Re: [math-fun] Quaternion eigenvectors, Part Deux
Henry Baker wrote ----- It is also well known that every 4D rotation can be expressed as the *independent* rotations of 2 planes orthogonal to one another, each with their own separate rotation. ----- 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. 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
Maybe ceil(n/2)? The statement appears to fail for R^3. On Sun, Nov 1, 2020 at 7:35 PM Dan Asimov <dasimov@earthlink.net> wrote:
Henry Baker wrote ----- It is also well known that every 4D rotation can be expressed as the *independent* rotations of 2 planes orthogonal to one another, each with their own separate rotation. -----
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.
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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Dan is correct. In R^3 every rotation is in some plane, and the unique direction left over is the axis of rotation. -- Gene On Sunday, November 1, 2020, 5:04:30 PM PST, Allan Wechsler <acwacw@gmail.com> wrote: Maybe ceil(n/2)? The statement appears to fail for R^3. On Sun, Nov 1, 2020 at 7:35 PM Dan Asimov <dasimov@earthlink.net> wrote:
Henry Baker wrote ----- It is also well known that every 4D rotation can be expressed as the *independent* rotations of 2 planes orthogonal to one another, each with their own separate rotation. -----
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.
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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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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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/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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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> 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/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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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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Hello again Jean (somewhat publically?) Correction (sigh!) --- the axis subspace is a co-line, not a co-plane. I must start by admitting that I just fired this comment off, without thinking very deeply about it. My reasoning is simply that the symmetry group of a quadratic geometry can be represented by the Clifford algebra with corresponding signature; and the versors of that algebra can be transformed into quasi-orthogonal matrices trivially via their action on basis vectors. Degenerate geometries (eg. real projective plane) involving zero components in the signature require special consideration involving limiting arguments as radius, angle of rotation approach infinity, zero. As I recall, Lorentz group matrices representing boosts just replace cos, sin by cosh, sinh in a 2x2 diagonal block. I'm not sure that I have parsed "What is" correctly here --- isometric transformations in projective n-space, represented via order n+1 matrices in homogeneous coordinates, were schoolroom material in my own youth, so I find it difficult to believe that you are unfamiliar with the technique! The translation vector occupies a single column juxtaposed to the usual orthogonal rotation matrix, along with a row of scale components 1,0,0,... On the other hand, I am intrigued by the improbable notion that such material is applicable to machine learning, so would appreciate your offer of a PDF. Regards, Fred On 11/3/20, Jean Gallier <jean@seas.upenn.edu> wrote:
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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Counter correction (groan!) --- ignore that, it was right first time. Brainstorm. Well, not quite: I did try hard to keep quiet about this topic, being well aware that I am simply "not current" in aircraft pilot jargon (besides being reluctant to assist in hijacking the thread). So regarding Dan's request for further background --- I must take greater care over any further contribution, for fear of succeeding only in promoting confusion (not to mention personal embarrassment). WFL On 11/4/20, Fred Lunnon <fred.lunnon@gmail.com> wrote:
Hello again Jean (somewhat publically?)
Correction (sigh!) --- the axis subspace is a co-line, not a co-plane. I must start by admitting that I just fired this comment off, without thinking very deeply about it.
My reasoning is simply that the symmetry group of a quadratic geometry can be represented by the Clifford algebra with corresponding signature; and the versors of that algebra can be transformed into quasi-orthogonal matrices trivially via their action on basis vectors. Degenerate geometries (eg. real projective plane) involving zero components in the signature require special consideration involving limiting arguments as radius, angle of rotation approach infinity, zero.
As I recall, Lorentz group matrices representing boosts just replace cos, sin by cosh, sinh in a 2x2 diagonal block.
I'm not sure that I have parsed "What is" correctly here --- isometric transformations in projective n-space, represented via order n+1 matrices in homogeneous coordinates, were schoolroom material in my own youth, so I find it difficult to believe that you are unfamiliar with the technique! The translation vector occupies a single column juxtaposed to the usual orthogonal rotation matrix, along with a row of scale components 1,0,0,...
On the other hand, I am intrigued by the improbable notion that such material is applicable to machine learning, so would appreciate your offer of a PDF.
Regards, Fred
On 11/3/20, Jean Gallier <jean@seas.upenn.edu> wrote:
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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Jean asked, as well he might --- "What is a projective transformation incorporating translation?". I was trying to say that Dan's observation can be extended to the full Euclidean group, ie. including rigid translations. Along with the Lorentz and Poincaré groups, and their generalisations to many dimensions: Möbius and Lie-sphere. Apologies to everybody. I'll shut up now until I can figure out how to keep my foot out of my mouth. WFL On 11/4/20, Fred Lunnon <fred.lunnon@gmail.com> wrote:
Counter correction (groan!) --- ignore that, it was right first time. Brainstorm. Well, not quite: I did try hard to keep quiet about this topic, being well aware that I am simply "not current" in aircraft pilot jargon (besides being reluctant to assist in hijacking the thread).
So regarding Dan's request for further background --- I must take greater care over any further contribution, for fear of succeeding only in promoting confusion (not to mention personal embarrassment).
WFL
On 11/4/20, Fred Lunnon <fred.lunnon@gmail.com> wrote:
Hello again Jean (somewhat publically?)
Correction (sigh!) --- the axis subspace is a co-line, not a co-plane. I must start by admitting that I just fired this comment off, without thinking very deeply about it.
My reasoning is simply that the symmetry group of a quadratic geometry can be represented by the Clifford algebra with corresponding signature; and the versors of that algebra can be transformed into quasi-orthogonal matrices trivially via their action on basis vectors. Degenerate geometries (eg. real projective plane) involving zero components in the signature require special consideration involving limiting arguments as radius, angle of rotation approach infinity, zero.
As I recall, Lorentz group matrices representing boosts just replace cos, sin by cosh, sinh in a 2x2 diagonal block.
I'm not sure that I have parsed "What is" correctly here --- isometric transformations in projective n-space, represented via order n+1 matrices in homogeneous coordinates, were schoolroom material in my own youth, so I find it difficult to believe that you are unfamiliar with the technique! The translation vector occupies a single column juxtaposed to the usual orthogonal rotation matrix, along with a row of scale components 1,0,0,...
On the other hand, I am intrigued by the improbable notion that such material is applicable to machine learning, so would appreciate your offer of a PDF.
Regards, Fred
On 11/3/20, Jean Gallier <jean@seas.upenn.edu> wrote:
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 > > _______________________________________________ > math-fun mailing list > math-fun@mailman.xmission.com > https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun >
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participants (6)
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Allan Wechsler -
Andy Latto -
Dan Asimov -
Eugene Salamin -
Fred Lunnon -
Jean Gallier