Notes from February 11


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1 Notes from February 11 Math 130 Course web site: Two lemmas Before proving the theorem which was stated at the end of class on February 8, we begin with two lemmas. The first lemma contains most of the work. It expresses the fact that two line whose direction vectors are not proportional will always meet in exactly one point. We continue to suppose that is an arbitrary field or skew field. ecall also that a vector w in 2 is proportional to a nonzero vector v if w = tv for some t in. Lemma 1. Let l = P + [v] and m = Q + [w] be lines. Suppose that v and w are not proportional. Then l and m meet in a single point. Proof. The points of l and m are the points of the form P + tv, Q + sw, t s respectively. To find a point of intersection of l and m (if any) we must solve for s and t the equation P + tv = Q + sw. Since both sides are vectors, this equation becomes two equations: we write P = (p 1, p 2 ) etcetera, and the two equations can be written p 1 + tv 1 = q 1 + sw 1 p 2 + tv 2 = q 2 + sw 2,
2 or equivalently tv 1 sw 1 = q 1 p 1 tv 2 sw 2 = q 2 p 2. The vector v, being a direction vector for the line l, is supposed to be nonzero. So at least one of v 1 and v 2 is nonzero. Let us suppose that v 1 is nonzero and proceed. We can multiply the first equation on the right by v1 1, and the two equations become t sw 1 v1 1 = q 1 v1 1 p 1 v1 1 tv 2 sw 2 = q 2 p 2. Subtract (first equation) v 2 from the second equation, and you arrive at the equivalent pair of equations t sw 1 v1 1 = (q 1 p 1 )v1 1 s(w 2 w 1 v1 1 v 2 ) = q 2 p 2 (q 1 p 1 )v1 1 v 2. There are now two cases to consider, according to whether the coefficient of s on the left in the second equation is zero or nonzero. Case 1: w 2 w 1 v 1 1 v 2 is nonzero. In this case, the second equation has the form sa = b, where a is nonzero. This equation has a unique solution for s, namely s = ba 1. Substituting this into the first equation gives also a unique solution for t. So in Case 1, the equations for s and t have a unique solution; which tells us that l and m intersect in exactly one point, as the lemma states. Case 2: w 2 w 1 v 1 1 v 2 is zero. In this case, we have w = (w 1, w 2 ) = (w 1, w 1 v 1 1 v 2 ) = (w 1 1, w 1 v 1 1 v 2 ) = (w 1 v 1 1 v 1, w 1 v 1 1 v 2 ) = (λv 1, λv 2 ), (λ = w 1 v 1 1 ) = λv, which tells us that w is proportional to v. The hypothesis of the lemma, however, is that v and w are not proportional. So Case 2 does not occur.
3 In conclusion, only Case 1 occurs; and l and m meet in a single point. emark. All we are doing in the proof above is examining two inhomogeneous linear equations in two unknowns s and t. Understanding the solutions of such linear equations is familiar linear algebra, at least when the field is. It is only because the might be a skew field that we need to take a little care. This accounts for the length and verbosity of the proof. While Lemma 1 deals with the case that v and w are not proportional, the next lemma deals with the opposite case: Lemma 2. Let l = P + [v] and m = Q + [w] be two lines again, but suppose now that v and w are proportional. Then there are two possibilities: (i) if Q lies on l, then the two lines l and m are the same; and (ii) if Q does not lie on l, then the lines l and m do not meet. Proof. We will prove (i) here. You will prove (ii) on the homework. So suppose v and w are proportional, and Q lies on l. Since the points of l have the form P + tv, this means that, for some t 1 in, we have Now we can describe the line m as m = Q + [w] Q = P + t 1 v. = Q + [v], (because v and w are proportional) = {Q + tv t } = { (P + t 1 v) + tv t } = { P + (t 1 + t)v t }. Now, as t runs through all the elements of, so t 1 + t also runs through all elements of. So, writing s = t 1 + t, we have So l = m as desired. m = { P + (t 1 + t)v t } = { P + sv s } = P + [v] = l.
