Completely positive factorizations associated with Euclidean distance matrices corresponding to an arithmetic progression
aa r X i v : . [ m a t h . SP ] S e p Completely positive factorizations associated with Euclideandistance matrices corresponding to an arithmetic progression
Damjana Kokol Bukovˇsek a,b,1 , Thomas Laffey c , Helena ˇSmigoc c a School of Economics and Business, University of Ljubljana, Slovenia b Institute of Mathematics Physics and Mechanics, Ljubljana, Slovenia c School of Mathematics and Statistics, University College Dublin, Ireland
Abstract
Euclidean distance matrices corresponding to an arithmetic progression have richspectral and structural properties. We exploit those properties to develop completelypositive factorizations of translations of those matrices. We show that the minimaltranslation that makes such a matrix positive semidefinite results in a completelypositive matrix. We also discuss completely positive factorizations of such matricesover the integers. Methods developed in the paper can be used to find completelypositive factorizations of other matrices with similar properties.
Keywords: completely positive matrices, Euclidean distance matrices
1. Introduction An n × n real symmetric matrix A is completely positive , if it can be written as A = BB T for some n × k entry-wise non-negative matrix B . The minimum numberof columns k in such a non-negative factor B is called the cp-rank of A . The set ofcompletely positive matrices forms a closed, convex cone, and better understandingof this cone is not only an interesting question, but it also plays an important rolein the field of copositive optimization , [8, 12, 9].It is immediate from the definition, that any completely positive matrix is non-negative and positive semidefinite. A matrix that is both non-negative and positivesemidefinite is said to be doubly non-negative . Not every n × n doubly non-negativematrix of order n ≥ Damjana Kokol Bukovˇsek acknowledges financial support from the Slovenian Research Agency(research core funding No. P1-0222). The work on the paper was done during her three-monthvisit to University College Dublin.
Preprint submitted to Linear Algebra Appl. September 30, 2019 ompletely positive is a hard problem that has been a topic of intense research overthe years. For a comprehensive survey on completely matrices we refer to [6].Research has shown that the question can be solved for different classes of ma-trices. Here we mention two representative results that are different in flavour. In[18] it is shown that every diagonally dominant non-negative matrix is completelypositive. Kogan and Berman [19] introduced a graph theoretic view on the problem.For a given n × n symmetric matrix A , let G ( A ) be a graph on n vertices with anedge ( i, j ) if and only if a ij = 0. A graph G is said to be completely positive ifevery doubly non-negative matrix with G ( A ) = G is completely positive. In [19] itis shown that a graph G is completely positive if and only if it does not contain anodd cycle of length greater than 4.Finding a factorization of a given completely positive matrix is a natural ques-tion in this context. A factorization algorithm that can factorize any matrix inthe interior of the cone of completely positive matrices is offered in [23], explicitfactorization of diagonally dominant matrices is presented in [18], and generalizedto matrices whose comparison matrix is positive definite in [11]. Factorization ofmatrices with conditions on their graph is studied for example in [4, 10]. Thosefactorizations are typically not optimal, the number of columns in the factorizationmatrix is larger than the minimal possible. Finding and optimizing the cp-rank addsadditional complexity to an already hard problem.Once we know that a factorization of a given matrix A exists, we may want toimpose further properties on the factors. A matrix A has a rational cp-factorization,if it can be decomposed as BB T where B the entries of B are non-negative andrational. Integer cp-factorization is defined in a similar way. Every rational matrixwhich lies in the interior of the cone of completely positive matrices has a rationalcp-factorization [13]. Integer question seems to be harder to study, and has onlyrecently been answered for 2 × n × n identity matrix will bedenoted by I n , and the n × n nilpotent Jordan block by J n . B ( i ) will denote thematrix obtained from B by deleting the i -th row and column. More generally, for I ⊆ { , , . . . , n } , we will denote by B ( I ) the matrix obtained from B by deletingthe row and columns indexed by I .
2. Euclidean Distance Matrices corresponding an arithmetic progression
For a set of distinct real numbers S = { a , a , ..., a n } , we define the Euclideandistance matrix to be the matrix EDM( S ) = ( a ij ) ni,j =1 with a ij = ( a j − a i ) . Such amatrix EDM( S ) is clearly entrywise non-negative, and it has rank 3, see [3].In this work we are interested in a family of Euclidean distance matrices thatcorrespond to an arithmetic progression, i.e. S = { a, a + d, ..., a + ( n − d } for somepositive real numbers a and d , and in particular for S = { , , ..., n } . We will denote A n := EDM( { , , ..., n } ) , hence ( A n ) ij = ( j − i ) . We can limit our discussion to this special case, as for S = { a, a + d, ..., a + ( n − d } we have EDM ( S ) = d A n , and all the propertiesof A n are easily adapted to this more general case. In our first result we explicitlycompute the eigenvalues and the eigenvectors of A n . Lemma 1.
