Affine transformations and convexity

Transcription

1 Affne transformatons and convexty The purpose of ths document s to prove some basc propertes of affne transformatons nvolvng convex sets. Here are a few onlne references for background nformaton: res/progeom/pgnotes02.pdf res/math133/metgeom.pdf Recall that an affne transformaton of R n s a map of the form (x) = b + A(x), where b E s some fxed vector and A s an nvertble lnear tranformaton of R n. Affne transformatons satsfy a weak analog of the basc denttes whch characterze lnear transformatons. LEMMA 1. Let as above be an affne transformaton, let x 0,, x k R n, and suppose that t 0,, t k R satsfy t = 1. Then = t (x ). Notaton. If t 0,, t k R satsfy t = 1 and x 0,, x k R n, then s sad to be an affne combnaton of the vectors x 0,, x k R n. Proof. Snce t = 1 we have t A x = A + whch s what we wanted prove. t b = + b = A t (A x + b) = + t b = t (x ) We also note the followng smple property of affne transformatons n R 2 : LEMMA 2. Let be an affne transformaton of R 2, and let x, y, z, w be ponts such that the lnes xy and zw are parallel. Then the lnes (x) (y) and (z) (w) are also parallel. Proof. Snce the two lnes are dsont and s 1 1, t follows that ther mages whch are also lnes because s an affne transformaton must also be dsont. CONVEX SETS. Here are the basc defntons we need for convexty: Defnton. If x, y R n, then the closed segment [xy] s the set of all vectors v such that where t R satsfes 0 < t < 1. v = t x + (1 t) y Ths corresponds to the ntutve noton of closed lne segment n elementary geometry. Defnton. A subset K R n s sad to be convex f x, y K mples that [xy] s contaned n K; n other words, x, y K and 0 t 1 mples that t x + (1 t) y K. 1

2 The followng result suggests that the notons of convexty and affne transformaton have some useful nterrelatonshps. LEMMA 3. Let K R n be convex, let x 0,, x m K, and suppose that t 0,, t m R satsfy t 0 and t = 1. Then K. Notaton. If t 0,, t m R satsfy t 0 and t = 1 and x 0,, x m R n, then s sad to be a convex combnaton of the vectors x 0,, x m R n. Proof. Snce a term t x makes no contrbuton to a sum f t = 0, t suffces to consder the case where each t s postve. The proof proceeds by nducton on m. If m = 1 the result s tautologcal, and f m = 2 the result follows from the defnton of convexty. Assume now that the result s true for m 2, and suppose we are gven scalars t 0,, t m+1 R satsfyng t > 0 and t = 1 together wth vectors x 0,, x m+1 K. Set σ equal to m t, and for 0 s m set s equal to t /σ. Then t follows that s > 0 and s = 1, so by nducton we know that y = s x s n K. By constructon we have 0 < σ < 1 and σ + t m+1 = 1, and therefore t follows that t x = t x + t m+1 x m+1 = whch s what we wanted to prove. COROLLARY 4. [A] s also convex. Proof. m σ y + t m+1 x m+1 K If s an affne transformaton of R n and A R n s convex, then the mage Suppose that x, y A and 0 t 1. Then Lemma 1 mples that (t x + (1 t) y) = t (x) + (1 t) (y) and hence the segment [ (x) (y)] s contaned n [A]. Snce every par of ponts n [A] can be expressed as (x) and (y) for some x, y A, the precedng sentence mples that [A] must be convex. Extreme ponts. Ths s a fundamental concept nvolvng convex sets. Defnton. A pont p n a convex set K s sad to be an extrme pont f t cannot be wrtten n the form p = t x + (1 t) y where x and y are dstnct ponts of K and 0 < t < 1; nformally speakng, ths means p s not between two other ponts of K. EXAMPLE 0. Let a < b R, and let X R be the closed nterval [a, b]. We clam that a and b are the extreme ponts of X. rst of all, f a < x < b and t = x a b a then 0 < t < 1 and x = (1 t)a + tb, so the two endponts are the only possble extreme ponts. To see that each s an extreme pont, suppose we are gven a pont x whch s NOT an extreme pont. Choose dstnct ponts u and v n [a, b] and t n the open nterval (0, 1) such that x = (1 t)u + tv; wthout loss of generalty we may as well assume u < v (note that t (0, 1) mples 1 t (0, 1) and 1 (1 t) = t). The nequaltes n the precedng sentence mply that u < x < v, and snce 2

