MAT 3272: Cartesian Planes and Transformations

Transcription

1 MAT 7: Cartesian Planes and Transformations March, 0 - (in honor of Hilbert s thirteen axioms for neutral geometry) are due by 5 p.m. on Wed., March, or electronically by 8 a.m. on Thursday, March. Ordered Fields, Incidence, and Betweenness Let Π(F) denote the Cartesian plane over a field F. (Recall that this means that the points of our plane Π(F) are ordered pairs of elements of F, the lines of Π(F) are the solution sets of linear equations, and incidence is given by set membership. With only this incidence relation defined, the Cartesian plane over F is often called the affine plane with coordinates in F. This is because any affine transformation (see next section) of the points preserves lines and incidence. Affine transformations are very useful for checking properties of the plane, as we have seen. As long as we don t care about congruence, we can use any affine transformation we like to reduce our situation to a simpler one. If the field is clear from context, we will omit it from the notation and just write Π.) Theorem. For any field F, Π(F ) satisfies Hilbert s Incidence Axioms -. Exercise. We have proven this theorem in class. As a review, verify that axioms I-I are satisfied. You may use either linear or elementary algebra, or a combination of both, as it suits you. If F is an ordered field, there is a natural way to define betweenness: Given a line with equation y = mx + b (so the coefficient of y is not zero) and three points on it, (x, mx + b), (x, mx + b), and (x, mx + b), define (x, mx + b) to be between (x, mx + b), and (x, mx + b) if (and only if) x is between x and x in the ordering of F. (It is trivial to show that, in this case, it is also the case that y = mx + b is between the other two y coordinates, and vice-versa.) In the remaining case, in which the line has equation x = x 0 (so the coefficient of y is zero), and given three points (x 0, y ), (x 0, y ) and (x 0, y ), define (x 0, y ) to be between (x 0, y ) and (x 0, y ) if (and only if) y is between y and y in the ordering of F. Theorem. The plane Π(F) satisfies Hilbert s Betweenness Axioms -4 if and only if F is an ordered field and the betweenness relation is given by the ordering of F (as described above). The proof that it is sufficient for F to be ordered, with the betweenness relation determined by the ordering of F, is fairly straightforward. BA follows directly from the definition. BA follows from the fact that, given any two elements of an ordered field, there is an element between them (take the average), an element less than both of them (subtract from the smaller of the two), and an element greater than both of them (add to the greater of the two). BA follows from the fact that, by definition of a linear order, in an ordered field any three distinct elements are pairwise comparable and ordered transitively in a unique way. BA4 is a bit trickier and is outlined below; it can be done without using affine transformations, but is an illustrative application of them. Note that the properties of a field are only needed for BA and BA4. BA4 can be proved directly for any line, but it is easier to note that this line may be transformed by an affine transformation so that it is the x-axis: y = 0. You will show that affine transformations

