Hermitian adjacency matrix of the second kind for mixed graphs
aa r X i v : . [ m a t h . C O ] F e b Hermitian adjacency matrix of the second kind for mixed graphs ∗ Shuchao Li † , Yuantian Yu Faculty of Mathematics and Statistics, Central China Normal University, Wuhan 430079, P.R. China
Abstract : This contribution gives an extensive study on spectra of mixed graphs via its Hermitianadjacency matrix of the second kind introduced by Mohar [21]. This matrix is indexed by the verticesof the mixed graph, and the entry corresponding to an arc from u to v is equal to the sixth root of unity ω = i √ (and its symmetric entry is ω = − i √ ); the entry corresponding to an undirected edge is equalto 1, and 0 otherwise. The main results of this paper include the following: Some interesting propertiesare discovered about the characteristic polynomial of this novel matrix. Cospectral problems amongmixed graphs, including mixed graphs and their underlying graphs, are studied. We give equivalentconditions for a mixed graph that shares the same spectrum of its Hermitian adjacency matrix of thesecond kind ( H S -spectrum for short) with its underlying graph. A sharp upper bound on the H S -spectral radius is established and the corresponding extremal mixed graphs are identified. Operationswhich are called three-way switchings are discussed–they give rise to a large number of H S -cospectralmixed graphs. We extract all the mixed graphs whose rank of its Hermitian adjacency matrix of thesecond kind ( H S -rank for short) is 2 (resp. 3). Furthermore, we show that all connected mixed graphswith H S -rank 2 can be determined by their H S -spectrum. However, this does not hold for all connectedmixed graphs with H S -rank 3. We identify all mixed graphs whose eigenvalues of its Hermitian adjacencymatrix of the second kind ( H S -eigenvalues for short) lie in the range ( − α, α ) for α ∈ (cid:8) √ , √ , (cid:9) . Keywords:
Mixed graph; Spectral radius; Characteristic polynomial; Switching equivalence; Cospectral-ity; H S -rankAMS subject classification: 05C35, 05C12
1. Background
Investigation on the eigenvalues of graphs has a long history. In 1965, G¨unthard and Primas [16]published a paper on the spectra of trees, which probably was the first one on eigenvalues of graphs.From then on, the eigenvalue of graphs was widely used in mathematical chemistry [19], combinatorics[5,6,10,13,25], code-designs theory [1,9] and theoretical computer science [3,12] and so on. For thedetails, one may be referred to Guo and Mohar’s contribution [15].In the mathematical literature, one may see that the eigenvalues on directed graphs (digraphs forshort) are scarce. One of the main reasons is that one can not choose a suitable matrix associated withthe digraph D such that this matrix would best reflect its structure properties by its spectrum. Inthe last century, the adjacency matrix for a digraph D of order n was introduced, defined as an n × n (0 , A ( D ) = ( a ij ) with a ij = 1 if and only if there is an arc from v i to v j . This matrix attractedmuch attention. For the advances on this matrix, we refer the reader to the survey [7]. In fact, one isnot satisfied with this matrix. Clearly, A ( D ) is not symmetric. So many nice properties of symmetricmatrix are lost for A ( D ). A more natural definition for the adjacency matrix of a digraph was proposed ∗ S.L. acknowledges the financial support from the National Natural Science Foundation of China (Grant No. 11671164). † Corresponding author.
Email addresses : [email protected] (S.C. Li), [email protected] (Y.T. Yu).
1y Cavers et al. [8]. It is called the skew-symmetric adjacency matrix S ( D ), in which the ( i, j )-entry is1 if there is an arc from v i to v j , and its symmetric entry is − Hermitian adjacencymatrix ( of the first kind ) for a mixed graph, in which the ( i, j )-entry is the imaginary unit i if there isan arc from v i to v j , − i if there is an arc from v j to v i , 1 if v i v j is an undirected edge, and 0 otherwise.This matrix is Hermitian and has many nice properties. Some basic theory on spectra of mixed graphswas established via its Hermitian adjacency matrix of the first kind in [15,20]. For the advances onthe Hermitian adjacency matrix of the first kind for mixed graphs, one may be referred to [18] and thereferences cited in.In 2020, Mohar [21] introduced the Hermitian adjacency matrix of the second kind for mixed graphs:each arc directed from v i to v j contributes the sixth root of unity ω = i √ to the ( i, j )-entry in thematrix and contributes ¯ ω = − i √ to the ( j, i )-entry; each undirected edge between v i and v j contributes1 to the ( i, j )- (resp. ( j, i )-) entry, and 0 otherwise. Clearly, this novel matrix is a Hermitian matrix.It has real eigenvalues. Mohar [21] [21] showed that the interlacing theorem holds for a mixed graphand its induced mixed subgraphs; he also showed that, for a mixed bipartite graph, its H S -spectrum isabout 0; he established some relationship between the H S -spectral radius and the largest eigenvalue ofthis new matrix.In this article, we investigate some basic properties of the Hermitian adjacency matrix of the secondkind, which may be viewed as a continuance of Mohar’s work [21]. The first natural problem is tostudy some interesting properties for the characteristic polynomial of this novel matrix. In particular,we interpret all the coefficients of this characteristic polynomial. Based on this result, we can findrecursions for the characteristic polynomial of some Hermitian adjacency matrices of the second kind.Furthermore, H S -cospectral problems among mixed graphs, including mixed graphs and their underlyinggraphs, are studied (see Section 3 in detail).The H S - spectrum of a mixed graph M is the multiset of the eigenvalues of its Hermitian adjacencymatrix of the second kind, where the maximum modulus is called the H S - spectral radius of M . A sharpupper bound on the H S -spectral radius is established and the corresponding extremal mixed graphs areidentified (see Section 4).Two mixed graphs are called H S - cospectral , if they have the same H S -spectrum. We mainly considerthe H S -cospectrality between two mixed graphs which have the same underlying graph here. Operationswhich are called three-way switchings are discussed–they give rise to a large number of H S -cospectralmixed graphs. Some equivalent conditions for a mixed graph that shares the same H S -spectrum withits underlying graph are deduced (see Section 5).It is interesting to study the rank of the Hermitian adjacency matrix of the second kind ( H S -rank forshort). We extract all the mixed graphs whose H S -rank equals 2 (resp. 3). Furthermore, we show thatall connected mixed graphs with H S -rank 2 can be determined by their H S -spectrum. However, thisdoes not hold for all connected mixed graphs with H S -rank 3. This kind of questions locates Section 6.Despite many unperceptive properties that the Hermitian adjacency matrix of the second kindexhibits, it is challenging to derive combinatorial structure of the mixed graph from its H S -eigenvalues.In Section 7, we find all mixed graphs whose H S -eigenvalues lie in the range ( − α, α ) for α ∈ (cid:8) √ , √ , (cid:9) .2 . Some definitions and preliminaries In this paper, we consider only simple and finite graphs. For graph theoretic notation and terminologynot defined here, we refer to [27].Let G = ( V ( G ) , E ( G )) be a graph with vertex set V ( G ) and edge set E ( G ). The number of vertices n and edges | E ( G ) | in a graph are called the order and size of G , respectively. We say that two vertices i and j are adjacent (or neighbours ) if they are joined by an edge and we write i ∼ j . A k - partite graph is a graph whose set of vertices is discomposed into k disjoint sets such that no two vertices within thesame set are adjacent. As usual, let P n , C n and K n denote the path, cycle and complete graph on n vertices, respectively. We use kG to denote the disjoint union of k copies of G .A mixed graph M G is obtained from a simple graph G , the underlying graph of M G , by orientingeach edge of some subset E ⊆ E ( G ). It is obvious that M G is a simple graph if E = ∅ while M G is andirected graph if E = E ( G ). Thus, mixed graphs are the generalizations of simple graphs and directedgraphs. A mixed graph M G ′ is a mixed subgraph of M G if G ′ is a subgraph of G and the directionof each edge in M G ′ coincides with that in M G . For a vertex subset V ′ of V ( G ) , M G [ V ′ ] is a mixedsubgraph of M G induced on V ′ . The order (resp. size ) of M G is exactly the order (resp. size) of G . Amixed graph is called to be connected if its underlying graph is connected.We write an undirected edge as { u, v } and a directed edge (or an arc) from u to v as −→ uv. Usually,we denote an edge of M by uv if we does not concern whether it is directed or not. Then M G − u (resp. M G − uv ) is the mixed graph obtained from M G by deleting the vertex u ∈ V ( G ) (resp. edge uv ∈ E ( M G )). This notation is naturally extended if more than one vertex or edge are deleted.The degree d G ( u ) of a vertex u (in a graph G ) is the number of edges incident with it. In particular,the maximum degree is denoted by ∆( G ). The set of neighbours of a vertex u is denoted by N G ( u ). Givena mixed graph M = ( V ( M ) , E ( M )) , let N M ( v ) = { u ∈ V ( M ) : { u, v } ∈ E ( M ) } , N + M ( v ) = { u ∈ V ( M ) : −→ vu ∈ E ( M ) } and N − M ( v ) = { u ∈ V ( M ) : −→ uv ∈ E ( M ) } . Clearly, N G ( v ) = N M ( v ) ∪ N + M ( v ) ∪ N − M ( v ),where G is the underlying graph of M . In our context, two vertices u, v in a mixed graph are called tobe adjacent if they are adjacent in its underlying graph and we also denote it by u ∼ v. The degree ofa vertex in a mixed graph M G is defined to be the degree of this vertex in the underlying graph G .The Hermitian adjacency matrix of the second kind , written as H ( M G ) = ( h st ), of a mixed graph M G was proposed by Mohar [21]. It is defined as h st = i √ , if −−→ u s u t is an arc from u s to u t ; − i √ , if −−→ u t u s is an arc from u t to u s ;1 , if { u s , u t } is an undirected edge;0 , otherwise , where ± i √ are the sixth roots of unity, i is imaginary unit. The sixth root of unity emerges realisticallyacross applications. It appears in the definition Eisenstein integers; in relation to matroid theory, thesixth root matroids play a special role next to regular and binary matroids; see [23, 28] for details.The H S - rank of M G is the rank of H ( M G ). The characteristic polynomial of H ( M G ) , P M G ( x ) =det( xI − H ( M G )), is also called the characteristic polynomial of M G , while its roots are just the H S -eigenvalues of M G .Note that H ( M G ) is Hermitian, that is, H ∗ ( M G ) = H ( M G ), where H ∗ ( M G ) denotes the conjugatetranspose of H ( M G ). Then its eigenvalues are real. The collection of H S -eigenvalues of M G (with3 (cid:13) (cid:13) (cid:13) PSfrag replacements (a) (b) (c) (d)Figure 1: The mixed cycles: (a) is positive; (b) is semi-positive; (c) is semi-negative and (d) is negative.repetition) is called the H S - spectrum of M G . We denote the H S -eigenvalues of M G by λ (= λ ( M G )) > λ (= λ ( M G )) > · · · > λ n (= λ n ( M G )) . Two mixed graphs are called H S - cospectral if they have the same H S -spectrum. The H S - spectral radius of M , written as ρ ( M ), is defined as ρ ( M ) = max {| λ | , | λ n |} . Let M be a mixed graph, and let M C = v v v · · · v l − v l v be a mixed cycle of M . Then the weightof M C is defined by wt ( M C ) = h h · · · h ( l − l h l , where h jk is the ( v j , v k )-entry of H ( M ). As h jk ∈ n , i √ , − i √ o if there is an edge between v j and v k , and ± i √ are the sixth roots of unity. We have wt ( M C ) ∈ ( , i √ , − i √ , − , − − i √ , − i √ ) , (2.1)the set of all the sixth roots of unity, we denote this set by S . Note that if, for one direction, the weight ofa mixed cycle is α , then for the reversed direction its weight is α , the conjugate of α . For convenience,for a mixed cycle M C , select a direction for it (clockwise or anticlockwise). Then its weight can bedetermined uniquely. For a mixed cycle M C , it is positive (resp. negative ) if wt ( M C ) = 1 (resp. − semi-positive if wt ( M C ) ∈ n i √ , − i √ o , whereas it is semi-negative if wt ( M C ) ∈ n − i √ , − − i √ o .An example of positive (resp. semi-positive, semi-negative, and negative) mixed cylce is depicted inFigure 1. Furthermore, we call a mixed graph M positive , if each mixed cycle of M is positive.A mixed cycle is called even (resp. odd ) if its order is even (resp. odd). An elementary mixed graph is a mixed graph such that every component is either an (oriented) edge or a mixed cycle. A spanningelementary subgraph of a mixed graph M is an elementary mixed subgraph such that it has the samevertex set as that of M . We define that the rank (resp. corank ) of a mixed graph M G is just the rank(resp. corank) of its underlying graph G . That is, r ( M G ) = n − c, s ( M G ) = m − n + c, where n, m and c are the order, size and number of components of M G , respectively.Further on we need the following preliminary results. Lemma 2.1.