The nonzero eigenvalues of A n are equal to: λ , = n ( n − ± q n ( n − n − , λ = − n ( n − ,λ > > λ > λ . The eigenvector corresponding to the eigenvalue λ is equal to w = ( w i ) ni =1 with w i := n + 1 − i for i = 1 , . . . , n . For j = 1 , . . . , n − let v ( j ) denote the vector with only four nonzero elements, defined as follows: v ( j ) j = 1 , v ( j ) j +1 = − , v ( j ) j +2 = 3 , v ( j ) j +3 = − . Then the vectors v ( j ) , j = 1 , . . . , n − ,form a basis of the null space of A n .Proof. We can prove that w is an eigenvector for A n corresponding to λ by directcomputation:( A n w ) i = n X j =1 ( j − i ) ( n + 1 − j ) = − n ( n − n + 1 − i ) . It is also straightforward to check that A n v ( j ) = 0, for j = 1 , . . . , n −
3. Since rankof A n is 3, we still need to compute two more nonzero eigenvalues.3o compute the remaining two eigenvalues, we note that the trace of A n is equalto zero, i.e. λ + λ + λ = 0, and we compute the trace of A n : λ + λ + λ = n X i =1 n X j =1 ( j − i ) = n ( n − n − . Taking into account that λ is known, we obtain the following quadratic equationfor λ and λ : 180 λ − n ( n − λ − n ( n − n −
4) = 0 . Solving this equation gives us: λ = n ( n −
1) + q n ( n − n − λ = n ( n − − q n ( n − n − . Finally, the inequality λ > λ > λ is straightforward to check. (cid:3) Since A n has two negative eigenvalues it is clearly not completely positive. Still,the aim of this work is to develop factorizations of matrices A n and A n + αI n thattake into account not only the nonnegativity of A n , but also as many of the followingproperties of A n as possible: symmetry, integer entries, pattern, and low rank of A n .We won’t be able to attend to all these properties with a single factorization, andwe will allow translations of A n to make it positive semidefinite.
3. Straightforward factorizations related to A n First we offer a factorization of A n , that is reminiscent of a completely positivefactorization, but has an additional factor with a negative element. Theorem 2.
We can write A n = L n RL Tn , where L n is a non-negative n × matrixand R = − . Proof.
Notice that the vectors v ( j ), j = 1 , . . . , n −
3, defined in Lemma 1 are thefirst n − I n − J Tn ) . It follows that first n − A n ( I n − J Tn ) are equal to zero, and, by symmetry, the first n − A n := ( I n − J n ) A n ( I n − J Tn ) are also zero. Hence, only the lower-right 3 × A n is non-zero, and it is straightforward to check that this corner is equal to R = − . It follows that A n = ( I n − J n ) − (0 n − ⊕ R )( I n − J Tn ) − , and L n is equal to thelast three columns of ( I n − J n ) − . The claim is proved by noting that ( I n − J n ) − = I n + J n + J n + ... + J n − n is a non-negative matrix. (cid:3) An interesting feature of the factorization in Theorem 2 is that the matrix R isindependent of n . Since we cannot hope to find a completely positive factorizationof A n , we focus on the factorizations of matrices of the form A n + g ( n ) I n . Clearly, A n + g ( n ) I n will become diagonally dominant, and hence completely positive,for all large enough g ( n ). The minimal g ( n ) for which A n is diagonally dominantis easily computed, and is equal to g D ( n ) := P n − j =1 j = n ( n − n − . On theother hand, the minimal g ( n ) for which A n + g ( n ) I n will become positive semidefinitecan be deduced from Lemma 1: f ( n ) = n ( n − g ( n ) for which A n + g ( n ) I n is completely positive therefore satisfies the following inequality:16 n ( n − n + 1) ≤ g ( n ) ≤ n ( n − n − . (1)The main result of this paper will show that the lower bound is achieved, but firstwe will improve the upper bound in (1) using an inductive approach. Informed by(1) we denote f ( n ) = n ( n − n + 1), and look for functions g ( n ) of the form g ( n ) = qf ( n ), for which we can prove that A n + g ( n ) I n is completely positive usinga straightforward inductive proof.Note that A n − = A n (1) = A n ( n ), and the most natural inductive approach tothe problem is to write A n + g ( n ) I n = ( A n − + g ( n − I n − ) ⊕ + R n , where v n − = (cid:0) ( n − ( n − . . . (cid:1) T and R n = (cid:18) ( g ( n ) − g ( n − I n − v n − v Tn − g ( n ) (cid:19) . If we already know that A n − + g ( n − I n − is completely positive by inductivehypothesis, then it is sufficient to prove that R n is completely positive to determine5he complete positivity of A n . Since the graph of R n contains no odd cycles of ordergreater than 3, the matrix R n is completely positive as soon as it is doubly