3 a and b are mnmal and maxmal ponts of the nterval X = [a, b] t follows that x a, b, whch means that a and b are extreme ponts of X. EXAMPLE 1. If a, b, c are noncollnear ponts and X s the sold trangular regon consstng of all convex combnatons of these vectors, then the extreme ponts of X are a, b, and c. rst of all, ths set s convex because Lemma 3 mples that a convex combnaton of convex combnatons s agan a convex combnaton. To prove the asserton about extreme ponts, note that f t a + u b + v c s a convex combnaton n whch at least two coeffcents are postve, then an argument lke the nductve step of Lemma 3 mples that ths convex combnaton s between two others, and therefore the only possble extreme ponts are the orgnal vectors. urthermore, f p = t x + (1 t) y where x and y are convex combnatons and 0 < t < 1, then one can check drectly that at least two barycentrc coordnates of p must be postve (ths s a bt messy but totally elementary). Therefore a pont that s not an extreme pont cannot be one of a, b, c and hence these must be the extreme ponts of X. EXAMPLE 2. Let X be the sold rectangular regon n R 2 gven by [0, p] [0, q] where 0 q p. In ths case we clam that X s convex and the extreme ponts are the vertces (0, 0), (p, 0), (0, q) and (p, q). Ths wll be a consequence of Example 0 and the followng result: PROPOSITION 5. Let K 1 and K 2 be convex subsets of R n and R m respectvely. Then K 1 K 2 R n R m = R n+m s convex. urthermore, a pont (p 1, p 2 ) s an extreme pont of K 1 K 2 f and only f p 1 s an extreme pont of K 1 and p 2 s an extreme pont of K 2 Proof. The frst step s to prove that K 1 K 2 s convex. Suppose that t (0, 1) and that (x 1, x 2 ) and (y 1, y 2 ) belong to K 1 K 2. Then (1 t) (x 1, x 2 ) + t (y 1, y 2 ) = ( (1 t) x 1 + t y 1, (1 t) x 2 + t y 2 ) and by convexty the frst and second coordnates belong to K 1 and K 2 respectvely. The statement about extreme ponts wll follow f we can prove the contrapostve: A pont p n K 1 K 2 s not an extreme pont f and only f at least one of ts coordnates s not an extreme pont of the correspondng factor. Wrte p = (p 1, p 2 ). If p s not an extreme pont then we have p = (p 1, p 2 ) = (1 t) (x 1, x 2 ) + t (y 1, y 2 ) where 0 < t < 1 and (x 1, x 2 ) and (y 1, y 2 ) are dstnct ponts of K 1 K 2. By the defnton of an ordered par, t follows that ether the frst or second coordnates of (x 1, x 2 ) and (y 1, y 2 ) are dstnct; f we choose = 1 or 2 such that the th coordnates are dstnct, then t follows that p cannot be an extreme pont of K. Conversely, suppose that one coordnate p of p s not an extreme pont of the correspondng convex set K. Wthout loss of generalty, we may as well assume that = 1 (f = 2, reverse the roles of 1 and 2 n the argument we shall gve to obtan the same concluson n that case). Choose x 1 y 1 K 1 and t (0, 1) such that p 1 = (1 t) x 1 + t y 1. Then we also have p = (p 1, p 2 ) = (1 t) (x 1, p 2 ) + t (y 1, p 2 ) and therefore p s not an extreme pont of K 1 K 2. The fnal result reflects the mportance of extreme ponts. THEOREM 6. Let A R n be a convex set, and suppose that s an affne transformaton of R n. Then maps the extreme ponts of A onto the extreme ponts of [A]. 3

4 Proof. We shall prove the followng contrapostve statement: If p A, then p s not an extreme pont of A f and only f (p) s not an extreme pont of [A]. Note that every pont q [A] s (p) for some p A. Suppose that p s not an extreme pont of A. Then p = t x + (1 t) y where x and y are dstnct ponts of A and 0 < t < 1. By Lemma 1 we then have (p) = t (x) + (1 t) (y) and snce s 1 1 t follows that (p) s not an extreme pont of [A]. To prove the converse, combne ths argument wth the fact that 1 s also affne. COROLLARY 7. If 0 p, q and 0 r, s and s an affne equvalence mappng [0, p] [0, q] onto [0, r] [0, s], then sends the vertces of the frst sold rectangular regon to the vertces of the second. Ths follows mmedately from the theorem and Example 2. Convex hulls. Gven a subset X n R n, the convex hull s defned so that t wll be the unque smallest convex subset contanng X. Defnton. If X R n, then the convex hull of X, wrtten Conv (X), s the set of all convex combnatons where x 0,, x m X and t 0,, t m R satsfy t 0 and t = 1. Here are some elementary propertes of convex hulls; they combne to prove that the convex hull s n fact the unque smallest convex subset of R n contanng X. LEMMA 8. The convex hull has the followng propertes: () If X R n, then Conv (X) s a convex subset of R n. () If X s convex, then X = Conv (X). () If X Y R n, then Conv (X) Conv (Y ). Proof. The thrd statement follows mmedately from the defnton, and the second follows mmedately from Lemma 3. To prove the frst statement, let y (where 1 n) be ponts of Conv (X), and let s 0 satsfy s = 1. We can then fnd fntely many x X such that for each we have y = t, x where each t, s nonnegatve and s y = t, = 1, and hence we also have the followng: s ( t, x ) = ( s t, ) x We clam that the sum of the coeffcents n the rght hand expresson s equal to 1; ths wll prove that the vector n queston belongs to Conv (X), whch s what we want to prove. Ths may be verfed as follows: ( s t, ) = s t, = s 1 = 1 4

5 As noted above, ths shows that Conv (X) s closed under takng convex combnatons and hence s convex. nally, the followng result s often very useful for studyng the effects of affne transformatons on geometrcal fgures, especally when combned wth Theorem 6. THEOREM 9. Conv ( [X]). If X R n and s an affne transformaton of R n, then maps Conv (X) onto Proof. We shall frst show that maps Conv (X) nto Conv ( [X]). To see ths, note that v Conv (X) mples that v = where x 0,, x m X and t 0,, t m R satsfy t 0 and t = 1, and snce s an affne transformaton we have = t (x ) Conv ( [X]). To see that every pont n Conv ( [X]) comes from a pont n Conv (X), note that a pont y n Conv ( [X]) has the form t (x ) for sutable t and x, and by Lemma 1 ths expresson s equal to ; snce the expresson nsde the parentheses les n Conv (X), t follows that y [Conv (X)] as requred. 5