Back to Cone

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# Back to Cone

## Back to Cone

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##### Presentation Transcript

1. 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}

2. 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

3. 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)

4. 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

5. 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 

6. 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.

7. (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.

8. (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 

9. 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.

10. 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. 

11. 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. 

12. Note that the decomposition K = S + (K  So) is not a unique representation. x2 a K K  So S x1 0 K = S + {ya: y0} also. So

13. 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’.

14. (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)

15. Ex) Consider the cone K = { xR3: 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.

16. Geometric view a1 a4 H4 a2 H3 a3 H1 H2 0

17. 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).

18. (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.)

19. (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 

20. 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.

21. 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.

22. 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.

23. (Ex-continued) Consider the cone K = { xR3: 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.

24. 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) )

25. Example) x2 K=K++ (1, 2) (-2, 1) (2, 1) x1 K+ (1, -2)

26. (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}.