On the positive semidenite polytope rank

Transcription

1 On the positive semidenite polytope rank Davíd Trieb Bachelor Thesis Betreuer: Tim Netzer Institut für Mathematik Universität Innsbruck February 16, 017

2 On the positive semidefinite polytope rank - Introduction 1 Introduction Note that this bachelor thesis largely follows []. If you want to do linear optimization over a polytope, the complexity of many algorithms depends on the size of the representation of the polytope. So, if you have a (complicated) polytope, the idea is to nd a simpler convex set of higher dimension which has the polytope as a linear image of it and then optimize over that instead. This motivates following denition: Denition 1.1. Let P R n be a polytope. For a closed convex cone C R m and an ane space L R m, C L is called a C-lift of P, if there exists a linear map π : R m R n such that P = π(c L). In this bachelor thesis, we will be interested in the cases C = R k + = {x R k x 1 0,..., x n 0} and especially C = S k + = {M R k k M symmetric, positive semidenite}. Note that R k + embeds into S k + via diagonal matrices. Now one is interested in nding the smallest cone in the families {R k +} and {S k +}, respectively, which allows a lift of a polytope P : Denition 1.. Let P R n be a polytope. (1) The nonnegative rank of a polytope is given by rank + (P ) := min({k N P has R k +-lift}). () The positive semidenite (psd) rank of a polytope is given by rank psd (P ) := min({k N P has S k +-lift}). In order to investigate the properties of the psd rank of a polytope, the following denitions are also needed: Denition 1.3. Let P R n be a n-dimensional polytope with vertex set {p 1,..., p v } and irredundant inequality representation P = {x R n b 1 a 1, x 0,..., b f a f, x 0}, where b j R and a j R n for j = 1,..., f. Then the slack matrix S P R v f is given by (S P ) ij = b j a j, p i. Short reminder: the polar dual of a cone C is given by C := {y R m x, y 0 x C}. Note that both S k + and R k + are self dual cones, i.e. C = C. For symmetric matrices A, B we will use the trace inner product A, B = Tr(AB). Denition 1.4. Let M R p q + be a nonnegative matrix and C a closed convex cone with its polar dual C. David Trieb

3 On the positive semidefinite polytope rank - Introduction (1) A pair of ordered sets a 1,..., a p C and b 1,..., b q C is called C-factorisation of M, if M ij = a i, b j for all i = 1,..., p and j = 1,..., q. () For C = S k + (respectively, R k +), a C-factorisation of M is called a psd (respectively, nonnegative) factorisation of M. (3) rank psd M := min({k N M has S k +-factorisation}) denotes the psd rank of M. rank + M := min({k N M has R k +-factorisation}) denotes the nonnegative rank of M. Note that in Denition 1.3 any positive scaling of a facet inequality of P can be used and therefore, if you positively scale the columns of a slack matrix S P of P, you get a slack matrix of P again. In the following, any such slack matrix is denoted by S P. Remark 1.5. These are some results we will use: rank psd P = rank psd S P and rank + P = rank + S P. [1], [4] rank + P and rank psd P are invariant under ane transformations. If P is the polar polytope of P and S P a slack matrix of P, then S P scaling) a slack matrix of P and therefore, rank psd P = rank psd P. is (up to row Since R k + embeds into S+ k via diagonal matrices it follows rank psd M rank + M for M R p q +. Together with the rst statement of this remark this yields rank psd P rank + P for any polytope P R n. Proof. We prove that rank + P = rank + S P. rank psd P = rank psd S P can be proven similarly. (1) rank + P rank + S P : Let P = {x R n 1 c 1, x 0,..., 1 c f, x 0} R n be a fulldimensional polytope with vertices p 1,..., p v and rank + P = k. Then there exists a linear map π : R k R n such that P = π(r k + L), where L is an ane subspace. Suppose that L = z + L 0, where L 0 is a linear subspace. Then P is given by P = {x R n x = π(w), w R k + (z + L 0 )}. Let π : R n R k be the adjoint of π and assume that R k + L 0 = {0} (we may do so, because P is bounded). Then, by strong conic duality: P = {x R n y π (x) R k +, y L 0, y, z = 1} Since y L 0, we have w i, y = 1 for any w i L 0 +z = L. Now dene a i := w i and b j := y π (c j ), where w i π 1 (p i ) R k + and y L 0 (Rk + + π (c j )) with y, z = 1. Such w i and y do exist and we get (S P ) ij = 1 c j, p i = 1 c j, π(w i ) = 1 π (c j ), w i = 1 w i, y b j = = w i, b j = a i, b j. As a i R k + and b j R k +, we have rank + S P k = rank + P. David Trieb 3

