## Reflections

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## Reflections

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Lecture Notes

- Pages:
- 2
- School:
- University of California, Riverside
- Course:
- Math 133 - Geometry

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NOTE ON REFLECTIONS One particularly important class of isometries for R 2 is given by plane reflections Geometrically such maps may be described as follows There is a line L the axis of symmetry such that every point in L is sent to itself but if x 6 L then x and its image x 0 are distinct and the line L is the perpendicular bisector of the segment xx 0 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 R 2 and let v be an orthonormal basis for V LEMMA Let A be the linear mapping of R 2 given by Ax 2 hx vi v x Then Ax x if x V and Ax x if x is perpendicular to v Proof Before starting we note that linearity is an immediate consequence of the definition and properties of the inner product h i If x V then x cv for some scalar c and Ax A cv c Av c 2hv 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 that we have described in geometric terms PROPOSITION Let L q V and A be as above and let T x Ax q Aq Then the 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 follows that T must be the reflection map we described above Proof i If x L then x q bv for some scalar b Then we have T x T q bv A q bv q Aq Aq b Av q Aq b Av q 1 2 and 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 which is perpendicular to V Therefore we have T 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 to x R x T

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