Journal of Applied Mathematics

Volume 2012 (2012), Article ID 276386, 28 pages

doi:10.1155/2012/276386

Research Article

Primitive Zero-Symmetric Sign Pattern Matrices with Zero Diagonal Attaining the Maximum Base

Ling Zhang,¹ Ting-Zhu Huang,¹ and Zhongshan Li2,3

¹School of Mathematical Sciences, University of Electronic Science and Technology of China, Chengdu, Sichuan 61173, China

²Department of Mathematics, North University of China, Taiyuan, Shanxi 030051, China

³Department of Mathematics and Statistics, Georgia State University, Atlanta, GA 30302-4110, USA

Received 1 March 2012; Accepted 1 December 2012

Academic Editor: Nicola Guglielmi

Copyright © 2012 Ling Zhang et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Abstract.

The base set of primitive zero-symmetric sign pattern matrices with zero diagonal is . In this paper, the primitive zero-symmetric sign pattern matrices with zero diagonal attaining the maximal base are characterized.

1. Introduction

A sign pattern matrix (or sign pattern) is a matrix whose entries are from the set . Notice that for a square sign pattern matrix , in the computation of (the signs of) the entries of the power , an ambiguous sign may arise when a positive sign is added to a negative sign. So a new symbol was introduced in [1] to denote such an ambiguous sign. The powers of a square sign pattern have been investigated to some extent, see, for example, [1–12]. In [1], the set is defined as the generalized sign set and the matrices with entries in the set are called generalized sign pattern matrices, and the addition and multiplication involving the symbol are defined as follows:

From now on, we assume that all the matrix operations considered in this paper are operations of the matrices over the set .

Definition 1.1. square generalized sign pattern matrix is called powerful if each power of contains no entry.

In [1], Li et al. introduced the concepts of base and period for (powerful) sign pattern matrices which are the generalizations of the concepts of “index of convergence” and period for square nonnegative matrices. These concepts are extended from (powerful) sign pattern matrices to (square) generalized sign pattern matrices by You et al. in [12] as follows.

Definition 1.2 (see [12]). Let be a square generalized sign pattern matrix of order and the sequence of powers of . (Since there are only different generalized sign patterns of order , there must be repetitions in the sequence.) Suppose is the first power that is repeated in the sequence, that is, is the least positive integer such that holds for some positive integer . Then is called the generalized base (or simply base) of , and is denoted by . The least positive integer such that holds for is called the generalized period (or simply period) of and is denoted by .

For a sign pattern matrix , we use to denote the matrix obtained from by replacing each nonzero entry by 1.

A nonnegative square matrix is primitive if some power . The least such as is called the primitive exponent (or simply exponent) of , denoted by . For convenience, a square sign pattern matrix is called primitive if is primitive, and in this case we define .

Definition 1.3 (see [3]). A square sign pattern matrix is called zero-pattern symmetric (abbreviated zero-symmetric, or simply ZS) if is symmetric.

It is well known that graph-theoretical methods are often useful in the study of the powers of square matrices, so we now introduce some graph-theoretical concepts.

Let be a digraph with vertex set and arc set (which permits loops but no multiple arcs). By assigning a sign of 1 or −1 to each arc of the digraph , we obtain a signed digraph . By a walk in the digraph (or the signed digraph ), we mean a sequence of vertices () such that is an arc of for . The number is called the length of the walk , denoted by . If the vertices are distinct, the walk is called a path, if , the path is called a cycle in (and in ). The sign of the walk in  , denoted by , is defined to be where is the sign of the arc .

Let be a sign pattern matrix of order . Then the associated digraph of is defined to be the digraph with vertex set and arc set . The associated signed digraph is obtained from by assigning the sign of to each arc in .

Definition 1.4 (see [10]). Let be a digraph (permitting loops but no multiple arcs). Digraph is called primitive if there is a positive integer such that for all ordered pairs of vertices and (not necessarily distinct) in , there exists a walk of length from to . The least such is called the primitive exponent of , denoted by .