4 On the positive semidefinite polytope rank - Introduction () rank + P rank + S P : Let P be a polytope as above. Assume that rank + S P = k. Then there exist a 1,..., a v, b 1,..., b f R k + such that (S P ) ij = a i, b j. Dene L := {(x, y) R n R k 1 x, c i = y, b j, j = 1,..., f}. Let L k R k denote the projection onto the second coordinate of L. Note that a i R k + L k and 0 / L k (0 L k x : 1 x, c i = 0 for all i = 1,..., f P is contained in the hyperplane {y x, y = 1}, which is a contradiction, because 0 P ). For all x R n for which there exist y R k + such that (x, y) L we have 1 x, c i 0, i = 1,..., f 1 x, y 0 y P x (P ) = P. Claim: y R k + L k :! x y R n : (x y, y) L. Proof of claim: The existence follows from the denition of L k. Now let x y and x y be two such points. Since (x y, y), (x y, y) L, we get 1 x y, c i = 1 x y, c i = y, b j, and thus we obtain for t R: 1 tx y + (1 t)x y, c i = 1 t x y, c i x y, c i + t x y, c i = y, b j, i.e., (tx y + (1 t)x y, y) L, which means that the line through x y and x y is contained in P, which is a contradiction, unless x y = x y. Therefore, the map R k + L k R n : y x y is well-dened. As this map is ane and 0 / L k, we may extend it to a linear map π : R k R n. We have already proven π(r k + L k ) P above. Since for all i = 1,..., v, a i R k + L k, it follows p i = π(a i ) π(r k + L k ) and because P is the convex hull of its vertices and π(r k + L k ) is convex, we obtain C π(r k + L k ). Thus C = π(r k + L k ) and we found a R k +-lift of P. Hence, rank + P rank + S P. Let P R n be a n-dimensional polytope with rank psd P = k. Then there exists a linear map π : S+ k R n such that P = π(s+ k L), where L S+ k is an ane subspace. Let π 1 : R n R n be an ane transformation. Then π 1 (P ) = z + π (π(s+ L)) k for some z R n and a linear function π. As the composition π π is also linear, this means that the polytope π 1 (P ) z has a S+-lift. k Since π 1 (P ) z can be mapped to P via another ane transformation, it follows rank psd π 1 (P ) z = k. From the rst statement of the remark, we know that rank psd π 1 (P ) z = = rank psd S π1 (P ) z. Now π 1 (P ) z has the same slack matrix as π 1 (P ) (see Lemma 1.6 below) and we get k = rank psd π 1 (P ) z = rank psd S π1 (P ) z = rank psd S π1 (P ) = rank psd π 1 (P ). The proof for rank + P works analogously. Lemma 1.6. Let P R n be a polytope with vertices p 1,..., p v and inequality representation P = {x R n b 1 a 1, x 0,..., b f a f, x 0}, and let z R n. Then S P = S P +z. David Trieb 4

