odd, +1 must be an eigenvalue.
This justifies why a rotation matrix Q has at most m 2D rotations if n = 2m or n = 2m + 1.
All this business of quaternions representing rotations has to do with Clifford algebras and the groups Spin(n), which have a special structure in low degrees (for example Spin(4) is isomorphic to Spin(3) x Spin (3) which is isomorphic to SU(2) x SU(2), with SU(2) the unit quaternions).
Best, -- Jean
--------------------------------------------------------------------------------------------------------------------- Hello everybody,
There is a spectral decomposition theorem about real normal matrices (AA^T = A^T A) which says that there is an orthogonal matrix P and a block diagonal matrix D such that A = P^ D P, where the blocks of D are either a scalar (real), or a 2 x 2 matrix of the form
a -b b a
with b > 0.
So indeed this amounts to finding an orthogonal basis and in some pairwise orthogonal planes, A behaves like a rotation composed with a positive scaling.
In the special case of an orthogonal matrix, the scalars are +1 or -1 and the 2 x 2 matrices are 2D rotation matrices
cos theta -sin theta sin theta cos theta
0 < theta < pi.
The corresponding vectors in P give you a plane. The angle theta correspond to the two eigenvalues cos theta + i sin theta and cos theta - i sin theta.
If det(A) = 1, then the number of -1 is even, and you can group them together in 2 x 2 matrices that correspond to 2D rotations of angle pi.
I don’t know how old the theorem is. It know that it is proven in Gantmaker (Theory of matrices), Berger (Geometry I), and I have a version in my recent book https://www.amazon.com/Algebra-Optimization-Applications-Machine-Learning/dp... <https://www.amazon.com/Algebra-Optimization-Applications-Machine-Learning/dp... <https://www.amazon.com/Algebra-Optimization-Applications-Machine-Learning/dp/9811207712>>
Please email me if you want a pdf.
On Nov 1, 2020, at 10:59 PM, Andy Latto <andy.latto@pobox.com> wrote:
On Sun, Nov 1, 2020 at 7:35 PM Dan Asimov <dasimov@earthlink.net> wrote:
Henry Baker wrote
-----
In fact any rotation of R^n (orientation-preserving isometry taking the origin to itself) is in fact the result of floor(n/2) rotations on mutually orthogonal 2-dimensional planes. If the angles are all distinct and not 0 or π, then this decomposition is unique.
What is meant by rotation on a 2-dimensional plane? Do you mean that there's a codimension-2 subspace that is fixed, and rotation in the other two coordinates, so that the matrix (in a suitable basis) looks like a 2x2 rotation matrix, and then 1's on the rest of the diagonal, and zeroes everywhere else?
Coxeter wrote a great paper on quaternions and
rotations and reflections of 4-dimensional space: "Quaternions and Reflections", the Monthly, Vol. 53, No. 3 (Mar., 1946), pp. 136-146.
—Dan
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