It is well known that a digraph is primitive if and only if is strongly connected and the greatest common divisor (gcd for short) of the lengths of all the cycles of is 1 (see [13]).

Definition 1.5 (see [3]). Let be a signed digraph of order . Then there is a sign pattern matrix of order whose associated signed digraph is . We say that is powerful if is powerful. Also we define .

We say that a sign pattern matrix has zero diagonal if for all . A digraph with vertex set and arc set is called symmetric provided that iff for all . It is clear that a sign pattern matrix is ZS iff its associated digraph is symmetric. For simplicity, we represent a symmetric (signed) digraph by its underlying graph.

The base set of primitive ZS sign pattern matrices and the base set of primitive ZS sign pattern matrices with zero diagonal are given, respectively, in [3, 10]. In [2], Cheng and Liu characterized the primitive ZS sign pattern matrices with the maximum base.

In this paper, we characterize the primitive sign pattern matrices with zero diagonal attaining the maximum base. Our main result is given in the following theorem.

Theorem 1.6. Let be an primitive zero-symmetric sign pattern matrix with zero diagonal. Then

and the equality holds if and only if is nonpowerful and skew symmetric, namely, for all , and the associated digraph is isomorphic to (see Figure 1).

Figure 1: The graph  . (All 2-cycles in are negative, is odd, ).

The proof of Theorem 1.6 will be given in Section 3.

2. Preliminary Results

In this section, we introduce some theorems, definitions, and lemmas which we need to use in the proof of our main result in Section 3.

In [1], Li et al. showed that if an irreducible sign pattern matrix is powerful, then . That is to say the study of the base for a primitive powerful sign pattern matrix is essentially the study of the base (i.e., exponent) for primitive matrices. Therefore, for a primitive powerful ZS sign pattern matrix with zero diagonal, Theorem 2.1 gives the base.

Theorem 2.1 (see [8]). Let be an primitive symmetric matrix with zero diagonal. Then and the primitive exponent set of primitive symmetric matrix with zero diagonal is , where is the set of odd numbers in .

Theorem 2.2 (see [10]). Let be an primitive ZS sign pattern matrix with zero diagonal. Then .

By Theorems 2.1 and 2.2, the sign pattern matrices with zero diagonal attaining this upper bound must be nonpowerful. So it remains to consider nonpowerful sign pattern matrices with zero diagonal.

Definition 2.3. Two walks and in a signed digraph are called a pair of SSSD walks, if they have the same initial vertex, same terminal vertex, and same length, but they have different signs.

Lemma 2.4 (see [8]). Let be a symmetric digraph. Then is primitive if and only if is strongly connected and there exists an odd cycle in .

Lemma 2.5 (see [1, 12]). If is a primitive signed digraph, then    is nonpowerful if and only if contains a pair of cycles and , with lengths and , respectively, satisfying one of the following conditions:

()  is odd and is even and ;() both and are odd and .

Lemma 2.6 (see [12]). Let be a primitive, nonpowerful signed digraph. Then we have the following.

(1)There is an integer such that there exists a pair of SSSD walks of length from each vertex to each vertex in .(2)If there exists a pair of SSSD walks of length from each vertex to each vertex , then there also exists a pair of SSSD walks of length from each vertex to each vertex in .(3)The minimal such (as in (1)) is just , the base of .

Lemma 2.7 (see [7]). Suppose that an sign pattern matrix is skew symmetric. Let be the associated signed digraph of . Let be an odd integer with (). Then if and only if is isomorphic to (see Figure 1).

3. Main Results

For an undirected walk of graph and two vertices , on , we denote by a shortest path from to on and by a shortest path from to on . For a cycle of , if and are two (not necessarily distinct) vertices on and is a path from to along , then denotes the path or cycle from to along obtained by deleting the edges of .

