Transformation Geometry Transformations of the Plane Transformations via Coordinates Instead of f (P)=P′, we could write f (x, y)=(x ′, y ′), where P has • A transformation of the plane is a function f to each point P in the coordinates (x, y) and P′ has coordinates (x ′, y ′). This is equivalent to plane another point P′ = f (P) so that having two functions f and f giving the x ′ and y ′ in terms of x and y : Compositions of Linear Transformations and • f is one-to-one, i.e., 1 2 The main idea behind the introduction of transformations in geometry is x ′ f x y Their Fixed Points to provide the means for comparing two figures. P �= Q ⇒ f (P) �= f (Q) = 1( , ) f : ′ • Two figures are congruent if there is a special transformation of the y = f2(x, y) plane (an isometry) taking the first figure onto the second. Sara Schleissmann Example. • Two figures are similar is there is a special transformation of the plane Q (a similarity) taking the first figure onto the second. x ′ =3x +2y − 1 State University of New York Q' f : ′ College at Oneonta Transformations or “motions”of the plane are used in Euclid’s P y = −x +4y +2 P' [email protected] “Elements”in Proposition I.4 (the SAS Postulate). Euclid assumes that if ′ ′ ′ two triangles satisfy the SAS Hypothesis, then one can “apply”one The image of the point P(2, 3) is the point P (x , y ), where x ′ =3· 2+2· 3 − 1 = 11 and y ′ = −2+4· 3+2=12. Student Research and Creative Activity Day triangle on top of the other to get a perfect match, thus proving the • f is onto, i.e., each point R in the plane has a preimage under f , The preimage of the point R(6, 9) is the point S(x, y) such that March 26, 2014 congruence of the given triangles. that is there exists a point S so that f (S)=R. 6=3x +2y − 1

--1 S=f (R) R 9=−x +4y +2

The only solution of the system above is x =1, y =2

S. Schleissmann Compositions and Fixed Points S. Schleissmann Compositions and Fixed Points S. Schleissmann Compositions and Fixed Points S. Schleissmann Compositions and Fixed Points

If, on the other hand P is not on either of the previous lines, then let’s • Linear transformations preserve noncollinearity. ←→ Linear Transformations look at line PC. Fixed Points of Transformations To prove this, let A, B, and C be three arbitrary, noncollinear points. We ←→ ←→ ←→ ←→ • If a transformation preserves collinearity, that is, if it maps any three a) If PC is not parallel to AB, then PC must intersect AB at Q. It would will show that A′ = f (A), B′ = f (B), C ′ = f (C) are noncollinear as well. collinear points to three collinear points, then the transformation is said ′ ←−→′ ′ ′ • A transformation f of the plane is said to have A as a fixed point if ′ ′ ′ then follow that f (Q)=Q belongs to A B = l and since C ∈ l, it to be linear. By contradiction, suppose that A , B , C are collinear, so they belong to f (A)=A. ′ ←−→′ ′ a line l. Now let P, different from A,B,C, be an arbitrary point in the would then follow that P = f (P) belongs to Q C = l. • If a given transformation fixes any point of the plane, then the C' plane. We will prove that P will always map to some point on line l. This transformation is called the identity mapping. B' f B' C will contradict being onto. B Example 1. The linear transformation f ←→ ←→ ←→ Q' If P lies on either of the lines AB, AC , or BC, then since f preserves Q P' B ←−→ ←−→ ←−→ P ′ ′ ′ ′ ′ ′ ′ C ′ collinearity, P = f (P) belongs to line A B = A C = B C = l. C' x = x +2y A A' ′ A A' y =3y ←→ ←→ • In terms of coordinates, the transformation B B' b) If If PC is parallel to AB, then let D ∈ AC be an arbitrary point. Line has (0, 0) as a fixed point. In fact, any point on the x−axis is a fixed ←→ ←→ ′ point since (x, 0) goes to (x +2· 0, 3 · 0) = (x, 0). ′ PD will then intersect line AB at some point T . Then, points T = f (T ) x = f1(x, y) P' Question. Can a linear transformation of the form P and D′ = f (D) must belong to line l and so P′ = f (P) must belong to l. ′ y = f2(x, y) C C' x′ = ax + by B' B ′ is linear if and only if the functions f1 and f2 are linear in x and y. That is A P y = cx + dy A' P' ′ C x = a1x + b1y + c1 C' D A D' ever have exactly one other fixed point besides (0, 0)? ′ T A' y = a2x + b2y + c2 T'

