Oops, I meant b is parallel to the z axis, not perpendicular to it. Tom Tom Karzes writes:
Here's a concrete counter-example:
Let R rotate the x axis into the y axis by angle t, so R is:
cos(t) -sin(t) 0 sin(t) cos(t) 0 0 0 1
Let b = (0, 0, 1). So b is perpendicular to the z axis, i.e. the axis of rotation.
Let s = 0.5. Then:
vf = (0, 0, 2)
is a fixed point:
R*vf = (0, 0, 2) s*R*vf = (0, 0, 1) f(vf) = s*R*vf + (0, 0, 1) = (0, 0, 2) = vf
Tom
Tom Karzes writes:
Hi Jean,
I think that would only be true if s = 1, but it's not. If 0 < s < 1, as in the problem statement, then I believe this doesn't apply. R is a proper rotation matrix, but A = s*R is a *scaled* proper rotation matrix. In addition to rotating and translating, it's also shrinking.
Tom
Jean Gallier writes:
If the translation vector b \not= 0 is parallel to the axis of rotation of your rotation, then there is no fixed point. These affine maps are known as screw motions. The fixed points of affine rigid motions can be completely classified. Look at Berger Geometry I (I think, may by II).
Best, -- Jean Gallier
On Feb 19, 2020, at 6:29 PM, Tom Karzes <karzes@sonic.net> wrote:
Does a function which is a scaled proper rotation plus an offset in R^n always have a fixed point?
More precisely, let R be a proper rotation matrix in R^n. Reflections are excluded. So R is an orthogonal matrix with determinant 1.
Let s be a scale factor, 0 < s < 1. Let A be the matrix s*R, combining the scaling and the rotation. Let b be a vector in R^n.
Then:
f(v) = A*v + b
It's clear that f can have at most one fixed point, since f changes the distances between points by a factor of s. Since the distance between two fixed points must remain constant, there can be at most one.
I want to know if f always has a fixed point, or if there are examples where it does not. I.e., can I always find a fixed point vf that satisfies:
vf = A*vf + b
So we need vf to satisfy:
(I - A)*vf = b
where I is the identity matrix.
If the determinant of (I - A) is zero, then there should be either no solution or multiple solutions. But we already know that there can be at most one solution, so I believe the question is equivalent to asking whether:
det(I - A) = 0
If so, then I believe there is no fixed point. Otherwise we have:
vf = inv(I - A)*b
Now consider the following. If b is the zero vector, then vf exists and is also the zero vector, for any A that satisfies the problem definition.
This implies that det(I - A) cannot be zero for any A that satisfies the problem definition. This is independent of the value of b. So (I - A) always has an inverse, and f always has exactly one fixed point.
Note that this argument works if R has determinant -1 as well, i.e. if R includes a reflection, so R can be any orthogonal matrix. The argument also works for s > 1.
Does this reasoning sound correct? Or did I miss something?
Thanks,
Tom
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