non-negative. Note that to determine that R n is positive semidefinite we only need toshow that det R n > g ( n ) = qf ( n ) intodet R n = ( g ( n ) − g ( n − n − (cid:0) g ( n )( g ( n ) − g ( n − − v Tn − v n − (cid:1) , and collecting ( g ( n )( g ( n ) − g ( n − − v Tn − v n − ) in terms of n , gives us q = 2 q as the optimal choice for q for this approach.Next we consider an induction step that takes us from n − n . Note that A n − = A n ( { , } ) = A n ( { n − , n } ) = A n ( { , n } ), and it turns out that developingour induction process from A n − = A n ( { , n } ) gives the best bound. Hence, wewrite A n + g ( n ) I = 0 ⊕ ( A n − + g ( n − I n − ) ⊕ + Q n , where Q n = g ( n ) u Tn − ( n − u n − ( g ( n ) − g ( n − I n − v n − ( n − v Tn − g ( n ) ,u n − = (cid:0) ... ( n − (cid:1) T , and v n − = (cid:0) ( n − ... (cid:1) T . As above, it issufficient to consider the complete positivity of Q n . The graph of Q n contains noodd cycles of order greater than 3, so Q n will be completely positive as soon as itwill be positive semidefinite. To study the positive definiteness of Q n , we considera matrix that is permutationally similar to Q n :ˆ Q n = ( g ( n ) − g ( n − I n − u n − v n − u Tn − g ( n ) ( n − v Tn − ( n − g ( n ) . Assuming g ( n ) > g ( n − Q n will be positive semidefinite if and only if det ˆ Q n ( n ) > Q n ≥
0. Using Schur complement to compute those determinants, andconsidering the leading coefficient in n , we determine that det ˆ Q n ( n ) > q > q , and that det ˆ Q n > (cid:18) δ n g ( n ) − u Tn − u n − δ n ( n − − v Tn − u n − δ n ( n − − v Tn − u n − δ n g ( n ) − v Tn − v n − (cid:19) > , where δ n = ( g ( n ) − g ( n − u Tn − u n − = v Tn − v n − = 130 ( n − n − n − (cid:0) n − n + 5 (cid:1) ,v Tn − u n − = 130 n ( n − n − (cid:0) n − n + 2 (cid:1) , g ( n ) = qf ( n ), we determine that q = q is optimal for this method. This givesus the following proposition that strengthens the bound given in (1). Proposition 3.
The matrix A n + q f ( n ) I n is completely positive.
4. Optimal g ( n ) At this point we would like to strengthen Proposition 3 to q = 1. To this endwe denote B n = A n + f ( n ) I n . Note that det B n = 0, hence we are operating at theboundary of positive semidefinite matrices. The following simple and well knownresult points out an additional property that the factors in the cp factorization of asemidefinite matrix must satisfy. Proposition 4.
Let A = BB T be a completely positive matrix with a nontrivialkernel N . Then B T v = 0 for every v ∈ N . The proof of the main result in this section will be rather technical, as we willexplicitly develop a completely positive factorization of B n . We will aim to write B n as the sum of rank one completely positive matrices, and a matrix with particularstructure that is made explicit in the following lemma. Lemma 5.
Let D ∈ M m ( R ) and D ∈ M k ( R ) be diagonal matrices, and let C bean m × k non-negative matrix. Then any matrix of the form A = (cid:18) D CC T D (cid:19) ∈ M m + k ( R ) , with an eigenvector w = (cid:0) w T − w T (cid:1) T corresponding to a non-negative eigenvalue,where w ∈ M m ( R ) and w ∈ M k ( R ) are strictly positive, is completely positive.Proof. First we observe that the matrix B = ( I m ⊕ − I k ) A ( I m ⊕ − I k ) has nonpositiveoff-diagonal elements, and a positive eigenvector ˆ w = (cid:0) w T w T (cid:1) T corresponding toa non-negative eigenvalue. We deduce that B ˆ w ≥
0, and hence B is an M -matrix,see for example [5]. Since B is also symmetric, it is positive semidefinite. This, inparticular, proves that D and D are non-negative. Since A is similar to B , we nowknow that A is doubly non-negative. Moreover, the graph of A is bipartite, hence A is completely positive, as it is proved in [4]. (cid:3) Proposition 4 and Lemma 5 bring to light, why it is convenient to know thesingular vector of B n explicitly when we are developing the completely positivefactorization of B n . In what follows let us denote the eigenvector correspondingto the eigenvalue λ given in Lemma 1 as w ( n ), or just w , when n is clear from7he context. (Clearly, w ( n ) is a singular vector for B n .) Part of the proof willbe split into an even and an odd case. One of the reasons for this is the slightdifference in the pattern of w ( n ) in those cases as w (2 m ) = (cid:0) v m , − v ′ m (cid:1) T and w (2 m + 1) = (cid:0) v m +1 , , v ′ m +1 (cid:1) T , where v n = (cid:0) n − n − . . . n − (2 ⌊ n ⌋ − (cid:1) T and v ′ n = (cid:0) n − (2 ⌊ n ⌋ − . . . n − n − (cid:1) T . Lemma 6.
