Mixed graphs with cut vertices having exactly two positive eigenvalues
aa r X i v : . [ m a t h . C O ] F e b Mixed graphs with cut vertices having exactly two positiveeigenvalues
Xiaocong He, Lihua Feng ∗ School of Mathematics and Statistics, HNP-LAMA, Central South University, Changsha, Hunan410083, P. R. China
Abstract : A mixed graph is obtained by orienting some edges of a simple graph. The positive inertiaindex of a mixed graph is defined as the number of positive eigenvalues of its Hermitian adjacency matrix,including multiplicities. This matrix was introduced by Liu and Li, independently by Guo and Mohar,in the study of graph energy. Recently, Yuan et al. characterized the mixed graphs with exactly onepositive eigenvalue. In this paper, we study the positive inertia indices of mixed graphs and characterizethe mixed graphs with cut vertices having positive inertia index 2.
Keywords : Mixed graph; Cut vertices; Hermitian adjacency matrix; Positive inertia index
We will start with introducing some background that will lead to our main results. Some significantpreviously established facts will also be presented.
Let G be a simple graph with vertex set V ( G ) = { v , v , . . . , v n } and edge set E ( G ). A mixed graph e G is obtained by orienting some edges of G . Accordingly, G is referred to as the underlying graph of e G . Consequently, a mixed graph e G is an ordered triple ( V ( G ) , E , E ), where E is the undirected edgeset and E is the directed edge (arc) set. It is obvious that e G is undirected if E = E ( G ) while e G isan oriented graph if E = ∅ . Thus, mixed graphs are the generalizations of simple graphs and orientedgraphs.The Hermitian adjacency matrix H ( e G ) = ( h st ) n × n of e G was proposed, independently, by Liu andLi [28], and Guo and Mohar [13]. It is defined as h st = , if v s v t is an undirected edge; i, if −−→ v s v t is an arc; − i, if −−→ v t v s is an arc;0 , otherwise,where i = √−
1. Note that H ( e G ) is Hermitian, that is, H ( e G ) ∗ = H ( e G ), where H ( e G ) ∗ denotes theconjugate transpose of H ( e G ). Thus the eigenvalues of H ( e G ) are real, called the H -eigenvalues of e G .The H -rank of e G , denoted by rk ( e G ), is the rank of H ( e G ), and the number of positive, negative andzero eigenvalues of H ( e G ) are defined as positive inertia index , negative inertia index and nullity of e G ,denoted by p ( e G ), n ( e G ) and η ( e G ), respectively. Obviously, p ( e G ) + n ( e G ) = rk ( e G ). The inertia index of e G is the triple In( e G ) = ( p ( e G ) , n ( e G ) , η ( e G )). Clearly, the Hermitian adjacency matrix of a mixed graph isa natural generalization of the adjacency matrix of a simple graph. Therefore the mixed graph attractsmore and more researchers’ attention; see [13, 21, 24, 25, 28, 35, 36, 52, 51, 50, 59, 60].In [13, 35], the authors considered the so-called four-way switching operation keeping the H -spectrumof a mixed graph unchanged. We describe the definition in the matrix-theoretic language below. Definition 1. [35] Let f G and f G be two mixed graphs. A four-way switching (FWS for short) is theoperation of changing f G into f G , if there exists a diagonal matrix Q with Q vv ∈ {± , ± i } such that H ( f G ) = Q − H ( f G ) Q . ∗ Corresponding author.
Email addresses : [email protected] (X.C. He), [email protected] (L.H. Feng)
1n fact, it is often difficult to check whether a mixed graph f G can be obtained from a mixed graph f G by a four-way switching. Because of this, Mohar [35] gave some special cases of four-way switchingcalled two-way switching , which can be described as a special similarity transformation. There are reallytwo different cases as follows.(1) Directed two-way switching . Reversing all edges in an edge cut consisting of directed edges only.(2)
Mixed two-way switching . If there is an edge cut which contains undirected edges (possibly empty)and directed edges (possibly empty) in one direction only, replacing the each directed edge in thecut by a single undirected edge, and replacing each former undirected edge by a single directed edgein the direction opposite to the direction of former directed edges.The converse e G ⊤ of a mixed graph e G is obtained from e G by reversing all directed edges in e G . Theyare cospectral since H ( e G ⊤ ) = H ( e G ) ⊤ . Two mixed graphs f G and f G are said to be switching equivalentif f G can be changed into f G by a FWS and taking a possible converse.The circle group, which is denoted by T = { z ∈ C : | z | = 1 } , is a subgroup of the multiplicative groupof all nonzero complex numbers C × . If v s v t ∈ E ( G ), we denote by e v s v t the oriented edge from v s to v t .Let −→ E be the set of oriented edges of G , i.e., −→ E = { e v s v t , e v t v s : v s v t ∈ E ( G ) } . A complex unit gain graph Φ = ( G, T , ϕ ) is a graph with a gain function ϕ : −→ E → T , such that ϕ ( e v s v t ) = ϕ ( e v t v s ) − = ϕ ( e v t v s ) forany v s v t ∈ E ( G ). Clearly, the complex unit gain graph is a natural generalization of a simple graph. So,the complex unit gain graph attracts much plenty of researchers’ attention; see [18, 24, 30, 31, 29, 34, 43,55, 58, 61].We usually write G ϕ for a complex unit gain graph Φ = ( G, T , ϕ ). The adjacency matrix of G ϕ is theHermitian matrix A ( G ϕ ) = ( a ϕst ) n × n , where a ϕst = (cid:26) ϕ ( e v s v t ) , if v s v t ∈ E ( G );0 , otherwise.The rank of G ϕ , denoted by rk ( G ϕ ), is the rank of A ( G ϕ ). The positive inertia index p ( G ϕ ), the negativeinertia index n ( G ϕ ) and the nullity η ( G ϕ ) of G ϕ are defined to be the number of positive eigenvalues,negative eigenvalues and zero eigenvalues of A ( G ϕ ), respectively. Obviously, p ( G ϕ ) + n ( G ϕ ) = rk ( G ϕ ).If ϕ ( −→ E ) ⊆ { , ± i } ⊆ T , then for any v s v t ∈ E ( G ), we define h st = (cid:26) ϕ ( e v s v t ) , if v s v t ∈ E ( G );0 , otherwise.Then ( h st ) n × n is the Hermitian adjacency matrix of a mixed graph e G and we have H ( e G ) = ( h st ) n × n = A ( G ϕ ). Hence, the complex unit gain graphs are the generalizations of mixed graphs. We can view amixed graph as a special case of a complex unit gain graph.Two complex unit gain graphs G ϕ and G ϕ ′ are switching equivalent if there is a mapping θ : V ( G ) → T such that ϕ ′ ( uv ) = θ ( u ) − ϕ ( uv ) θ ( v ). In this case, the mapping ϕ ′ can be written as ϕθ . It leads to that A ( G ϕ ′ ) = diag ( θ ( v ) , θ ( v ) , . . . , θ ( v n )) − A ( G ϕ ) diag ( θ ( v ) , θ ( v ) , . . . , θ ( v n )) . Therefore, G ϕ and G ϕ ′ share the same spectrum. It is clear that the switching equivalence is an equiva-lence relation.We write C n , S n , K n for the cycle, star and complete graph of order n , respectively. A mixed graphis called a mixed cycle, mixed star, mixed complete graph , etc., if its underlying graph is a cycle, star,complete graph, etc., respectively. A cut vertex in G is a vertex whose removal increases the numberof connected components of G . For v ∈ V , v is called a pendant vertex if its degree d G ( v ) = 1. Let K n ,n ,...,n k denote the complete k -partite graph with V , V , . . . , V k being the partition class such that | V j | = n j for 1 ≤ j ≤ k . We follow [2] for notations and terminologies not defined here.A subgraph of e G (resp. G ϕ ) is a subgraph of G in which each edge preserves the original direction(resp. gain fuction) in e G (resp. G ϕ ). Let v ∈ V ( G ), we write e G − v (resp. G ϕ − v ) for the inducedsubgraph obtained from e G (resp. G ϕ ) by deleting the vertex v and all edges incident with v . For an2nduced subgraph e R (resp. R ϕ ) of e G (resp. G ϕ ), denote by e G − e R (resp. G ϕ − R ϕ ), the subgraphobtained from e G (resp. G ϕ ) by deleting all vertices of e R (resp. R ϕ ) and all incident edges. For a