5 On the positive semidefinite polytope rank - Hadamard square roots Proof. The polytope P + z has vertices p 1 + z,..., p v + z and it holds P + z = {x R n b 1 a 1, x + a 1, z 0,..., b f a f, x + a f, z 0}. Therefore, we obtain (S P +z ) ij = b j a j, p i + z + a j, z = b j a j, p i a j, z + a j, z = (S P ) ij. The following proposition compares psd rank with usual rank: Proposition 1.7. [3] Let M R p q +. Then rank M rank psd M min({p, q}). Proof. (1) Let rank psd M = k and assume that A 1,..., A p, B 1,..., B q give an S+-factorisation k of M. For N S+ k dene ( vec(n) := N 11, N,..., N kk, N 1, N 13,..., N 1k, N3,..., N k,..., N 34,..., ) N (k 1)k. Then vec(n) R ( k+1 ) and vec(a i ), vec(b j ) = = k m=1 n=1 k l=1 Thus rank M ( ) k+1 and it follows rank M (k + 1)k k 1 A i ll Bj ll + k m=1 n=m+1 A i mnb j mn = k A i mnbmn j = Tr(A i B j ) = A i, B j = M ij. = 0 k + k rank M = k rank M. () Let e j denote the jth standard unit vector. Since diag(m i ), diag(e j ) = = Tr (diag(m i ) diag(e j )) = M i, e j = M ij, there exists an S q + -factorisation of M. Similarly, one gets a S p + -factorisation of M. Hadamard square roots In this section we will use Hadamard square roots to study the psd rank of nonnegative matrices. Denition.1. Any matrix whose (i, j)-entry is a square root of the (i, j)-entry of M is called a Hadamard square root of M and is denoted by M. In particular, let + M be the all-nonnegative Hadamard square root of M. The square root rank of M is dened as rank M := min({rank M}). David Trieb 5

6 On the positive semidefinite polytope rank - Hadamard square roots Proposition.. Let M R p q + be a nonnegative matrix. Then rank psd M rank M. Proof. If M is a Hadamard square root ( of M of rank r, then there exist vectors M ) a 1,..., a p and b 1,..., b q R r such that = a i, b j. We obtain ij ( r ) r r M ij = a i, b j = (a i b j) = a ik b jk = a ik a il b jk b jl = k=1 k=1 l=1 a i1 a i1 a i a i1 a ir b a i1 a i a j1 b j1 b j b j1 b jr i a i a ir b j1 b j b j b j b jr = Tr = a i1 a ir a i a ir a ir b j1 b jr b j b jr b jr ( ) = Tr a i a i b jb j = a i a i, b jb j and thus get a S r +-factorisation of M. Hence, rank psd M r. Example.3. This simple example shows that the upper bound in. can be strict: Consider M := It holds rank M = rank M = 3, but rank psd M =, as the follwing S+-factorisation of M shows: ( ) ( ) ( ) A :=, A :=, A :=, ( ) ( ) ( ) B 1 0 :=, B 0 0 :=, B 3 1 := One can easily check that M ij = A i, B j. Lemma.4. k = rank M is the smallest k that admits a S+- k factorisation for M R p q + where all factors have rank one. Proof. (1) If k = rank M, then there exists a Hadamard square root M of M such that rank M = k and the proof of proposition. gives a S k +-factorisation of M where all factors have rank one. () k rank M: Let a 1 a 1,..., a pa p S k + and b 1 b 1,..., b qb q S k + be a S k +-factorisation of M with rank one factors. Then M ij = a i a i, b jb j = a i, b j and the matrix M with ( M ) ij = a i, b j is a Hadamard square root of M with rank M k. David Trieb 6