Let K n ∈ M n ( R ) be the matrix with ones on the anti-diagonal and zeroselsewhere. Then K n = I n , K n A n K n = A n and K n v n = v ′ n . Lemma 7.
Let n = 2 m . There exists a non-negative matrix U satisfying U T w ( n ) =0 , diagonal matrices D, D ′ ∈ M m ( R ) and a non-negative matrix C ∈ M m ( R ) , suchthat B n = U U T + (cid:18) D CC T D ′ (cid:19) . Proof.
As suggested by the result we will view B n as a 2 × B n = (cid:18) A m + f ( n ) I m H m H Tm A m + f ( n ) I m (cid:19) , where H m = ( h ij ) ∈ M m ( R ) is defined by h ij = ( m + j − i ) . We will construct U byfinding two matrices V and V ′ satisfying U U T = V V T + V ′ V ′ T , where each columnof V corresponds to an off-diagonal pair of entries in the upper-left block of B n andeach column of V ′ corresponds to an off-diagonal pair of entries in the lower-rightblock of B n .Explicitly, V := P m − i =1 P mj = i +1 u ij u Tij , where u ij := ( j − i )( e i + e j + α ij e m + k ) , k := ⌊ m + j ⌋ + 1 − i, and α ij := k − (2 m − i − j + 1) . Note that the matrix u ij u Tij has exactly nine nonzero entries, three on the maindiagonal, one at the position ( i, j ), two in the upper-right block and symmetricallythree more below the main diagonal. The entry in ( i, j ) position is equal to ( i − j ) ,and hence equal to the ( i, j )-th entry of B n . Furthermore, once k is chosen, thevalue of α ij is determined from the condition u Tij w ( n ) = 0. (While there is somechoice in what k can be, we need to make sure that we choose it in such a way thatthe off-diagonal blocks remain non-negative after subtraction from B n .) Thus thematrix V V T has the form V V T = (cid:18) A m + E SS T E (cid:19) , where E and E are diagonal matrices, and satisfies the condition V T w ( n ) = 0.8e define V ′ := K n V . Using block partition K n = (cid:18) m K m K m m (cid:19) and Lemma 6 it is easy to determine that V ′ V ′ T = (cid:18) E ′ S ′ S ′ T A m + E ′ (cid:19) and V ′ T w ( n ) = 0 , where E ′ = K m E K m , E ′ = K m E K m and S ′ = K m S T K m . In particular, S ′ xy = S m +1 − y,m +1 − x . To complete the proof, we need to show that the matrix C = H m − S − S ′ is non-negative. Let us first consider the ( x, y )-entry in S : s xy = P ≤ i 1) + 1 − m . In the second case we have j = x and i = ⌊ m + x ⌋ + 1 − y . So, at most three columns of matrix V V T contribute to theentry s xy . We will refer to this contributions as matrices S I , S II and S III , whosenonzero entries are defined as follows:( S I ) xy = ( u x, x + y − − m u Tx, x + y − − m ) x,y + m for 1 ≤ x < x + y − − m ≤ m ( S II ) xy = ( u x, x + y − − m u Tx, x + y − − m ) x,y + m for 1 ≤ x < x + y − 1) + 1 − m ≤ m ( S III ) xy = ( u ⌊ m + x ⌋ +1 − y,x u T ⌊ m + x ⌋ +1 − y,x ) x,y + m for 1 ≤ ⌊ m + x ⌋ + 1 − y < x ≤ m. The computation of relevant entries of u ij u Tij and reordering of the bounds, gives usthe following formulas for the nonzero entries of S I , S II and S III :( S I ) xy = y − ( x + 2 y − m − (3 m − x − y + 3) for m + 3 − y ≤ x ≤ m + 1 − y, ( S II ) xy = y − ( x + 2 y − m − (3 m − x − y + 2) for m + 2 − y ≤ x ≤ m − y, ( S III ) xy = y − ( x − ⌊ m + x ⌋ + y − (2 m − ⌊ m + x ⌋ + y − x )for max { m + 3 − y, y − m − } ≤ x ≤ m. Note that outside the indicated regions the entries of those matrices are equal tozero. To simplify the calculations we estimate the matrix S III by a matrix ˆ S III withnonzero entries in positions ( x, y ) satisfying m + 3 − y ≤ x ≤ m and defined by:( S III ) xy ≤ ( ˆ S III ) xy = y − ( x − m + x − + y − (2 m − m + x − + y − x ) . Since