Definition. If f and g are invertible affine transformations of R^{n}, the conjugate of f by g is the transformation g \circ f \circ g^{-1}.
When the context is clear, one sometimes abbreviates notation for conjugation by writing
| g . f = g \circ f \circ g^{-1} . |
One views this as g “acting on” f. That is, an affine transformation of R^{n} not only describes a motion of R^{n} but also a motion on the set of affine transformations of R^{n}: the action of the set of affine transformations on itself by conjugation.
Theorem 1. If f(x) = U x + v is an affine transformation of R^{n} and tau(x) = x + a is translation by the vector a, then f . tau is translation by the vector U a.
Proof. f^{-1}(x) = U^{-1} x - U^{-1} v, and, therefore y = tau \circ f^{-1}(x) = U^{-1} x - U^{-1} v + a. Then (f . tau)(x) = U y + v = x + U a.
Warning. Although when a translation is conjugated by an arbitrary affine transformation, the result is a translation, it is not true that the result of conjugating an isometry by an affine transformation is always an isometry.
For example, if f is the simple linear transformation (x, y) -> (2x, y) of R^{2} and rho is rotation through the angle pi/2 about the origin, i.e., (x, y) -> (-y, x), then f . rho is the linear map (x, y) -> (-2 y, x/2), which is not an isometry.
On the other hand, it is clear that the result of conjugating an isometry by an isometry is an isometry. Moreover:
Proposition 1. If f and g are affine transformations of R^{n} and the point c is a fixed point of f, then the point g(c) is a fixed point of g . f = g \circ f \circ g^{-1}.
Proof. The proof is left as an exercise.
Proposition 2. If f and g are affine transformations of R^{n}, then f is orientation-preserving if and only if g . f = g \circ f \circ g^{-1} is orientation-preserving.
Proof. If f has matrix part U, i.e., f(x) = U x + v for some v, then the matrix part of g . f is the matrix B U B^{-1} where B is the matrix part of g. Therefore, the matrix parts of f and of g . f have the same determinant, and so the question of orientation-preservation, which depends on the sign of the determinant, is the same for f and g . f.
Theorem. Each of the four classes of isometries of the plane is carried to itself under conjugation by an isometry.
Proof. For isometries of the plane the four classes are determined by two property switches: (1) whether or not there is a fixed point and (2) whether or not the isometry is orientation-preserving. The two propositions above show that these questions are not affected by conjugation.
Find a rotation of R^{2} that fails to be conjugated into an isometry by the affine transformation (x, y) -> (x + y, y).
Find a reflection of R^{2} that fails to be conjugated into an isometry by the affine transformation (x, y) -> (x + y, y).
Prove Proposition 1 above.
Show that if a rotation of the plane is conjugated by an isometry of the plane, then the resulting rotation either involves the same angle or its negative.
Inasmuch as each translation of R^{n} is given by a unique vector, the set of all translations of R^{n} may be regarded as an abstract vector space in which vector addition is composition of translations and for each scalar c and each translation tau the translation c tau is defined as the map x -> (1 - c) x + c tau(x). Explain why, for a given affine transformation f, the map tau -> f . tau is a linear transformation of the vector space of translations.