7 On the positive semidefinite polytope rank - Hadamard square roots Next, we show a method to increase the psd rank of a matrix by one. Proposition.5. ( ) Let M R p q + with rank psd M = k, w R q M 0 +, α > 0 and M :=. w α Then rank psd M = k + 1. Furthermore, in any S+ k+1 -factorisation of M, the factor associated to the last column of M is of rank one. Proof. (1) First, we show that rank psd M cannot be smaller than k + 1: Assume that there is a S k +-factorisation of M with factors A 1,..., A p, A S k + associated to its rows and B 1,..., B q, B S k + associated to its columns. From A, B = α 0 we get A 0 B. Let r = rank B > 0. Then there exists an orthogonal matrix U such that U 1 BU = diag(λ 1,..., λ r, 0,..., 0) =: D, where λ 1,..., λ r are the positive eigenvalues of B. Let A i := U 1 A i U S k + i = 1,..., p. Then i = 1,..., p: D, A i = U 1 BU, U 1 A i U = Tr(U 1 BA i U) = Tr(BA i ) = B, A i = 0. Because the diagonal entries of A i are nonnegative and those of D positive, the rst r diagonal entries of A i must be zero, since Tr(DA i ) = 0. Since A i is psd, it follows that the rst r rows and columns of A i are all zero (if this were not the case, there would be at least one negative eigenvalue of A i ). For B j := U 1 B j U, j = 1,..., q we have A i, B j = Tr(U 1 A i UU 1 B j U) = Tr(U 1 A i B j U) = Tr(A i B j ) = A i, B j = M ij i = 1,..., p and j = 1,..., q. Since A i only has nonzero entries in its bottom right (k r) (k r)-block, it follows that M ij = A i, B j, where A i and B j are the bottom right (k r) (k r)-submatrices of A i and B j respectively. Now we have found a S+ k r -factorisation of M which is a contradiction to rank psd M = k. Therefore, rank psd M k + 1. () If A 1,..., A p, B 1,..., B q S+ k is an S+-factorisation k of M, then A 1,..., A p, A, B 1,..., B q, B S+ k+1 is an S+ k+1 -factorisation of M, where ( ) ( ) ( ) ( ) A Ai 0 i :=, B j Bj 0 :=, A 0 0 :=, B 0 0 := w j α This proofs that rank psd M = k + 1. (3) Now let B be the matrix associated to the last column of M in a S+ k+1 -factorisation of M. Since α > 0, B 0. So, let r = rank B > 0. By applying the same arguments as above, we get a S+ k+1 r -factorisation of M. Because of rank psd M = k and r > 0, r = rank B must be one. David Trieb 7

8 On the positive semidefinite polytope rank - Psd ranks of polytopes Example.6. If M R n n + is a diagonal matrix with positive entries, then rank psd M = n. Furthermore, each factor of any S+-factorisation n of M is of rank one. Proof. The proof is by induction on n. For n = 1, the rst statement is trivially true. If rank psd M = n for a diagonal matrix M R+ n n with positive entries, then rank psd M = ( ) M 0 n + 1, where M = and α > 0 by Proposition.5. That each factor of a 0 α S+-factorisation n of M must have rank one follows from the second part of Proposition.5 applied to M and M. 3 Psd ranks of polytopes For this section, let P R n be a n-dimensional polytope. Lemma 3.1. rank S P = n + 1. Proof. If the vertices of P are p 1,..., p v and P = {x R n b 1 a 1, x 0,..., b f a f, x 0} is an irredundant inequality representation, then (S P ) ij = b j a j, p i and 1 p 1 ( ) b1 b S P = f... 1 p a 1 a f v Call the rst factor F and the second one G. Since P is a n-dimensional polytope, there exist n+1 vertices of P such that no vertex lies in the linear span of the other n vertices, implying that rank F n + 1 and thus, because F R v (n+1), rank F = n + 1. Now assume that rank G n. Then there exists a nonzero c R n+1 such that c G = (0,..., 0). If c 1 0, then normalise c such that c 1 = 1 and set c := (c,..., c n+1 ). Then, for all j: b j a j, c = 0, implying that c lies on every facet of P, which is a contradiction. If c 1 = 0, it follows a j, c = 0 (c as above) for all j = 1,..., f. This means that c (as a vector) is parallel to every facet of P, which is a contradiction, since P is fulldimensional. Thus, rank G = n + 1. As both factors have rank n + 1, it also holds rank S P = n + 1. Proposition 3.. rank psd P n + 1. If equality holds, every S+ n+1 -factorisation of S P has only rank one factors. Proof. The proof is by induction on the dimension n. For n = 1, P is a line segment with vertices p 1, p and facets f 1, f, where p 1 = f and p = f 1. Because of (S P ) 1 = b 1 a 1 p = b 1 a 1 f 1 = 0, and, analogously, (S P ) 1 = 0, the slack matrix S P of P is a diagonal matrix with positive entries. From Example.6 we get that rank psd P = rank psd S P = and any S+-factorisation only uses rank one factors. David Trieb 8