Lemma 3.1. Let be an primitive nonpowerful ZS sign pattern matrix with zero diagonal. If all the 2 cycles in are positive, then .

Proof. Since is primitive, it follows from Lemma 2.4 that is strongly connected and there is an odd cycle in such that . Since has zero diagonal, there are no loops in and so . Without loss of generality, we assume that is an odd cycle with the least length in .

Case  1. There exists at least one negative even cycle in .

Let be a negative even cycle in . Without loss of generality, we assume that is a negative even cycle with the least length in . Since all 2 cycles in are positive, .

Subcase  1.1. and have no common vertices.

Let be the shortest path from to . Suppose intersects at vertex and intersects at vertex and there are vertices on where . Let . Let and be any two (not necessarily distinct) vertices in . Suppose that is the shortest path from to and intersects at vertex and is the shortest path from to and intersects at vertex . Then ,  , where and . By the proof of Case  1 of Lemma  4.5 in [3], there exists a pair of SSSD walks from to of length . Therefore, by Lemma 2.6, .

Subcase  1.2. and have at least one common vertex.

Let . Let and be any two (not necessarily distinct) vertices in . Suppose that is the shortest path from to and intersects at vertex and is the shortest path from to and intersects at vertex . Denote by . Assume has vertices, where , then

If , then is a path with vertex set and edge set . In this case, if , then there exists another odd cycle with length , which contradicts our assumption that is an odd cycle with the least length in . Therefore, and if , then .

Subcase  1.2.1. and .

Subcase  1.2.1.1. . See Figure 2(a).

We consider two subcases: the subcase and the subcase .

First, we consider the subcase . Without loss of generality, we assume that . Let . If is even, set

Otherwise, set

Let

where is a positive 2-cycle that contains vertex . Since is odd, and have different parity. Therefore, both and are even. Then, if is even,

Otherwise,

Since both and are even, it follows that . We see that the pair , is a pair of SSSD walks with even length. Therefore, there exists a pair of SSSD walks from to with length .

Then, we consider the subcase . Without loss of generality, we assume that . Let

If is even, set

Otherwise, set

Let

where is a positive 2-cycle that contains vertex . Since is an odd cycle, and have different parity. Therefore, both and are even. Then, if is even,

Otherwise,

Since both and are even, it follows that . We see that the pair , is a pair of SSSD walks with even length. Therefore, there exists a pair of SSSD walks from to with length .

Subcase  1.2.1.2. . See Figure 2(b).

The proof for this subcase is similar to that of the Subcase and is omitted.

Subcase  1.2.1.3. . See Figure 2(c).

Let . If is even, set

Otherwise, set

Let

where is a positive 2-cycle that contains vertex . Since is odd, we see that both and are even. Then, if is even,

Otherwise,

Since both and are even, it follows that . We see that the pair , is a pair of SSSD walks with even length. Therefore, there exists a pair of SSSD walks from to with length .

Subcase  1.2.2. and .

Subcase  1.2.2.1. and . See Figure 3(a).

Without loss of generality, we assume that . Let . If is even, set

Otherwise, set

Let

where is a positive 2-cycle that contains vertex . Since is odd, and have different parity. Therefore, both and are even. Noting that and are paths of , we have ,  . Then

Since both and are even, it follows that . We see that the pair , is a pair of SSSD walks with even length. Therefore, there exists a pair of SSSD walks from to with length .

Subcase  1.2.2.2. and . See Figure 3(b).

Without loss of generality, we assume that . Then we consider two subcases: the subcase , and and the subcase and .

First, we consider the subcase , and . Let . If is even, set

Otherwise, set

Let

where is a positive 2-cycle that contains vertex . Since is odd, and have different parity. Therefore, both and are even. Noting that and are two paths of , then we have ,  . Then

Since both and are even, it follows that . We see that the pair , is a pair of SSSD walks with even length. Therefore, there exists a pair of SSSD walks from to with length .