Let M be an elementary mixed graph with order n . If the components of M consist of N edges, N even mixed cycles and N odd mixed cycles. Then r ( M ) ≡ N + N ( mod . Proof.
If there are l mixed cycles of length c l , then the equation 2 N + P lc l = n shows that N ≡ n ( mod r ( M ) = n − c = n − N − ( N + N ) ≡ N + N (mod 2) . M be a mixed graph with connected components M , M , . . . , M t . Then H ( M ) can be writtenas H ( M ) = H ( M ) . . . H ( M t ) . Hence the following result is clear.
Lemma 2.2.
Let M be a mixed graph with connected components M , M , . . . , M t . Then P M ( x ) = t Y j =1 P M j ( x ) . Lemma 2.3.
Let M G be an n -vertex mixed graph of size m and let λ ≥ λ ≥ · · · ≥ λ n be its H S -eigenvalues. Then P nj =1 λ j = 2 m. Proof.
Let H = H ( M G ). Since H is Hermitian and has only entries 0 , , and ± i √ , we have H uv H vu = H uv H uv = 1whenever H uv = 0. This implies that the ( u, u )-diagonal entry in H is the degree of u in G . Hence n X j =1 λ j = tr ( H ) = X u ∈ V ( H ) uu = X u ∈ V d G ( u ) = 2 m, as desired.Suppose that λ ≥ λ ≥ · · · ≥ λ n and µ ≥ µ ≥ · · · ≥ µ n − t (where t ≥ λ l (1 ≤ l ≤ n ) and µ j (1 ≤ j ≤ n − t ) interlace iffor every s = 1 , . . . , n − t , we have λ s ≥ µ s ≥ λ s + t . The following interlacing theorem is well-known.
Theorem 2.4 ([15]) . If H is a Hermitian matrix and B is a principal submatrix of H , then theeigenvalues of B interlace those of H . Theorem 2.4 implies that the H S -eigenvalues of any induced mixed subgraph interlace those of themixed graph itself. Corollary 2.5 ([21]) . The H S -eigenvalues of an induced mixed subgraph interlace the H S -eigenvaluesof the mixed graph.
3. The characteristic polynomial of mixed graphs
In this section, we study the determinant of H ( M ) and interpret the coefficients of P M ( x ), which aremotivated by those of Hermitian adjacency matrices of the first kind (see [20]). Furthermore, we givesome consequences of H S -cospectra and recurrence relations on P M ( x ).5 heorem 3.1. Let M be a mixed graph with vertex set V = { v , v , . . . , v n } and let H = H ( M ) . Then det H = X M ′ ( − r ( M ′ )+ l s ( M ′ )+ l n ( M ′ ) · l p ( M ′ )+ l n ( M ′ ) , where the summation is over all spanning elementary subgraphs M ′ of M and l p ( M ′ ) , l n ( M ′ ) , l s ( M ′ ) are the number of positive, negative, semi-negative cycles in M ′ , respectively.Proof. According to the definition of determinant, we havedet H = X π sgn ( π ) h π (1) h π (2) · · · h nπ ( n ) , where the summation is over all permutations π of 1 , , . . . , n . More formally, consider a term h π (1) h π (2) · · · h nπ ( n ) , which equals zero if h kπ ( k ) = 0, i.e., there is no edge between v k and v π ( k ) for some k ∈{ , , . . . , n } . Thus, if the term is non-zero, in the cycle decomposition of π , each cycle ( jk ) of length 2corresponds to the factor h jk h kj , and signifies an edge v j v k . Each cycle ( pqr · · · t ) of length greater than2 corresponds to the factor h pq h qr · · · h tp , and signifies a mixed cycle v p v q · · · v t v p in M ′ . Consequently,each non-zero term in the determinant expansion gives rise to an elementary mixed subgraph M ′ of M ,with V ( M ′ ) = V ( M ). That is, M ′ is a spanning elementary subgraph of M . The sign of a permutation π is ( − N e , where N e is the number of even cycles (i.e. cycles with even length) in π . Clearly, N e = N + N , where N and N are the number of edge components and even cycle components in M ′ , respectively. By Lemma 2.1, the sign of π is equal to ( − r ( M ′ ) .Each spanning elementary subgraph M ′ gives rise to several permutations π for which the corre-sponding term in the determinant expansion is non-zero. The number of such π arising from a given M ′ is 2 s ( M ′ ) , since for each mixed cycle-component in M ′ there are two ways of choosing the correspondingcycle in π . Furthermore, if for some direction of a permutation π , a mixed cycle-component has weight i √ (or − i √ ), then for the other direction the mixed cycle-component has weight − i √ (or i √ )and vice versa. Thus, the mixed cycle-component has weight in each direction on average. If for somedirection of a permutation π , a mixed cycle-component has weight − i √ (or − − i √ ), then for theother direction the mixed cycle-component has weight − − i √ (or − i √ ) and vice versa. Thus, themixed cycle-component has weight − in each direction on average. Similarly, if for some direction of apermutation π , a mixed cycle-component has weight 1 (or − −
1) too.As s ( M ′ ) = l p ( M ′ ) + l n ( M ′ ) + l s ( M ′ ) + l s ′ ( M ′ ), where l s ′ ( M ′ ) is the number of semi-positive cyclesin M ′ . Thus each M ′ contributes( − r ( M ′ ) · s ( M ′ ) · ( 12 ) l s ′ ( M ′ ) · ( −
12 ) l s ( M ′ ) · ( − l n ( M ′ ) = ( − r ( M ′ )+ l s ( M ′ )+ l n ( M ′ ) · l p ( M ′ )+ l n ( M ′ ) to the determinant and the result follows.Given an n -vertex mixed graph M , we give a description of all the coefficients of the characteristicpolynomial P M ( x ). For convenience, let P M ( x ) = x n + c x n − + c x n − + · · · + c n , (3.1)where c , . . . , c n are real. 6 heorem 3.2. Let M be a mixed graph of order n , then the coefficients of the characteristic polynomial P M ( x ) in (3.1) are given by c k = X M ′ ( − − k + r ( M ′ )+ l s ( M ′ )+ l n ( M ′ ) · l p ( M ′ )+ l n ( M ′ ) , where the summation is over all elementary subgraphs M ′ of M with k vertices and l p ( M ′ ) , l n ( M ′ ) , l s ( M ′ ) are the number of positive, negative, semi-negative cycles in M ′ , respectively.Proof. According to the expansion of det( xI − H ( M )), where I is the unit matrix of order n . Thenumber ( − k c k is the sum of all principal minors of H ( M ) with k rows and columns. Each such minoris the determinant of the Hermitian-adjacency matrix of the second kind for an induced subgraph of M with k vertices. Any elementary subgraph with k vertices is contained in precisely one of these inducedsubgraphs, and so, by applying Theorem 3.1 to each minor, we obtain the required result.From Theorem 3.2 we can deduce that c = 0 and c = −| E ( M ) | for each mixed graph M . As M has no elementary subgraph of order 1, and has | E ( M ) | elementary subgraphs of order 2, each of whichis an edge and hence contributes − c . In [21], Mohar showed that if M is a mixed graph whoseunderlying graph is bipartite, then the H S -spectrum of M is symmetric about 0. This can be easilyseen from Theorem 3.2, as M has no elementary subgraph of odd order, c k = 0 if k is odd.According to Theorem 3.2, we also have the following corollaries. Corollary 3.3.
