Abstract
An matrix is called an -matrix if all its specified principal minors are nonpositive. In the context of partial matrices, a partial matrix is called a partial -matrix if all its specified principal minors are nonpositive. In this paper we characterize the existence of an -matrix completion of a partial -matrix whose associated graph is a directed cycle.
1. Introduction
A partial matrix is a rectangular array, some of whose entries are specified while the remaining unspecified entries are free to be chosen (from a certain set). In this paper we are going to work on the set of the real numbers and to assume that all diagonal entries are prescribed. A completion of a partial matrix is the matrix resulting from a particular choice of values for the unspecified entries. A completion problem asks if we can obtain a completion of a partial matrix with some prescribed properties.
The technics to obtain this completion depend on the pattern of the partial matrix which can be combinatorially symmetric (i.e., is specified if and only if is) or noncombinatorially symmetric. Here we are going to work with this second class of partial matrices.
A natural way to describe an partial matrix is via a graph , where the set of vertices is and there is an arc from to if and only if position of is specified. In general, a directed graph (resp., nondirected graph) is associated with a noncombinatorially symmetric (resp., combinatorially symmetric) partial matrix. Since all main diagonal entries are specified we omit loops.
A cycle in a directed graph is a sequence of arcs , where for all .
In the last years many completions problems have been analysed. The completion problem for partial -matrices, -matrices, -matrices, , has been studied by Johnson, Hogben, Urbano, Mendes, , among others. See, for instance [1–9] and the references therein.
As a class of square real matrices that contains the -matrices we define the -matrices, real matrices , where all its principal minors are nonpositive. Since -matrices are preserved by principal submatrices we define a partial -matrix as a partial matrix whose completely specified principal submatrices are -matrices.
In general it is not always true that a partial -matrix has an -matrix completion as the following matrix shows (see [7]): is a partial -matrix that has an -matrix completion that leads us to analyze the -matrix completion problem depending on the pattern of the partial matrix. We have studied in [7] when a combinatorially symmetric partial -matrix with no null main diagonal entries such that the graph of its specified entries is a 1-chordal graph or a cycle has an -matrix completion.
In this paper we study the mentioned problem for partial -matrices that can have zeros at the main diagonal and whose associated graph is a directed cycle of length equal to the order of the matrix. In this case we may suppose without loss of generality that these matrices have the form: where denotes a specified entry and an unspecified one.
Observe that when we say “, , where the subscripts are expressed module ”, we are using the congruence module ; that is, we are considering the entries .
Given a matrix of size the submatrix lying in rows and columns , , is denoted by and the principal submatrix is abbreviated to .
We denote if and if .
In the next section we introduce necessary and sufficient conditions in order to guarantee the existence of an -matrix completion of a partial -matrix whose associated graph is a directed cycle.
2. Completion of Partial -Matrices
It is easy to prove that -matrices as well as partial -matrices satisfy the following properties.
Proposition 1. Let be an -matrix. (1)If is a permutation matrix, then is an -matrix. (2)If is a positive diagonal matrix, then and are -matrices. (3)If is a nonsingular diagonal matrix, then is an -matrix. (4)If , , then , for all , . (5)Every principal submatrix of is an -matrix.
In [7] the authors proved that any -matrix with no null diagonal entries is diagonally similar to an -matrix in the set: But since there are -matrices with some entries equal to zero we need to introduce the following set:
We also extend the definition of matrices to partial matrices; that is, consists of the partial matrices such that if then , for all specified entry , , . The following matrix is an example of a matrix of ,
The following results, consequence of Proposition 1, allow us to transform a partial -matrix , whose associated graph is a directed cycle, into a matrix whose diagonal nonzero values are −1; the nonzero elements of the first upperdiagonal are 1 and the element in position is .
Proposition 2. Let be an partial -matrix. There exists a positive diagonal matrix such that matrix is also an -partial matrix with equal to −1 or to zero, for all .
Proof. It suffices to consider defined if and if .