vertex v of V ( G ) \ V ( R ), we write e R + v (resp. R ϕ + v ) to denote the induced subgraph of e G (resp. G ϕ ) withvertex set V ( R ) S { v } . A vertex of e G (resp. G ϕ ) is called a pendant vertex if it is of degree one in G . Avertex of e G (resp. G ϕ ) is called a cut vertex if it is a cut vertex of G .The spectral-based graph invariants are widely investigated in the literature, such as inertia index[9, 12, 15, 23, 39, 37, 42, 52, 44, 49, 56, 57, 61] of simple graphs, signed graphs, mixed graphs andcomplex unit gain graphs, nullity [3, 7, 6, 8, 10, 14, 38, 41, 45, 56, 57, 64] of simple graphs and signedgraphs, rank [1, 11, 46, 47, 48, 54] of simple graphs, signed graphs and complex unit gain graphs, the H -rank [4, 5, 50] of mixed graphs and skew-rank [20, 22, 32, 33, 26, 27, 40, 53, 62] of oriented graphs.The spectral parameter “inertia index” can determine the structure of a graph to some extent andhas attracted much attention recently. Smith [44] characterized all (undirected) graphs with positiveinertia index 1. Fan et al. [9] determined sharp bounds for the positive and negative inertia index ofa graph. Oboudi [42] completely characterized the (undirected) graphs with exactly two non-negativeeigenvalues. Yu et al. [56] determined the signed graphs with positive inertia index 1 and the signedgraphs containing pendant vertices with positive inertia index 2. Wang et al. [49] extended the abovework to the signed graphs containing cut vertices with positive inertia index 2. Gregory et al. [15]investigated the subadditivity of the inertia indices and gave some properties of Hermitian rank that candetermine the biclique decomposition number. Gregory et al. [12] studied the inertia indices of a partialjoin of two graphs and provided certain connections between biclique decompositions of partial joins ofgraphs and the inertia indices. Geng et al. [16] studied characterizations of graphs with given inertiaindex achieving the maximum diameter.Recently, much attention has been paid to the inertia index of mixed graphs. Wissing and van Dam[52] characterized all mixed graphs with negative inertia index 1. He et al. [17] studied the negative andpositive inertia indices of mixed unicyclic graphs and gave the upper and lower bounds of the negativeand positive inertia indices for mixed graphs. Zheng et al. [63] gave some eigenvalue inequalities betweentwo mixed graphs by means of inertia indices. Wei et al. [51] investigated relations between the inertiaindices of a mixed graph and those of its underlying graph. Yuan et al. [60] provided a characterizationof mixed graphs with positive inertia index 1 and studied the determination of some mixed graphs bytheir H -spectra.Motivated by the results in [13, 28, 35, 51, 52, 49, 50, 56, 57, 60], in this paper, we consider the mixedgraphs with connected underlying graph and characterize all mixed graphs with cut vertices which havepositive inertia index 2. For more general setting, we leave it for further research. Before announcing the main theorem, we introduce one class mixed graph. Let −→ C ( t , t , t ) be themixed complete tripartite graph shown in Figure. 1, where | A | = t , | B | = t , | C | = t . An odd triangle is a mixed triangle containing odd number of directed edges. In particular, −→ C (1 , ,
1) is exactly an oddtriangle. Similarly, a mixed triangle containing even number of directed edges is called an even triangle . (cid:13)(cid:13) (cid:13)(cid:13) PSfrag replacements
AB C
Figure 1: −→ C ( t , t , t )We now list our main results in the following. 3 heorem 1.1. Let e G be a connected mixed graph with pendant vertices. Then p ( e G ) = 2 if and only if e G is obtained by adding some undirected edges (possibly empty) and directed edges (possibly empty) witharbitrary direction between the center of a mixed star and some vertices of a mixed graph e F , where e F isswitching equivalent to a complete multipartite graph or to some −→ C ( t , t , t ) . A most interesting problem attracting us is to determine all mixed graphs without pendant verticeswhose positive inertia indices are 2. It seems difficult for us to give a complete solution now. Note thata mixed graph with a pendant vertex must have a cut vertex. In the following, we extend the result ofTheorem 1.1 to consider the mixed graphs with cut vertices which have positive inertia index 2. In orderto accomplish our goal, we should have the aid of the following graph transformations and terminologies.Let f G and f G be two mixed graphs. Let f G • v • f G be the mixed graph obtained by identifying onevertex v ′ of f G with one vertex v ′′ of f G (we denote the new vertex by v ).Let K ( q , . . . , q r ; n , . . . , n k ; p ) be a simple graph obtained from K q ,...,q r and K n ,...,n k by adding anew vertex v and adding edges such that v is adjacent to all vertices of K q ,...,q r and adjacent to allvertices of S pj =1 V j , where 1 ≤ p ≤ k and V , V , . . . , V k are the partition classes of V ( K n ,n ,...,n k ) suchthat | V j | = n j for 1 ≤ j ≤ k (see Figure. 2). Similarly, let K (0; n , . . . , n k ; p ) be a simple graph obtainedfrom K n ,...,n k by adding a new vertex v and adding edges such that v is adjacent to all vertices of S pj =1 V j , where 1 ≤ p ≤ k . If K q ,...,q r and K n ,...,n k are complete graphs K r and K k , respectively, thenwe denote K ( q , . . . , q r ; n , . . . , n k ; p ) by K ( r ; k ; p ). Similarly, if K n ,...,n k is a complete graph K k , thenwe denote K (0; n , . . . , n k ; p ) by K (0; k ; p ). (cid:13)(cid:13) PSfrag replacements v Figure 2: K (3 ,
2; 2 , ,
3; 2)For nonnegative integers a, b, c and d satisfying 1 ≤ a + b + c + d ≤ k , let K ( q , . . . , q r ; n , . . . , n k ; a i , b − i , c , d − , k − a − b − c − d ) = ( K ( q , . . . , q r ; n , . . . , n k ; a + b + c + d ) , T , ϕ )be a complex unit gain graph with underlying graph K ( q , . . . , q r ; n , . . . , n k ; a + b + c + d ) and gainfunction ϕ : ϕ ( e vu ) = ϕ ( e uv ) − = i, if u ∈ S aj =1 V j ; ϕ ( e vu ) = ϕ ( e uv ) − = − i, if u ∈ S a + bj = a +1 V j ; ϕ ( e vu ) = ϕ ( e uv ) − = 1 , if u ∈ S a + b + cj = a + b +1 V j ; ϕ ( e vu ) = ϕ ( e uv ) − = − , if u ∈ S a + b + c + dj = a + b + c +1 V j ; ϕ ( e uw ) = ϕ ( e wu ) − = 1 , if uw is any other edge of K ( q , . . . , q r ; n , . . . , n k ; a + b + c + d ).Similarly, we can define K (0; n , . . . , n k ; a i , b − i , c , d − , k − a − b − c − d ). If K ( q , . . . , q r ; n , . . . , n k ; a + b + c + d ) is K ( r ; k ; a + b + c + d ), then we denote K ( q , . . . , q r ; n , . . . , n k ; a i , b − i , c , d − , k − a − b − c − d )by K ( r ; k ; a i , b − i , c , d − , k − a − b − c − d ). Similarly, if K (0; n , . . . , n k ; a + b + c + d ) is K (0; k ; a + b + c + d ),then we denote K (0; n , . . . , n k ; a i , b − i , c , d − , k − a − b − c − d ) by K (0; k ; a i , b − i , c , d − , k − a − b − c − d ).Let f C n be a mixed cycle with vertex set { u , u , . . . , u n } . The value of f C n , denoted by h ( f C n ), is definedas h h . . . h n . Clearly, if in one direction the value of f C n is h , then its value is h (the conjugate number4f h ) for the reverse direction. Thus, we refer to the mixed cycle f C n with value 1 (resp. −
1) as a positive(resp. negative) mixed cycle regardless of the direction of f C n . For some orientation of f C n , the signature of f C n , denoted by σ ( f C n ), is defined as the difference between the number of its forward and backwarddirected edges in f C n . Hence, a mixed cycle is positive (resp. negative) if and only if its signature withrespect to an arbitrary direction is congruence to 0 (resp. 2) modulo 4. A mixed graph e G is said to be positive if all its mixed cycles are positive, otherwise it is called non-positive . Theorem 1.2.
