i=1 α ip i, where s The analogue of subspaces

Transcription

1 Definition: Let X = {P 1,...,P s } be an affine basis for A. If we write P = s i=1 α ip i, where s i=1 α i = 1 then the uniquely determined coefficients, α i, are called the barycentric coordinates of P with respect to the affine basis X. Example: 1) Let X = {P 1 = (1,1,0),P 2 = (0,0,1),P 3 = (1,0,1),P 4 = (0, 1,2)} and let A = Aff(X). The vectors P 1 P 2 = (1,1, 1),P 3 P 2 = (1,0,0),P 4 P 2 = (0, 1,1) are easily seen to NOT be linearly independent, so these four points are not an affine basis for A (just notice that (1,1, 1) + (0, 1,1) = (1,0,0) and so P 1 P 2 and P 4 P 2 are a basis for D(A) and hence Y = {P 1,P 2,P 4 } is an affine basis for A. The vector 5P 1 2P 2 2P 4 = 5(1,1,0) 2(0,0,1) 2(0, 1,2) = (5,7, 6) is in A and has barycentric coordinates (5, 2, 2) with respect to the affine basis Y of A. The point 5P 1 2P 2 2P 4 + P 4 = 5P 1 2P 2 P 4 = (5,6, 4) / A. 2) We can take P 0 = (0,0,...,0), P 1 = (1,0,...,0),P 2 = (0,1,...,0),...,P n = (0,0,...,0,1) as an affine basis for R n (considered as an affine set). It is called the standard affine basis for R n. Notice that, {P i P 0 = e i,i = 1,...,n} is the standard basis for R n, which is the giacitura of R n, considered as a vector space. Observation: With the discussion above, it is now clear that every affine subset A of R n can be generated by a finite subset of A. If dim A = s, it is enough to take a basis for the giacitura of A, call that basis v 1,...,v s and then take any point P 0 A. It will follow that the set is actually an affine basis for A. P 0,P 1 = P 0 + v 1,...,P s = P 0 + v s The analogue of subspaces There is one more simple concept I want to generalize from the notions you considered when studying vector spaces. This is the concept of an affine subset of an affine set, which generalizes the notion of a subspace of a subspace. Definition: Let A R n be an affine subset. We say that A is an affine subset of A if two things are true: 1

2 i) A is a subset of A; ii) A is an affine subset of R n. Example: Given A R n an affine subset, write A = P + D(A). Then, as we have seen, P A and D(A) is a subspace of R n. Now, let W D(A) be a subspace of D(A) and let P be any point in A, then Claim: A = P + W is an affine subset of A. Pf. It will be enough to show that A satisfies the two conditions, and those are obvious. In fact, the subsets of A described in the example are the ONLY affine subsets of A. That is the content of the following: Proposition: If A is an affine subset of the affine set A in R n then A = P + W where W is a subspace of D(A), P A. Proof: Inasmuch as A is an affine set we have that A = P + V, where P A and V is a subspace of R n. Since A is a subset of A, we must have that P A, and that is the last statement of the Proposition. On the other hand, since A is a subset of A and any v V can be written in the form v = u 1 u 2 with u i A it follows that u 1 u 2 D(A) as well. Thus, V D(A) and hence is a subspace of D(A). A Characterization of Affine Subsets of R n Let A = be an affine subset of R n. If P 1,P 2 are any two distinct points in A then X = {P 1,P 2 } is an affinely independent set of points. If we let Aff(X) be the affine subset spanned by X then Aff(X) = P 1 + P 2 P 1 i.e. the points of Aff(X) are the points P such that P = P 1 + t(p 2 P 1 ) = tp 2 + (1 t)p 1, t R. These points P are obviously in A. 2

3 It is a natural question to ask if: Suppose that X is a subset of R n with the property that: whenever P 1,P 2 X then Aff(P 1,P 2 ) X, is it then true that X is an affine subset of R n? Interestingly enough, the answer is Yes!! Theorem: Let X be a subset of R n with the property that: whenever P 1,P 2 X then the line Aff(P 1,P 2 ) X. Then X is an affine subset of R n. Proof: We need to show that X = P + V where P X and V is a subspace of R n. So, we pick a specific element in X and call it P. We will consider that fixed element for the rest of the proof. Lemma: Let V = {Q P Q X }. Then V is a subspace of R n. Proof of Lemma: We will first show that V is closed under scalar multiplication. So, pick an element in V (call it Q P) and an element λ R. We want to show that λ(q P) is back in V. But, by hypothesis, every point in Aff(P,Q) X, so, for λ as above, we must have N = (1 λ)p + λq X (where λ is as given above). Rewriting this last equation we have N = 1 P λp + λq, i.e. N P = λ(q P) and we have written λ(q P) in the form N P for some N X. We now want to show that V is closed under vector addition. So, let v and v be in V, i.e. v = M P and v = M P where M,M X we want to show that v + v is in V. 3