Let M G be a positive mixed graph, then M G and G are H S -cospectral.Proof. Let M G ′ be an elementary subgraph of M G . Then G ′ is an elementary subgraph of G and viceversa. Hence( − r ( M G ′ )+ l s ( M G ′ )+ l n ( M G ′ ) · l p ( M G ′ )+ l n ( M G ′ ) = ( − r ( M G ′ ) · l p ( M G ′ ) = ( − r ( G ′ ) · l p ( G ′ ) , which implies that P M G ( x ) = P G ( x ), and so M G , G are H S -cospectral. Corollary 3.4.
Let G be a simple graph with cut edges, and let M , M be the mixed graphs with theunderlying graph G , and differ only on some cut edges of G . Then M and M are H S -cospectral.Proof. Let S be the set of cut edges that differ in M and M . If an elementary subgraph M ′ of M contains no edge from S , then M ′ is also an elementary subgraph of M . Obviously, we have( − r ( M ′ )+ l s ( M ′ )+ l n ( M ′ ) · l p ( M ′ )+ l n ( M ′ ) = ( − r ( M ′ )+ l s ( M ′ )+ l n ( M ′ ) · l p ( M ′ )+ l n ( M ′ ) . If an elementary subgraph M ′ = M G ′ of M contains some edges S ′ from S , then correspondingly thereis an elementary subgraph M ′′ = M G ′′ of M that satisfies G ′ ∼ = G ′′ and differs from M ′ only on S ′ andvice versa. As cut edges contained in no cycle, we have( − r ( M ′ )+ l s ( M ′ )+ l n ( M ′ ) · l p ( M ′ )+ l n ( M ′ ) = ( − r ( M ′′ )+ l s ( M ′′ )+ l n ( M ′′ ) · l p ( M ′′ )+ l n ( M ′′ ) . Thus, c k ( M ) = c k ( M ) for all integer k , and so M , M are H S -cospectral.As we know, each edge of a forest is a cut edge, hence we have Corollary 3.5. If M T is a mixed forest, then M T and T are H S -cospectral.
7n the following, we will give two reduction formulas for P M ( x ), which are similar to those ofadjacency matrices of simple graphs [11, Section 2] and those of Hermitian adjacency matrices of thefirst kind for mixed graphs [2]. Theorem 3.6.
Let M be a mixed graph, and let u be a vertex of M . Then P M ( x ) = xP M − u ( x ) − X v ∼ u P M − v − u ( x ) − X Z ∈ C ( u ) (cid:16) wt ( Z ) + wt ( Z ) (cid:17) P M − V ( Z ) ( x ) , (3.2) where C ( u ) is the set of mixed cycles containing u , wt ( Z ) is the weight of Z in a direction.Proof. We prove our result by defining a one-to one correspondence between elementary subgraphs M ′ that contribute to a coefficient on the left-hand side of (3.2), and those M ′′ that contribute to acoefficient on the right-hand side. We distinguish three possible cases for an elementary subgraph M ′ of M on k vertices:(i) if u / ∈ V ( M ′ ), then M ′′ = M ′ , regarded as a subgraph of M − u ;(ii) if u lies in a component of an edge uv of M ′ , then M ′′ = M ′ − u − v, regarded as a subgraph of M − u − v ;(iii) if u lies in a mixed cycle Z of M ′ , then M ′′ = M ′ − V ( Z ) , regarded as a subgraph of M − V ( Z ).Now, by applying Theorem 3.2 to each of the graphs that play an essential role in (3.2), we can showthat if M ′ contributes c to the coefficient of x n − k on the left, then M ′′ contributes c to the coefficientof x n − k on the right.In case (i), M ′′ contributes c to the coefficient of x n − − k in P M − u ( x ), hence contributes c to thecoefficient of x n − k in xP M − u ( x ). Note that M ′′ does not contribute to the coefficient of x n − k in theremaining terms, hence M ′′ contributes c to the coefficient of x n − k on the right.In case (ii), M ′′ is an elementary subgraph of M − u − v with v ∼ u . Its contribution to the coefficientof x ( n − − ( k − (= x n − k ) in P M − v − u ( x ) is( − k − · ( − r ( M ′′ )+ l s ( M ′′ )+ l n ( M ′′ ) · l p ( M ′′ )+ l n ( M ′′ ) = − ( − k · ( − r ( M ′ )+ l s ( M ′ )+ l n ( M ′ ) · l p ( M ′ )+ l n ( M ′ ) = − c. As M ′ and M ′′ have the same mixed cycles and r ( M ′ ) − r ( M ′′ ) = 1. Moreover, M ′′ does not contributeto the coefficient of x n − k in the remaining terms, and so M ′′ contributes c to the coefficient of x n − k onthe right.In case (iii), M ′′ is an elementary subgraph of M ′′ − V ( Z ) with Z ∈ C ( u ). If | V ( Z ) | = r , then thecontribution of M ′′ to the coefficient of x ( n − r ) − ( k − r ) (= x n − k ) in P M − V ( Z ) ( x ) is( − k − r · ( − r ( M ′′ )+ l s ( M ′′ )+ l n ( M ′′ ) · l p ( M ′′ )+ l n ( M ′′ ) . We know that r ( M ′ ) − r ( M ′′ ) = ( | V ( M ′ ) | − c ( M ′ )) − ( | V ( M ′′ ) | − c ( M ′′ )) = r − . If Z is a positive cycle, then l p ( M ′ ) − l p ( M ′′ ) = 1 and l n ( M ′ ) = l n ( M ′′ ) , l s ( M ′ ) = l s ( M ′′ ) . Hence,( − k − r · ( − r ( M ′′ )+ l s ( M ′′ )+ l n ( M ′′ ) · l p ( M ′′ )+ l n ( M ′′ ) = − ( − k ·
12 ( − r ( M ′ )+ l s ( M ′ )+ l n ( M ′ ) · l p ( M ′ )+ l n ( M ′ ) = − c. M ′′ to the coefficient of x n − k in (cid:16) wt ( Z ) + wt ( Z ) (cid:17) P M − V ( Z ) ( x ) is − c . Similarly, we can prove that if Z is a negative, semi-positive or semi-negative cycle, then thecontribution of M ′′ to the coefficient of x n − k in (cid:16) wt ( Z ) + wt ( Z ) (cid:17) P M − V ( Z ) ( x ) is also − c . Besides, M ′′ does not contribute to the coefficient of x n − k in the remaining terms, hence M ′′ contributes c to thecoefficient of x n − k on the right.This completes the proof. Theorem 3.7.
Let M be a mixed graph, and let uv be a mixed edge of M . Then P M ( x ) = P M − uv ( x ) − P M − v − u ( x ) − X Z ∈ C ( uv ) (cid:16) wt ( Z ) + wt ( Z ) (cid:17) P M − V ( Z ) ( x ) , where C ( uv ) is the set of mixed cycles containing uv , wt ( Z ) is the weight of Z in a direction.Proof. The proof is similar to the proof of Theorem 3.6, and we omit it here.
Corollary 3.8.
Let M be a mixed graph with uv being a cut edge of its underlying graph, and let M , M be two components of M − uv with u ∈ V ( M ) , v ∈ V ( M ) . Then P M ( x ) = P M ( x ) P M ( x ) − P M − u ( x ) P M − v ( x ) . Proof.
According to Theorem 3.7, we have P M ( x ) = P M − uv ( x ) − P M − v − u ( x ) , as uv is contained in no mixed cycle of M . By Lemma 2.2, P M − uv ( x ) = P M ( x ) P M ( x ) , P M − v − u ( x ) = P M − u ( x ) P M − v ( x ) . This completes the proof.This result is the same as the corresponding result for the simple graphs which has been proved in[11, Section 2] by another method. More reduction formulas for P M ( x ) which are the same as the caseof simple graphs can be seen in [11, Section 2].A set G of graphs is called a class of cospectrum if all the graphs in G have the same spectrum. Thenext theorem gives the number of classes of H S -cospectrum of mixed graphs with a same underlyinggraph G , where G is a unicyclic graph (i.e., a connected graph with only one cycle). Theorem 3.9.
Let G be a unicyclic simple graph, let G be the set of all mixed graphs whose underlyinggraph is G . Then G can be partitioned into four classes of H S -cospectrum.Proof. For each M ∈ G , let M C be the unique mixed cycle of M , and let uv be an edge on M C . Thenby Theorem 3.7, we have P M ( x ) = P M − uv ( x ) − P M − v − u ( x ) − (cid:16) wt ( M C ) + wt ( M C ) (cid:17) P M − V ( M C ) ( x ) . As M − uv, M − v − u and M − V ( M C ) are all mixed forests. By Corollary 3.5, for all M, M ′ ∈ G , onehas P M − uv ( x ) = P M ′ − uv ( x ) , P M − v − u ( x ) = P M ′ − v − u ( x ) , P M − V ( M C ) ( x ) = P M ′ − V ( M ′ C ) ( x ) . P M ( x ) only depends on the value of wt ( M C ) + wt ( M C ). According to (2.1), we know that wt ( M C ) + wt ( M C ) ∈ {± , ± } , i.e., wt ( M C ) + wt ( M C ) has four possible values. This implies that for all M ∈ G , there are four possible characteristic polynomials for M , each characteristic polynomial gives aclass of H S -cospectrum.This completes the proof.In view of the proof of Theorem 3.9, the following result is clear. Corollary 3.10.
All the positive (resp. negative, semi-positive, semi-negative) cycles of order n are H S -cospectral, which constitute just four different classes of H S -cospectrum of mixed graphs with underlyinggraph C n . The graphs depicted in Figure 1 clearly lie in different classes of H S -cospectrum with underlyinggraph C .