Proposition 3. Let be an partial -matrix, whose associated graph is a directed cycle, such that is −1 or 0, for all . Then there exists a diagonal matrix such that is a partial -matrix with for all , or zero for all and .
Proof. It suffices to consider .
Therefore, if is an partial -matrix whose associated graph is a directed cycle, we will assume, without loss of generality, that has the following structure: −1 or zeros on the main diagonal, 1’s or zeros in the first upper diagonal and in position .
The following theorem characterizes the matrices as an intermediate step to obtain the desired completion. It can be easily obtained from the transformations of Propositions 2 and 3.
Theorem 4. Let be an partial matrix, even (resp., odd), whose associated graph is a directed cycle. If all entries , , where the indices are expressed module , are nonzero the matrix is diagonally similar to an element of if and only if (resp., ).
Now we analyze the existence of an -matrix completion of a partial -matrix with an associated directed cycle, by distinguishing between matrices with no null main diagonal entries and matrices with some null values in the main diagonal.
Theorem 5. Let be an partial -matrix, with nonzero main diagonal entries such that its associated graph is a directed cycle. The following statements are equivalent: (1) if is even ( if is odd),(2) is diagonally similar to an element of ,(3)there exists an -matrix completion of .
Proof. Observe that from (4) of Proposition 1, we have that all the specified entries are nonzero. Then, from commentary after Proposition 3, we assume that all the elements in the main diagonal are −1 and the first upper diagonal is formed by 1’s.
Let us suppose that is even; the case odd is analogous. Since the upper diagonal and the element in position are nonzero, by applying Theorem 4, the condition is equivalent to item 2.
Now, we assume that the second statement is true. We consider , where if is a specified value of , for , where subscripts are expressed module and . Then is an partial -matrix, with nonzero main diagonal entries such that its associated graph is a nondirected cycle. Theorem 4.3 of [7] assures that , and therefore has an -matrix completion.
Finally, from the note after Proposition 1, the third statement implies the second one.
Now, it arises the question about establishing an analogous result to Theorem 5, when zero entries appear in the main diagonal. The answer is negative since if we admit a zero diagonal element and a zero entry in the upper diagonal, there exist matrices in , is even, such that is negative, but that admits an -matrix completion. For example, matrix defined is diagonally similar to an element of by using and it has an -matrix completion, , although is negative,
The following results characterize this type of matrices. Note that, if there are some null main diagonal entries, the existence of an -matrix completion implies that if , then . So, we add this condition as a hypothesis. In addition, recall that we are going to assume that or zero for all ; the entries in the first upper diagonal are 1 or zero and the value of the element in position is .
Theorem 6. Let be an , even (resp., odd), , partial -matrix with some null main diagonal entries, whose associated graph is a directed cycle. Let one suppose that if for all , where the indices are expressed module , then . If there exists , , where the indices are expressed module , then there exists an -matrix completion.
Proof. Let be, with for all , the corresponding indices to the negative diagonal values of matrix and , where and for all , the corresponding ones to the zero diagonal entries. Since there is , , such that , the diagonal similarity allows us to assume, without loss of generality, that if is even and that if is odd. Let be the matrix where is the permutation matrix , being the canonical vector for all . Consider partitioned in a block matrix, where is of size and is .
Note that elements of the first upper diagonal , , are moved to blocks , , depending on the value of and : if both entries are zero, after the permutation will be in ; if and the element will be in ; if and then will be in and in other cases will be in . This is shown in Table 1(a). In Table 1(b) we can see the position that the element of will occupy after the permutation.
Then we can be sure that each of the first lines of the permutated matrix will have as maximum a nonzero value and the last ones will have exactly one −1 and only another nonzero value as maximum.
We complete with zeros the unspecified entries of blocks , , and . In order to complete we distinguish two cases.(a)The element of is at position . If and , then and, after the permutation, it will be in submatrix . If then position of the permutated matrix will be unspecified. Now we partially complete as follows: and for all , and by Theorem 5 we get that there exists an -completion of named .(b)The element of is not at position . We complete with 1’s and −1’s in order to obtain a matrix with all their diagonals formed alternatively by 1’s and −1’s. All the principal minors of this new matrix will be zero.
