aa r X i v : . [ m a t h . SP ] S e p THE INTEGER CP-RANK OF × MATRICES
THOMAS LAFFEY, HELENA ˇSMIGOC
Abstract.
We show the cp-rank of an integer doubly nonnegative2 × Key words: completely positive matrices, doubly nonnegative ma-trices, integer matrices. MSC[2010] 15B36, 15B48.
Dedicated to Charles R. Johnson. Introduction A n × n matrix A is said to be completely positive , if there existsa (not necessarily square) nonnegative matrix V such that A = V V T .Completely positive matrices have been widely studied, and they playan important role in various applications. It is a subject to which C.R.Johnson has made an important contribution [10, 9, 8, 7]. For furtherbackground on completely positive matrices, we refer the reader to thefollowing works and citations therein [3, 2, 5, 4].Clearly, any completely positive matrix is nonnegative and positivesemidefinite. We call the family of matrices that are both nonnegativeand positive semidefinite doubly nonnegative . Doubly nonnegative ma-trices of order less than 5 are completely positive [14]. However, thisis no longer true for matrices of order larger than or equal to 6 [12].Any n × n completely positive matrix A has many cp-factorizations of the form A = V V T , where V is an n × m matrix. Note that m is also not unique. We define the cp-rank of A to be the minimalpossible m . If we demand that V has rational entries, then we saythat A has a rational cp-factorization . We define the rational cp-rank correspondingly. In this note we will study integer cp-factorizations ,where we demand V to be an integer nonnegative matrix, and integercp-rank , the minimal number of columns in the integer cp-factorizationof a given matrix.Every rational matrix which lies in the interior of the cone of com-pletely positive matrices has a rational cp-factorization [11], but thequestion is still open for rational matrices on the boundary of the re-gion. On the other hand, for n ≥ n × n integer completely positive matrices that do not have an integercp-factorization. In [13] the authors answered a question posed in [1],by proving that for n = 2 every integer doubly nonnegative matrix hasan integer cp-factorization. An alternative proof of this result can be found in [6]. Neither of those proofs offer a bound on the integer cp-rank of such matrices. In this note we prove that the integer cp-rankof 2 × Main Result
The question of determining the completely positive integer rank fora given n × n completely positive matrix is not trivial, even in the casewhen n = 1. In this case the answer is given by Lagrange’s Four-SquareTheorem. Theorem 2.1. [Lagrange’s Four Square Theorem] Every positive inte-ger x can be written as the sum of at most four squares. If x is not ofthe form (1) x = 4 r (8 k + 7) for some nonnegative integers r and k , then x is the sum of at mostthree squares. With Theorem 2.1, rank one matrices are easy to analyse.
Lemma 2.1.
An integer doubly nonnegative matrix A of rank iscompletely positive, and has the integer cp-rank equal to the integercp-rank of the greatest common divisor of its diagonal elements.Proof. Let A = ( a ij ) be an integer doubly nonnegative matrix of rank1, and let d := gcd( a , a , . . . , a nn ). Since a ij = √ a ii a jj , d also dividesall the off-diagonal elements of A . Hence, A = dB , where B = ( b ij )satisfies gcd( b , b , . . . , b nn ) = 1.We claim that each diagonal element in B is a perfect square. If thisis not true, then b i i = pc i , where p is a product of distinct primes.Now b i j = b i i b jj = pc i b jj . Hence b jj = pc j for some positive integer c j and for all j . This implies that p divides b jj for all j . As thiscontradicts our assumption, the claim is proved. (cid:3) Proposition below not only gives the first bound on the integer cp-rank of 2 × Proposition 2.1. A × integer doubly nonnegative matrix has aninteger cp-factorization, and an integer cp-rank less than or equal to .Proof. Let A = (cid:18) a bb c (cid:19) be an integer doubly nonnegative matrix. First we prove that we canreduce our problem to the case when a ≥ b and c ≥ b . To this end we HE INTEGER CP-RANK OF 2 × assume b > c , and write b = αc + b , where α ≥ b ∈ { , . . . , c − } . Let S ( α ) := (cid:18) − α (cid:19) . We claim that A = S ( α ) AS ( α ) T = (cid:18) a − αb + α c b − αcb − αc c (cid:19) = (cid:18) a b b c (cid:19) is a doubly nonnegative matrix. From det S ( α ) = 1, we deduce det A =det A ≥