Let e G be a connected mixed graph with a cut vertex v and without pendant vertices. Then p ( e G ) = 2 if and only if e G is switching equivalent to one of the following mixed graphs: (i) e K m ,m ,...,m x • v • e K m ′ ,m ′ ,...,m ′ y , where e K m ,m ,...,m x (resp. e K m ′ ,m ′ ,...,m ′ y ) is not a mixed star, and e K m ,m ,...,m x (resp. e K m ′ ,m ′ ,...,m ′ y ) is positive or switching equivalent to some −→ C ( t , t , t ) . (ii) K ( q , . . . , q r ; n , . . . , n k ; p ) , where r ≥ , k ≥ , p ≥ , satisfying one of the following conditions: (1) p = 1 and either n = 1 and k − p ≥ or n ≥ ; (2) p ≥ and k − p ≤ ; (3) p ≥ , k − p ≥ and r + p + k − p − ≥ . (iii) K ( q , . . . , q r ; n , . . . , n k ; a i , b − i , , − , s ) , where r ≥ , k ≥ , a ≥ b ≥ , s = k − a − b , satisfying a = b = 1 and r = 2 . (iv) K ( q , . . . , q r ; n , . . . , n k ; a i , − i , c , − , s ) , where r ≥ , k ≥ , a ≥ c ≥ , s = k − a − c , satisfyingone of the following conditions: (1) a = c = 1 and either s = 0 or s = 1 ; (2) a = c = 1 , s = 2 and either r = 3 or r = 4 ; (3) a = c = 1 , s = 3 and r = 3 ; (4) a = c = 1 , s ≥ and r = 2 ; (5) a = c = 2 , s = 0 and ≤ r ≤ ; (6) a = c = 2 , s = 1 and r = 2 ; (7) a = 3 , s = 0 and r = c = 2 ; (8) a = 4 , c = r = 2 and s = 0 ; (9) a ≥ , c = 1 and as − a + s ≤ r − . (cid:13)(cid:13) (cid:13) (cid:13) (cid:13)(cid:13) (cid:13) (cid:13)(cid:13)(cid:13) (cid:13)(cid:13)(cid:13) (cid:13) (cid:13)(cid:13) (cid:13) (cid:13) PSfrag replacements e Gv v · · ·· · ·· · ·· · · f G ′ Figure 3: Two mixed graphs e G and f G ′ with positive inertia index 2. n n n n (= 1)...... q q q Example 1.3. e G = K (3 ,
2; 3 ,
1; 1 i , − i , , − , satisfies Theorem 1.2 (iii) and f G ′ = K ( q , q , q ; n , n , n , n ; 2 i , − i , , − , satisfies Theorem 1.2 (iv) (9) (see Figure. 3). It is easy to see that p ( e G ) = p ( f G ′ ) = 2 . In the rest of this section we recall some known results. In Section 2 we give the proof of Theorem1.1. In Section 3 we first establish some technical lemmas that help us characterize the mixed graphswith cut vertices having exactly two positive eigenvalues and then present the proof of Theorem 1.2.5 .3 Preliminaries
In this subsection, we introduce some helpful lemmas and known results.
Lemma 1.4. [19] (Sylvester’s law of inertia) If two Hermitian matrices M and N are congruent, thenthey have the same positive (resp. negative) inertia indices. Lemma 1.5. [28] Let e G be a mixed graph. Then the following conditions are equivalent. (1) Every induced mixed cycle of e G is positive. (2) e G is positive. (3) e G is switching equivalent to its underlying graph. Lemma 1.6. [51, 60] Let v be a pendant vertex of mixed graph e G with neighbor v . Then p ( e G ) = p ( e G − v − v ) + 1 , n ( e G ) = n ( e G − v − v ) + 1 , η ( e G ) = η ( e G − v − v ) . Lemma 1.7. [51] Let e G be a mixed graph with u ∈ V ( G ) . Then p ( e G ) − ≤ p ( e G − u ) ≤ p ( e G ) , n ( e G ) − ≤ n ( e G − u ) ≤ n ( e G ) . Lemma 1.8. [60] Let v be a cut vertex of a mixed graph e G , and let f G , f G , . . . , f G l be all the componentsof e G − v . (1) If there exists a component f G i satisfying rk ( f G i + v ) = rk ( f G i ) + 2 , then In( e G ) = In( e G − v ) + (1 , , −
1) = l X j =1 In( f G j ) + (1 , , − . (2) If there exists a component f G i satisfying rk ( f G i + v ) = rk ( f G i ) , then In( e G ) = In( e G i ) + In( e G − e G i ) . In particular, if rk ( f G i + v ) = rk ( f G i ) for each i = 1 , , . . . , l , then In( e G ) = In( e G − v ) + (0 , ,
1) = l X j =1 In( f G j ) + (0 , , . Lemma 1.9. [50] Let f C n be a mixed cycle of order n . Then η ( f C n ) = , if n is odd, σ ( f C n ) is odd; , if n is odd, σ ( f C n ) is even; , if n is even, σ ( f C n ) is odd; , if n is even, σ ( f C n ) is even, n + σ ( f C n ) ≡ mod ; , if n is even, σ ( f C n ) is even, n + σ ( f C n ) ≡ mod . Lemma 1.10. [60] Let e G be a mixed graph. Then p ( e G ) = 1 if and only if the subgraph induced byall non-isolated vertices of e G is switching equivalent either to a complete multipartite graph or to some −→ C ( t , t , t ) . Proof for Theorem 1.1Proof of Theorem 1.1:
The sufficiency can be verified by Lemma 1.6, and we only prove necessity. “Necessity”. If e G is a mixed graph with pendant vertices, without loss of generality, let v be apendant vertex of e G with neighbor v . Then by Lemma 1.6, we have p ( e G ) = p ( e G − v − v ) + 1. Thecondition p ( e G ) = 2 implies p ( e G − v − v ) = 1. Thus, by Lemma 1.10, the subgraph, say e F , induced byall non-isolated vertices of e G − v − v is switching equivalent either to a complete multipartite graph orto some −→ C ( t , t , t ). Hence, p ( e F ) = 1. It is easy to see that e G is obtained by adding some undirectededges (possibly empty) and directed edges (possibly empty) with arbitrary direction between the centerof a mixed star and some vertices of e F . This completes the proof. In this subsection we present a few technical lemmas aiming to provide some fundamental charac-terizations of mixed graphs with cut vertices and without pendant vertices having exactly two positiveeigenvalues.Two vertices u, w ∈ V ( e G ) are called twins if e G is switching equivalent to f G ′ such that the rows of H ( f G ′ ) indexed by u and w are same. The relation of being twins is an equivalence relation on vertex setof e G . Denote by [ u ] the equivalence class containing u . The twin reduction graph of e G , written as T e G , isa mixed graph whose vertices are the equivalence classes and [ u ][ w ] ∈ E ( T e G ) if and only if uw ∈ E ( e G )and −−−→ [ u ][ w ] ∈ E ( T e G ) if and only if −→ uw ∈ E ( e G ). Lemma 3.1. [35] Let e G and f G ′ be mixed graphs with the same underlying graph. Then they are switchingequivalent if and only if T e G and T f G ′ are switching equivalent. Lemma 3.2. [50] Adding or removing twins preserves the H -rank of a mixed graph, and also preservesthe positive and negative inertia index of a mixed graph. Lemma 3.3. rk ( −→ C ( t , t , t )) = 2 .Proof. It is easy to see that T −→ C ( t ,t ,t ) = −→ C (1 , ,
1) and rk ( −→ C (1 , , rk ( −→ C ( t , t , t )) = rk ( −→ C (1 , , . Lemma 3.4. [50] Let e G be a connected mixed graph. Then rk ( e G ) = 3 if and only if T e G is an eventriangle. Lemma 3.5.
Let e G be a positive complete k -partite mixed graph with one partition class containingexactly one vertex. If k ≥ and v ∈ V ( e G ) is in one partition class containing exactly one vertex, then rk ( e G ) = rk ( e G − v ) + 1 .Proof. By Lemma 1.5, we have e G is switching equivalent to its underlying graph. Then the vertices in thesame partition class of e G are twins. Thus, by Lemma 3.1, we have T e G is switching equivalent to K k . UsingLemma 3.2, we have rk ( e G ) = rk ( T e G ) = rk ( K k ) = k . Similarly, we have rk ( e G − v ) = rk ( K k − ) = k − rk ( e G ) = rk ( e G − v ) + 1. Lemma 3.6.