9 On the positive semidefinite polytope rank - Psd ranks of polytopes Now suppose the rst statement holds up to dimension n 1. Let P R n be a fulldimensional polytope. Let F be a facet of P with vertices p 1,..., p s and facets f 1,..., f t and slack matrix S F. Assume that f j corresponds to the facet F j of P for j = 1,..., t. Then, by induction hypothesis, rank psd F n. Suppose that, if p / F is a vertex of P, the top left (s + 1) (t + 1) submatrix of S P is indexed by p 1,..., p s, p in the rows and F 1,..., F t, F in the columns. We will call that submatrix S F. S F has the form ( ) S F SF 0 =, α where α > 0. From Proposition.5 we get that rank psd S P rank psd S F n + 1. If rank psd P = n + 1, then there exists a S+ n+1 -factorisation of S P, and therefore of S F. By Proposition.5, the factor corresponding to the facet F must be of rank one. Applying this argumentation to all facets F of P, it follows that all factors indexed by the facets of P have rank one. Recall that S P is, up to row scaling, a slack matrix of the polar dual polytope P and therefore, all factors corresponding to the vertices of P also have rank one. Corollary 3.3. If a facet of a polytope P has psd rank k, then rank psd P k + 1. Especially, if P R n is a n-dimensional polytope with psd rank n + 1, then, for every i-dimensional face F of P, rank psd F = i + 1. Theorem 3.4. rank psd P = n + 1 if and only if rank S P = n + 1. Proof. By remark 1.5, rank psd P = rank psd S P. If rank S P = n+1, then, by Proposition., rank psd P = rank psd S P rank S P = n + 1 and rank psd P n + 1 (Proposition 3.) implies rank psd = n + 1. On the other hand, if rank psd P = n + 1, then there exists a S+ n+1 -factorisation of S P, whose factors are by Proposition 3. all of rank one. By Lemma.4, rank S P n + 1. Since rank psd S P rank S P, rank S P = n + 1. Example 3.5. Consider the pentagon P with vertices p 1 = (0, 0), p = (1, 0), p 3 = (, 1), p 4 = (1, ), p 5 = (0, 1) and inequality representation { ( ) ( ) ( ) x P = R 0 x, y 1 y ( ) ( ) ( ) 1 x 1 3, 0, 1, 1 y 1 ( 1 0, 1 1 ( ) x 0, y ( 1 0 ) ( ) x, 0, y ) ( ) x, y } 0. David Trieb 9

10 On the positive semidefinite polytope rank - Psd ranks of polytopes Then we obtain the slack matrix S P = (the inequalities have been multiplied by, 4, 4, 4, and, respectively). Theorem 4.5 will show that the psd rank of P is at least four. Due to the following S+-factorisation 4 of P we have rank psd P = 4: It holds (S P ) ij = A i, B j with A i, B j as follows: A 1 := , A := , A3 := A 4 := , A5 := , B 1 := , B := , B3 := , B 4 := , B5 := Now let us look at rank S P. Let 0 a b c d e f S := g 0 0 h i j k 0 0 l. m n o 0 0 Then there exists a Hadamard square root of S P with rank 4 if and only if there exists a solution to this system of equations: det(s) = 0, a = 4, b = 1, c = 4,..., o = 8. Using a computer one can easily check that this system of equations has no solution and it follows rank S P = 5. Thus, the psd rank of a polytope may be smaller than the square root rank of its slack matrix., David Trieb 10