Then, we consider the subcase and . Let . If is even, set

Otherwise, set

Let

where is a positive 2-cycle that contains vertex . Since is odd, and have different parity. Therefore, both and are even. Noting that and are two paths of , we have ,  . Then

Since both and are even, it follows that . We see that the pair , is a pair of SSSD walks with even length. Therefore, there exists a pair of SSSD walks from to with length .

Subcase  1.2.2.3. and . See Figure 3(c).

The proof for this subcase is similar to that of the Subcase and is omitted.

Subcase  1.2.2.4. and . See Figure 3(d).

Let . If is even, set

Otherwise, set

Let

where is a positive 2-cycle that contains vertex . Since is odd, and have different parity. Therefore, both and are even. Then if is even,

Otherwise,

Since both and are even, it follows that . We see that the pair , is a pair of SSSD walks with even length. Therefore, there exists a pair of SSSD walks from to with length .

Subcase  1.2.2.5. and . See Figure 3(e).

The proof for this subcase is similar to that of Subcase , so we omit it.

Subcase  1.2.2.6. and . See Figure 3(f).

Without loss of generality, we assume that . Let . If is even, set

Otherwise, set

Let

where is a positive 2-cycle that contains vertex . Since is odd, and have different parity. Therefore, both and are even. Noting that and are two paths of , then we have and . Then, if is even,

Otherwise,

Since both and are even, it follows that . We see that the pair , is a pair of SSSD walks with even length. Therefore, there exists a pair of SSSD walks from to with length .

Subcase  1.2.3. and . We assume that .

Subcase  1.2.3.1. and . See Figure 4(a).

Let . If is even, set

Otherwise, set

Let

where is a positive 2-cycle that contains vertex . Since is odd, both and are even. Then, if is even,

Otherwise,

Since both and are even, it follows that . We see that the pair , is a pair of SSSD walks with even length. Therefore, there exists a pair of SSSD walks from to with length .

Subcase  1.2.3.2. , , and . See Figure 4(b).

Let . If is even, set

Otherwise, set

Let

where is a positive 2-cycle that contains vertex . Since is odd, and have different parity. Therefore, both and are even. Noting that and are two paths of , we have ,  . Then,

Since both and are even, it follows that . We see that the pair , is a pair of SSSD walks with even length. Therefore, there exists a pair of SSSD walks from to with length .

Subcase  1.2.3.3. and . See Figure 4(c).

The proof for this subcase is similar to that of the Subcase and is omitted.

Thus in each of the above subcases, there exists a pair of SSSD walks from to with length . Therefore, we get by Lemma 2.6.

Case  2. There exists no negative even cycle in .

Since is primitive, nonpowerful and there exist no negative even cycles in , it follows from Lemma 2.5 that there exist two odd cycles and with different signs in . We assume that and . Since the trace of is zero, there are no loops in and so ,  . Without loss of generality, we assume .

Subcase  2.1. and have no common vertices.

Let be the shortest path from to . Suppose intersects at vertex and intersects at vertex and there are vertices on where . Let . Let and be two arbitrary (not necessarily distinct) vertices in . Suppose that is the shortest path from to and intersects at vertex and is the shortest path from to and intersects at vertex . Then

Since and have diffident signs, if there exists an even walk with length , then and have the same length and different signs. As the proof of Subcase  1.1, we can construct an even walk with length . So there exists a pair of SSSD walks from to of length . It remains to consider the following subcase.

Subcase  2.2. and have at least one common vertex.

Let . Let and be two arbitrary (not necessarily distinct) vertices in . Suppose that is the shortest path from to and intersects at vertex and is the shortest path from to and intersects at vertex . Denote by . Assume has vertices, where , then

If , then is a path with vertex set and edge set . If , since there exists no negative even cycles in , then and have the same sign, a contradiction. Therefore, .

Subcase  2.2.1. and . See Figure 3(a).