2 preserve betweenness as well as incidence relationships, so verifying BA4 may be reduced to the case of the x-axis, which is fairly trivial, based on the fact that the betweenness relationships depend only on the signs of the y coordinates of the points. The exercises below take you through the necessary steps.. (Concrete example) Consider the points (, ), (, 4), and (4, 8). Verify that these points are collinear. Let T (, ) be the translation defined by (x, y) (x +, y ). Show that the images of these points under this translation are collinear by giving an equation for the line on which they lie. Show in addition that the betweenness relationship among the points is preserved after translation.. (General result) Let b = (b, b ) be a vector with coordinates in F. Denote by T b the translation defined by (x, y) (x, y) + b = (x + b, y + b ). Show that if (x, y ), (x, y ), and (x, y ) are collinear, then T b (x, y ), T b (x, y ), and T b (x, y ) are collinear by giving an equation for the line on which they lie in terms of the coefficients for the equation satisfied by (x, y ), (x, y ), and (x, y ). Show in addition that the betweenness relationship among the points is preserved. A Crash Course in Linear Algebra in Two Dimensions! Consider F F as a vector space, with addition and scalar multiplication defined coordinate by coordinate in the obvious way. (A vector space over a field F is just a set in which addition and multiplication by elements of F, which are called scalars, are defined and obey the natural distributive properties, and which contains an additive identity element and all additive inverses.) In general, a transformation L of a vector space is linear if, given vectors v and w and given scalars a, b F, L(av + bw) = al(v) + bl(w). For our two-dimensional vector space F F, this means that a linear map is completely determined by its outputs for (, 0) and (0, ), since (x, y) = x(, 0) + y(0, ). I invite you to check that if L(, 0) = (v, v ) and L(0, ) = (w, w ), then L may be computed by matrix multiplication: [ ] [ ] v w L(x, y) = x w y v Conversely, any transformation defined by matrix multiplication is linear. An affine transformation is any composition of linear transformations and translations. 4. (Concrete example) Consider again the points (, ), (, 4), and (4, 8), thinking of them as vectors. Let L be any linear transformation of the vector space F F. Show that L(, ), L(, 4), and L(4, 8) are collinear, and give an equation for the line on which they lie. Show in addition that the betweenness relationship among the points is preserved. 5. (General result) Given three collinear points and a linear transformation L, show that the images of these points under L are collinear, and give an equation for the line on which these images lie, in terms of the equation for the line through the original points. Show in addition that the betweenness relationship among the points is preserved. 6. Given a linear transformation L, show that L T b = T L(b) L. In addition, show that T a T b = T a+b. Finally, show that the composition of two linear transformations is linear. Conclude that any affine transformation may be written as a (single) linear transformation followed by a (single) translation. 7. Give the matrix for an invertible linear transformation that transforms the y-axis into the x-axis. (There is not a single correct answer.) Give the matrix for the inverse of this transformation. 8. Give the matrix for an invertible linear transformation that transforms the line y = mx into the x-axis. (Again, there is not a single correct answer.) Give the matrix for the inverse of this transformation.

3 9. Give the formula for an invertible affine transformation that transforms the line x = x 0 into the x-axis. Give the formula for the inverse of this transformation. 0. Give the formula for an invertible affine transformation that transforms the line y = mx + b into the x-axis. Give the formula for the inverse of this transformation.. Let l be any line, let P and Q be any points not lying on l, and let A be any invertible affine transformation. Show that P and Q are on opposite sides (resp. the same side) of l if and only if A(P ) and A(Q) are on opposite sides (resp., the same side) of A(l).. Let P and Q be any two points that do not lie on the x-axis. Show that P and Q are on opposite sides (resp., the same side) of the x-axis if and only if their y coordinates have opposite signs (resp., the same sign).. Prove that Betweenness Axiom 4 holds in the Cartesian plane over an ordered field. The proof of the converse, that if BA-4 are satisfied in a Cartesian plane, then the field must ordered and the betweenness relation of the plane must correspond to that order, is considerably harder. Frankly, we glossed over the proof in class. In fact, a generally excellent source for this material, Geometry: Euclid and Beyond, by the algebraic geometer Robin Hartshorne, contains an incorrect argument for this result! I will explain! Let us think carefully about the hypothesis of this proposition. We have a Cartesian plane over some field: the points are ordered pairs of elements in this field, the lines are solution sets of linear equations. We are assuming that incidence is given by set membership: a point is on a line if it satisfies the equation of that line (that is, if it is in the solution set of that equation). Although we are assuming there is a betweenness relation among the points in our plane, we are not defining it. We are only assuming it satisfies BA-4. Since we haven t assumed our field is ordered, we cannot define a betweenness relation based on such an ordering! We are also not defining a congruence relation, nor even assuming there is one. We could define a congruence relation in the usual way, but our purpose is to show that the betweenness axioms alone are sufficient, with or without a congruence relation, to imply that our field is ordered and this ordering determines the betweenness relation. In particular, we cannot find points with particular coordinates by marking off congruent segments, because we are not assuming any notion of congruence. (This is where Hartshorne commits his error.) So what do we have to work with? We know whether or not two lines are parallel, since we can determine if their equations have a common solution. If two lines are parallel, all the points on one are on the same side of the other one. Recall also the trick that if P, Q, and R lie on a line l, and m is a distinct line through Q, then P Q R P and R are on opposite sides of m. (Sides of a line are defined, since we have a betweenness relation.) These tools alone are sufficient to reach the desired conclusion! Before proceeding with the proof, let us consider how to prove a field is ordered. Recall that we need to define a set P of positive elements. The set of positive elements induces an order with the definition that a < b b a P. The requirement that a < b a + c < b + c is automatically met, since (b + c) (a + c) = b a. To ensure that the other requirements of an ordered field are met, P must satisfy the following properties: 0. It goes without saying that 0 P. This property ensures anti-reflexivity, since any element minus itself is zero.. For any field element a 0, either a P or a P. This ensures that any two field elements are comparable (since a b = (b a)).. P is closed under addition. This ensures transitivity (since c a = (c b) + (b a)).. P is closed under multiplication. This ensures that if a < b and c > 0, ac < bc. (Short exercise: show this!) Define P to be the set of x-coordinates for points on the x-axis that are on the same side of (0, 0) as (, 0). (That is, (0, 0) does not lie between these points and (, 0); equivalently, they are on the same

