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