Let f M be a mixed graph with a vertex v ′ and f M be another mixed graph with a vertex v ′′ , and let e G = f M • v • f M . Then p ( f M − v ′ ) + p ( f M − v ′′ ) ≤ p ( e G ) ≤ p ( f M ) + p ( f M ) . roof. By Lemma 1.7, we obtain p ( f M − v ′ ) ≤ p ( f M ) ≤ p ( f M − v ′ ) + 1 . Similarly, p ( f M − v ′′ ) ≤ p ( f M ) ≤ p ( f M − v ′′ ) + 1 . Lemma 1.7 applying to the induced subgraph e G − v = ( f M − v ′ ) S ( f M − v ′′ ) of e G yields that p ( f M − v ′ ) + p ( f M − v ′′ ) ≤ p ( e G ) ≤ p ( f M − v ′ ) + p ( f M − v ′′ ) + 1 . If p ( f M ) = p ( f M − v ′ ) + 1, then p ( e G ) ≤ p ( f M − v ′ ) + p ( f M − v ′′ ) + 1 ≤ p ( f M ) + p ( f M ) . If p ( f M ) = p ( f M − v ′′ ) + 1, we also have p ( e G ) ≤ p ( f M − v ′ ) + p ( f M − v ′′ ) + 1 ≤ p ( f M ) + p ( f M ) . Suppose now p ( f M ) = p ( f M − v ′ ) and p ( f M ) = p ( f M − v ′′ ) and consider the Hermitian adjacency matrix H ( e G ) of e G : H ( e G ) = H ( f M − v ′ ) α α ∗ β ∗ β H ( f M − v ′′ ) . The condition p ( f M ) = p ( f M − v ′ ) implies that rk ( f M ) ≤ rk ( f M − v ′ ) + 1, thus the equation H ( f M − v ′ ) X = α has a solution, say X . Similarly, suppose Y is a solution of the equation H ( f M − v ′′ ) Y = β and let S = I − X
00 1 00 − Y I . Then S ∗ H ( e G ) S = H ( f M − v ′ ) 0 00 − α ∗ X − β ∗ Y
00 0 H ( f M − v ′′ ) . From p ( f M ) = p ( f M − v ′ ) and p ( f M ) = p ( f M − v ′′ ) it follows that − α ∗ X ≤ − β ∗ Y ≤
0. Thusby Lemma 1.4, p ( e G ) = p ( S ∗ H ( e G ) S ) = p ( f M − v ′ ) + p ( f M − v ′′ ) = p ( f M ) + p ( f M ). This completes theproof. Lemma 3.7.
Let e K m ,m ,...,m x and e K m ′ ,m ′ ,...,m ′ y be mixed complete multipartite graphs without pendantvertices. Let v ′ (resp. v ′′ ) be a vertex of e K m ,m ,...,m x (resp. e K m ′ ,m ′ ,...,m ′ y ). If p ( e K m ,m ,...,m x ) = 1 and p ( e K m ′ ,m ′ ,...,m ′ y ) = 1 , then p ( e K m ,m ,...,m x • v • e K m ′ ,m ′ ,...,m ′ y ) = 2 .Proof. Since e K m ,m ,...,m x and e K m ′ ,m ′ ,...,m ′ y are mixed complete multipartite graphs without pendantvertices, we have K m ,m ,...,m x and K m ′ ,m ′ ,...,m ′ y are not mixed stars. Thus, e K m ,m ,...,m x − v ′ (resp. e K m ′ ,m ′ ,...,m ′ y − v ′′ ) has at least one edge. It follows that p ( e K m ,m ,...,m x − v ′ ) ≥ p ( e K m ′ ,m ′ ,...,m ′ y − ′′ ) ≥
1. By Lemma 1.7, we have p ( e K m ,m ,...,m x − v ′ ) = 1 and p ( e K m ′ ,m ′ ,...,m ′ y − v ′′ ) = 1. In view ofLemma 3.6, we have 2 = p ( e K m ,m ,...,m x − v ′ ) + p ( e K m ′ ,m ′ ,...,m ′ y − v ′′ ) ≤ p ( e K m ,m ,...,m x • v • e K m ′ ,m ′ ,...,m ′ y ) ≤ p ( e K m ,m ,...,m x ) + p ( e K m ′ ,m ′ ,...,m ′ y )=2 . This completes the proof.Let J n × m and n × m be respectively the all-one and the all-zero n × m matrices. Let n = J n × . Weoften simply write J , and , respectively, if the order of these matrices is clear from the context. Lemma 3.8.
For r ≥ , k ≥ , a ≥ b, a ≥ , and s = k − a − b , the positive inertia index of e G = K ( r ; k ; a i , b − i , , − , s ) is 2 if and only if one of the following conditions holds: (1) a = 1 , b = 0 ; (2) a = b = 1 and r = 2 ; (3) a ≥ , b = 0 and s ≤ ; (4) a ≥ , b = 0 , s ≥ and r + a + s − ≥ .Proof. The Hermitian adjacency matrix of e G = K ( r ; k ; a i , b − i , , − , s ) can be written as H ( e G ) = H ( K r ) ⊤ i ⊤ − i ⊤ − i H ( K a ) J J0 i H ( K b ) J0 0 J J H ( K s ) . Case 1. a = 1 and b = 0. If s = 1, then e G has a pendant vertex, by Theorem 1.6, it is easy to seethat p ( e G ) = 2. If s ≥
2, then H ( e G ) = H ( K r ) ⊤ i − i ⊤ H ( K s ) . Let C = I r − r −
00 0 − s − I s . Then C ∗ H ( e G ) C = H ( K r ) − rr − i − i − ss −
00 0 0 H ( K s ) .
9t is easy to see that the eigenvalues of − rr − i − i − ss − ! are both negative, therefore p ( C ∗ H ( e G ) C ) = p ( H ( K r )) + p ( H ( K s )) = 2 . Thus in this case, the positive inertia index of K ( r ; k ; a i , b − i , , − , s ) must be 2. Case 2. a = b = 1. If s = 0, then H ( e G ) = H ( K r ) ⊤ i − i − i i . Let C = I r − r − − i i . Then C ∗ H ( e G ) C = H ( K r ) − rr − . Thus p ( e G ) = 2 if and only if r = 2.If s = 1, then H ( e G ) = H ( K r ) ⊤ i − i − i i . Let C = I r − r − − − i − . Then C ∗ H ( e G ) C = H ( K r ) − rr − i − i − . p ( e G ) = 2 if and only if − rr − i − i − ! . has no positive eigenvalue, if and only if r = 2.If s ≥
2, let C = I r − r −
00 0 − s − − s − I s . Then C ∗ H ( e G ) C = H ( K r ) − rr − i − i − i − ss − − s − i − s − − ss −
00 0 0 0 H ( K s ) . If r >
2, since the determinant of − rr − i − i − i − ss − − s − i − s − − ss − is positive, the positive inertia index of this matrix is at least 1, and thus p ( e G ) ≥
3. If r = 2, it is easyto see that the above matrix has 0 as an eigenvalue and the other two eigenvalues are negative, and thus p ( e G ) = 2. Consequently, when a = b = 1, p ( e G ) = 2 if and only if r = 2. Case 3. a ≥
2. If k = a = 2, then by Lemma 3.6,2 = p ( K r ) + p ( K k ) ≤ p ( e G ) ≤ p ( K r +1 ) + p ( ^ K k +1 ) = 2 , and thus p ( e G ) = 2. Suppose k ≥
3. Let U n be the strictly upper triangular matrix of order n whose( i, j )-entries are 1 for all 1 ≤ i < j ≤ n . Let D = diag ( − a − , − a , · · · , − k − ). Set C = I r
00 0 P , where P is the following matrix P = I a J D I k − a + U k − a D ! . C ∗ H ( e G ) C is the following matrix: H ( K r ) · · · · · · ⊤ i ⊤ (1 − a ) ia − − a ) ia · · · (1 − a ) ia + b − b − a ) ia + b − · · · ( b − a ) ik − − i H ( K a ) 0 0 · · · · · · (2 a − ia − − aa − · · · · · · (2 a − ia − ( a +1) a · · · · · · · · · ... (2 a − ia + b − · · · − ( a + b − a + b − · · · ( a − b ) ia + b − · · · − ( a + b ) a + b − · · · · · · ... ( a − b ) ik − · · · · · · − ( k − k − . Let C = I r − r − . . . . . . . . . . . . ia − I a . . . . . . (2 a − ia . . . . . . (2 a − ia +1 . . . . . . . . . ... (2 a − ia + b − . . . . . . ( a − b ) ia + b . . . . . . . . . ... ( a − b ) ik − . . . . . . . Then C ∗ C ∗ H ( e G ) C C is the following matrix: H ( K r ) . . . . . . ρ . . . . . . H ( K a ) 0 0 . . . . . . − aa − . . . . . . − ( a +1) a . . . . . . . . . ... . . . − ( a + b − a + b − . . . . . . − ( a + b ) a + b − . . . . . . ... . . . . . . − ( k − k − , where ρ = (2 a − (cid:18) a − − a + b − (cid:19) + ( a − b ) (cid:18) a + b − − k − (cid:19) − rr − − aa − b (2 a − ( k −
1) + s ( a − b ) ( a − a − a + b − k − − rr − − aa − . φ = b (2 a − ( k − s ( a − b ) ( a − a − a + b − k − . Clearly, p ( e G ) = 2 if and only if ρ ≤
0, if and only if φ ≤ rr − aa − . Clearly, φ ≥ b (2 a − ( a − a + b − . If b ≥
2, since 2 a − ≥ a + b − φ ≥ a − a − > ≥ rr − aa − , and thus p ( e G ) >
2. If b = 1, then φ ≥ (2 a − a ( a − > ≥ rr − aa − , leading to p ( e G ) >
2. If b = 0, then φ = a s ( a − a + s − . If s = 0, then φ = 0. If s = 1, then φ = aa − .In both cases, we have φ ≤ rr − + aa − and thus p ( e G ) = 2. If s ≥
2, the condition φ ≤ rr − + aa − isequivalent to 1 r + 1 a + 1 s − ≥ . This completes the proof.