11 On the positive semidefinite polytope rank - Psd ranks of polytopes Example 3.6. Consider a pentagonal pyramid P P R 3 with the pentagon P from Example 3.5 as the base. The vertices of P P are p 1 = 0, p = 0, p 3 = 1, p 4 =, p 5 = 1, p 6 = and it holds P P = {(x, y, z) R 3 y z 0, 1 x + y z 0, 3 x y z 0, 1 + x y z 0, x z 0, z 0}. Since P is a facet of P P and rank psd P = 4 (see above), we get from Corollary 3.3 that rank psd P P 5. A slack matrix of P P is given by S P P = The top left 5 5 submatrix of S P P is equal to S P and thus, by Proposition.5, rank psd P P = rank psd S P P = 5. Example 3.7. Let H R be a regular hexagon with vertices ( ) ( ) ( ) p 1 =, p 0 = 3, p 3 = 3, p 4 = ( ) ( ) 1 1, p 0 5 = 3, p 6 = ( 1 3 and inequality representation { ( ) ( ) ( ) ( ) ( ) x H = R y / x 0, 0, 3 y 4/ x, 0, 3 y ( ) ( ) ( ) ( ) / x, 0, 3 y / x, 0, 3 y ( ) ( ) ( ) ( ) } 0 4/ x, 0, 3 y / x, 0. 3 y The slack matrix S H of H is then given by S H = ) David Trieb 11

12 On the positive semidefinite polytope rank - Polytopes of minimum psd rank It holds rank + S H = 5, while the following Hadamard square root of H has rank four: This example shows that it is not enough to compute rank + S P rank S P. in order to get Example 3.8. Let H be as in Example 3.7. In Proposition 3. and Theorem 3.4 we have shown that if P is an n-dimensional polytope with rank psd P = n + 1, then rank psd P = rank P and all S+ n+1 -factorisations of S P have only rank one factors. From Theorem 4.5 we have that rank psd H 4, hence we get from Theorem 3.4 that the square root rank of S H is at least four and since we have seen a Hadamard square root of S H of rank four in the example above, it holds rank S H = 4. We also get from Theorem 4.5 that rank psd H 4 and the following S+-factorisation 4 proofs that rank psd H = 4: A 1 := , A := , A3 := , A 4 := , A5 := , A6 := , B 1 := , B := , B3 := , B 4 := , B5 := , B6 := Note that rank A i = for all i = 1,..., 6. Thus, there can be S rank psd P + -factorisations of a slack matrix S P of a polytope P, where there are factors with rank greater than one even if rank psd P = rank P. David Trieb 1

13 On the positive semidefinite polytope rank - Polytopes of minimum psd rank 4 Polytopes of minimum psd rank For a n-dimensional polytope P R n, rank + P n + 1 and the only n-dimensional polytopes that achieve the lower bound of n + 1 are simplices. In this section, we will see that there are several classes of polytopes that achieve the minimum psd rank of n + 1. Denition 4.1. A full fulldimensional polytope P R n is called -level if its slack matrixes entries are all 0 or 1. Remark 4.. A polytope P is -level if and only if for every facet F of P, all vertices of P lie on the union of the facet F and one hyperplane parallel to the hyperplane containing F. For example, any n-dimensional cube is -level. Corollary 4.3. If P is a n-dimensional -level polytope, then rank psd P = n + 1 and every S+ n+1 - factorisation of P only has factors of rank one. Proof. Let S P be a slack matrix of a n-dimensional -level polytope P. We have rank S P rank + S P = rank S P = n+1 and therefore, by Proposition., rank psd S P rank S P n+1 and thus rank psd P = rank psd S P = n+1. The second statement follows from Proposition 3.. Theorem 4.4. For any fulldimensional polytope P R n with n + vertices we have rank psd P = n + 1. Proof. Let P be a polytope with n + vertices and f facets. Then S P R (n+) f + and rank S P = n + 1. Since rank S P = n + 1, there exist a i R (not all of them equal to zero) such that n+ i=1 a i(s P ) i = (0,..., 0). Because every vertex lies on at least n dierent facets, each column of S P has at least n zeros. Thus, considering above equation componentwise, we have that for all j = 1,..., f there exist i 1, i {1,..., n + } such that a i1 (S P ) i1 j + a i (S P ) i j = 0. Now let b i := sgn(a i ) a i. Then b i1 (SP ) i1 j + b i (SP ) i j = 0. Since this is true for all components, we get that n+ i=1 b i (SP ) i = (0,..., 0) and thus rank + S P = n + 1 = rank S P. This together with Theorem 3.4 proves the statement. Theorem 4.5. Let P R be a -dimensional polytope. Then rank psd P = 3 P has at most four vertices. Proof. " ": Follows from 4.4. " ": Let P R be a polytope with at least 5 vertices. Since the psd rank is invariant under ane transformations we can assume that two facets of P are given by f 1 = {(x, y) R x 0} and f = {(x, y) R y 0} with vertices v 1 = (0, 0), v = (0, 1) and v 3 = (1, 0). Furthermore, let v 4 = (a, b) and v 5 = (c, d) be the vertices sharing an edge with (0, 1) and (1, 0), respectively. These two edges are given David Trieb 13