Without loss of generality, we assume that . Let . If is odd, set

Otherwise, set

Let

where is a positive 2-cycle that contains . Since is odd, and have different parity. Therefore, both and are even. Then, if is odd,

Otherwise,

Since both and are even, it follows that . We see that the pair , is a pair of SSSD walks with even length. Therefore, there exists a pair of SSSD walks from to with length .

Subcase  2.2.2. and . See Figure 3(b).

Let . If is odd, set

Otherwise, set

Let

where is a positive 2-cycle that contains . Since is odd, and have different parity, Therefore, both and are even. Then, if is odd,

Otherwise,

Since both and are even, it follows that . We see that the pair , is a pair of SSSD walks with even length. Therefore, there exists a pair of SSSD walks from to with length .

Subcase  2.2.3. and . See Figure 3(c).

Without loss of generality, we assume that . Let . If is odd, set

Otherwise, set

Let

where is a positive 2-cycle that contains . Since is odd, and have different parity. Therefore, both and are even. Noting that and are paths of , then we have ,  . Therefore,

Since both and are even, it follows that . We see that the pair , is a pair of SSSD walks with even length. Therefore, there exists a pair of SSSD walks from to with length .

Subcase  2.2.4. and . See Figure 3(d).

The proof for this subcase is similar to that of Subcase , so we omit it.

Subcase  2.2.5. and . See Figure 3(e).

The proof for this subcase is similar to that of Subcase , so we omit it.

Subcase  2.2.6. and . See Figure 3(f).

The proof for this subcase is similar to that of Subcase , so we omit it.

From all the above subcases, there exists a pair of SSSD walks from to with length . Therefore, we get by Lemma 2.6.

(a)

(b)

(c)

Figure 2: Illustrate for Subcase of Lemma 3.1.

(a)

(b)

(c)

(d)

(e)

(f)

Figure 3: Illustration for Subcases   and 2.2 of Lemma 3.1.

(a)

(b)

(c)

Figure 4: Illustrate for Subcase of Lemma 3.1.

Lemma 3.2. Let be an primitive nonpowerful ZS sign pattern matrix with zero diagonal. If there exist a negative 2-cycle and a positive 2-cycle in , then .

Proof. Since is primitive, it follows from Lemma 2.4 that is strongly connected and there is an odd cycle in with length . Since has zero diagonal, there are no loops in and so . Without loss of generality, we assume that is an odd cycle with the least length in . Since is ZS and contain a positive 2-cycle and a negative 2-cycle, there exists a positive 2-cycle and a negative 2-cycle such that . Let .

Let be the shortest path from to . Let and be two arbitrary (not necessarily distinct) vertices in . Suppose there are vertices on and intersects at , where . Suppose is the shortest path from to and intersects at and is the shortest path from to and intersects at where ,  . We consider the following three cases.

Case  1. and .

By the Subcase 2.1 of Lemma  4.2 in [3], there exists a pair of SSSD walks from to of length .

Case  2. Only one of and belongs to .

By the Subcase of Lemma  4.2 in [3], there exists a pair of SSSD walks from to of length .

It remains to consider the following case.

Case  3. and . See Figure 5.

Subcase  3.1. .

Let . If is even, set

Otherwise, set

Let

Since is odd, and have different parity. Therefore, both and are even. Then, if is odd,

Otherwise,

Since both and are even, it follows that . We see that the pair , is a pair of SSSD walks with even length. Therefore, there exists a pair of SSSD walks from to with length .

Subcase  3.2. .

Without loss of generality, we assume that . Let and . Let , If is odd, set

Otherwise, set

Let

Since is odd, and have different parity. Therefore, both and are even. Then, if is odd,

Otherwise,

Since both and are even, it follows that . We see that the pair , is a pair of SSSD walks with even length. Therefore, there exists a pair of SSSD walks from to with length .

Therefore, by Lemma 2.6.

Figure 5: Illustration for Case 3 of Lemma 3.2.