4 But, let N = 1 2 M M. This is a point on the line Aff(M,M ) and so is in X by hypothesis. Thus N P X. But, N P = 1 2 (M P) (M P) = 1 2 (v + v ) V. But, now we can use the fact that V is closed under scalar multiplication to conclude that v + v is in V. That completes the proof that V is a subspace of R n. We now prove that X = P + V and that will finish the proof of the theorem. Notice that Q X implies that Q P V and hence Q = P + (Q P) P + V. Thus, we have X P + V. For the reverse inclusion, every element in P + V is of the form P + v where v V, i.e. is of the form P + (Q P) with Q X. But, P + Q P = Q and so we have that P + V X. That completes the proof. The Functions Between Affine Sets in R n As I have said on several occasions, once one has a structure it is then very important to look at the appropriate functions on that structure. In our case, we are dealing with affine subsets of R n and I want to now begin to think about what are the appropriate functions between them. We ve seen that if A is an affine subset of R n and X = {P 0,P 1,...,P s } generates A (whether or not they are an affine basis for A) then every element, P, in A can be written in the form s a i P i = P i=0 We use this fact to define our functions. where s a i = 1. Definition: Let A 1 R n and A 2 R m be two affine sets. Let f : A 1 A 2 be a function. We say that f is an affine transformation if and only if, given any subset P 1,...,P r of A 1 and any constants a 1,...,a r R for which r i=1 a i = 1, then if P = r i=1 a ip i we want i=0 f(p) = a 1 f(p 1 ) + + a r f(p r ). 4

5 Remarks: Notice that this definition is not so easy to apply, i.e. it is not a trivial task to see if a given function is an affine transformation. One has to check the condition on EVERY finite subset of A 1 and on EVERY collection of coefficients. It almost looks impossible to verify if anything is an affine transformation!! So, part of our task is to tame this definition. We will see that it is not, in fact, very hard to use and we will be able to find lots of affine transformations between affine spaces. Indeed, we will be able to find them all. But, first we can make some easy observations. Observations: 1) If A 1 = V 1 is a subspace of R n and A 2 = V 2 is a subspace of R m and f : V 1 V 2 is a linear transformation, then f is an affine transformation. In fact, if P = a 1 P a r P r then we automatically have f(p) = a 1 f(p 1 ) + + a r f(p r ), even without the restriction on the sum of the a i s. 2) However, not every affine transformation between vector spaces is a linear transformation. E.g.let A 1 = A 2 = R 2 and let P = (1,1) R 2. Define the function f : R 2 R 2 as follows: f(v) = v + (1,1). Claim: This f is an affine transformation (and it is called a translation of R 2 by the vector P = (1,1)) which is not a linear transformation. Pf: It is obvious that f is not a linear transformation since f((0,0)) = (1,1) and so f does not carry the origin of R 2 into itself, as it would have to if it were a linear transformation. Let s now try to use the definition to show that f is an affine transformation. So, we pick P 1,...,P r any r points in R 2 and we pick a 1,...,a r, any real numbers with the property that r i=0 a i = 1 and look at the point P = a 1 P a r P r. 5

6 We need to verify that f(p) = a 1 f(p 1 ) + + a r f(p r ). But, f(p) = (1,1) + a 1 P a r P r while, a 1 f(p 1 ) + + a r f(p r ) = a 1 ((1,1) + P 1 ) + + a r ((1,1) + P r ) = (a a r )(1,1) + a 1 P a r P r. Since a a r = 1, we are done. There are some other things that are easy to show: 1) Let A 1, A 2, A 3 be three affine sets and suppose f 1 : A 1 A 2 and f 2 : A 2 A 3 are two affine transformations. Then f 2 f 1 : A 1 A 3 (the composition of f 1 and f 2 ) is again an affine transformation. 2) The identity map, 1 A : A A is an affine transformation for any affine set A. 3) Let A 1 and A 2 be any two affine sets and suppose that P A 2. The function f P : A 1 A 2 defined by f P (u) = P for all u A 1. is an affine transformation. We call this the constant affine transformation with constant value P. 6