14 On the positive semidefinite polytope rank - Polytopes of minimum psd rank by f 3 = {(x, y) R y b 1 a x + 1} = {(x, y) R a + (b 1)x ay 0} and f 4 = {(x, y) R y d c 1 (x 1)} = {(x, y) R (c 1)y d(x 1) 0}. Then, the 5 4 submatrix S P of S P indexed by these vertices and facets is given by 0 0 a d S P c 1 + d = 1 0 a + b 1 0 a b 0 cb b da + d. c d bc c ad + a 0 We now show that rank S P 4. Theorem 3.4 then yields rank psd P > 3, which proves the statement. It is enough to show that any Hadamard square root H of the top left 4 4 submatrix of S P has rank four. Assume that H = 0 0 ± a ± d 0 ±1 0 ± c 1 + d ±1 0 ± a + b 1 0 ± a ± b 0 ± cb b da + d has rank three. It is obvious that the rst three rows are independent, hence we can write the fourth row as a linear combination of the rst three and it must hold H 4 = ± ah 3 ± bh a + b 1H 1 = ( = ± a, ± b, 0, d(a + b 1) ± ) b(d + c 1). Let α := b(d + c 1) and β := d(a + b 1). Thus H 44 = ± cb b da + d = ± α β! = ± α β, and this implies α = β bc b = da d a 1 = c 1. Hence, the slopes of the lines between (a, b) and (1, 0), and (c, d) and (1, 0) are the same, meaning that (a, b), (c, d) and (1, 0) are collinear and cannot be all vertices, except when (a, b) = (c, d). Lemma 4.6. Let P R 3 be a fulldimensional polytope with psd rank four and p be a vertex of P of degree four. Then the four faces incident to p are all triangles. Proof. Assume that one of the four faces is not a triangle. By Corollary 3.3, this face has psd rank 3 and hence, by Theorem 4.5, it is a quadrilateral. Thus, P contains the conguration b d p 1 f 3 p 5 f 1 p f 4 p 4 f p p 3 David Trieb 14