4. An upper bound for the H S -spectral radius In this section, we show that ρ ( M G ) is bounded above by ∆( G ) and when G is connected, we characterizethe mixed graphs attaining this bound. Recall that S = ( , i √ , − i √ , − , − − i √ , − i √ ) , which will be used in this section and the subsequent section. Theorem 4.1.
Let M be an n -vertex mixed graph whose underlying graph is G . Then ρ ( M ) ∆( G ) .When G is connected, the equality holds if and only if G is ∆( G ) -regular and one can partition V ( M ) into six (possibly empty) parts V , V − , V i √ , V − i √ , V − i √ , V − − i √ such that one of the followingholds: (i) For j ∈ S , the induced mixed graph M [ V j ] contains only undirected edges and each of the rest edgesin E ( M ) \ ( S j ∈ S E ( M [ V j ])) is an arc −→ uv satisfying u ∈ V j and v ∈ V − i √ · j for some j ∈ S ; seeFig. . (ii) For j ∈ S , the induced mixed graph M [ V j ] is an independent set; every undirected edge { u, v } of M satisfies u ∈ V j and v ∈ V − j for some j ∈ S , and every arc −→ uv of M satisfies u ∈ V j and v ∈ V − i √ · j for some j ∈ S ; see Figure .Proof. Let H = H ( M ) and let x = ( x , x , . . . , x n ) T be an eigenvector corresponding to the eigenvalue λ of H . Associate a labeling of vertices of M (with respect to x ) in which x i is a label of v i . Withoutloss of generality, let | x | = max {| x i | : 1 i n } . On the one hand, we consider the first entry of H x :( H x ) = X v i ∈ N M ( v ) x i + 1 + i √ X v j ∈ N + M ( v ) x j + 1 − i √ X v k ∈ N − M ( v ) x k . On the other hand, from H x = λ x , we obtain( H x ) = λx . (4.1)10 (cid:13)(cid:13) (cid:13) (cid:13)(cid:13) (cid:13)(cid:13)(cid:13) (cid:13) (cid:13)(cid:13) PSfrag replacements V V V i √ V i √ V − i √ V − i √ V − V − V − − i √ V − − i √ V − i √ V − i √ Figure 2: Cases (i) and (ii) of Theorem 4.1.Thus, | λx | = | ( H x ) | = (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) X v i ∈ N M ( v ) x i + 1 + i √ X v j ∈ N + M ( v ) x j + 1 − i √ X v k ∈ N − M ( v ) x k (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) X v i ∈ N M ( v ) | x i | + (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) i √ (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) X v j ∈ N + M ( v ) | x j | + (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) − i √ (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) X v k ∈ N − M ( v ) | x k | (4.2) X v i ∈ N M ( v ) | x | + X v j ∈ N + M ( v ) | x | + X v k ∈ N − M ( v ) | x | (4.3)= d G ( v ) | x | ∆( G ) | x | . (4.4)Hence, | λ | ∆( G ). Note that λ is an arbitrary H S -eigenvalue of M , and by the definition of H S -spectralradius, we have ρ ( M ) ∆( G ). In what follows, we characterize all the mixed graphs attaining thisbound if the underlying graph G is connected.Note that ρ ( M ) = ∆( G ) holds if and only if equalities above must hold through out. We see thatthe equality in (4.4) holds if and only if d G ( v ) = ∆( G ), whereas the equality in (4.3) holds if and onlyif | x k | = | x | for all v k ∈ N G ( v ). (4.5)Since the choice of v is arbitrary among all vertices attaining the maximum absolute value in x , wemay apply this same discussion to any vertex adjacent to v in G . Therefore, G is ∆( G )-regular.Note that G is connected. One has | x k | = | x | for all v k ∈ V ( M ). We may normalize x such that x = 1. Hence, | x i | = 1 for i ∈ { , , . . . , n } . The inequality in (4.2) follows from the triangle inequalityfor sums of complex numbers, and so equality holds if and only if every complex number in the followingset W has the same argument, where W = (cid:8) x i : v i ∈ N M ( v ) (cid:9) ∪ ( i √ x j : v j ∈ N + M ( v ) ) ∪ ( − i √ x k : v k ∈ N − M ( v ) ) . (4.6)11igure 3: The semi-negative triangle.Now we consider (4.1). The equality in (4.1) holds if and only if every complex number in W hasthe same argument as λx . There are three cases for λ : λ = 0 , λ > λ <
0. Since we are boundingabove the H S -spectral radius, and the only mixed graph with ρ ( M ) = 0 is the empty graph, it sufficesto consider the following two cases. Case 1. λ > . In this case, if ρ ( M ) = ∆( G ), then together with (4.1) we have ( H x ) = ∆( G ) x .Combining with (4.6) we deduce that every complex number in W is just x and is thus equal to 1. Weconclude that x i = , if v i ∈ N M ( v ); − i √ , if v i ∈ N + M ( v ); i √ , if v i ∈ N − M ( v ) . Repeating the argument at a vertex v j such that x j = i √ gives x i = i √ , if v i ∈ N M ( v j );1 , if v i ∈ N + M ( v j ); − i √ , if v i ∈ N − M ( v j ) . Similar argument can be applied to x j = − i √ , − , − − i √ or − i √ . From this we conclude that V ( M ) is partitioned into V ∪ V − ∪ V i √ ∪ V − i √ ∪ V − i √ ∪ V − − i √ according to the value of x j , and so condition (i) holds. Case 2. λ < . In this case, if ρ ( M ) = ∆( G ), then together with (4.1) we have ( H x ) = − ∆( G ) x .Combining with (4.6) we obtain that every complex number in W is just − x and thus equals −
1. Weconclude that x i = − , if v i ∈ N M ( v ); − i √ , if v i ∈ N + M ( v ); − − i √ , if v i ∈ N − M ( v ) . Repeating the discussion at a vertex v j with x j = i √ yields x i = − − i √ , if v i ∈ N M ( v j ); − , if v i ∈ N + M ( v j ); − i √ , if v i ∈ N − M ( v j ) . x j = − i √ , − , − − i √ and − i √ , respectively, gives that V ( M ) hasa partition V ∪ V − ∪ V i √ ∪ V − i √ ∪ V − i √ ∪ V − − i √ , which is based on the value of x j , and so condition (ii) holds.Now, we consider the converse for the two cases of the theorem. Let M be a mixed graph whoseunderlying graph is k -regular. Assume that V ( M ) has a partition S j ∈ S V j satisfying condition (i) or(ii).Let x be the vector indexed by the vertices of M such that x i = j if v i ∈ V j , where j ∈ S . Then it iseasy to see that for every vertex v i we have ( H x ) i = kx i (by item (i)) or ( H x ) i = − kx i (by item (ii)).Thus x is an eigenvector of H with eigenvalue k or − k , and so ρ ( M ) = k = ∆( G ). Then the bound istight as claimed.For simple graphs, ρ ( G ) is always larger or equal to the average degree. However, for mixed graphs, ρ ( M G ) can be smaller than the minimum degree in G . An example is the semi-negative triangle shownin Figure 3, whose characteristic polynomial is x − x + 1 (based on Theorem 3.2). Clearly, its H S -spectral radius is less than 2, while the minimum degree of its underlying graph is 2. Of course, thisanomaly is also confirmed by Theorem 4.1, since the semi-negative triangle shown in Figure 3 does nothave the structure as depicted in Figure 2.
5. Switching equivalence and H S -cospectrality In this section, we focus on properties of mixed graphs that are H S -cospectral and introduce someoperations on mixed graphs that preserve the H S -spectrum. In particular, we are inspired to studymixed graph operations that preserve the H S -spectrum and conserve the underlying graph. We tryto demonstrate the spectral information about the underlying graph by looking at some H S -spectrumpreserving operations that do not change the underlying graph.In view of the characteristic polynomial, Corollaries 3.3-3.5 reveal the H S -cospectrality betweenmixed graphs having the same underlying graph. In fact, all of them can be generalized by a similaritytransformation (based on the structure of Theorem 4.1(i)).Suppose that the vertex set of M is partitioned into six (possibly empty) sets, V ( M ) = V ∪ V − ∪ V i √ ∪ V − i √ ∪ V − i √ ∪ V − − i √ . (5.1)An arc −→ xy or an undirected edge { x, y } is said to be of type ( j, k ) for j, k ∈ S if x ∈ V j and y ∈ V k . Thepartition is said to be admissible if both of the following two conditions hold:(i) Each undirected edge is one of the type ( j, j ) , (cid:16) j, i √ · j (cid:17) for j ∈ S ;(ii) Each arc is one of the type ( j, j ) , (cid:16) j, − i √ · j (cid:17) or (cid:16) j, − − i √ · j (cid:17) for j ∈ S . A three-way switching with respect to an admissible partition (5.1) is the operation of changing M intothe mixed graph M ′ by making the changes in what follows (see Fig. 4):(i) replacing each undirected edge of type (cid:16) j, i √ · j (cid:17) with an arc directed from V j to V i √ · j for j ∈ S ; 13 . Switching equivalence and -cospectrality In this section, we focus on properties of mixed graphs that are -cospectral and introduce someoperations on mixed graphs that preserve the -spectrum. In particular, we are inspired to studymixed graph operations that preserve the -spectrum and conserve the underlying graph. We tryto demonstrate the spectral information about the underlying graph by looking at some -spectrumpreserving operations that do not change the underlying graph.In view of the characteristic polynomial, Corollaries 3.3-3.5 reveal the -cospectrality betweenmixed graphs having the same underlying graph. In fact, all of them can be generalized by a similaritytransformation (based on the structure of Theorem 4.1(i)).Suppose that the vertex set of is partitioned into six (possibly empty) sets,) =
1+ 1+ (5.1)An arc −→ xy or an undirected edge x, y is said to be of type j, k ) for j, k if and . Thepartition is said to be admissible if both of the following two conditions hold: V V V i √ V i √ V − i √ V − i √ V − V − V − − i √ V − − i √ V − i √ V − i √ Figure 4: Three-way switching with respect to an admissible partition.(i) Each undirected edge is one of the type ( j, j j, for(ii) Each arc is one of the type ( j, j j, or j, for three-way switching with respect to an admissible partition (5.1) is the operation of changing intothe mixed graph by making the changes in what follows (see Fig. 4):(i) replacing each undirected edge of type j, with an arc directed from to for(ii) replacing each arc of type j, with an undirected edge for(iii) reversing the direction of each arc of type j, for Theorem 5.1.