the matrices H m and S + S ′ are symmetric with respect to the counter-diagonal it is enough to prove the nonnegativity of elements ( H m − S − S ′ ) xy that9atisfy x + y ≤ m + 1. Inside this region we will consider four subregions, that aredetermined by the location of nonzero entries of matrices S I , S II , S III , S ′ I , S ′ II and S ′ III . Before we define the regions, we observe that ( S II ) xy = 0 for x + y ≥ m + 1,hence ( S ′ II ) xy = 0 for x + y ≤ m + 1 and S ′ II doesn’t need to be considered. Similarly,( S I ) xy = 0 for x + y ≥ m + 2, so ( S ′ I ) xy only contribution is when x + y = m + 1. Atthis point we also note that ( S ) m = ( S ′ ) m = 0. Now let us define and study eachof the four regions:Region (1): { ( x, y ); x ≤ m + 1 − y } , only ( ˆ S ′ III ) xy = 0, hence C xy = ( H m − S ′ III ) xy ≥ ( H m − ˆ S ′ III ) xy . Introducing new variables u = y − ≥ v = m + 1 − x − y ≥ C xy ≥ u + 129 u + 81 u + 2 v + 52 uv + 66 u v + 12 uv u + 2 v ) > . Region (2): { ( x, y ); x = m + 2 − y, y ≥ } , ( S II ) xy = 0 and ( ˆ S ′ III ) xy = 0, hence C xy ≥ ( H m − S II − ˆ S ′ III ) xy , which we express in terms of u = y − ≥ C xy ≥ 320 + 1184 u + 1558 u + 855 u + 162 u u )(5 + 4 u ) > . Region (3): { ( x, y ); m + 3 − y ≤ x ≤ m − y } , ( S I ) xy = 0, ( S II ) xy = 0, ( ˆ S III ) xy = 0,( ˆ S ′ III ) xy = 0, and C xy ≥ ( H m − S I − S II − ˆ S III − ˆ S ′ III ) xy . Change ofvariables u = x + 2 y − m − ≥ v = m − x − y ≥ C xy ≥ p ( u, v )4(5 + 2 u + 2 v )(7 + 2 u + 4 v ) > , where p ( u, v ) = 1261 + 1487 u + 601 u + 122 u + 12 u + 3629 v + 3306 uv ++ 849 u v + 72 u v + 3565 v + 2351 uv + 312 u v + 1371 v ++ 516 uv + 162 v . Region (4): { ( x, y ); x = m +1 − y, y ≥ } , ( S I ) xy = ( S ′ I ) xy = 0, ( ˆ S III ) xy = ( ˆ S ′ III ) xy =0, and C xy ≥ ( H m − S I − S III ) xy , which we express in terms of u = y − ≥ C xy ≥ u + 12 u + 3 u u ) > . Lemma 8. Let n = 2 m + 1 . There exists a non-negative matrix U satisfying U T w ( n ) = 0 , α ≥ , diagonal matrices D, D ′ ∈ M m ( R ) and a non-negative ma-trix C ∈ M m ( R ) , such that B n = U U T + D C α C T D ′ . Proof. The proof will go along the same lines as the proof of Lemma 7. This timewe will view B n as a 3 × B n = A m + f ( n ) I m b m H m b Tm f ( n ) ( b ′ m ) T H Tm b ′ m A m + f ( n ) I m , where b m = (cid:0) m ( m − . . . (cid:1) T , b ′ m = K m b m and H m = ( h ij ) ∈ M m ( R ) isdefined by h ij = ( m + 1 + j − i ) . We will write U as: U U T = V V T + V ′ V ′ T + tt T + ZZ T , where (as in the proof of Lemma 7) each column of V will correspond to an off-diagonal pair of entries in the upper-left block of B n , and each column of V ′ willcorrespond to an off-diagonal pair of entries in the lower-right block of B n . Inthis case we will need to consider separately the position ( m − , m ) in the upper-left block, and the position ( m + 2 , m + 3) in the lower-right block. To deal withthose positions we introduce a rank one matrix tt T . Finally, each column of Z willcorrespond to an entry in the upper-middle block of B n .As in the even case, we define V := P m − i =1 P mj = i +1 u ij u Tij , where u ij := ( j − i )( e i + e j + α ij e m +1+ k ) , k := ⌊ m + j +12 ⌋ − i,α ij := k (2 m − i − j + 2) , and V ′ := K n V . There are two positions in the diagonal blocks left to cover, andthis is done by tt T , where: t := e m − + e m + e m +2 + e m +3 . It is easy to verify that V T w ( n ) = 0, V ′ T w ( n ) = 0, t T w ( n ) = 0, and V V T + V ′ V ′ T + tt T = A m + ˆ E S S A m + ˆ E , E and ˆ E are diagonal matrices. Finally, we