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## Back to Cone

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**Back to Cone**• Motivation: From the proof of Affine Minkowski, we can see that if we know generators of a polyhedral cone, they can be used to describe the polyhedron in Rn. • Which generators are important for generating a polyhedral cone? • Def: Given cone K, then for a K \ {0}, the half line { ya: y 0} is called a ray of cone K. We term a 0 as a ray of K, but think { ya : y 0}**Def: A ray of cone K is called an extreme ray if it cannot**be written as a proper (weights are > 0) conical combination of two other distinct rays of K, i.e. a K \ {0} is an extreme ray when a = y1a1 + y2a2, y1, y2> 0, and a1, a2 K \ {0} either z1 > 0 s.t. a1 = z1a or z2 > 0 s.t. a2 = z2a ex) Note that the cone is generated by extreme rays. Then can we say that all cones are generated by extreme rays? O X O O 0**Ex) Consider cone K = { (x1, x2) : x2 0 }**Consider vector (2, 0) below, (2, 0) = 3(1, 0) +1(-1, 0), i.e. (2, 0) is a positive scalar multiple of (1, 0) and it is impossible to express (2, 0) as proper conical combination without using a vector having the same direction as (2, 0), hence (2, 0) ( and (1, 0) ) is an extreme ray • Note: • Not both of (1, 0) and (-1, 0) need to be positive multiple of (2, 0). • Extreme rays are (-1, 0) and (1, 0), but this K is not generated by these extreme rays. x2 K x1 0 (2, 0) (1, 0) (-1, 0)**Def: The lineality space of cone K is K(-K), where (-K)**= { -a : a K}, i.e. K (-K) = { a : a K, (-a) K }. It is a subspace. Observe that for K = { x : Ax 0}, have lineality subspace { x : Ax = 0} The lineality of K is the rank of K (-K) Cone K is said pointed provided K (-K) = {0} x2 K K K(-K) K(-K) x1 0 -K -K Pointed cone**Prop: For pointed cones, a K \ {0} is an extreme ray**when a = y1a1 + y2a2, y1, y2> 0, and a1, a2 K \ {0} both z1 > 0 s.t. a1 = z1a and z2 > 0 s.t. a2 = z2a Pf) Suppose a = y1a1 + y2a2, y1, y2 > 0, a1, a2 K \ {0} and z1 > 0 with a1 = z1a. Then a = y1z1a + y2a2 , i.e. a( 1 – y1z1 ) = y2a2 (1) 1 – y1z1 = 0 either y2 = 0 or a2 = 0, contradiction. (2) 1 – y1z1 < 0 -a = { y2 / ( y1z1-1) } a2 -a K, contradiction. From (1), (2) (1 – y1z1) > 0, i.e. a2 = z2a **Thm: Suppose K is pointed, nontrivial ( {0}, )**polyhedral cone. Then K has finitely many extreme rays, say A = { a1, …, am }, and K is generated by A. Pf) Minkowski’s theorem guarantees K is finitely generated, say by A = { a1, …, al }. Suppose this set of generators is minimal in that A \ {ak} does not generate K. We will show A, A are the same set up to positive multiplication. When l = 1, K is a half line and conclusion clear. Suppose l > 1. We first show A A. Pick any extreme ray ak A. Since ak K and A generates K, we have y1, …, yl 0 s.t. ak = y1a1 + … + ylal and some yj > 0 (since ak A ak 0), hence ak = yjaj + ( i j yiai ). Now, if i j yiai = 0, then ak = yjaj and if i j yiai 0, then ak = yjaj + a ( a K \ {0} ) So, from ak extreme ray aj = zak for some z > 0. Hence a positive multiple of ak appears in A. Thus A A and so A is a finite set.**(continued)**Now need to show A A. Show this by supposing ak A \ A (i.e. ak is a generator but not an extreme ray), and derive contradiction. Since ak A ak is not an extreme ray. So b1, b2 K \ {0}, s.t. ak = z1b1 + z2b2, z1, z2 > 0, b1, b2 ak. Since b1, b2 K si , ti 0 s.t. b1 = i=1l siai , b2 = i=1l tiai Hence ak = i=1l (z1si + z2ti ) ai ( 1 - z1sk - z2tk ) ak = i k (z1si + z2ti ) ai 3 cases: (1) (1 - z1sk - z2tk ) > 0 ak is positive combination of {ai : i k}. A \ {ak} generates K A not minimal, contradiction. (2) (1 - z1sk - z2tk ) < 0 -ak = i k {(z1si + z2ti ) / (z1sk + z2tk – 1) } ai -ak K ak lineality space of K K not pointed, contradiction.