S. Schleissmann Compositions and Fixed Points S. Schleissmann Compositions and Fixed Points S. Schleissmann Compositions and Fixed Points S. Schleissmann Compositions and Fixed Points

Linear Algebra Approach Geometry Approach • Linear transformations send midpoints to midpoints. Proof. Let AC be an arbitrary segment and let M be its midpoint. Let Finding the fixed points of the given linear transformation is equivalent to ′ ′ • If a linear transformation fixes two points A, B, then any point on the Now let’s show that l � m . A′, C ′, M′ be the images of A, C, M via the linear transformation f . We finding the solutions of the system ′ ′ ′ ′ ′ segment AB is fixed. Indeed, by contradiction, suppose that l ∩ m = P . Since P ∈ l , it will prove that M′ is the midpoint of A′C ′. To do so, construct a ′ To prove this, we need some additional results. follows that P must come from some point on l. Similarly, since parallelogram ⋄ABCD. Since parallel lines go to to parallel lines, the x = ax + by ′ ′ ′ • Linear transformations send parallel lines to parallel lines. P ∈ m , it follows that P must come from some point on m. But then, quadrilateral ⋄A′B′C ′D′ is also a parallelogram with diagonals A′C ′ and ′ ′ −1 ′ y = cx + dy, Proof. Indeed, suppose l � m and let’s first show that l �= m . f (P ) ∈ l ∩ m, which is impossible since l ∩ m = ∅. B′D′ meeting at the midpoint N of A′C ′ and B′D′. But M lying on both AC and BD maps to M′ on both A′C ′ and B′D′. So M′ = N =midpoint l'=m' or equivalently l l' m' ′ ′ m A' l of A C . (a − 1)x + by =0 A m f C' D D' C' cx +(d − 1)y =0. f P' C B B' Such a system has solutions other than (0, 0) iff C M f = a − 1 b ′ ′ N M' det =0. By contradiction, suppose l = m . Let A, B ∈ l and C ∈ m. Since l � m, cd− 1 it follows that A, B, C are noncollinear. But then A   B' A′ = f (A) ∈ l ′, B′ = f (B) ∈ l ′, C ′ = f (C) ∈ m′ = l ′ are collinear. This B A' If this is the case, then, in fact, the system has an entire “line worth”of contradicts the fact that linear transformations preserve noncollinearity. solutions.

S. Schleissmann Compositions and Fixed Points S. Schleissmann Compositions and Fixed Points S. Schleissmann Compositions and Fixed Points S. Schleissmann Compositions and Fixed Points

Back to the proof More fixed points Isometries: Special Linear Transformations • If a linear transformation fixes three noncollinear points, then it fixes • If a linear transformation f fixes points A and B, then f fixes any point every point of the plane and, hence, it is the identity transformation. on the segment AB. Observation. The same result can be proven by means of coordinates. Proof. Let A, B, and C be three arbitrary, noncollinear fixed points and Proof. Let P ∈ AB be arbitrary. Since f is linear, it takes the midpoint Definition. A linear transformation f is an isometry if and only if for any For example, consider the transformation f given by ′ ′ let P be an arbitrary point in the plane. Construct △ABC. If ′ ′ M1 of AB to the midpoint of A B = AB. But this means that M1 is two points P and Q with P = f (P) and Q = f (Q) we have that P ∈△ABC, then P is fixed based on the previous result. If P ∈△/ ABC, ′ ′ fixed. Now, without loss of generality assume P ∈ AM1 and look at the PQ = P Q . x ′ =3x +2y − 1 then construct lines l and m passing through P and intersecting the sides midpoint M2 of AM1. As before, M2 is fixed. If P ∈ AM2, we consider ′ of △ABC. y = −x +4y +2 M3 as the midpoint of AM2. If, on the other hand, P ∈ M1M2, we let M3 Q' Q be the midpoint of M1M2. A l f and the points A(2, 3) and B(−4, −3). The midpoint of AB is M(−1, 0). m ′ ′ ′ P2 f A A B B M M P1 P sends to (11, 12), to (−19, −6), and to (−4, 3). As we B ′ ′ ′ A M2 P M3 M1 can see, M turns out to be the midpoint of A B . P4 The same kind of argument can be used for any linear transformation f P P' B P and any two points A(x1, y1), B(x2, y2). 3 C This way, we construct an entire sequence of (fixed) midpoints {Mn}n≥1 converging to P. But then, by continuity of linear transformations, It follows from the definition and the previous result that the only Since P, P1, P2 ∈ m, it follows that f (P), f (P1), f (P2) belong to some isometry that has three noncollinear fixed points is the identity mapping. m′ P′ P P m′ P P f (P)=f ( lim Mn) = lim f (Mn) = lim Mn = P, line . But this means that , 1, 2 belong to since 1, 2 are n→∞ n→infty n→∞ ′ fixed. But since P1, P2 ∈ m, then m = m . Similarly, P′, P , P belong to line l ′ = l. But since so P is fixed. 3 4 P′ ∈ l ′ ∩ m′ = l ∩ m = P, it follows that P′ = f (P)=P, so P is fixed.