If a partition in (5.1) is admissible, then the mixed graph obtained from by thethree-way switching is -cospectral with M. Figure 4: Three-way switching with respect to an admissible partition.(ii) replacing each arc of type (cid:16) j, − i √ · j (cid:17) with an undirected edge for j ∈ S ;(iii) reversing the direction of each arc of type (cid:16) j, − − i √ · j (cid:17) for j ∈ S . Theorem 5.1.
If a partition in (5.1) is admissible, then the mixed graph M ′ obtained from M by thethree-way switching is H S -cospectral with M. Proof.
We use a similarity transformation with the diagonal matrix D whose ( v, v )-entry D v is equal to j ( ∈ S ) if v ∈ V j . Let H = H ( M ). The entries of the matrix H ′ = D − HD are given by H ′ uv = D − u H uv D v . It is clear that H ′ is Hermitian whose non-zero elements are in S , the set of all the sixth roots of unity.Admissibility is needed here so that H ′ has no entry in n − , − ± i √ o . To see it, note that theentries within the parts in (5.1) remain unchanged. If { u, v } is an undirected edge of type (cid:16) j, i √ · j (cid:17) for some j ∈ S , then H ′ uv = j − H uv · i √ · j = 1 + i √ . Similarly, H ′ uv = j − H uv · − i √ j = 1 , if −→ uv is an arc of type ( j, − i √ · j ) for some j ∈ S ; j − H uv · − − i √ j = − i √ , if −→ uv is an arc of type ( j, − − i √ · j ) for some j ∈ S .It turns out that H ′ is the Hermitian adjacency matrix of the second kind for M ′ . As H ′ is similarto H , M ′ is H S -cospectral with M. There is a special case of the three-way switching in which four of the six sets are empty: Let V ( M ) = V k ∪ V l be a partition which has undirected edges (possibly empty) and directed edges (possiblyempty) in one direction only between V k and V l . This special three-way switching replaces each directededge between V k and V l by an undirected edge, and replaces each undirected edge between V k and V l by a directed edge in the direction opposite to the direction of former directed edges (see Fig. 5), where( k, l ) ∈ { (1 , − i √ ) , ( − i √ , − − i √ ) , ( − − i √ , − , ( − , − i √ ) , ( − i √ , i √ ) , ( i √ , } . roof. We use a similarity transformation with the diagonal matrix whose ( v, v )-entry is equal to) if . Let ). The entries of the matrix HD are given by uv uv It is clear that is Hermitian whose non-zero elements are in , the set of all the sixth roots of unity.Admissibility is needed here so that has no entry in . To see it, note that theentries within the parts in (5.1) remain unchanged. If u, v is an undirected edge of type j, for some then uv uv uv uv = 1 if −→ uv is an arc of type ( j, ) for some uv if −→ uv is an arc of type ( j, ) for someIt turns out that is the Hermitian adjacency matrix of the second kind for . As is similarto is -cospectral with M. There is a special case of the three-way switching in which four of the six sets are empty: Let) = be a partition which has undirected edges (possibly empty) and directed edges (possiblyempty) in one direction only between and . This special three-way switching replaces each directededge between and by an undirected edge, and replaces each undirected edge between andby a directed edge in the direction opposite to the direction of former directed edges (see Fig. 5), where k, l ∈ { (1 1)
1+ 1+ 1+ 1+ V k V k V l V l Figure 5: A special case of the three-way switching.Given a mixed graph , let be its converse (the graph obtained by reversing all the arcs of). It is immediate from the definition of the Hermitian adjacency matrix of the second kind that) = . This implies the following result.
Theorem 5.2.
A mixed graph and its converse are -cospectral.
Two mixed graphs and are switching equivalent if one can be obtained from the other by asequence of three-way switchings and operations of taking the converse.Our next result characterizes the mixed graph -cospectral to its underlying graph, which ismotivated by [22].
Theorem 5.3.
Let be a connected simple graph of order and let be a mixed graph whoseunderlying graph is a spanning subgraph of . Then the following statements are equivalent: (a) and are -cospectral. (b) ) = 14
Figure 5: A special case of the three-way switching.Given a mixed graph M , let M c be its converse (the graph obtained by reversing all the arcs of M ). It is immediate from the definition of the Hermitian adjacency matrix of the second kind that H ( M c ) = H ( M ) T . This implies the following result. Theorem 5.2.
A mixed graph M and its converse are H S -cospectral. Two mixed graphs M and M are switching equivalent if one can be obtained from the other by asequence of three-way switchings and operations of taking the converse.Our next result characterizes the mixed graph H S -cospectral to its underlying graph, which ismotivated by [22]. Theorem 5.3.
Let G be a connected simple graph of order n and let M = M G be a mixed graph whoseunderlying graph G is a spanning subgraph of G . Then the following statements are equivalent: (a) G and M are H S -cospectral. (b) λ ( G ) = λ ( M ) . (c) G = G , and the vertex set of M has a partition S j ∈ S V j such that the following holds: For j ∈ S ,the induced subgraph M [ V j ] contains only undirected edges; each of the rest edges uv of M is anarc −→ uv with u ∈ V j and v ∈ V − i √ · j for some j ∈ S . (d) G and M are switching equivalent.Proof. Clearly, (a) implies (b), and (d) implies (a). By the definition of three-way switching, (c) implies(d) directly. Hence, it suffices to show that (b) implies (c).Assume that (b) holds. Let H = H ( G ) , H ′ = H ( M ) and let x be a normalized eigenvector of H corresponding to λ ( G ). This means that H x = λ ( G ) x , x T x = 1 and x T H x = λ ( G ). By Perron-Frobenius Theorem (see [4]), we may assume that x is real and positive. Then it is uniquely determined.Similarly, let y ∈ C n be a normalized eigenvector of H ′ corresponding to λ ( M ) and let z ∈ R n bedefined by z i = | y i | , i ∈ { , . . . , n } . Then, λ ( M ) = y T H ′ y = n X j =1 n X k =1 ( H ′ ) jk y j y k (5.2) n X j =1 n X k =1 | ( H ′ ) jk | z j z k (5.3) n X j =1 n X k =1 ( H ) jk z j z k (5.4) n X j =1 n X k =1 ( H ) jk x j x k (5.5)= λ ( G ) . (5.6)15ote that λ ( G ) = λ ( M ). Hence, equalities must hold throughout. The equality in (5.5) holds if andonly if z = x , since x is the unique positive normalized vector that attains the maximum x T H x . Theequality in (5.4) holds if and only if ( H ) jk = | ( H ′ ) jk | for all j, k ∈ { , . . . , n } , which is equivalent to saythat no edge has been removed, i.e., G = G . Finally, the equality in (5.3) holds if and only if( H ′ ) jk y j y k = | ( H ′ ) jk | z j z k = | ( H ′ ) jk || y j || y k | (5.7)for every edge v j v k . Since y = 0, without loss of generality, we can assume that y ∈ R + , one has y / | y | = 1. Then in view of Eq. (5.7), we can see, if v k ∈ N M ( v ) , H ′ k = 1, then y k / | y k | = 1; if v k ∈ N + M ( v ) , H ′ k = i √ , then y k / | y k | = − i √ ; if v k ∈ N − M ( v ) , H ′ k = − i √ , then y k / | y k | = i √ . Note that G is connected. Then repeating the above argument shows that y k / | y k | ∈ S for k ∈{ , . . . , n } . Let V j = { v k ∈ V ( M ) | y k / | y k | = j } , j ∈ S . Then they construct a partition of V ( M ). It isstraightforward to check that the edges within and between the parts are as claimed in (c).This completes the proof.
6. Characterizing mixed graphs with H S -rank or H -rank of a mixed graph, we mean the rank of its Hermitian adjacency matrix of thefirst kind, and when we say the H S -rank of a mixed graph, we mean the rank of its Hermitian adjacencymatrix of the second kind.Mohar [22] determined all the mixed graphs with H -rank 2, and constructed a class of mixed graphswhich can not be determined by their Hermitian spectra. Wang, Yuan and Li [26] determined all themixed graphs with H -rank 3, and they also showed that all connected mixed graphs with H -rank 3are determined by their Hermitian spectra. Inspired directly from [22, 26], we focus on determiningall mixed graphs with H S -rank 2 and 3, respectively. Furthermore, we show that all connected mixedgraphs with H S -rank 2 can be determined by their H S -spectrum. However, this does not hold for allconnected mixed graphs with H S -rank 3.Let M be a mixed graph of order n , the H S -rank of M is denoted by ξ ( M ), and the nullity ofthe Hermitian adjacency matrix of the second kind for M is denoted by η ( M ). Then it is clear that η ( M ) = n − ξ ( M ). Thus we can use nullity instead of H S -rank in some cases.It is well known that η ( T ) = n − µ ( T ) for any tree T of order n , where µ ( T ) is the matching numberof T . Since for any mixed forest, its H S -spectrum is the same as the adjacency spectrum of its underlyinggraph, we immediately get the following two lemmas, which are the same as the corresponding resultsfor Hermitian adjacency matrices of the first kind for mixed graphs [26]. Lemma 6.1. If M = M T is a mixed tree of order n , then η ( M ) = n − µ ( T ) , where µ ( T ) is thematching number of T . Lemma 6.2.
Let M P be a mixed path of order n . Then η ( M P ) = (cid:26) , if n is odd , , if n is even . (6.1) Lemma 6.3.
Let M be a mixed graph containing a pendent edge uv , and let M ′ = M − u − v . Then η ( M ) = η ( M ′ ) . The proof of this lemma is the same as [26, Lemma 3.3], we omit it here.16 emma 6.4.
Let M C be a mixed cycle of order n . Then η ( M C ) = , if n is odd , , if n ≡ and M C is negative , , if n ≡ and M C is positive, semi-positive or semi-negative , , if n ≡ and M C is positive , , if n ≡ and M C is negative, semi-positive or semi-negative . (6.2) Proof.