define Z := P mi =1 z i z Ti , where z i := ( m + 1 − i )( e i + e m +1 + e m +2 − i ) . We have z Ti w n = 0, and ZZ T = F b m F K m b Tm β ( b ′ m ) T K m F b ′ m K m F K m , where F is a diagonal matrix.To complete the proof, we need to show that α := f ( n ) − β > 0, and that thematrix C := H m − ˆ S − F K m is non-negative. We have β = m X i =1 i = 16 m ( m + 1)(2 m + 1) < f ( n ) = 23 m ( m + 1)(2 m + 1) , proving α > tt T influences only the lower-left 2 × C . For n = 5 wehave C = (cid:18) (cid:19) , and for n ≥ × C is equal to (cid:18) (cid:19) ≥ . The ( x, y ) entry of the matrix F K m is nonzero only when x + y = m + 1, and inthis case ( F K m ) xy = y .We will consider V V T and V ′ V ′ T separately, and to this end we denote by S the top-right corner of V V T and by S ′ = K m S T K m the top-right corner of V ′ V ′ T .Contributions of S and S ′ are dealt with as in Lemma 7, with only minor changes inindexing. Considering nonzero entries of the matrix u ij u Tij at position ( x, m + 1 + y ),we get ( x, y ) = ( i, k ) or ( x, y ) = ( j, k ). This results in three possibilities, as follows.First we can have i = x and j = 2 x + 2 y − − m , and we define the correspondingmatrix S I by ( S I ) xy = ( u x, x +2 y − − m u Tx, x +2 y − − m ) x,m +1+ y for 1 ≤ x < x + 2 y − − m ≤ m, x ≤ m − , and zero for ( x, y ) outside this range.Similarly, for i = x and j = 2 x + 2 y − m we have S II with nonzero entries( S II ) xy = ( u x, x +2 y − m u Tx, x +2 y − m ) x,m +1+ y ≤ x < x + 2 y − m ≤ m, x ≤ m − 2, and for j = x and i = ⌊ m + x +12 ⌋ − y wedefine S III with nonzero entries( S III ) xy = ( u ⌊ m + x +12 ⌋− y,x u T ⌊ m + x +12 ⌋− y,x ) x,m +1+ y for 1 ≤ ⌊ m + x +12 ⌋ − y < x ≤ m, ⌊ m + x +12 ⌋ − y ≤ m − u ij u Tij and reordering of the bounds, gives usthe following formulas for the nonzero entries of S I , S II and S III :( S I ) xy = y ( x + 2 y − m − (3 m − x − y + 3) for m + 2 − y ≤ x ≤ m − y, ( S II ) xy = y ( x + 2 y − m ) (3 m − x − y + 2) for m + 1 − y ≤ x ≤ m − y, y ≥ , ( S III ) xy = y ( x − ⌊ m + x +12 ⌋ + y ) (2 m − ⌊ m + x +12 ⌋ + y − x + 2)for max { m + 2 − y, y − m } ≤ x ≤ m, ( x, y ) = ( m, . We estimate the matrix S III by a matrix ˆ S III , ˆ S III ≥ S III , with nonzero entries inpositions ( x, y ) satisfying m + 2 − y ≤ x ≤ m, ( x, y ) = ( m, 1) and defined by:( S III ) xy ≤ ( ˆ S III ) xy = y ( x − m + x + y ) (2 m − m + x + y − x + 2) . The matrices H m , S + S ′ and F K m are symmetric with respect to the counter-diagonal, so it is enough to prove the nonnegativity of elements ( H m − F K m − S − S ′ ) xy that satisfy x + y ≤ m + 1. Inside this region we will consider four subregions,that are determined by the location of nonzero entries of matrices S I , S II , S III , S ′ I , S ′ II , S ′ III and F K m . Since ( S ′ I ) xy = ( S ′ II ) xy = 0 for x + y ≤ m + 1, matrices S ′ I and S ′ II don’t need to be considered. The four regions are defined as follows:Region (1): { ( x, y ); x ≤ m − y } , only ( S ′ III ) xy = 0, hence C xy = ( H m − S ′ III ) xy ≥ ( H m − ˆ S ′ III ) xy . Introducing new variables u = y − ≥ v = m − x − y ≥ C xy ≥ 24 + 220 u + 258 u + 81 u + 8 v + 104 uv + 66 u v + 12 uv u + v ) > . Region (2): { ( x, y ); x = m + 1 − y, y ≥ } , ( S II ) xy = 0 and ( S ′ III ) xy = 0, hence C xy ≥ ( H m − S II − ˆ S ′ III ) xy , which we express in terms of u = y − ≥ C xy ≥ 305 + 691 u + 435 u + 81 u u ) > . { ( x, y ); m + 2 − y ≤ x ≤ m − y } , ( S I ) xy = 0, ( S II ) xy = 0, ( S III ) xy = 0,( S ′ III ) xy = 0, and C xy ≥ ( H m − S I − S II − ˆ S III − ˆ S ′ III ) xy . Change ofvariables u = x + 2 y − m − ≥ v = m − x − y ≥ C xy ≥ p ( u, v )8(2 + u + v )(3 + u + 2 v ) , where p ( u, v ) = − − u + 25 u + 30 u + 6 u + 361 v + 568 uv + 241 u v ++ 36 u v + 909 v + 825 uv + 156 u v + 531 v + 258 uv + 81 v . We get C xy > u, v ) ∈ { (0 , , (1 , } . For ( u, v ) ∈ { (0 , , (1 , } we get ( x, y ) ∈ { ( m − , , ( m − , } . In those two cases we compute C m − , = 3 > C m − , = > { ( x, y ); x = m + 1 − y, y ≥ } , ( F K m ) xy = 0, ( S III ) xy = ( S ′ III ) xy = 0,and C xy ≥ ( H m − F K m − S III ) xy , which we express in terms of u = y − ≥ C xy ≥ 68 + 116 u + 52 u + 7 u u ) > . (cid:3) Theorem 9. The matrix B n = A n + f ( n ) I is completely positive for every n .Proof. Suppose first that n = 2 m is even. By Lemma 7 we write B n as B n = U U T + (cid:18) D CC T D ′ (cid:19) , and since the matrix (cid:18) D CC T D ′ (cid:19) satisfies conditions of Lemma 5, the proof is com-plete in this case.Suppose now that n = 2 m + 1. By Lemma 8 we can write B n as B n = U U T + D C α C T D ′ . Now (cid:18) D CC T D ′ (cid:19) is completely positive by Lemma 5, since its eigenvector correspond-ing to 0 is just w (2 m + 1) with the middle entry (which is equal to zero) omitted.14n addition, we have proved that α > 0, and this allows us to conclude that D C α C T D ′ is completely positive. (cid:3) 5. Integer completely positive factorization In this section we want to take into account that A n has integer entries, and weask the question, for what values of g ( n ) does the matrix A n + g ( n ) I have an integercompletely positive factorization. It turns out, that for small n , g ( n ) = f ( n ) works,but for n ≥ g ( n ) > f ( n ). Proposition 10. Let B n := A n + f ( n ) I n . Then:1. B n has an integer completely positive factorization for n ≤ .2. B does not have an integer completely positive factorization.3. B + I has an integer completely positive factorization.Proof. To prove the first item and the third item, we find integer completely positivefactorizations explicitly, as follows: B = (cid:18) (cid:19) = (cid:18) (cid:19) (cid:0) (cid:1) ,B = = (cid:0) (cid:1) + 3 (cid:18) (cid:19) ,B = 10 1 4 91 10 1 44 1 10 19 4 1 10 = + 8 (cid:0) (cid:1) , = 20 1 4 9 161 20 1 4 94 1 20 1 49 4 1 20 116 9 4 1 20 = ++ 3 . Furthermore: B + I = A + 36 I = 36 1 4 9 16 251 36 1 4 9 164 1 36 1 4 99 4 1 36 1 416 9 4 1 36 125 16 9 4 1 36 == (cid:0) (cid:1) + 2 ++ 3 (cid:18) (cid:19) + 6 . To prove the second item, we assume that B = 35 1 4 9 16 251 35 1 4 9 164 1 35 1 4 99 4 1 35 1 416 9 4 1 35 125 16 9 4 1 35 has an integer completely positive factorization: B = U U T . Let u i , i = 1 , . . . , r ,denote the columns of U . First we note, that all columns of U need to be orthogonal16o w = (cid:0) − − − (cid:1) T by Proposition 4. Next we look at conditions thatare coming from the fact that the superdiagonal of B has all elements equal to1. In particular, this implies, that if two consecutive entries of any column of U are nonzero, they both have to be equal to 1. Moreover, since ( B ) = 1,one of the columns in U has to have the first two entries both equal to one, and,without loss of generality, we may assume that this holds for the first column of U : u = (cid:0) ∗ ∗ ∗ ∗ (cid:1) T . Taking into account all the conditions on u that wehave listed so far, u has to be equal to one of the following three vectors: v = (cid:0) (cid:1) T ,v ′ = (cid:0) (cid:1) T ,v ′′ = (cid:0) (cid:1) T . In the first case we have B − u u T = U U T = 34 0 3 8 15 240 34 0 3 8 153 0 34 0 3 