Corollary 3.9.
For r ≥ , k ≥ and p ≥ , the positive inertia index of e G = K ( r ; k ; p ) is 2 if and onlyif one of the following conditions holds: (1) p = 1 ; (2) p ≥ and k − p ≤ ; (3) p ≥ , k − p ≥ and r + p + k − p − ≥ .Proof. Using mixed two-way switching of mixed graphs we have e G = K ( r ; k ; p ) is switching equivalentto the mixed graph e G = K ( r ; k ; p i , − i , , − , k − p ). Thus, the result follows from Lemma 3.8. Lemma 3.10.
For r ≥ , k ≥ , a ≥ c, a ≥ and s = k − a − c , the positive inertia index of e G = K ( r ; k ; a i , − i , c , − , s ) is 2 if and only if one of the following conditions holds: (1) a = 1 , c = 0 ; (2) a = c = 1 and either s = 0 or s = 1 ; (3) a = c = 1 , s = 2 and either r = 3 or r = 4 ; (4) a = c = 1 , s = 3 and r = 3 ; (5) a = c = 1 , s ≥ and r = 2 ; (6) a = c = 2 , s = 0 and ≤ r ≤ ; (7) a = c = 2 , s = 1 and r = 2 ; (8) a = 3 , s = 0 and r = c = 2 ; (9) a = 4 , c = r = 2 and s = 0 ; (10) a ≥ , c = 1 and as − a + s ≤ r − ; (11) a ≥ , c = 0 and either s = 0 or s = 1 ; (12) a ≥ , c = 0 , s ≥ and r + a + s − ≥ .Proof. The Hermitian adjacency matrix of e G = K ( r ; k ; a i , − i , c , − , s ) can be written as H ( e G ) = H ( K r ) ⊤ i ⊤ ⊤ − i H ( K a ) J J0 1 J H ( K c ) J0 0 J J H ( K s ) . ase 1. a = 1 and c = 0. The result follows from Lemma 3.8 (1). Thus in this case, the positiveinertia index of K ( r ; k ; a i , − i , c , − , s ) must be 2. Case 2. a = c = 1. If s = 0, then H ( e G ) = H ( K r ) ⊤ i − i . Let T = I r − r − − i . Then T ∗ H ( e G ) T = H ( K r ) − rr − . Thus p ( e G ) = 2 since r ≥
2. When s = 1, then H ( e G ) = H ( K r ) ⊤ i − i . Let T = I r − r − − − − . Then T ∗ H ( e G ) T = H ( K r ) − rr − i − − i − − . Thus p ( e G ) = 2 if and only if − rr − i − − i − − ! r ≥
2. When s ≥
2, let T = I r − r −
00 0 − s − − s − I s . Then T ∗ H ( e G ) T = H ( K r ) − rr − i − i − ss − − s − − s − − ss −
00 0 0 0 H ( K s ) . If r ≥
3, since the determinant of − rr − i − i − ss − − s − − s − − ss − (3.1)is r ( s − − s ( r − s − . If s ≥
4, then r ( s − − s ( r − s − ≥ s − r − s − >
0. Thus the positive inertia index of matrix (3.1)is at least 1, and thus p ( e G ) ≥
3. If s = 3 and r = 3, matrix (3.1) has eigenvalues 0, − and − p ( e G ) = 2. If s = 3 and r >
3, then r ( s − − s ( r − s − = r − r − >
0. Thus the positive inertia index of matrix (3.1)is at least 1, and thus p ( e G ) ≥
3. If s = 2 and r >
4, then r ( s − − s ( r − s − = r − r − >
0. Thus the positive inertiaindex of matrix (3.1) is at least 1, and thus p ( e G ) ≥
3. If s = 2 and r = 4, matrix (3.1) has eigenvalues 0, − √ and − −√ . If s = 2 and r = 3, matrix (3.1) has eigenvalues − . − . − . p ( e G ) = 2.If r = 2, the characteristic polynomial of matrix (3.1) is f ( λ ) = λ + s − s − λ + s +1)( s − λ + s − . Wehave f ′ ( λ ) = 3 λ + s − s − λ + s +1)( s − . It is easy to see that f ′ ( λ ) > λ ≥
0. Thus f ( λ ) is strictlyincreasing in [0 , + ∞ ). Since f (0) = s − >
0, we have matrix (3.1) has no positive eigenvalue, and thus p ( e G ) = 2. Case 3. a ≥
2. If k = a = 2, then by Lemma 3.6,2 = p ( K r ) + p ( K k ) ≤ p ( e G ) ≤ p ( K r +1 ) + p ( ^ K k +1 ) = 2 , and thus p ( e G ) = 2. Suppose k ≥
3. Similarly, as in the proof of Case 3 of Lemma 3.8, set T = I r
00 0 P . T ∗ H ( e G ) T is the following matrix: H ( K r ) · · · · · · ⊤ i ⊤ ( a − − aia − a − − aia · · · ( a − − aia + c − − c − aia + c − · · · − c − aik − − i H ( K a ) 0 0 · · · · · · ( a − aia − − aa − · · · · · · ( a − aia − ( a +1) a · · · · · · · · · ... ( a − aia + c − · · · − ( a + c − a + c − · · · − c + aia + c − · · · − ( a + c ) a + c − · · · · · · ... − c + aik − · · · · · · − ( k − k − . Let T = I r − r − · · · · · · · · · · · · ia − I a · · · · · · ( a − aia · · · · · · ( a − aia +1 · · · · · · · · · ... ( a − aia + c − · · · · · · − c + aia + c · · · · · · · · · ... − c + aik − · · · · · · . Then T ∗ T ∗ H ( e G ) T T is the following matrix: H ( K r ) . . . . . . ξ . . . . . . H ( K a ) 0 0 . . . . . . − aa − . . . . . . − ( a +1) a . . . . . . . . . ... . . . − ( a + c − a + c − . . . . . . − ( a + c ) a + c − . . . . . . ... . . . . . . − ( k − k − , where ξ = [( a − + a ] (cid:18) a − − a + c − (cid:19) + ( a + c ) (cid:18) a + c − − k − (cid:19) − rr − − aa − c [( a − + a ]( k −
1) + s ( a − a + c )( a − a + c − k − − rr − − aa − . γ = c [( a − + a ]( k − s ( a − a + c )( a − a + c − k − . Clearly, p ( e G ) = 2 if and only if ξ ≤
0, if andonly if γ ≤ rr − aa − . If c ≥
4, we have γ ≥ c [( a − + a ]( a − a + c − ≥ a − + a ]( a − a −
1) = 4(2 a − a + 1)2 a − a + 1 > ≥ rr − aa − , and thus p ( e G ) > c = 3 and a ≥
4, then we have γ ≥ c [( a − + a ]( a − a + c −
1) = 3[( a − + a ]( a − a + 2) > ≥ rr − aa − , and thus p ( e G ) > c = a = 3, then γ = + s s ) , where k = a + c + s . If γ ≤ rr − + aa − = rr − + , then we have + s s ) ≤ rr − . Hence, we obtain 2 < rr − , i.e. r <
2, a contradiction. Thus, p ( e G ) > c = 2 and a ≥
4, we now consider three cases. If r ≥
3, then f R (see Fig. 4) is an induced subgraphof e G . If s ≥
1, then f R (see Figure. 4) is an induced subgraph of e G . By direct calculation, we have p ( f R ) = p ( f R ) = 3. Hence, p ( e G ) = 2 implies r = 2 and s = 0. Let γ = c [( a − + a ]( a − a + c − = a − a +1) a − ≤ rr − + aa − = 2 + aa − . We have a = 4, r = 2 and s = 0. Hence, in this case, p ( e G ) = 2 if and only if a = 4 , r = 2 and s = 0. (cid:13) (cid:13)(cid:13) (cid:13)(cid:13) PSfrag replacements vv f R f R Figure 4: Forbidden subgraph f R and f R .If c = 2 and a = 3, then γ = + s s ) , where k = a + c + s . If γ ≤ rr − + aa − = rr − + , thenwe have (3 + 4 s ) r ≤ s . Note that r ≥