The completion of , , is an -matrix since (a)the principal minors lying rows and columns with indices in are zero, as we can prove by developing by the nonzero elements (there is as maximum one nonzero entry by line);(b)the principal minors lying rows and columns with indices in are less than or equal to zero, since is an -matrix completion either of is at position or not;(c)the principal minors with rows and columns with indices in and are zero, because of they have a row of zeros or, as before, by developing by the only nonzero element of each row, the minor is reduced to one of submatrix .
So, matrix has an -matrix completion.
Theorem 7. Let be an , even (resp., odd), , partial -matrix with some null main diagonal entries, whose associated graph is a directed cycle. Let one suppose that if for all where the indices are expressed module , then . If for all , where the indices are expressed module , then the condition for even (resp., for odd) is necessary and sufficient for the existence of an -matrix completion.
Proof. Let be even. If some changes in the proof of the above theorem gives the desired completion. Specifically, choose the zero diagonal entry with less index. If since matrix can be transformed by diagonal similarity in a matrix such that if is even, and are equal to a positive value and if is odd, and are equal to a negative value. By permutation similarity we get a block-matrix analogous to the previous result and by a similar reasoning we get that there exists an -matrix completion of . If we transform by the permutation , where is the canonical vector for all and we proceed in a similar way to the previous result.
Now, we are going to prove the necessity of the condition by induction on ; that is, if there exists an -completion of , , then if is even and if is odd. We denote .
For , if we suppose that by analyzing the seven different cases that arise depending on the number of zeros, one, two, or three in the main diagonal, we get that , that is a contradiction. Then .
Let be. Suppose is even (if is odd the process is analogous). We are showing that in the mentioned conditions if there exists an -completion of , then . Let be an -completion of .
Let us suppose by hypothesis of induction (HI) that the statement is true for all . Since is an -matrix, we obtain, from the principal minors, that all the entries of the first under diagonal are greater than or equal to zero; that is, for all .
In addition, if we consider with , , we get that if is odd and if is even for , , .
From the nonpositivity of for all , , we obtain that the upper diagonals follow the same rule of signs of the under diagonals, alternatively positive and negative, but with the option of zero; that is, if is odd and if is even for , , .
Now we study the case in which there exists such that .(a)If we consider . The submatrix of can be considered as a completion of a partial matrix of size strictly smaller than with all the first upper diagonal formed by 1’s. If has at least a zero diagonal entry, taking into account that if is odd and if is even, the hypothesis of induction allows us to assure that the entry in position of ; that is, is positive. In the other case, if all the diagonal entries are nonzero we get the same conclusion by Theorem 5. This ends the proof in this case.(b)If and with , we analyze two cases: (b.1) or and (b.2) and . In the first one, we get from . If , and and for all we get from . If , and there exists such that we get also the same result from .
In this case for all if some entry of the upper triangular part of is nonzero, we also obtain that by using and HI.
Therefore, it remains to analyze the case in which all the upper triangular part except the first upper diagonal is zero. As we will see now, most of the cases can not be given.
Let be the nonzero diagonal entry with less index. If from or if and from we get a contradiction. Now we study the remaining cases depending on the values of and .
If from we get (if to show it we also use ).
If and from we obtain (if to show it we also use and if ). If , and from we get (if to show it we also use and if ); if and leads a contradiction.
If and from we get (if to show it we also use and and if , the nonpositivity of . If , and from and we get . In this case if the nonpositivity of and leads to a contradiction.
Finally, if and we get a contradiction by using and also if , or if the nonpositivity of and .
We sum up the results of Theorems 6 and 7 in Table 2. One can consider a similar one for odd.
Conflict of Interests
The authors declare that there is no conflict of interests regarding the publication of this paper.
Acknowledgments
The authors would like to render thanks to the anonymous reviewers for their comments and suggestions that have improved the readability of the paper. This paper has been partially supported by Ministerio de Ciencia y Tecnologia MTM2011-28636-C02-02.