0. Now, the inequalities det A = a c − b ≥ c = c > a = ( a − αb − αc ) ≥ S ( α ) − >
0, any completely positive factorization of A : A = B B T , gives us a completely positive factorization of A : A = ( S ( α ) − B )( S ( α ) − B ) T . Clearly, B and ( S ( α ) − B ) have the same number of columns, hencethe two factorizations give the same bound on the cp-rank. With thiswe have proved, that to find an integer completely positive factorizationfor A it is sufficient to solve the problem for A , that satisfies a ≥ b and c ≥ b . If b > a , we can repeat the above argument, with theroles of the diagonal elements reversed.From now on we may assume that our given matrix A satisfies a ≥ b and c ≥ b . Under this assumption we can write: A = (cid:18) b bb b (cid:19) + (cid:18) a − b c − b (cid:19) (2) = b (cid:18) (cid:19) + (cid:18) a − b
00 0 (cid:19) + (cid:18) c − b (cid:19) . (3)By Lemma 2.1 each of the rank 1 matrices in the above sum have theinteger cp-rank at most 4, so the integer cp-rank of A is at most 12. (cid:3) To reduce the bound for the cp-rank to 11 we look more closely atthe family of integers that cannot be written as a sum of less than foursquares.
Lemma 2.2.
Let x be a positive integer of the form (1) . Then x − , x − , x + 2 and x + 6 are not of the form (1) .Proof. Let x = 4 r (8 k + 7) for some nonnegative integers r and k . Then: x ≡ r = 0 ,x ≡ r = 1 ,x ≡ r ≥ . THOMAS LAFFEY, HELENA ˇSMIGOC
In each case, it is straightforward to check that x − x − x + 2, x + 6are not equivalent to 7, 4 or 0 modulo 8, so they cannot be of the form(1). (cid:3) Theorem 2.2.
Let A = (cid:18) a bb c (cid:19) be an integer doubly nonnegative matrix. Then A has an integer cp-factorization, and an integer cp-rank less than or equal to .Proof. From (2) and Theorem 2.1 it is clear that the bound 12 will notbe reached unless b , a − b and c − b are all of the form (1). In particular,the result holds for b ≤
6, and for a − b ≤ c − b ≤
6. So we assume b , a − b and c − b are all greater than or equal to 7, and that they allrequire four squares in Theorem 2.1.First let us consider the case when a − b a − b − ≡ a − b − ≡ a − b − A = (cid:18) a − b − c − b + 2 (cid:19) + ( b − (cid:18) (cid:19) + (cid:18) (cid:19) (cid:0) (cid:1) . Under our assumption, we can write each a − b − c − b + 2 and b − A is at most 3 + 3 + 3 + 1 = 10. The case,when c − b a − b ≡ c − b ≡ A = (cid:18) a − b − c − b + 2 (cid:19) + ( b − (cid:18) (cid:19) + (cid:18) (cid:19) (cid:0) (cid:1) . Since a − b − ≡ c − b + 2and b − A is at most 3 + 3 + 3 + 1 = 10. (cid:3) Next example shows that the integer cp-rank of a 2 × × Example 2.1.
Let A = (cid:18) a c (cid:19) , where a and c are positive integers. Then any integer cp-factorizationof A must involve (cid:18) (cid:19) (cid:0) (cid:1) , so the integer cp-rank of A is p + q ,where p , q are the least number of squares of integers needed to represent a − , c − , respectively. In particular, A has the integer cp-rank if a and c are both divisible by . HE INTEGER CP-RANK OF 2 × Example 2.2.
Let B = (cid:18) a c (cid:19) with integers c ≥ a ≥ . The decomposition B = (cid:18) a − c − (cid:19) + 2 (cid:18) (cid:19) shows that the integer cp-rank of B is at most , unlessboth a − and c − are of the form (1) . But if a − is of the form (1) a − is not, by Lemma 2.2. In this case the decomposition B = (cid:18) a − c − (cid:19) + (cid:18) (cid:19) (cid:0) (cid:1) shows that the integer cp-rank of B is at most . Weconclude that the integer cp-rank of all such B is at most . References [1] Abraham Berman. Completely positive matrices – real, rational and integral.
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