2, we obtain 2(3 + 4 s ) ≤ s . Thus, we have s = 0. Inaddition, 3 r ≤ r = 2. Thus, in this case, p ( e G ) = 2 if and only if r = 2 and s = 0.If c = 2 and a = 2, then γ = s )+8 s s ) , where k = a + c + s . If γ ≤ rr − + aa − = rr − + 2, then wehave (3 + 9 s ) r ≤
12 + 12 s . Note that r ≥
2, we obtain 2(3 + 9 s ) ≤
12 + 12 s . Thus, we have s = 0 or 1. If s = 0, we have 2 ≤ r ≤
4. If s = 1, we obtain r = 2. Thus, in this case, p ( e G ) = 2 if and only if (i) s = 0and 2 ≤ r ≤ s = 1 and r = 2.If c = 1, then γ = [( a − + a ]( k − s ( a − a +1) a ( a − k − = [( a − + a ]( a + s )+ s ( a − a +1) a ( a − a + s ) , where k = a + c + s .The condition γ ≤ rr − + aa − is equivalent to as − a + s ≤ r − .If c = 0, then γ = a s ( a − k − = a s ( a − a + s − , where k = a + c + s . If s = 0, then γ = 0. If s = 1, then γ = aa − . In both cases, we have γ ≤ rr − + aa − and thus p ( e G ) = 2. If s ≥
2, the condition γ ≤ rr − + aa − is equivalent to 1 r + 1 a + 1 s − ≥ . This completes the proof. 17 emma 3.11.
Let f F be a connected mixed graph with p ( f F ) = 1 . Let f F be a mixed graph obtained from f F by adding a new vertex v and adding some mixed edges between v and some vertices of f F . If rk ( f F ) = rk ( f F ) + 1 and p ( f F ) = 2 , then f F is switching equivalent to some complete k -partite graph K n ′ ,n ′ ,...,n ′ k and f F is switching equivalent to some complex unit gain graph K (0; n ′′ , n ′′ , . . . , n ′′ k ; a i ∗ , b − i ∗ , c ∗ , d − ∗ , k − a ∗ − b ∗ − c ∗ − d ∗ ) .Proof. Since p ( f F ) = 1, by Lemma 1.10, we have f F is switching equivalent to a complete multipartitegraph or some −→ C ( t , t , t ). In view of Lemma 3.3, we have rk ( −→ C ( t , t , t )) = 2. If f F is switchingequivalent to some −→ C ( t , t , t ), then combining rk ( −→ C ( t , t , t )) = 2 and rk ( f F ) = rk ( f F ) + 1, we have rk ( f F ) = 3. By Lemma 3.4, we have T f F is an even triangle. Since f F is a subgraph of f F , hence f F contains an odd triangle. By Lemma 1.9, odd triangle is not equivalent to even triangle. Thus, we obtain T f F is not an even triangle, a contradiction. Hence, we have f F is switching equivalent to a completemultipartite graph.Without loss of generality, assume that f F is switching equivalent to K n ′ ,n ′ ,...,n ′ k . Then f F is switchingequivalent to a complex unit gain graph F ϕ with F ϕ − v = K n ′ ,n ′ ,...,n ′ k , where ϕ ( −−−→ E ( F )) ⊆ {± , ± i } .Now we consider F ϕ . Let V , V , . . . , V k be the partition classes of V ( K n ′ ,n ′ ,...,n ′ k ) such that | V j | = n ′ j for j = 1 , , . . . , k . Since rk ( f F ) = rk ( f F ) + 1, we have v is not an isolated vertex. In the following, we willprove the following two facts. Fact 1.
Let ≤ j ≤ k . If there exists one edge connecting v and a vertex of V j in F , then v is adjacentto all vertices of V j in F . Proof of Fact 1
Suppose on the contrary, without loss of generality, let u, w ∈ V j such that vu ∈ E ( F )and vw / ∈ E ( F ). Without loss of generality, choose arbitrary vertex v t ∈ V t for 1 ≤ t ≤ k, t = j . TheHermitian adjacency matrix of the complex unit gain graph induced by vertex subset { v, u, w } [ (cid:16) [ ≤ t ≤ k,t = j { v t } (cid:17) of F ϕ can be written as N = ǫ η ∗ ⊤ ǫ ⊤ η k − − I k − , where ǫ ∈ {± , ± i } . Set Q = −
00 0 0 I k − , then Q ∗ N Q is the following matrix: − ǫ ǫ η ∗ − ǫ ǫ ⊤ η k − − I k − . Note that the equation ⊤ k − − I k − ! X = (cid:18) ǫη (cid:19) (cid:18) µϑ (cid:19) . Let Q = − µ − ϑ k − . Then Q ∗ Q ∗ N Q Q is the following matrix: − ǫµ − η ∗ ϑ − ǫ − ǫ ⊤ k − − I k − . Thus rank ( N ) = rank ( Q ∗ Q ∗ N Q Q )= rank − ǫµ − η ∗ ϑ − ǫ − ǫ ! + rank ⊤ k − − I k − ! = 2 + k = 2 + rk ( e F ) . Since rk ( e F ) = rk ( F ϕ ) ≥ rank ( N ), we have rk ( e F ) ≥ rk ( e F ), a contradiction. Fact 2. ∀ ≤ j ≤ k , A ( F ϕ ) vu = A ( F ϕ ) vw for all { u, w } ⊆ V j . Proof of Fact 2
Let 1 ≤ j ≤ k . If there are no edges between v and V j , then it is easy to see that A ( F ϕ ) vu = A ( F ϕ ) vw = 0 for all { u, w } ⊆ V j . If there are edges between v and V j , then by Fact 1 , wehave vu and vw are edges of F for all { u, w } ⊆ V j . Suppose A ( F ϕ ) vu = A ( F ϕ ) vw for some { u, w } ⊆ V j .Without loss of generality, choose arbitrary vertex v t ∈ V t for 1 ≤ t ≤ k and t = j . The Hermitianadjacency matrix of the complex unit gain graph induced by vertex subset { v, u, w } [ ( [ ≤ t ≤ k,t = j { v t } )of F ϕ can be written as M = A ( F ϕ ) vu A ( F ϕ ) vw ζ ∗ A ( F ϕ ) vu ⊤ A ( F ϕ ) vw ⊤ ζ k − − I k − . Set Q = −
00 0 0 I k − , then Q ∗ M Q is the following matrix: A ( F ϕ ) vu − A ( F ϕ ) vw A ( F ϕ ) vw ζ ∗ A ( F ϕ ) vu − A ( F ϕ ) vw A ( F ϕ ) vw ⊤ ζ k − − I k − . ⊤ k − − I k − ! X = (cid:18) A ( F ϕ ) vw ζ (cid:19) has a solution, say (cid:18) νδ (cid:19) . Let Q = − ν − δ k − . Then Q ∗ Q ∗ N Q Q is the following matrix: − νA ( F ϕ ) vw − ζ ∗ δ A ( F ϕ ) vu − A ( F ϕ ) vw A ( F ϕ ) vu − A ( F ϕ ) vw ⊤ k − − I k − . Since A ( F ϕ ) vu = A ( F ϕ ) vw , we obtain rank ( M ) = rank ( Q ∗ Q ∗ M Q Q )= rank − νA ( F ϕ ) vw − ζ ∗ δ A ( F ϕ ) vu − A ( F ϕ ) vw A ( F ϕ ) vu − A ( F ϕ ) vw ! + rank ⊤ k − − I k − ! = 2 + k = 2 + rk ( f F ) . Since rk ( f F ) = rk ( F ϕ ) ≥ rk ( M ), we have rk ( f F ) ≥ rk ( f F ), a contradiction. This completes theproof of Fact 2.Combining Facts 1 and 2, we have f F is switching equivalent to some complex unit gain graph K (0; n ′′ , n ′′ , . . . , n ′′ k ; a i ∗ , b − i ∗ , c ∗ , d − ∗ , k − a ∗ − b ∗ − c ∗ − d ∗ ) . Now, we are ready to prove the second main result of this paper.
