Response from Prof. B.

"Reflection" is in common usage to denote mirror imagery, and that would lead one to think that there is only one concept of reflection for each dimension, I agree. But there is a somewhat extended notion more prevalent in mathematics that allows you to reflect over anything that you can project onto. Usually that means reflecting across a hyperplane (the case you considered) or reflecting across a line in a space of dimension greater than two, or reflecting across a plane in four-space etcetera. Using a bit of the terminology of linear algebra, we say that a mapping is a "projection" if doing it twice is no different from doing it once, so for every vector X, we have P(X) = P(P(X)). To reflect across the space we are projecting into, we want the projection P(X) to be half way between X and its reflection R(X) so P(X) = 1/2( X + R(X) ) and therefore R(X) = 2P(X) - X. In Chapter 8, we will be able to express this sort of mapping in terms of coordinates, or just by "adding mappings". It is convenient to introduce the "identity mapping" I which sends every vector to itself, so I(X) = X for all X. We can then express the defining property of a reflection as R(X) = 2P(X) - I(X) for all X, or R = 2P - I. It helps to have things in this form if you are going to express them to a computer.

Somewhat more technically, we express the projection of a vector into a line along a unit vector Z by using the dot product: P(X) = (X€Z)Z. Then R(X) = 2(X€Z)Z - X. More generally, if we have a plane determined by two perpendicular unit vectors Z and Y, then the projection is defined by adding the projections to the two lines, so P(X) = (X€Z)Z + (X€Y)Y and the reflection is defined accordingly. Of course when someone gives you something to project into or reflect across, they usually don't furnish you with an "orthonomal basis" like Y and Z, so part of linear algebra is showing how to extract that description from any of the many other ways that we can define a "subspace" to project into.

Does this make sense? There should be a diagram to accompany this of course, but perhaps you can make one for yourself? Or work one out to go with this comment?