**(continued) ( 1 - z1sk - z2tk ) ak = i k (z1si +**z2ti ) ai (3) (1 - z1sk - z2tk ) = 0 Observe that j k s.t. z1sj + z2tj > 0. Why? else si = ti = 0 i k b1 = skak, b2 = tkak b1, b2 are positive multiplication of ak, contradiction. Then if j is unique, we have 0 = (1 - z1sk - z2tk)ak = (z1sj + z2tj ) aj Note that (z1sj + z2tj ) > 0, aj 0. So (z1sj + z2tj ) aj 0, contradiction. Hence (1 - z1sk - z2tk)ak - (z1sj + z2tj ) aj = i j, k (z1si + z2ti ) ai -aj = i j, k {(z1si + z2ti ) / (z1sj + z2tj )} ai Note that (z1si + z2ti ) 0, (z1sj + z2tj ) > 0. So - aj K aj lineality space of K K not pointed, contradiction **The proof shows that the set of extreme rays is the unique**(up to positive scalar multiplication) minimal set of generators for nontrivial, polyhedral, pointed cone K, i.e. we have shown For nontrivial, polyhedral cone K, pointedness of K {extreme rays} = {generators} How about the converse? i.e. for K nontrivial, polyhedral cone, does K not pointed K not generated by extreme rays? • For example, consider a line through the origin, which is not a pointed cone. Then there exist 2 extreme rays and these do generate all K. Hence, the converse of the Theorem is false.**But consider K nontrivial, polyhedral, not a line. Then**does K not pointed K not generated by extreme rays? (yes) Pf) Since K is not pointed, a 0 in the lineality space of K, i.e. a, -a K. Consider any x K \ {0}, but not in { ya: y R}. Then x is not a positive scaling of either a or –a. Hence, neither are x+a, x-a. But x+a, x-a K and x = ½(x+a) + ½(x-a) So this x is not an extreme ray of K. Therefore, a, -a are the only candidates for extreme rays of K, yet a, -a cannot generate K. **Prop (Cone decomposition):**Let K be a convex cone with lineality space S. Then K = S + (K So) and (K So) is pointed. Pf) x K x’ S, x’’ So s.t. x = x’ + x’’ (from HW) But x’ S - x’ K x + (-x’) = x’’ K. Hence x’ S, x’’ K So Alternatively, if x* S K and x** K So, then x* + x** K since both x*, x** K. Together, have K = S + (K So) To see that (K So) is pointed, suppose a (K So) and -a (K So) a, -a K a S, hence a (S So) a = 0 Hence (K So) is pointed. **Note that the decomposition K = S + (K So) is not a**unique representation. x2 a K K So S x1 0 K = S + {ya: y0} also. So**Finding Generators of Cone K**• Suppose given K = { x: Ax 0}, polyhedral cone, want to find generators of K. Then lineality space of K is S = { x : Ax = 0} First find a basis (rows of) B for S (using G-J elimination) ( recall that if rows of A generates a subspace, S = Ao = { x: Ax = 0}. Suppose the columns of A are permuted so that AP = [ B : N ], where B is m m and nonsingular. By elementary row operations, obtain EAP = [ Im : EN ], E = B-1. Then the columns of the matrix D constitute a basis for S.) Here the basis matrix B = D’.**(continued)**Then find extreme rays of K So, say rows of matrix C. Have K = { y’B + z’C: y Rp, z R+q}, where rows of B are basis for S and rows of C are extreme rays of pointed cone K So. Once B known, then So = { x : Bx = 0} K So = { x : Ax 0, Bx 0, -Bx 0} ( recall that for S = { y’A: y Rm}, T = { x Rn: Ax = 0}, then have So = T, To = S ) Finding generators of K reduces to finding extreme rays of a pointed cone. Observe that { x : Ax = 0, Bx = 0, -Bx = 0} = {0} because K So is pointed. Hence rank of matrix is n (full column rank)**Ex) Consider the cone K = { xR3: x1 + x2 + x3 0, x2**+ 2x3 0} Hence, basis for S ( = {x: Ax = 0}, null space of A) is (1, -2, 1)’ So = { x : x1 – 2x2 + x3 = 0} (So is the row space of A). Hence, K So = { x R3: x1 + x2 + x3 0, x2 + 2x3 0, x1 – 2x2 +x3 0, - x1 + 2x2 - x3 0 } • Yet we don’t know how to identify generators (extreme rays) of pointed cone K So.**Geometric view**a1 a4 H4 a2 H3 a3 H1 H2 0**Characterizing Extreme Rays**• Thm: Suppose K = { x: Ax 0}, where A: m n has rank n. Let x K and reorder rows of A as A = , where Ux = 0, Vx < 0. Then x generates extreme ray of K U has rank n-1. Pf) Prove contraposition in both directions. ) Suppose rank(U) n-1. If rank(U) = n x = 0 (since Ux = 0) x not an extreme ray. If rank(U) < n-1 rank < n vector a 0 s.t. Ua = 0 and xa = 0 (latter a not a multiple of x) Consider x + a, x - a, for > 0 and small. Then U( x a) = 0. Also Vx < 0, so for small > 0, V(x + a) < 0, V(x - a) < 0, i.e, A(x a) 0. So (x a) K (for small > 0).