Proof of Theorem 1.2 “Sufficiency”. If e G is a mixed graph switching equivalent to a mixed graphsatisfying (i), then by Lemmas 1.10 and 3.7, the result is naturally obtained. If e G is a mixed graphswitching equivalent to a mixed graph satisfying (ii), then by Lemma 3.1, T e G is switching equiva-lent to K ( r ; k ; p ) satisfying one of the three conditions of Corollary 3.9. By Corollary 3.9, we have p ( K ( r ; k ; p )) = 2. In view of Lemmas 3.1 and 3.2, we have p ( e G ) = p ( T e G ) = p ( K ( r ; k ; p )) = 2. If e G is a mixed graph switching equivalent to a mixed graph satisfying (iii), then by Lemma 3.1, T e G isswitching equivalent to K (2; k ; 1 i , − i , , − , s ) satisfying the condition (2) of Lemma 3.8. By Lemma3.8, we have p ( K (2; k ; 1 i , − i , , − , s ) = 2. In view of Lemmas 3.1 and 3.2, we have p ( e G ) = p ( T e G ) = p ( K ( r ; k ; a i , b − i , , − , s ) = 2. If e G is a mixed graph switching equivalent to a mixed graph satisfying(iv), then by Lemma 3.1, T e G is switching equivalent to K ( r ; k ; a i , − i , c , − , s ) satisfying one of theconditions (2)-(10) of Lemma 3.10. By Lemma 3.10, we have p ( K ( r ; k ; a i , − i , c , − , s )) = 2. In viewof Lemmas 3.1 and 3.2, we have p ( e G ) = p ( T e G ) = p ( K ( r ; k ; a i , − i , c , − , s )) = 2. “Necessity”. We proceed by establishing some claims.20 laim 1. e G − v has exactly two components. Proof of Claim 1.
Suppose that e G − v = f G ∪ f G ∪ . . . ∪ f G t is the disjoint union of different componentsof e G − v . Since v is a cut vertex of e G , we have t ≥
2. Since e G has no pendant vertices, we have f G i has atleast one edge for all i = 1 , , . . . , t . Hence, p ( f G i ) ≥ i = 1 , , . . . , t . From 2 = p ( e G ) ≥ p ( e G − v ) = p ( e G ) + . . . + p ( e G t ) it follows that t = 2. Hence, e G − v has exactly two components f G and f G . Claim 2. p ( f G i ) = 1 for each i = 1 , . Proof of Claim 2.
Noting that p ( f G ) ≥ p ( f G ) ≥
1, then from2 = p ( e G ) ≥ p ( f G ) + p ( f G ) ≥ p ( f G i ) = 1 for each i = 1 , f G (resp. f G ) is switching equivalent to a complete multipartitegraph or some −→ C ( t , t , t ). Suppose, without loss of generality, that p ( f G + v ) ≤ p ( f G + v ). Claim 3. p ( f G + v ) = 1 . Proof of Claim 3.
We distinguish the following three cases.
Case 1.
There exists f G i such that rk ( f G i + v ) = rk ( f G i ) + 2 . By Lemma 1.8 (1), we have p ( e G ) = p ( e G − v ) + 1 = p ( f G ) + p ( f G ) + 1 = 3, a contradiction. Case 2.
There exists f G i such that rk ( f G i + v ) = rk ( f G i ) . Subcase 1. rk ( f G + v ) = rk ( f G ) . By Lemma 1.8 (2), we have p ( e G ) = p ( f G ) + p ( e G − f G ) = p ( f G ) + p ( f G + v ). Hence, 1 ≤ p ( f G + v ) ≤ p ( f G + v ) = 1. It follows that p ( f G + v ) = p ( f G + v ) = 1. Subcase 2. rk ( f G + v ) = rk ( f G ) . By Lemma 1.8 (2), p ( e G ) = p ( f G )+ p ( e G − f G ) = p ( f G )+ p ( f G + v ). Hence, p ( f G + v ) = 1. Furthermore,from rk ( f G + v ) = rk ( e G ), we have p ( f G + v ) = p ( f G ) = 1. Case 3. rk ( f G i + v ) = rk ( f G i ) + 1 for each i = 1 , . Consider the Hermitian adjacency matrix of e G : H ( e G ) = H ( f G ) τ τ ∗ ψ ∗ ψ H ( f G ) . The condition rk ( f G + v ) = rk ( f G ) + 1 implies that the equation H ( f G ) X = τ has a solution, say X .Similarly, suppose Y is a solution of the equation H ( f G ) Y = ψ and let W = I − X
00 1 00 − Y I . Then W ∗ H ( e G ) W = H ( f G ) 0 00 − τ ∗ X − ψ ∗ Y
00 0 H ( f G ) . p ( e G ) = p ( f G ) + p ( f G ) and Lemma 1.4, we have − τ ∗ X − ψ ∗ Y ≤
0. Note that − τ ∗ X = 0 and − ψ ∗ Y = 0. Hence, at least one of − τ ∗ X and − ψ ∗ Y is less than 0. If − ψ ∗ Y >
0, then − τ ∗ X < p ( f G + v ) = p ( f G ) = 1. If − ψ ∗ Y <
0, we have p ( f G + v ) = p ( f G ) = 1. Hence, by thecondition of p ( f G + v ) ≤ p ( f G + v ), we have p ( f G + v ) = 1.Combining the proof of Cases 1 -- Observation 1. If rk ( f G i + v ) = rk ( f G i ) + 1 for each i = 1 , , and − ψ ∗ Y > , where ψ and Y aredefined in Case 3 of Claim 3, then p ( f G + v ) = 2 . Otherwise, p ( f G + v ) = 1 . If rk ( f G i + v ) = rk ( f G i ) for each i = 1 ,
2, then combining Claim 3 and Observation 1, we have p ( f G + v ) = p ( f G + v ) = 1. In view of Lemma 1.10, we obtain f G + v (resp. f G + v ) is switchingequivalent to complete multipartite graph or some −→ C ( t , t , t ). Furthermore, if f G + v (resp. f G + v ) isswitching equivalent to complete multipartite graph, then v is in the partition class containing at leasttwo vertices of V ( G + v ) (resp. V ( G + v )).If rk ( f G i + v ) = rk ( f G i ) and rk ( f G j + v ) = rk ( f G j ) + 1 for i = j . Without loss of generality, assume rk ( f G + v ) = rk ( f G ) and rk ( f G + v ) = rk ( f G ) + 1. By Claim 3 and Observation 1, we have p ( f G + v ) = p ( f G + v ) = 1. In view of Lemma 1.10, we obtain f G + v is switching equivalent to complete multipartitegraph or some −→ C ( t , t , t ). Furthermore, if f G + v is switching equivalent to complete multipartitegraph, then by rk ( f G + v ) = rk ( f G ), we have v is in the partition class containing at least two verticesof V ( G + v ). Combining rk ( f G + v ) = rk ( f G ) + 1 and Lemma 1.10, we obtain f G + v is switchingequivalent to some complete k ′ -partite graph satisfying k ′ ≥ v is in the partition class containingexactly one vertex of V ( G + v ).If rk ( f G i + v ) = rk ( f G i ) + 1 for each i = 1 ,
2, and − ψ ∗ Y <
0, then by Claim 3 and Observation 1, wehave p ( f G + v ) = p ( f G + v ) = 1. Combining Lemma 1.10 and rk ( f G i + v ) = rk ( f G i ) + 1 for each i = 1 , f G + v (resp. f G + v ) is switching equivalent to some complete k -partite (resp. k -partite)graph satisfying k ≥ k ≥
3) and v is in the partition class containing exactly one vertex of V ( G + v ) (resp. V ( G + v )).If rk ( f G i + v ) = rk ( f G i ) + 1 for each i = 1 ,
2, and − ψ ∗ Y >
