NOTE ON REFLECTIONSOne particularly important class of isometries for R2is given by plane reflections. Geometri-cally, such maps may be described as follows: There is a line L (the axis of symmetry) suchthat every point in L is sent to itself, but if x 6∈ L then x and its image x0are distinct, and theline L is the perpendicular bisector of the segment [xx0]. The purpose of this note is to give a formula for such a mapping in terms of linear algebra.Write L = q + V , where q ∈ L and V is a 1-dimensional vector subspace of R2, and let {v} bean orthonormal basis for V .LEMMA. Let A be the linear mapping of R2given by Ax = 2 · hx, vi v − x. Then Ax = xif x ∈ V and Ax = −x if x is perpendicular to v.Proof. Before starting, we note that linearity is an immediate consequence of the definitionand properties of the inner product h , i.If x ∈ V , then x = cv for some scalar c, andAx = A(cv) = c(Av) =c2hv, vi − v) = c(2 · v − v) = cv = x .On the other hand, if x is perpendicular to V , then hx, vi = 0 and therefore we have Ax = −x.The preceding lemma leads directly to an purely algebraic formula for the reflection map thatwe have described in geometric terms.PROPOSITION. Let L = q + V and A be as above, and let T (x) = Ax + (q − Aq). Thenthe following hold:(i) T (x) = x if x ∈ L.(ii) For all x 6∈ L we have T (x) 6= x.(iii) If x 6∈ L then the line joining T (x) and x meets L at the midpoint of the segment[x T (x) ], and the lines L and x T (x) are perpendicular.Since T has the geometrical propeties of the reflection map in the previous discussion, it followsthat T must be the reflection map we described above.Proof. (i) If x ∈ L then x = q + bv for some scalar b. Then we haveT (x) = T (q + bv) = A(q + bv) + q − Aq =Aq + b Av + q − Aq = b(Av) + q12and since Av = v this implies that T (x) = q + bv = x.(ii) If x 6∈ L then x − q = y + z where y = bv for some scalar b and z is a nonzero vector whichis perpendicular to V . Therefore we haveT (x) = A(q + bv + z) + q − Aq =Aq + A(bv) + Az + q − Aq = Aq + bv − z + q − Aq = q + bv − z .It follows that x − T (x) = 2 · z and is nonzero, so that x 6= T (x).(iii) Continuing in the setting of the previous part, we see that the line x T (x) is equal tox + R ·x − T (x)= x + R · (2z) = x + R · z(since R · (2z) = R · z), and this line is perpendicular to L because z is perpendicular to v.Finally, we have12x + T (x)= q + bv which is also a point of L, and hence L meets x T (x) atthe midpoint of [x T (x)
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