**(continued)**But x = ½( x + a) + ½( x - a), and neither ½( x + a) nor ½( x - a) is a positive multiple of x. Hence x not extreme in K. ) Suppose x does not generate an extreme ray of K. Then either x = 0, in which case U = A (definition of U) and rank(A) = n rank(U) = n n-1. So suppose x 0 and not extreme in K, then must have x = y1a1 + y2a2 with y1, y2 > 0, a1, a2 K \ {0} and neither a1 nor a2 is a positive multiplication of x. Observe that a1, a2 are linearly indep., else contradiction. ( Why? a1 = a2 with > 0 x = ’a2, ’> 0, contradiction to above. If < 0 cases (1) x = 0 (contradiction) (2) x = ’a2, ’> 0 (contradiction) (3) – x K K not pointed, contradiction.)**(continued)**Also 0 = Ux = y1(Ua1) + y2(Ua2) Note that y1 > 0, Ua1 0, y2 > 0, Ua2 0, so Ua1 = Ua2 = 0 Ua1 = Ua2 = 0 rank(U) n-2, i.e. Rank(U) n-1 **So to find extreme rays of a pointed cone, list all**submatrices of A which have rank n-1. Suppose that those submatrices are Bi , i I, | I | < +. Now rank Bi = n-1 rank {x: Bix = 0} = 1. The solution set is a line, i.e. of form Li { ybi : y R} for some bi Rn ( find bi from Bi by G-J elimination.) Now 3 things can happen ( Observe Li K because K pointed) Consider Li = Li+ Li- , where Li+ = {ybi : y 0} Li- = {y(-bi) : y 0} (1) Li+ K, then bi generates extreme ray of K. (2) Li- K, then - bi generates extreme ray of K. (3) neither (1) nor (2), then bi or – bi fails to generate extreme ray of K.**a1**a4 • Ex) H4 a2 H3 a3 H1 H2 0 H1 and H2 generate a2, H1 and H4 generate a1. But H1 and H3 intersect outside of K, hence fail to generate extreme ray of K.**In summary, given K = { x : Ax 0}** K = S + K So ( S = { x : Ax = 0} ) Find basis matrix B of lineality space S using G-J. So = { x : Bx = 0 } ( So is subspace generated by rows of A) Let K’ = K So K’ = { x: Ax 0, Bx 0, - Bx 0 } = { x: Ax 0, Bx = 0} To find extreme ray of K’, choose n-1 independent rows of the above constraints to obtain Bi matrix. Then find +, - basis of the null space of Bi, i.e. { x: Bix = 0} using G-J Plug in +, - basis to K’ to check if the vector is in K’. If the vector is in K’, it is an extreme ray.**(Ex-continued)**Consider the cone K = { xR3: x1 + x2 + x3 0, x2 + 2x3 0} basis for S ( = {x: Ax = 0}, null space of A) is (1, -2, 1)’ So = { x : x1 – 2x2 + x3 = 0} (So is the row space of A). K So = { x R3: x1 + x2 + x3 0, x2 + 2x3 0, x1 – 2x2 +x3= 0} Consider the combinations 1st –2nd (unnecessary), 1st-3rd, 2nd-3rd constraints.**Conversely, how can we find the constrained form of a cone**given that generators of the cone are known? Use K = K++ iff K is finitely generated nonempty cone. Recall that for K = { y’A: y 0}, L = { x: Ax 0}, we have K+ = L, L+ = K. Hence, given K = { y’A: y 0}, we first construct K+ which is given by K+ = L = { x: Ax 0}. We then find generators of K+ using previous results. Since K+ now is described by generators, we take the polar cone of K+ again to get K++ (= K) which is now described as a constrained system. (Note that in K = S + (K So), S is described by linear combinations of a basis. By taking basis as generators of S, we can describe K as conical combinations of basis of S and generators of the pointed cone (K So) )**Example)**x2 K=K++ (1, 2) (-2, 1) (2, 1) x1 K+ (1, -2)**(ex-continued)**Cone K is generated by two vectors (1, 2) and (2, 1). Hence its polar is K+ = { x: x1 + 2x2 0, 2x1 + x2 0}. Lineality space of K+ = {0}, so K+ is a pointed cone and its extreme rays are generators of K+. To find generators of K+, we set n-1 = 1 of its constraints at equality and find the one dimensional line satisfying the equality. From x1 + 2x2 = 0, we get two vectors (2, -1) and (-2,1). Among these two vectors, (-2, 1) is in the cone, hence is an extreme ray of K+. Similarly, we get extreme ray (1, -2) from 2x1 + x2 = 0. These two vectors are all extreme rays of K+. Hence its polar cone is described as K++ = K = { x: -2x1 + x2 0, x1 – 2x2 0}.