0, then by Claim 3 and Observation 1, wehave p ( f G + v ) = 1 and p ( f G + v ) = 2. Combining Lemma 1.10 and rk ( f G + v ) = rk ( f G )+1, we have f G + v is switching equivalent to some complete ( r + 1)-partite graph K q ,q ,...,q r , satisfying r ≥ v is in thepartition class containing exactly one vertex. In view of Lemmas 1.10 and 3.11, we obtain f G is switchingequivalent to some complete k -partite graph K n ′ ,n ′ ,...,n ′ k and f G + v is switching equivalent to somecomplex unit gain graph K (0; n ′′ , . . . , n ′′ k ; a ∗ i , b ∗− i , c ∗ , d ∗− , k − a ∗ − b ∗ − c ∗ − d ∗ ). Thus e G is switchingequivalent to some complex unit gain graph K ( q , . . . , q r ; n ′′′ , . . . , n ′′′ k ; a i , b − i , c , d − , k − a − b − c − d ). Inthe following, we will prove at least two of a, b, c, d are equal to 0. Otherwise, we have abc = 0, abd = 0, acd = 0 or bcd = 0.Let H denote the Hermitian adjacency matrix of K ( q , . . . , q r ; n ′′′ , . . . , n ′′′ k ; a i , b − i , c , d − , k − a − b − c − d ). Suppose abc = 0 and note that r ≥
2, then H = i − i
10 0 − i i is a submatrix of H . By a direct calculation, we have p ( H ) = 3. Thus, p ( e G ) = p ( H ) ≥ p ( H ) = 3,a contradiction. Similarly, we also have abd = 0 , acd = 0 and bcd = 0. Thus, we have at least two of a, b, c, d are equal to 0.If d = 0, then K ( q , . . . , q r ; n ′′′ , . . . , n ′′′ k ; a i , b − i , c , − , k − a − b − c ) is a mixed graph. Hence, we have e G is switching equivalent to mixed graph K ( q , . . . , q r ; n ′′′ , . . . , n ′′′ k ; a i , b − i , c , − , k − a − b − c ).22f d >
0, we proceed by considering four cases.
Case 1. a, b, c are all equal to 0.In this case, there exists a diagonal matrix D with ( D ) uu = − u ∈ V ( G + v ). Other-wise, ( D ) uu = 1. It is easy to see that D − HD is the adjacency matrix of some simple graph K ( q , . . . , q r ; n ′′′ , . . . , n ′′′ k ; d ). Thus e G is switching equivalent to K ( q , . . . , q r ; n ′′′ , . . . , n ′′′ k ; d ). Case 2. a > b = c = 0.In this case, there exists a diagonal matrix D with ( D ) uu = i if u ∈ V ( G + v ). Otherwise,( D ) uu = 1. It is easy to see that D − HD is the Hermitian adjacency matrix of some mixed graph K ( q , . . . , q r ; n , . . . , n k ; d i , − i , a , − , k − a − d ) . Thus e G is switching equivalent to some mixed graph K ( q , . . . , q r ; n , . . . , n k ; d i , − i , a , − , k − a − d ) . Case 3. b > a = c = 0.In this case, there exists a diagonal matrix D with ( D ) uu = − i if u ∈ V ( G + v ). Otherwise,( D ) uu = 1. It is easy to see that D − HD is the Hermitian adjacency matrix of some mixed graph K ( q , . . . , q r ; n , . . . , n k ; 0 i , d − i , b , − , k − b − d ) . Thus e G is switching equivalent to some mixed graph K ( q , . . . , q r ; n , . . . , n k ; 0 i , d − i , b , − , k − b − d ) . Case 4. c > a = b = 0.In this case, there exists a diagonal matrix D with ( D ) uu = − i if u ∈ V ( G + v ). Otherwise,( D ) uu = 1. It is easy to see that D − HD is the Hermitian adjacency matrix of some mixed graph K ( q , . . . , q r ; n , . . . , n k ; c i , d − i , , − , k − c − d ) . Thus e G is switching equivalent to some mixed graph K ( q , . . . , q r ; n , . . . , n k ; c i , d − i , , − , k − c − d ) . Since simple graphs are mixed graphs, according to the above analysis, without loss of generality,assume that e G is switching equivalent to some mixed graph K ( q , . . . , q r ; n , . . . , n k ; a i , b − i , c , − , k − a − b − c ), where a, b, c are nonnegative integers such that at least one of them is equal to 0 and 1 ≤ a + b + c ≤ k .Note that [ v, V ( G )] is an edge cut of K ( q , . . . , q r ; n , . . . , n k ; a i , b − i , c , − , k − a − b − c ). We canperform a sequence of two-way switching and operations of taking the converse so that e G is switchingequivalent to one of the following mixed graphs:(i) K ( q , . . . , q r ; n , . . . , n k ; p );(ii) K ( q , . . . , q r ; n , . . . , n k ; a i , b − i , , − , k − a − b ), where a ≥ b ≥ K ( q , . . . , q r ; n , . . . , n k ; a i , − i , c , − , k − a − c ), where a ≥ c ≥ e G is switching equivalent to K ( q , . . . , q r ; n , . . . , n k ; p ), then according to Lemmas 3.1, 3.2 andCorollary 3.9, T e G is switching equivalent to K ( r ; k ; p ), where r ≥ , k ≥ , p ≥
1, satisfying one of thefollowing conditions:(1) p = 1;(2) p ≥ k − p ≤ p ≥ , k − p ≥ r + p + k − p − ≥
1. 23y the definition of twin reduction graph and G without pendant vertices, it is easy to see that e G is switching equivalent to K ( q , . . . , q r ; n , . . . , n k ; p ), where r ≥ , k ≥ , p ≥
1, satisfying one of theconditions stated in Theorem 1.2 “only if ” part (ii).If e G is switching equivalent to K ( q , . . . , q r ; n , . . . , n k ; a i , b − i , , − , k − a − b ), where a ≥ b ≥ T e G is switching equivalent to K ( r ; k ; a i , b − i , , − , s ), where r ≥ , k ≥ , a ≥ b ≥ , s = k − a − b , satisfying a = b = 1 and r = 2. By the definition of twin reductiongraph and G without pendant vertices, it is easy to see that e G is switching equivalent to K ( q , . . . , q r ; n , . . . , n k ; a i , b − i , , − , s )where r ≥ , k ≥ , a ≥ b ≥ , s = k − a − b, satisfying a = b = 1 and r = 2.If e G is switching equivalent to K ( q , . . . , q r ; n , . . . , n k ; a i , − i , c , − , k − a − c ), where a ≥ c ≥ T e G is switching equivalent to K ( r ; k ; a i , − i , c , − , s ), with r ≥ , k ≥ , a ≥ c ≥ , s = k − a − c , satisfying one of the following conditions:(1) a = c = 1 and either s = 0 or s = 1;(2) a = c = 1 , s = 2 and either r = 3 or r = 4;(3) a = c = 1, s = 3 and r = 3;(4) a = c = 1, s ≥ r = 2;(5) a = c = 2 , s = 0 and 2 ≤ r ≤ a = c = 2 , s = 1 and r = 2;(7) a = 3 , s = 0 and r = c = 2;(8) a = 4 , c = r = 2 and s = 0;(9) a ≥ , c = 1 and as − a + s ≤ r − .By the definition of twin reduction graph and G without pendant vertices, it is easy to see that e G isswitching equivalent to K ( q , . . . , q r ; n , . . . , n k ; a i , − i , c , − , s )with r ≥ , k ≥ , a ≥ c ≥ , s = k − a − c , satisfying one of the conditions stated in Theorem 1.2 “onlyif ” part (iv). This completes the proof. Acknowledgement
This work was supported by NSFC (Grant Nos. 11871479, 12071484), Hunan Provincial NaturalScience Foundation (Grant Nos. 2018JJ2479, 2020JJ 4675).
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