4 emember that two lines are said to be parallel if they are either one and the same line or if they do not intersect. The second lemma can be summarized by saying that lines with the proportional direction vectors are parallel. Meanwhile, the first lemma says that lines with nonproportional direction vectors are not parallel and meet at exactly one point. Putting it all together, we have: Corollary 3. Lines l and m in 2 are parallel if and only if their direction vectors are proportional. Lines that are not parallel meet in exactly one point. Checking that the axioms hold We can now verify that three axioms (A1), (A2) and (A3) for an affine plane do hold in the coordinate plane 2. (These axioms appear on the first handout, the Notes from January 30.) Axiom (A1) holds. We must show that, if P and Q are two distinct point, there exists a unique line passing through P and Q. To verify the existence of such a line, consider the line l = P + [v], where v = Q P. This line l does indeed contain P and Q (the latter because Q = P + v). Could there be a different line m, also passing throughp and Q? No, because by Corollary 3, we know that two different lines can only intersect in either zero points (if they are parallel) or exactly one point (if they are not parallel). This completes the verification of Axiom (A1). Axiom (A2) holds. Given a line m = Q + [w] and a point P, we must show that there exists exactly one line l through P that is parallel to m. According to Corollary 3, a line l is paralllel to m if and only if its direction vector is proportional to w. So the one and only line through P parallel to m is the line l = P + [w]. Axiom (A3) holds. We must show that there are three noncollinear points in 2. The points (0, 0), (1, 0) and (0, 1) are three such points (as you will be asked to confirm in the next homework).
5 Desargues Theorems in the coordinate plane Desargues Theorem II We will show that Desargues Theorem II holds in the affine coordinate plane 2. We still allow to be an arbitrary field or skew field here. The setup for Desargues Theorem II is that we have three distinct concurrent lines l, m and n, meeting at a point P. (Previously, this point was called O; but now we will use P because P may not be the origin (0, 0) in the coordinate plane.) We have points A, A on l, points B, B on m and C,C on n. None of these six points should coincide with P. The assumptions of Desargues Theorem are that AB and BC are parallel to A B and B C respectively. We must prove that AC is parallel to A C. The vector u = A P is a direction vector for the line l, so we can write l = P + [u] m = P + [v] n = P + [w] where A = P + u B = P + v C = P + w.
6 Because A lies on l, it has the form P + t 1 u for some t 1 in. So we have A = P + t 1 u B = P + t 2 v C = P + t 3 w. No two of the vectors u, v and w can be proportional, because if u and v were proportional (for example) then the lines l and m would be the same line. A direction vector for the line AB is the vector B A, which is equal to v u. Similarly, the line A B has direction vector t 2 v t 1 u. We are given that these two lines are parallel, so t 2 v t 1 u is proportional to v u. In other words, for some s in, we have (t 2 v t 1 u) = s(v u), or in other words (t 2 s)v = (t 1 s)u. However, the vectors u and v are not proportional; so the last equation can hold only if both the coefficients t 2 w and t 1 s are zero. It follows that t 2 and t 1 are equal. We have just seen that t 2 = t 1 follows from our assumption that AB and A B are parallel. Similarly, because BC and B C are parallel, it follows in the same way that t 2 = t 3. Putting these together, we conclude that t 1 = t 3. Now w u is a direction vector for the line AC, and t 3 w t 1 u is a direction vector for A C. Because t 3 and t 1 are equal, the direction vectors w u and t 3 w t 1 u are proportional. So AC and A C are parallel also. Desargues Theorem I We now turn to the first version of Desargues Theorem. In this version, we have three distinct parallel lines l, m and n and points A, A, B, B and C,C as shown. The hypothesis is that AB is parallel to A B and BC to B C. We must show that AC is parallel to A C.
7 Since the lines are parallel, we can take the same direction vector for all three: call it v. Further, since A, B and C lie on l, m and n, we can write l = A + [v] m = B + [v] n = C + [v] In particular, this means that for some t 1, t 2 and t 3 in. A = A + t 1 v B = B + t 2 v C = C + t 3 v Now let us see what the hypothesis that AB is parallel to A B amounts to. It tells us that B A is proportional to B A, so B A = s(b A) for some s in. Substituting from the formulae above, we see that this means t 2 v t 1 v + (B A) = s(b A) or in other words (t 2 t 1 )v = (s 1)(B A). The vector B A cannot be proportional to v (for otherwise we would have B = A + tv for some t, which would tell us that B lies on l, which is not the case). So s must equal 1, for if s 1 then we would have (B A) = (s 1) 1 (t 2 t 1 )v i.e. B A would be proportional to v. Knowing that s = 1, we now see that from the displayed equation above that t 2 t 1 must also be zero. So t 1 = t 2.
8 As in the proof of Desargues II previously, we have now seen: Similarly, AB A B = t 1 = t 2. BC B C = t 2 = t 3. Putting these together, we conclude that t 1 = t 3. This tells us that C A = C A ; so the lines AC and A C are also parallel, as claimed.
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