Sufficient conditions to be exceptional

Abstract A copositive matrix A is said to be exceptional if it is not the sum of a positive semidefinite matrix and a nonnegative matrix. We show that with certain assumptions on A−1, especially on the diagonal entries, we can guarantee that a copositive matrix A is exceptional. We also show that the only 5-by-5 exceptional matrix with a hollow nonnegative inverse is the Horn matrix (up to positive diagonal congruence and permutation similarity).


Introduction
All of the matrices considered will be symmetric matrices with real entries. We will say a matrix is a nonnegative matrix if all of its entries are nonnegative, and likewise for a vector. A symmetric matrix A ∈ R n×n is positive semidefinite (positive definite) if x T Ax ≥ 0 for all x ∈ R n (x T Ax > 0 for all x ∈ R n , x ≠ 0). A symmetric matrix A ∈ R n×n is called copositive (strictly copositive) if x T Ax ≥ 0 for all x ∈ R n , x ≥ 0 (x T Ax > 0 for all x ∈ R n , x ≥ 0, x ≠ 0). We will let e i ∈ R n denote the vector with ith component one and all other components zero. A permutation matrix is an n-by-n matrix whose columns are e 1 , ..., en in some order. For n ≥ 2, an n-by-n matrix is said to irreducible [9] if under similarity by a permutation matrix, it cannot be written in the form with A 11 and A 22 square matrices of order less than n. We call an n-by-n matrix hollow if all of its diagonal entries are zero.

When the inverse is nonnegative and hollow
The results in this paper grew out of a question that arose from studying symmetric, nonnegative, hollow, invertible matrices in [4]. Theorem 1, despite its short proof and the fact that we will extend it in Section 3, is the core theorem of this paper. Theorem 1. Suppose A ∈ R n×n is symmetric, invertible, and that A −1 is nonnegative and hollow. If A is of the form A = P + N, with P positive semidefinite and N nonnegative, then P is zero.

Proof
The assumption e T i A −1 e i = 0, for all i, 1 ≤ i ≤ n, can be rewritten e T i A −1 AA −1 e i = 0. Then if A = P + N, i Px i = 0, for all i, 1 ≤ i ≤ n, but then Px i = 0, for all i, so P = 0. The conclusion of Theorem 1, stated as "For P nonzero, then A is not of the form P + N", is where our main interest lies. In this contrapositive form, we note that A being copositive is not an assumption of the theorem. Diananda [7] proved that for n = 3, and n = 4, copositivity coincides with being of the form P + N. So from Theorem 1 if A −1 is any 3-by-3 or 4-by-4 hollow, nonnegative matrix then A cannot be copositive with P nonzero when written as P + N. An example of a matrix meeting the hypotheses of Theorem 1 is A = Here, not only is A not of the form P + N, it is not copositive either (note the central 3-by-3 block).
A copositive matrix, known as the Horn matrix, is An example suggesting we cannot improve on Theorem 1 by having n − 1 zero diagonal It would also appear to be not possible to improve on Theorem 1 by A −1 having all zero diagonal entries and not requiring A −1 to be nonnegative, by considering and this is also of the form P + N.

Theorem 2. Suppose A ∈ R n×n is invertible. Both A and A −1 are nonnegative if and only if A is the product of a permutation matrix and a diagonal matrix with positive diagonal entries.
Since Theorem 1 is only concerned with symmetric matrices, Theorems 1 and 2 imply that the only way an invertible matrix A of the form A = P + N, can have all zeroes on the diagonal of its nonnegative inverse is if P = 0, n is even, and A consists of blocks on the diagonal of A, in which each diagonal block is a product of a symmetric permutation matrix and a positive diagonal matrix.
A simple observation is that if P is a positive semidefinite matrix and N is nonnegative, then A = P + N is a copositive matrix. It is well-known (see [7], [8], [10], [12]) that copositive matrices do not have to be of this form, an example of which is the 5-by-5 matrix H (from above) that we called the Horn matrix in [12]. In fact the Horn matrix is extreme [10], i.e. it cannot be written nontrivially as the convex sum of two copositive matrices. In [12] we called copositive matrices exceptional if they are not the trivial sum of a positive semidefinite matrix and a nonnegative matrix. Otherwise, we call them non-exceptional.
The proof of Theorem 3 will use the property proved in [11] (or see [13], [15]) that for any copositive matrix A, if x ≥ 0 and x T Ax = 0, then Ax ≥ 0. In [2], [3], Baumert studied copositive matrices that had a weak form of extremity, namely, copositive matrices that are not of the form C + N (nontrivially), in which C is copositive, and N is nonnegative with all zeroes on its diagonal. Baumert gave a characterization for such matrices in [1], which included an error, later corrected in [5]. In [5], the authors called such matrices irreducible with respect to the nonnegative cone. Obviously, if a matrix is not of the form C + N, then it is not of the form P + N. For Theorem 3 we need the assumption that n ≥ 3, since in the proof we will write A −1 in block form with a specified (1, 2) entry, as well as another nonzero column to the right of it.
Theorem 3. For n ≥ 3, suppose that A ∈ R n×n is symmetric, irreducible, invertible, and A −1 is nonnegative and hollow. If A is of the form C + N, in which C is copositive and N is nonnegative and hollow, then N is zero.
Proof Our method of proof will be to show, with the stated assumptions, that if A = C + N, we must have that N is diagonal and therefore N = 0.
We proceed now to show that N is diagonal. Choose a permutation matrix R, so that if N has a nonzero off-diagonal entry n ij , we have n ij in the (1, 2) position of R T NR. In other words, we may assume n 12 )︃ , with B 1 as a 2-by-2 matrix and the other blocks of conforming dimensions.
Next, let Q be the permutation matrix given by Q = permutation matrix chosen so that However, N 1 b is the (n − 1)-by-1 matrix with first two components n 11 b 1n + n 12 b 2n + · · · = 0 and n 12 b 1n + n 22 b 2n +· · · = 0. Since all entries of N 1 and b are nonnegative, this forces n 12 = 0, which is a contradiction.
Thus, the only way a copositive matrix A can satisfy the assumptions of Theorem 3 is for A to be "irreducible with respect to the nonnegative hollow cone". Again, the Horn matrix provides an example of such a matrix.