4 side of the y-axis as (, 0).) Pictorially, we will represent points on the x-axis that are on the same side as (, 0) as being to the right of the y-axis, and we will represent the points on the y-axis that are on the same side as (0, ) as being above the x-axis, in the usual manner. Property (0) above is clearly satisfied. The following exercises lead you through the proofs of the remaining three properties. (You are, of course, welcome to develop arguments different from the ones suggested.) 4. First show that (a, 0) and (, 0) are on the same side of the y-axis if and only if (0, a) and (0, ) are on the same side of the x-axis. (In other words, the points on the same side of the x-axis as (0, ) are exactly those points with y-coordinate in P.) Consult the following picture, which illustrates that (0, 0) (, 0) (a, 0) (0, 0) (0, ) (0, a). The proof that (0, 0) (a, 0) (, 0) (0, 0) (0, a) (0, ) is similar. By BA, exactly one of these situations must hold, since (0, 0) is not in the middle. (0,a) (0,) (0,0) (,0) (a,0) 5. More generally, the method of the previous exercise shows that (a, 0) (b, 0) (c, 0) (0, a) (0, b) (0, c). 6. To show that P is closed under multiplication (Property ), assume b P (and hence that (0, b) and (0, ) are on the same side of the x-axis), and show that (a, 0) and (ab, 0) are on the same side of the y-axis.. Consult the following illustration. (0,b) (0,) (0,0) (a,0) (ab,0) 7. Either as an algebraic consequence of the previous result, or using the illustration below, show that P is closed under multiplicative inverses.