Denote the characteristic polynomial of H ( M C ) by P M C ( x ) = P nj =0 c j x n − j . To prove η ( M C ) = 0(resp. 2), it is sufficient to prove that c n = 0 (resp. c n − = 0 and c n − = c n = 0). By Theorem 3.2, itis easily verified.The following lemma is similar to [22, Lemma 5.1]. Lemma 6.5.
Suppose that M is a mixed graph and M ′ is an induced mixed subgraph of M . Then the H S -rank of M is greater than or equal to the H S -rank of M ′ . H S -rank Lemma 6.6.
Suppose that M is a mixed graph with H S -rank . Then M has the following properties: (a) M consists of one connected component with more than one vertex together with some isolatedvertices. (b) Every induced subgraph of M has H S -rank or . (c) The underlying graph of M contains no induced path on at least vertices and no induced cycle oflength at least . The proof of this Lemma is similar to that of [22, Lemma 5.2], we omit it here.
Theorem 6.7. If M = M G is a connected mixed graph with H S -rank , then G is a complete bipartitegraph.Proof. According to Lemmas 6.4 and 6.5, we know that G contains no odd cycle, hence G is bipartite.A shortest path between any two nonadjacent vertices in opposite parts of the bipartition would inducea path on at least 4 vertices. Since M has no induced P (based on Lemma 6.6), there are no suchnonadjacent vertices. Since it contains at least one edge, it is necessarily a complete bipartite graph.Two vertices u, v ∈ V ( M ) are twins if M is switching equivalent to a mixed graph M ′ in which N M ′ ( u ) = N M ′ ( v ), N + M ′ ( u ) = N + M ′ ( v ) and N − M ′ ( u ) = N − M ′ ( v ). It is easy to see that by removing oradding twins the H S -rank remains the same (but the H S -spectrum changes), and the relation of beinga twin of each other is an equivalence relation on V ( M ). Let [ u ] denote the equivalence class containingthe vertex u . Mohar [22] defines the twin reduction graph of M , denoted by T M , to be a graph whosevertices are the equivalence classes and [ u ][ v ] ∈ E ( T M ) if uv ∈ E ( M ′ ). Note that M ′ is determined onlyup to switching equivalence, and thus also T M is determined only up to switching equivalence. So T M has no twins. The following observation is easy to obtain, and enables us to assume that there are notwins when one classifies mixed graphs of a fixed H S -rank. Lemma 6.8.
Let M and M be two mixed graphs with the same underlying graph. Then they areswitching equivalent if and only if T M and T M are switching equivalent, M and T M have the same H S -rank. heorem 6.9. Let M = M G be a mixed graph of order n whose H S -rank is equal to and let ρ be itspositive H S -eigenvalue. Then M is switching equivalent to K a,b ∪ tK , where t ≥ . Moreover, we have n = a + b + t and ρ = ab . (6.3) Proof.
By Lemma 6.6, M has t ≥ t = 0, so that M is connected. By Theorem 6.7, G is a complete bipartitegraph K a,b with parts A, B, where | A | = a, | B | = b and b ≥ a ≥
1. By Lemma 6.8, we may assumethat M has no twins, i.e., M = T M .If a = b = 1, then M is switching equivalent to K , which gives the outcome. Suppose now that b >
1. Then H ( M ) can be written as H ( M ) = NN T ! , where the first a rows (columns) are indexed by the vertices in A , while the last b rows (columns) areindexed by the vertices in B . Then it is clear that N has more than one column, and the column vectorsof N are pairwise linearly independent. For a vertex x ∈ V ( M ), let H x denote its column in H ( M ).Let u ∈ A and let v, v ′ ∈ B . Then H u , H v and H v ′ are linearly independent, hence the rank of H ( M )is at least 3, a contradiction. This proves the first part of the theorem. n = a + b + t is clear. Let ρ and λ be two non-zero H S -eigenvalues of M , then it is clear that ρ + λ = 0 and ρ + λ = 2 | E ( M ) | = 2 ab ,which implies that λ = ρ = ab and this proves the second part of the theorem. Theorem 6.10.
All connected mixed graphs of order n with H S -rank are determined by their H S -spectrum.Proof. Let M be a connected mixed graph of order n with H S -rank 2. Then M is switching equivalentto K a,b ( a ≥ b ). If there exists a connected mixed graph M ′ with the same H S -spectrum to M , then M ′ is switching equivalent to K a ′ ,b ′ ( a ′ ≥ b ′ ). By (6.3), we have a + b = a ′ + b ′ , ab = a ′ b ′ , which implies a = a ′ , b = b ′ , i.e., M is switching equivalent to M ′ . Hence M is determined by its H S -spectrum.In Theorem 6.10, if condition “connected” is omitted, then M is not determined by its H S -spectrum.For example, K , ∪ ( n − K is H S -cospectral with K , ∪ ( n − K . Note that if K a,b ∪ ( n − a − b ) K is H S -cospectral with K a ′ ,b ′ ∪ ( n − a ′ − b ′ ) K , then K ta,sb ∪ ( n − ta − sb ) K is H S -cospectral with K ta ′ ,sb ′ ∪ ( n − ta ′ − sb ′ ) K for every integer t, s ≥
1. This implies the following proposition.
Proposition 1.
There are infinitely many mixed graphs with H S -rank which are not determined bytheir H S -spectrum. .2. Mixed graphs with H S -rank Lemma 6.11.
Suppose that M is a mixed graph with H S -rank . Then M has the following properties: (a) M consists of one connected component with more than one vertex together with some isolatedvertices. (b) Every induced subgraph of M has H S -rank , or . (c) The underlying graph of M contains no induced path on at least vertices and no induced cycle oflength at least .Proof. Since the trace of H ( M ) is 0, this holds for each connected component. The component withmore than one vertex contributes at least 2 to the rank of H ( M ), so we immediately see that (a)holds. Note that, in particular, no mixed graph has H S -rank 1. Thus, Lemma 6.5 implies (b). Finally,combining Lemma 6.2, 6.4 and 6.5, (c) holds. Lemma 6.12.
Let M = M G be a connected mixed graph of order . Then ξ ( M ) = 3 if and only if M is switching equivalent to one of the mixed graphs as depicted in Fig. 6. Proof.
Let be a connected mixed graph of order with -rank 2. Then is switching equivalentto a,b ). If there exists a connected mixed graph with the same -spectrum to , thenis switching equivalent to ,b ). By (6.3), we have , ab which implies , b , i.e., is switching equivalent to . Hence is determined by its-spectrum.In Theorem 6.10, if condition “connected” is omitted, then is not determined by its -spectrum.For example, 13) is -cospectral with 12) . Note that if a,b is -cospectral with ,b , then ta,sb ta sb is -cospectral with ta ,sb ta sb for every integer t, s
1. This implies the following proposition.
Proposition 1.
There are infinitely many mixed graphs with -rank which are not determined bytheir -spectrum.
Suppose that is a mixed graph with -rank . Then has the following properties: (a) consists of one connected component with more than one vertex together with some isolatedvertices. (b)
Every induced subgraph of has -rank or (c)
The underlying graph of contains no induced path on at least vertices and no induced cycle oflength at leastProof.
Since the trace of ) is 0, this holds for each connected component. The component withmore than one vertex contributes at least 2 to the rank of ), so we immediately see that (a)holds. Note that, in particular, no mixed graph has -rank 1. Thus, Lemma 6.5 implies (b). Finally,combining Lemma 6.2, 6.4 and 6.5, (c) holds.
Lemma 6.12.
Let be a connected mixed graph of order . Then ) = 3 if and only ifis switching equivalent to one of the mixed graphs as depicted in Fig. 6. (a) (b) (c) (d) (e) (f)Figure 6: Some 4-vertex mixed graphs.
Proof.
If is a mixed tree or a mixed cycle of order 4, then ) = 2 or 4 by Lemmas 6.1 and 6.4.So, we can assume that contains a mixed triangle with vertex set , v , v . Let be in \ { , v , v . If ) = 1 then assume that ) = . By Lemma 6.3,) = ) = ) = 0 (6.4)which implies ) = 4. So we only need to consider the following two cases.18
Figure 6: Some 4-vertex mixed graphs.
Proof. If M is a mixed tree or a mixed cycle of order 4, then ξ ( M ) = 2 or 4 by Lemmas 6.1 and 6.4.So, we can assume that M contains a mixed triangle M C with vertex set { v , v , v } . Let v be in V ( M ) \ { v , v , v } . If d G ( v ) = 1 , then assume that N G ( v ) = { v } . By Lemma 6.3, η ( M ) = η ( M − v − v ) = η ( K ) = 0 , (6.4)which implies ξ ( M ) = 4. So we only need to consider the following two cases. Case 1. d G ( v ) = 2. In this case, assume N G ( v ) = { v , v } . Then G is obtained from K by deletingan edge. Denote the characteristic polynomial of H ( M ) by P M ( x ) = x + c x + c x + c x + c . Hence, ξ ( M ) = 3 is equivalent to c = 0 and c = 0. By Lemmas 6.4 and 6.5, ξ ( M ) ≥
3, and so if c = 0, one can easily obtain that c = 0. So in order to complete the proof in this case, it suffices toshow c = 0.In fact, by Theorem 3.2 one has c = X M ′ ( − r ( M ′ )+ l s ( M ′ )+ l n ( M ′ ) · l p ( M ′ )+ l n ( M ′ ) , M ′ of M . Note that there are exactlytwo perfect matchings in G and one spanning mixed cycle (say M C ) in M . Then c = 2 + ( − l s ( M C )+ l n ( M C ) · l p ( M C )+ l n ( M C ) = 0if and only if M C is a positive cycle, i.e., M is switching equivalent to (a), (b), (c) or (d) depicted inFig. 6. Case 2. d G ( v ) = 3. In this case, G is isomorphic to K . Denote the characteristic polynomial of H ( M ) by P M ( x ) = x + c x + c x + c x + c . Similar to Case 1, it suffices to show c = 0. By Theorem 3.2, c = X M ′ ( − r ( M ′ )+ l s ( M ′ )+ l n ( M ′ ) · l p ( M ′ )+ l n ( M ′ ) , (6.5)where the summation is over all spanning elementary subgraphs M ′ of M . Note that there are exactly3 perfect matchings, say E , E , E , in G and 3 spanning mixed cycles, say M C , M ′ C , M ′′ C , in M . Let M = { M E , M E , M E } , C = { M C , M ′ C , M ′′ C } . Then (6.5) gives c = X M ′ ∈M ( − r ( M ′ )+ l s ( M ′ )+ l n ( M ′ ) · l p ( M ′ )+ l n ( M ′ ) + X M ′ ∈C ( − r ( M ′ )+ l s ( M ′ )+ l n ( M ′ ) · l p ( M ′ )+ l n ( M ′ ) = 3 + X M ′ ∈C ( − r ( M ′ )+ l s ( M ′ )+ l n ( M ′ ) · l p ( M ′ )+ l n ( M ′ ) . Hence, c = 0 if and only if there are three semi-positive mixed cycles or two positive and one semi-negative mixed cycles in C . It is easy to check that there does not exist mixed graph M K containingthree semi-positive spanning cycles. Furthermore, all the mixed graphs M K containing two positiveand one semi-negative spanning cycles are switching equivalent to (e) or (f) as depicted in Fig. 6.This completes the proof. Lemma 6.13.