Extending Theorem 1
Our next theorem (and its proof) reduces to Theorem 1 when the matrix B of Theorem 4 is the identity matrix. Theorem 4 improves on Theorem 1, since the signs of the entries, including the diagonal entries, of A −1 are not restricted to being nonnegative. This may be seen from the examples of exceptional matrices from [11] and [12] following the theorem.

Theorem 4. Let A ∈ R n×n be symmetric and invertible. Suppose there exists an invertible matrix B ∈ R n×n such that A −1 B is nonnegative, and B T A −1 B is hollow. If A is of the form A = P + N, with P positive semidefinite and N nonnegative, then P is zero. Moreover, whether or not A is of the form P + N, if A is copositive then B is nonnegative.
Proof Suppose A can be written as A = P + N, with P positive semidefinite and N nonnegative. Then, with the assumptions on the matrix B, and letting This implies for each i, 0 = e T i C T PCe i . Then PCe i = 0 for all i, so P = 0. For the "Moreover" part of the statement of the theorem, since for each i we have e T i C T ACe i = 0, and A is copositive, then ACe i ≥ 0, from the property of copositive matrices stated in Section 2. Therefore B = AC ≥ 0.
An example of a matrix A to illustrate Theorem 4 is the Hoffman-Pereira matrix [11], as we called it in [12], which is copositive. This exceptional A along with its inverse is Another illustration of the same theorem is the 7-by-7 extension of the Horn matrix given in [12], which is the exceptional matrix A, along with A −1 given by Using similar reasoning to that given in Theorem 8 of [12] we also have Theorem 5. Proof Suppose 0 = e T i A −1 e i , for i = 1, 2, 3. Then, as in the proof of Theorem 1, we have when i = 1 that 0 = e T 1 A −1 NA −1 e 1 , which means that the (n − 1) -by-(n − 1) block of N obtained by deleting row and column 1 is zero. Arguing in the same way for i = 2, and i = 3, we have that N = 0.

The 5-by-5 case
In this section, we will use a theorem from [5], which we state as Theorem 6, to show that the only 5-by-5 exceptional matrix with a hollow nonnegative inverse is the Horn matrix, up to positive diagonal congruence and permutation similarity. Let −cos θ 1 cos(θ 1 + θ 2 ) cos(θ 4 + θ 5 ) −cos θ 5 −cos θ 1 1 −cos θ 2 cos (θ 2 + θ 3 ) cos(θ 5 + θ 1 ) cos(θ 1 + θ 2 ) −cos θ 2 1 −cos θ 3 cos(θ 3 + θ 4 ) cos(θ 4 + θ 5 ) cos(θ 2 + θ 3 ) −cos θ 3 1 −cos θ 4 −cos θ 5 cos(θ 5 + θ 1 ) cos(θ 3 + θ 4 ) −cos θ 4 1 Theorem 6 appears at the end of [5], where they use C 5 , S 5 + and N 5 , respectively, to denote the copositive, positive semidefinite, and nonnegative matrices, in R 5×5 . Theorem 6. Let A ∈ C 5 − (S 5 + + N 5 ). Then, up to permutation similarity and positive diagonal congruence, A can be written as A = S + N, for some hollow N ∈ N 5 , where θ i ≥ 0, for 1 ≤ i ≤ 5, and Let now A be a 5-by-5 exceptional matrix that has a hollow nonnegative inverse. Theorem 6 implies that, up to permutation similarity and positive diagonal congruence, A can be written as A = S + N, where N is hollow and nonnegative. We would like to apply Theorem 3, but we need to first check that A is irreducible. If A is reducible, it is permutation similar to a matrix with irreducible diagonal blocks. We note that if A is reducible this does not necessarily imply S is reducible. If A had a 1-by-1 diagonal block (under permutation similarity), then its inverse could not be hollow. If A had a 2-by-2 diagonal block, then this 2-by-2 block, when inverted, must be nonnegative with both diagonal entries being zero. Then the (not inverted) 2-by-2 block of A would also be nonnegative with both diagonal entries being zero, but S has all ones on the diagonal, in which case we could not have A = S + N (under permutation similarity or positive diagonal congruence). Now applying Theorem 3, since A has a hollow nonnegative inverse, we know that N = 0. We next determine the values of the θ i 's, for 1 ≤ i ≤ 5, that ensure S has a hollow inverse. In effect, we will show that the θ i 's are all equal to zero, whereupon S becomes the Horn matrix. Let us examine the 4-by-4 principal minors of S. A computer algebra system can used to show that the top left 4-by-4 principal minor of S, namely det(S [1,2,3,4]), satisfies det(S[1, 2, 3, 4]) = − [︁ cos(θ 1 + θ 2 + θ 3 ) + cos(θ 4 + θ 5 ) ]︁ 2 sin 2 θ 2 .