5 (0,a) (0,) (0,0) (/a,0) (,0) 8. Show that P is close under addition (Property ()). Consult the following illustration, and then apply the previous results. (Hint: Assume, without loss of generality, that (0, 0) (a, 0) (b, 0). Then (0, 0) and (0, b) are on opposite sides of (0, a)(b, 0). Since (0, 0) and (a, 0) are on the same side of ( ) ab (0, a)(b, 0), (0, b) and (a, 0) are on opposite sides, by BA4. Hence (0, b) a+b, ab a+b (a, 0), ( ) from which it follows that and (a, 0) are on the same side of the y-axis. Note that ab a+b, ab a+b BA and Prop.. are used on several occasions. Finish by using the fact that a vertical line is parallel to the y-axis, so all points on it are on the same side of the y-axis.) (0,b) (0,a) (ab/(a+b),ab/(a+b)) (0,0) (ab/(a+b),0) (a,0) (b,0) 9. Finally, show that if a 0 and a / P, then a P (Property ()). (Hint: An argument similar to the one used in the previous exercise shows that if (a, 0) and (b, 0) are both on the opposite side of the y-axis from (, 0), then so is their sum. Hence, if neither a nor a is in P, we would obtain the obvious contradiction that (0, 0) is on the opposite side of the y-axis from (, 0). Now that we know our coordinate field F is ordered, it is not hard to show that this ordering gives the same betweenness relation that we assumed to begin with. After all, the ordering of the field elements is determined by the betweenness relation of the plane! It suffices to show that betweenness and ordering are the same for the x-axis (proof left to the reader for extra credit!): a < b < c or a > b > c a b < 0 < c b or a b > 0 > c b a b is negative and c b is positive, or a b is positive and c b is negative, (a b, 0) and (c b, 0) are on opposite sides of the y axis a and c are on opposite sides of the line x = b (proof left to the reader for extra credit!) (a, 0) (b, 0) (c, 0). Ordered Pythagorean Fields and Congruence Observe that we must have a betweenness relation in order to introduce congruence, because without it we cannot define segments, rays, or angles! (An angle consists of two rays emanating from a common point, so without rays, there are no angles.) There wouldn t be any objects that could be congruent! Once we have a betweenness relation, we can define congruence in a Cartesian plane using the dot product. In what follows, the dot product will be denoted by. It is simplest to think of points as vectors, so that we can subtract them to get the vector from one point to another. (That way we avoid putting little hats over pairs of points to denote vectors, as we did in class. Mathematics is a process

6 of continual refinement and improvement!) We then make the following precise definition of congruence for segments: Definition. Segment P Q is congruent to segment RS if (and only if) (P Q) (P Q) = (R S) (R S). 0. Show that the dot product is linear in each coordinate; that is, if u, v, and w are vectors, and a and b are scalars (elements of F), then u (av+bw) = a(u v)+b(u w) and (au+bv) w = a(u w)+b(v w). We summarize these properties by saying that the dot product is bilinear. (Hint: it saves time to first show that the dot product is commutative: u v = v u.). Show that congruence, as defined above, is an equivalence relation. In particular, be sure to show that (P Q) (P Q) = (Q P ) (Q P ), which is necessary to prove reflexivity, since by BA segment P Q is equal to segment QP. We showed in class that Congruence Axiom is satisfied if and only if the field F is Pythagorean. Here are some concrete examples that put the Pythagorean property into practice, along with some further exercises on affine transformations.. Show that a point with positive x-coordinate exists at distance from the origin on the line y = x if and only if 4 F. (You showed on the first exam that, if F is Pythagorean, then 4 is indeed an element of F.). What are the coordinates of both points at distance from the point (, ) on the line y = x+ 7? 4. Using straightedge and compass, construct a Pythagorean spiral: Choose a unit segment. First construct a right triangle with legs of length. Its hypotenuse has length. Using the hypotenuse of this first triangle as one leg, construct a right triangle adjacent to the first whose other leg has length. The hypotenuse of the second triangle will have length. You can, of course, continue in this fashion indefinitely. Construct the first five triangles in the spiral. (Remark: In the Cartesian plane over a Pythagorean field, all of these points would be constructible. But if the field is not also Euclidean, there would be some straightedge and compass constructions that could not be done; that is, the points of intersection between some lines and circles would not really exist because their coordinates would not be in the field.) 5. The figure shows a unit square grid and some points, P, Q, and R. On graph paper, carefully draw the image of this grid and the images, P, Q, and R, of P, Q, and R under each linear transformation whose matrix is given below. P (a) R [ ] 0 Q [ ] (b) (c) [ 6. Which of the linear maps in Exercise 5 preserve congruence of segments? ] (d) [ 7. Which of the linear maps in Exercise 5 preserve congruence of angles? ] (e) [ ] 8. Let A = (, ), B = (4, ), and C = (, ). Let A = (, ), B = (, 4), and C = (, ). It should be evident that ABC = A B C. Give a formula for the affine transformation that takes A to A, B to B, and C to C. (We will show that this transformation is unique.)