Let M K n be a mixed graph on n ≥ vertices. Then ξ ( M K n ) ≥ .Proof. By Lemma 6.5, it is enough to prove ξ ( M K ) ≥
4. Let M = M K . Assume ξ ( M ) ≤
3. Thencombine with Lemmas 6.5 and 6.12 we know that each induced subgraph on 4 vertices of M is switchingequivalent to (e) or (f) as depicted in Fig. 6. Hence, it is straightforward to check that M can beswitching equivalent only to the mixed graph M as depicted in Fig. 7. By a direct computation, onehas ξ ( M ) = 5, a contradiction to the assumption ξ ( M ) ≤
3. This completes the proof.
Figure 7: The mixed graph whose underlying graph is
Theorem 6.14.
Assume ) = 3 . If the underlying graph is connected, then is either a completetripartite graph or a complete -partite graph.Proof.
By Lemma 6.13, we know that contains no 5-clique. Suppose that has no triangle. Thenis bipartite since otherwise, a shortest odd cycle of length at least 5 would be induced in , whichcontradicts the second part of Lemma 6.11(c). Hence, the -spectrum of is symmetric about 0,which implies that the -rank of is even, a contradiction. So in what follows, we suppose thatcontains a triangle uvw
Assume firstly that contains no 4-clique. Let be the largest induced complete tripartite subgraphof containing , and let
A, B, C be the parts of such that
A, v and . Ifthen there is a vertex ) that is adjacent to some vertices in . We may assume that isadjacent to some vertices in . As has no , together with Lemmas 6.5 and 6.12, is adjacent toprecisely two vertices in , say u, v . Consider any triangle uvw with . Then is not adjacentto , otherwise, contains a . Considering all triangles vw and uv for A, v , we seeis adjacent to every and . Adding to , we can get a complete tripartite graph biggerthan , a contradiction.Assume now that contains a 4-clique with ) = u, v, w, z . Let be the largest inducedcomplete 4-partite subgraph of containing , and let
A, B, C, D be the parts of such that
A, v B, w and . If , then there is a vertex ) that is adjacent to somevertices in . We may assume that is adjacent to some vertices in . As contains no 5-clique as asubgraph, together with Lemmas 6.5 and 6.12, is adjacent to precisely three vertices, say u, v, w, inFor any 4-clique with vertex set u, v, w, z is not adjacent to , otherwise, contains asa induced subgraph. Considering all 4-cliques with vertex set , v, w, z u, v , w, z and u, v, w , z for
A, v and , we see that , x , x . Adding to yields a complete4-partite graph, which contradicts the maximality of , a contradiction.By a similar discussion as the proof of [26, Lemma 4.5], we obtain the following lemma.
Lemma 6.15.
Let be a mixed graph whose underlying graph is a tripartite (resp -partite) graph,and let be Hermitian adjacency matrix of the second kind for . For all vertices , letdenotes the th row in . If and are linearly independent for two vertices u, v in the samepartite of , then
Recall that, for a mixed graph is the twin reduction graph of
Theorem 6.16.
Let be a connected mixed graph. Then ) = 3 if and only if is either a mixedtriangle or switching equivalent to or as depicted in Fig. 6.Proof.
As ) = ), the sufficiency follows by Lemmas 6.4 and 6.12. Now we prove the necessity.By Theorem 6.14, the underlying graph of is a complete tripartite graph or complete 4-partite graph,so is , say a,b,c,d , where a, b, c >
A, B, C, D , where a, b, c, . If = 1 and = 0, we have the first outcome; if20
Figure 7: The mixed graph M whose underlying graph is K .20 heorem 6.14. Assume ξ ( M G ) = 3 . If the underlying graph G is connected, then G is either a completetripartite graph or a complete -partite graph.Proof. By Lemma 6.13, we know that G contains no 5-clique. Suppose that G has no triangle. Then G is bipartite since otherwise, a shortest odd cycle of length at least 5 would be induced in G , whichcontradicts the second part of Lemma 6.11(c). Hence, the H S -spectrum of M is symmetric about 0,which implies that the H S -rank of M is even, a contradiction. So in what follows, we suppose that G contains a triangle T = uvw .Assume firstly that G contains no 4-clique. Let Q be the largest induced complete tripartite subgraphof G containing T , and let A, B, C be the parts of Q such that u ∈ A, v ∈ B and w ∈ C . If G = Q ,then there is a vertex z ∈ V ( G ) \ V ( Q ) that is adjacent to some vertices in Q . We may assume that z isadjacent to some vertices in T . As G has no K , together with Lemmas 6.5 and 6.12, z is adjacent toprecisely two vertices in T , say u, v . Consider any triangle uvw ′ with w ′ ∈ C . Then z is not adjacentto w ′ , otherwise, G contains a K . Considering all triangles u ′ vw and uv ′ w for u ′ ∈ A, v ′ ∈ B , we see z is adjacent to every u ′ ∈ A and v ′ ∈ B . Adding z to Q , we can get a complete tripartite graph biggerthan Q , a contradiction.Assume now that G contains a 4-clique F with V ( F ) = { u, v, w, z } . Let K be the largest inducedcomplete 4-partite subgraph of G containing F , and let A, B, C, D be the parts of K such that u ∈ A, v ∈ B, w ∈ C and z ∈ D . If G = K , then there is a vertex x ∈ V ( G ) \ V ( K ) that is adjacent to somevertices in K . We may assume that x is adjacent to some vertices in F . As G contains no 5-clique as asubgraph, together with Lemmas 6.5 and 6.12, x is adjacent to precisely three vertices, say u, v, w, in F .For any 4-clique with vertex set { u, v, w, z ′ } , z ′ ∈ D , x is not adjacent to z ′ , otherwise, G contains K asa induced subgraph. Considering all 4-cliques with vertex set { u ′ , v, w, z } , { u, v ′ , w, z } and { u, v, w ′ , z } for u ′ ∈ A, v ′ ∈ B and w ′ ∈ C , we see that x ∼ u ′ , x ∼ v ′ , x ∼ w ′ . Adding x to K yields a complete4-partite graph, which contradicts the maximality of K , a contradiction.By a similar discussion as the proof of [26, Lemma 4.5], we obtain the following lemma. Lemma 6.15.
Let M be a mixed graph whose underlying graph is a tripartite (resp. -partite) graph,and let H be Hermitian adjacency matrix of the second kind for M . For all vertices x ∈ V ( M ) , let H x denote the x th row in H . If H u and H v are linearly independent for two vertices u, v in the same partiteof M , then ξ ( M ) ≥ . Recall that, for a mixed graph M , T M is the twin reduction graph of M . Theorem 6.16.
Let M be a connected mixed graph. Then ξ ( M ) = 3 if and only if T M is either a mixedtriangle or switching equivalent to ( e ) or ( f ) as depicted in Fig. 6.Proof. As ξ ( M ) = ξ ( T M ), the sufficiency follows by Lemmas 6.4 and 6.12. Now we prove the necessity.By Theorem 6.14, the underlying graph of M is a complete tripartite graph or complete 4-partite graph,so is T M , say K a,b,c,d , where a, b, c > d ≥
0. Denote the four parts in T M by A, B, C, D , where | A | = a, | B | = b, | C | = c, | D | = d . If a = b = c = 1 and d = 0, we have the first outcome; if a = b = c = d = 1, Lemma 6.12 gives the second outcome. For all v ∈ V ( T M ), denote by H v the rowvector indexed by v in H ( T M ).If d = 0, then the underlying graph of T M is a complete tripartite graph. Let T = xyz be a trianglein T M , where x ∈ A, y ∈ B, z ∈ C . Note that T is an induced subgraph of T M and ξ ( T ) = 3. Then21 x , H y and H z are linearly independent. If a ≥
2, there is a vertex x ′ = x in A . We can assert that H x and H x ′ are linearly independent. Otherwise, there exists a constant k such that H x ′ = kH x , and thus x and x ′ have exactly the same neighborhood under a three-way switching. In other word, x ′ is a twinof x , a contradiction. On the other hand, if H x and H x ′ are linearly independent, by Lemma 6.15, onehas ξ ( T M ) ≥
4, a contradiction. Thus, a = 1. By a similar discussion, we obtain b = c = 1.If d = 0, then the underlying graph of T M is a complete 4-partite graph. By a similar discussion, wehave a = b = c = d = 1. This completes the proof. Remark 1.
Wang et al. [26] characterized all mixed graphs with H -rank 3, and show that all connectedmixed graphs with H -rank 3 can be determined by their H -spectrum. Here we identify all connectedmixed graphs with H S -rank 3. However, not all connected mixed graphs with H S -rank 3 are determinedby their H S -spectrum. For example, K , , is not switching equivalent to M K , , whose twin reductiongraph is a semi-positive triangle, whereas both of them are H S -cospectral.