S. Schleissmann Compositions and Fixed Points S. Schleissmann Compositions and Fixed Points S. Schleissmann Compositions and Fixed Points S. Schleissmann Compositions and Fixed Points

Reflections On the other hand, the line y = x is the line of reflection. To see this, let Composing Linear Transformations Translations as Compositions of Reflections ′ ′ Definition. If a fixed line l is the perpendicular bisector of PP′ for any P = f (P) and note that the slope of PP equals P P′ f P f point , where = ( ), then the linear transformation is said to be y ′ − y x − y a reflection in line l. m ′ = = = −1. PP x ′ − x y − x Example: Consider the reflections f and g in the parallel lines y = x and If f and g are any two linear transformations, then the composition of g P y = x +2. with f is defined as (g ◦ f )(P)=g(f (P)). Similarly, the composition of f This says that the segment PP′ and the line y = x are perpendicular. x ′ = y ′ with g is defined as (f ◦ g)(P)=f (g(P)). In general, g ◦ f �= f ◦ g. f Also, the midpoint M of PP belongs to the line y = x since : ′ ′ ′ y = x. l x+x y+y x+y y+x M M ′ ′ ′′ ( 2 , 2 )= ( 2 , 2 ). P −→ f (P)=P −→ g(f (P)) = g(P )=P x ′′ = y ′ − 2 P g : P' ′′ ′ • If f is a reflection in line l, then f 2 = f ◦ f is the identity mapping, that y = x +2. 2 M is it fixes any point in the plane. When composing g with f , we obtain x ′′ = y ′ − 2=x − 2 and Example: The following linear transformation has an entire line of fixed Proof. P f P P′ l y ′′ = x ′ +2=y +2. In this manner, (g ◦ f )(x, y)=(x − 2, y + 2). points and is a reflection in that line. P' Let be arbitrary and let ( )= , where is the l perpendicular bisector of PP′. It then follows that f (P′)=P and so Any linear transformation of the form x′ = x + a, y ′ = y + b is called a 2 ′ ′ f (P)=f (f (P)) = f (P )=P. translation which maps the origin to the point (a, b) and displaces every x = y 2 ′ other point by an equal amount. y = x.

′ ′ • An isometry with an entire line of fixed points (and no others) is a Indeed, the fixed points will be solutions of the equation (x , y )=(x, y), reflection in that line. meaning that x = y. So any point on the line y = x is fixed.

S. Schleissmann Compositions and Fixed Points S. Schleissmann Compositions and Fixed Points S. Schleissmann Compositions and Fixed Points S. Schleissmann Compositions and Fixed Points

Rotations as Compositions of Reflections Final Observation Similarly, when composing f with g, we obtain x ′′ = y ′ = x + 2 and From the point of view of fixed points, reflections, rotations, and y ′′ = x ′ = y − 2. In this manner, (f ◦ g)(x, y)=(x +2, y − 2), and so translations behave as follows: f ◦ g is a translation, once more, which happens to be the inverse • Any rotation is the composition of two reflections in lines that are transformation of g ◦ f . intersecting each other. P 4 m

2

rotation: single fixed point O P' reflection: line worth of fixed points l 2 P'' • Any translation can be viewed as the composition of two reflections in parallel lines. • The only isometry with exactly one fixed point is a rotation.

translation: no fixed points

S. Schleissmann Compositions and Fixed Points S. Schleissmann Compositions and Fixed Points S. Schleissmann Compositions and Fixed Points