Lemma 3.3. Let be an primitive nonpowerful ZS sign pattern matrix with zero diagonal. If , then all 2-cycles in are negative.

Proof. Assume that . By Lemma 3.1, has at least one negative 2-cycle. It then follows from Lemma 3.2 that all 2-cycles in are negative.

Lemma 3.4. Let be an primitive nonpowerful ZS sign pattern matrix with zero diagonal. Suppose that . Then there exists an odd cycle in and is the only cycle of length at least 3 in .

Proof. Since is primitive, it follows from Lemma 2.4 that is strongly connected and there is an odd cycle in with length . Since has zero diagonal, there is no loop in and so . Consider the following two directed cycles:

Since there exists no positive 2-cycle in by Lemma 3.3, the arcs and have different signs. Thus by the fact that is odd. Without loss of generality, we assume that is an odd cycle with the least length in .

If , suppose there exists another cycle with length , we consider the following two cases.

Case  1. and have no common vertices.

Let be the shortest path from to . Suppose intersects at vertex and intersects at vertex and there are vertices on where . Let . Let and be two arbitrary (not necessarily distinct) vertices in . Suppose that is the shortest path from to and intersects at vertex and is the shortest path from to and intersects at vertex . Then

We consider the following six subcases.

Subcase  1.1. and . See Figure 6(a).

Let . If is odd, set

Otherwise, set

Let

Since is odd, and have different parity. Therefore, both and are even. Then, if is odd,

Otherwise,

Since both and are even, it follows that . We see that the pair , is a pair of SSSD walks with even length. Therefore, there exists a pair of SSSD walks from to with length .

Subcase  1.2. and . See Figure 6(b).

Let . If is odd, set

Otherwise, set

Let

Since is odd, and have different parity. Therefore, both and are even. Then, if is odd,

Otherwise,

Since both and are even, it follows that . We see that the pair , is a pair of SSSD walks with even length. Therefore, there exists a pair of SSSD walks from to with length .

Subcase  1.3. and . See Figure 6(c).

Let . If is odd, set

Otherwise, set

Let

Since is odd, and have different parity. Therefore, both and are even. Then, if is odd,

Otherwise,

Since both and are even, it follows that . We see that the pair , is a pair of SSSD walks with even length. Therefore, there exists a pair of SSSD walks from to with length .

Subcase  1.4. and . See Figure 6(d).

Without loss of generality, we assume that . Let . If is odd, set

Otherwise, set

Let

Therefore, both and are even. Then, if is odd,

Otherwise,

Since both and are even, it follows that . We see that the pair , is a pair of SSSD walks with even length. Therefore, there exists a pair of SSSD walks from to with length .

Subcase  1.5. and . See Figure 6(e).

Let , if is odd, set

Otherwise, set

Let

Since is odd, both and are even. Then, if is odd,

Otherwise,

Since both and are even, it follows that . We see that the pair , is a pair of SSSD walks with even length. Therefore, there exists a pair of SSSD walks from to with length .

Subcase  1.6. and . See Figure 6(f).

Let . If is odd, set

Otherwise, set

Let

Since is odd, both and are even. Then if is odd,

Otherwise,

Since both and are even, it follows that . We see that the pair , is a pair of SSSD walks with even length. Therefore, there exists a pair of SSSD walks from to with length .

Therefore, by Lemma 2.6, a contradiction.

Case  2. and have at least one common vertex.

Let . Let and be two arbitrary (not necessarily distinct) vertices in . Suppose that is the shortest path from to and intersects at vertex and is the shortest path from to and intersects at vertex . Denote by . Assume has vertices, where , then

If , then is a path with vertex set and edge set . If is odd, by the minimality of . If is even, suppose , then there exists an odd cycle with which contradicts our assumption that is an odd cycle with the least length in . Therefore, if , then . We consider the following three subcases.

Subcase  2.1. and .