88 3 0 34 0 315 8 3 0 34 024 15 8 3 0 34 . If u = v ′ , then some column of U (suppose u ) has 1 in the second and the thirdposition. This forces the first entry of u to be zero, and the condition w T u = 0leaves us with only one choice for u : u = (cid:0) (cid:1) T , thus B − u u T − u u T = U U T = 34 0 3 8 15 240 33 0 2 7 153 0 33 0 2 88 2 0 33 0 315 7 2 0 33 024 15 8 3 0 34 . If u = v ′′ , we notice that some column of U (suppose u ) has 1 in the lasttwo positions, and since the first two entries of u cannot both be equal to one, wededuce that u = (cid:0) (cid:1) T . Since ( B ) = 1, we still need a column,say u , in U that has 1 in both the second and the third position. Again, we are left17ith only one option: u = (cid:0) (cid:1) T , and B − u u T − u u T − u u T = U U T = 33 0 1 6 15 230 33 0 0 8 151 0 25 0 0 66 0 0 25 0 115 8 0 0 33 023 15 6 1 0 33 . In all three cases we are left with a matrix U i U Ti that has all the entries on thesuperdiagonal equal to zero. In particular, the columns of U i have at least everyother entry equal to zero. Moreover, one of the columns in U i has to have the firstand the third entry nonzero, so it is of the form: ˆ v = (cid:0) a b c d (cid:1) T , where cd = 0. Looking at the values of U i U Ti in positions (1 , , 5) and (2 , 6) we get ab ≤ bc ≤ 3, and bd ≤ 8. This, together with the condition w T ˆ v = 0, gives usonly one option for ˆ v : ˆ v = (cid:0) (cid:1) T . In the third case we also need bc = 0, so this choice immediately leads to contradic-tion. In the remaining two cases U i U Ti − ˆ v ˆ v T still has the (1 , U i , i = 1 , 2, of the same form. However, in bothcases U i U Ti − vv T has a negative entry in (3 , 5) position, leading to a contradiction. (cid:3) The ad hoc approach, that we used to find completely positive factorizations inthe proof above, is not well suited for generalisation to matrices of arbitrary size.Instead, we develop a more general technique, that can be applied to a class ofToeplitz matrices, but that does not produce the optimal q for distance matrices. Lemma 11. Let n ∈ N , i ∈ { , . . . , n − } , and E i := I + J in + J in + ... + J ( n − in + ( J Tn ) i + ( J Tn ) i + ... + ( J Tn ) ( n − i . Any matrix of the form P n − i =1 a i E i , a i ∈ N ∪ { } , has an integer completely positivefactorization.Proof. Clearly, it is sufficient to prove the statement for E i , i = 1 , . . . , n − 1. So,let us choose i ∈ { , . . . , n − } , and write n = qi + r where r ∈ { , . . . , i − } . Wedefine: U Ti = (cid:0) I i I i ... I i D r (cid:1) , where D r is an i × r (possibly null) matrix with 1’s on the main diagonal and 0’selsewhere. Equality E i = U i U Ti can be checked by a straightforward calculation. (cid:3) Example 12. The matrix of the form E + E + . . . + E n − = ( n − I + n − X i =1 τ ( i )( J in + ( J Tn ) i ) has an integer completely positive factorization, where τ ( i ) := P d | i d , the numberof all divisors of i . The previous example shows a straightforward application of Lemma 11, where a i are all chosen to be one. To determine the values for a i , i = 1 , . . . , n − 1, sothat the matrices A n and P n − i =1 a i E i agree outside the diagonal, we need the Jordantotient function: J ( k ) := k Y p | kp prime (1 − p ) . Theorem 13. The matrix A n + g J ( n ) I has an integer completely positive factoriza-tion for g J ( n ) := P n − k =1 J ( k ) .Proof. The claim will follow from Lemma 11, after we show that: A n + g J ( n ) I n = n − X i =1 J ( i ) E i . 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