15 On the positive semidefinite polytope rank - Polytopes of minimum psd rank where p 1,..., p 5 are vertices of P. According to Lemma 3.1, rank S P = 4. Since rank psd P = 4, by Theorem 3.4, there exists a Hadamard square root S P of S P of rank four. Let f 1, f, f 3 and f 4 be the faces of P containing the vertices p, p 1, p ; p, p, p 3 ; p, p 1, p 5 and p, p 3, p 4, p 5, respectively and M be the submatrix of S P indexed by p, p 1, p, p 3, p 4 in the rows and f 1, f, f 3, f 4 in the columns. Then, by scaling the columns of S P accordingly, it holds M = a 1 0 b 0, c d e 0 where a, b, c, d, e > 0. Since the four rows of S P (and S P ) corresponding to the rst four rows of M are clearly independent, we can write the row of S P (and S P ) corresponding to the fth row of M as a linear combination of the other four. This yields the conditions d+ae = abc (for S P ) and d +a e = a b c (for S P ), implying d +a e = d +a e + ade and thus ade = 0, which is a contradiction to a, b, c, d, e > 0. Denition 4.7. Let F(P ) denote the set of all faces of a polytope P. Two polytopes Q and Q are called combinatorially equivalent, if there exists an isomorphism f : F(Q) F(Q ) which preserves the partial order given by inclusion. Remark 4.8. Roughly speaking, two polytopes that are combinatorially equivalent have the same number of faces of each dimension arranged in the same way. For example, every three-dimensional polytope with six two-dimensional quadrilateral faces is combinatorially equivalent to a cube. Theorem 4.9. A fulldimensional polytope P R 3 with rank psd P = 4 is combinatorially equivalent to an octahedron, bisimplex, simplex, quadrilateral pyramid, triangular prism, or cube. Proof. Let v, e, f be the number of vertices, edges, and faces of P, respectively. Further, v t := #{v v vertex of P, deg v = 3}, v q := #{v v vertex of P, deg v = 4}, f t := #{f f triangular face of P } and f q := #{f f quadrangular face of P }. By counting the edges on each face, e = 3f t + 4f q, and thus (by considering P ), we also have e = 3v t + 4v q. Since v e + f = (Euler's formula) and v = v t + v q, f = f t + f q, we get e = v t + v q + f t + f q 4 and thus: 3v t + 4v q = v t + v q + f t + f q 4 v t = f t + f q 4 v q, and analogously, f t = v t + v q 4 f q, from which we get that v t and f t are even and v t + f t = v t + f t 8 v t + f t = 8. Therefore, the only polytopes we have to consider are polytopes, where (v t, f t ) equals (0, 8), (, 6), (4, 4), (6, ) and (8, 0). (1) (v t, f t ) = (0, 8): In this case, every vertex of P has degree four. From Lemma 4.6 we get that all faces of P must be triangular. An octahedron is the only polytope in R 3 that satises these conditions. David Trieb 15

16 On the positive semidefinite polytope rank - Polytopes of minimum psd rank () (v t, f t ) = (, 6): Here, there must be at least one vertex p of P of degree four, and thus, by Lemma 4.6, P contains the following structure: p 1 p 4 p p p 3 (a) Since v t =, at least two of the vertices p 1, p, p 3, p 4 have degree four. Now assume, that two of those degree four vertices (let's say, p 1 and p ) are adjacent. Then, again by Lemma 4.6, p 1 and p are contained in four faces which are triangles, implying that p 1 and p each lie on two triangular faces not shown in the gure. Since p 1 already shares a face with p, they can only share at most one of the four extra triangular faces, meaning that there must be at least 7 triangular faces, which is a contradiction to f t = 6. Thus, only two out of p 1, p, p 3, p 4 have degree four and those two are nonadjacent. Again, both lie on two extra triangular faces not shown in the gure, and because of f t = 6, they must share both of them. Consequently, P is a bisimplex (3) (v t, f t ) = (4, 4): If v q = 0, then there are only four vertices in total and the polytope must be a simplex. If v q > 0, then, by Lemma 4.6, P contains structure (a). Since f t = 4, p i must have degree three, i = 1,, 3, 4 (if one of the p i 's had the degree four, then there would be too many triangular faces). Therefore, the polytope is a quadrilateral pyramid. (4) (v t, f t ) = (6, ): Then, for the polar dual of P we have (v t, f t ) = (, 6), hence, the polar dual of P is a bisimplex and thus, P is a combinatorial triangular prism. (5) (v t, f t ) = (8, 0): Again, for the polar dual of P : (v t, f t ) = (0, 8). Thus P is the polar dual of an octahedron, i.e. a cube. References [1] Pablo A. Parrilo João Gouveia and Rekha R. Thomas. Lifts of convex sets and cone factorisations. Mathematics of OR, 38:4864, 013. [] Richard Z. Robinson João Gouveia and Rekha R. Thomas. Polytopes of minimum positive semidenite rank, 01. [3] Richard Z. Robinson. The positive semidenite rank of matrices and polytopes, 014. [4] Mihalis Yannakakis. Expressing combinatorial optimization problems by linear programs. J. Comput. System Sci., 43:441466, David Trieb 16