Let . If is odd, set

Otherwise, set

Let

Since is odd, and have different parity. Therefore, both and are even. Then, if is odd,

Otherwise,

Since both and are even, it follows that . We see that the pair , is a pair of SSSD walks with even length. Therefore, there exists a pair of SSSD walks from to with length .

Subcase  2.2. and .

Without loss of generality, we assume that . Let . If odd, set

Otherwise, set

Let

Since is odd, and have different parity. Therefore, both and are even.

Then, if is odd,

Otherwise,

Since both and are even, it follows that . We see that the pair , is a pair of SSSD walks with even length. Therefore, there exists a pair of SSSD walks from to with length .

Subcase  2.3. and .

The verification for this subcase is similar to that of the Subcase 2.2 and is omitted.

Therefore, by Lemma 2.6, a contradiction.

(a)

(b)

(c)

(d)

(e)

(f)

Figure 6: Illustration for Lemma 3.4.

Lemma 3.5. Let be an primitive nonpowerful ZS sign pattern matrix with zero diagonal. If , then is isomorphic to (see Figure 1).

Proof. Let be an primitive nonpowerful ZS sign pattern matrix with zero diagonal. Since is primitive, it follows from Lemma 2.4 that is strongly connected and there is an odd cycle in with length . Since has zero diagonal, there is no loop in and so . Without loss of generality, we assume that is an odd cycle with the least length in . If , then by Lemma 3.3, there exists no positive 2-cycle in . Then is isomorphic to by Lemma 2.7. We will give anther proof of the theorem.

Denote the vertex set of by and the vertex set of by . By Lemmas 3.4 and 3.3, the cycle is the only cycle of length at least 3 in and there exists no positive 2-cycle in . Consider the two directed cycles and . Since there exists no positive 2-cycle in , the arcs and have different signs. Thus by the fact that is odd. Let and be two arbitrary (not necessarily distinct) vertices in . Suppose that is the shortest path from to and intersects at vertex and is the shortest path from to and intersects at vertex . If , is isomorphic to an odd cycle of length . If , it is enough to show that there exists a vertex in such that where is the shortest path from to and intersects at . Suppose not, then

Now we consider the following three cases.

Case  1. and .

Let . If is odd, set

Otherwise, set

Let

Since is odd, and have different parity. Therefore, both and are even. Then, if is odd,

Otherwise,

Since both and are even, it follows that . We see that the pair , is a pair of SSSD walks with even length. Therefore, there exists a pair of SSSD walks from to with length .

Case  2. and .

In this case, . Let . If is odd, set

Otherwise, set

Let

Since is odd, and have different parity. Therefore, both and are even. Then, if is odd,

Otherwise,

Since both and are even, it follows that . We see that the pair , is a pair of SSSD walks with even length. Therefore, there exists a pair of SSSD walks from to with length .

Case  3. and .

In this case, and . Let . If is odd, set

Otherwise, set

Let

Since is odd, and have different parity. Therefore, both and are even. Then, if is odd,

Otherwise,

Since both and are even, it follows that . We see that the pair , is a pair of SSSD walks with even length. Therefore, there exists a pair of SSSD walks from to with length .

From all the above cases, there exists a pair of SSSD walks from to with length , therefore, by Lemma 2.6, a contradiction.

Proof of Theorem 1.6. We get from Theorem 2.2. Suppose that is nonpowerful and skew symmetric and is isomorphic to . Then by Lemma 2.7.

Conversely, suppose that . We need to prove that is nonpowerful and skew symmetric and is isomorphic to . In [1], Li et al. showed that if an irreducible sign pattern matrix is powerful, then . Therefore, if is powerful, then by Theorem 2.1, which contradicts . Hence is nonpowerful. Consequently, is skew symmetric by Lemma 3.3 and is isomorphic to by Lemma 3.5.

Acknowledgments

This paper is supported by NSFC (61170311), Chinese Universities Specialized Research Fund for the Doctoral Program (20110185110020), and Sichuan Province Sci. & Tech. Research Project (12ZC1802).

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