7. Mixed graphs with small H S -spectral radius In this section, using interlacing theorem, we characterize all mixed graphs whose H S -eigenvalues havesmall absolute values. We characterize all the mixed graphs whose H S -spectra are contained in ( − α, α )for α ∈ (cid:8) √ , √ , (cid:9) .Recall that the H S -spectral radius of an n -vertex mixed graph M is defined as ρ ( M ) = max {| λ | , | λ n |} , where λ (resp. λ n ) is the largest (resp. smallest) H S -eigenvalue of M . Thus the H S -spectrum of M contains in ( − α, α ) if and only if ρ ( M ) < α . H S -spectral radius is less than √ H S -eigenvalues are equal to 1 or −
1. Then we characterize all the mixedgraphs whose H S -spectrum is contained in ( −√ , √ . Theorem 7.1.
A mixed graph M has the property that λ ∈ {− , } for each H S -eigenvalue λ if andonly if M is switching equivalent to tK for some t . The proof of this theorem is the same as [15, Theorem 9.1], which is omitted here.By Corollary 3.5 we know that all the mixed paths on n vertices are H S -cospectral. By Corollary 3.10,all the positive (resp. semi-positive, negative, semi-negative) cycles on n vertices are H S -cospectral. Wedenote by C n , M C n , M C n , M C n the n -vertex mixed cycles having no arc, just one arc, just two consec-utive arcs with the same direction and just three consecutive arcs with the same direction, respectively.Then they are positive, semi-positive, semi-negative, negative cycles on n vertices, respectively. Thefollowing fact is well-known (see [10, Section2.6]). Lemma 7.2 ([10]) . The characteristic polynomials of the paths satisfy the recurrence relation P P n ( x ) = xP P n − ( x ) − P P n − ( x ) with P P ( x ) = 1 and P P ( x ) = x . And the spectrum consists of simple eigenvalues λ j = 2 cos πjn + 1 , j = 1 , . . . , n. H S -eigenvalues of mixed graphs with C as the underlying graph Mixed graph Characteristic polynomial Eigenvalues C x − x − , − (2) M C x − x − . , − . , − . M C x − x + 1 1 . , . , − . M C x − x + 2 1 (2) , − H S -eigenvalues of mixed graphs with C as the underlying graph Mixed graph Characteristic polynomial Eigenvalues C x − x ± , (2) M C x − x + 1 ± . , ± . M C x − x + 3 ±√ , ± M C x − x + 4 ±√ (2) Theorem 7.3.
For a mixed graph M , the following are equivalent: (a) ρ ( M ) < √ ρ ( M ) ≤ Every component of M is ether an undirected edge, an arc or an isolated vertex.Proof. One may see that (b) implies (a) trivially. Note that, if (c) holds, then together with Theorem7.1 one has that (b) holds immediately. In order to complete the proof, it suffices to show that (a)implies (c).In fact, consider a mixed graph M on n vertices with H S -eigenvalues λ > λ > · · · > λ n . Assumethat −√ < λ n λ < √
2. Let M ′ be an induced subgraph of M on three vertices and let µ > µ > µ be the H S -eigenvalues of M ′ . By Corollary 2.5, we have −√ < µ i < √ i = 1 , , . We confirm that M ′ is not connected. Otherwise, M ′ is switching equivalent to P , C , M C , M C or M C . By Lemma 7.2, we have ρ ( P ) = √
2. By a direct calculation, the H S -spectra of C , M C , M C and M C are obtained (see Table 1), each of which contradicts that of (a). So M ′ is unconnected. Bythe arbitrary of M ′ , we know that every component of M is either an undirected edge, an arc or anisolated vertex. Hence, (c) holds.This completes the proof. Theorem 7.4.
Let M be an n -vertex mixed graph, then ρ ( M ) < √ if and only if every component of M is switching equivalent to P , P , P , P or M C .Proof. “Necessity”: Let M be a mixed graph on n vertices with H S -eigenvalues λ > λ > · · · > λ n .Suppose that λ < √ λ n > −√
3. Note that 2 cos πn +1 is increasing as n tends to infinity, and2 cos π = √
3. Hence, by Corollary 2.5 and Lemma 7.2 we know M contains no induced path with order23able 3: H S -eigenvalues of mixed graphs with C as the underlying graph Mixed graph Characteristic polynomial Eigenvalues C x − x + 5 x − , . (2) , − . (2) M C x − x + 5 x − . , , . , − . , − . M C x − x + 5 x + 1 1 . , . , − . , − , − . M C x − x + 5 x + 2 1 . (2) , − . (2) , − M contains no induced mixed cycle with order no less than 6. Refer to Tables 1 , M contains only M C as an induced mixed cycle.If M contains a vertex v with d M ( v ) ≥
3, then M contains either an induced mixed star on 4 verticesor a mixed triangle. Notice that M contains no mixed triangle. Hence, M must contain an inducedmixed star on 4 vertices. As every mixed star on 4 vertices is switching equivalent to its underlyinggraph K , , and by a direct calculation we know that ρ ( K , ) = √ . By Corollary 2.5, this cannothappen for M . Thus d M ( v ) ≤ v ∈ V ( M ). Therefore, every component of M is switchingequivalent to P , P , P , P or M C .“Sufficiency”: It is straightforward to check that if every component of M is switching equivalentto P , P , P , P or M C , then ρ ( M ) < √
3, as desired. H S -spectral radius is less than H S -spectral radius is smaller than 2. A T -shapetree Y a,b,c is a tree with exactly one vertex of degree greater than two such that the removal of thisvertex gives rise to paths P a , P b and P c . This tree has a + b + c + 1 vertices and contains a uniquevertex of degree 3 if a, b, c are all positive. The following lemma is well known (see also Smith [24] andLemmens and Seidel [17]). Lemma 7.5.
The largest adjacency eigenvalue of a connected simple graph is smaller than if and onlyif the graph is either a path or the graph Y a,b, for some a ≥ b ≥ , where either b = 1 and a ≥ , or b = 2 and ≤ a ≤ . As a mixed tree is H S -cospectral with its underlying graph, whose spectral radius is equal to itslargest eigenvalue. We have Corollary 7.6.
Let M be a mixed forest. Then ρ ( M ) < if and only if each component of the underlyinggraph of M is either a path or the graph Y a,b, for some a ≥ b ≥ , where either b = 1 and a ≥ , or b = 2 and ≤ a ≤ . In the following, we consider the case when M contains at least one mixed cycle. The spectral radiusof C n is 2 for n ≥
3, which follows from the following result.
Lemma 7.7 ([4]) . For n ≥ , the spectrum of C n consists of eigenvalues λ j = 2 cos 2 jπn , j = 1 , . . . , n. Lemma 7.8.
For every n ≥ , the characteristic polynomials of H ( C n ) , H ( M C n ) , H ( M C n ) and H ( M C n ) satisfy the following: P C n ( x ) = P P n ( x ) − P P n − ( x ) − P M Cn ( x ) = P P n ( x ) − P P n − ( x ) − P M Cn ( x ) = P P n ( x ) − P P n − ( x ) + 1; P M Cn ( x ) = P P n ( x ) − P P n − ( x ) + 2 . Lemma 7.9. If n ≥ is odd, then λ is an eigenvalue of C n if and only if − λ is an H S -eigenvalue of M C n .Proof. Let P P n ( x ) = x n + c x n − + · · · + c n − x + c n , P P n − ( x ) = x n − + c ′ x n − + · · · + c ′ n − x + c ′ n − . As P n is bipartite, c j − = c ′ k − = 0 , j ∈ { , , . . . , n +12 } , k ∈ { , , . . . , n − } . Hence P P n ( x ) and P P n − ( x )are odd functions in x . By Lemma 7.8, one has P C n ( λ ) = 0 ⇔ P P n ( λ ) − P P n − ( λ ) = 2 ⇔ P P n ( − λ ) − P P n − ( − λ ) = − ⇔ P M Cn ( − λ ) = 0 . This completes the proof.Combine with Lemmas 7.7 and 7.9, we obtain that the H S -spectral radius of M C n is 2 for odd n .The following result follows directly from Corollary 2.5. Corollary 7.10. If M is a mixed graph with ρ ( M ) < , then M contains no induced positive or oddnegative cycle. For the other types of mixed cycles, we can also show that their H S -spectral radii are strictly lessthan 2, which reads as the following result. Lemma 7.11. If M C is a semi-positive, semi-negative cycle or negative cycle of even order, then the H S -spectral radius of M C is strictly less than .Proof. From (5.2)-(5.6), we have λ ( M C ) ≤ λ ( C ). Similarly, replacing λ ( M C ) by | λ n ( M C ) | in (5.2)-(5.6) gives | λ n ( M C ) | ≤ λ ( C ), i.e., ρ ( M C ) ≤ ρ ( C ) = 2. It is sufficient to show that neither 2 nor − H S -eigenvalue of M C if M C is one of those mixed cycles. This follows directly by substituting2 and − n is even, then P C n (2) = P C n ( −
2) = 0; if n is odd, then P C n (2) = P M Cn ( −
2) = 0.)
Lemma 7.12.
Let M be a mixed graph. If ρ ( M ) < , then M contains no positive quadrangle andevery negative (resp. semi-positive, semi-negative) quadrangle in M forms an induced mixed subgraphof M .Proof. Note that the H S -spectral radius of every induced positive quadrangle is 2. Hence, if there is apositive quadrangle in M on consecutive vertices v , v , v , v , we may assume without loss of generalitythat v and v are adjacent. We proceed by considering whether v and v are adjacent or not.We consider firstly that v v . Since every triangle in M must be semi-positive or semi-negative,it is easy to see that the mixed graph induced on { v , v , v , v } contains either two semi-positivetriangles (they are switching equivalent, and call one of these mixed graphs Q ), or two semi-negativetriangles (they are switching equivalent, and call one of these mixed graphs Q ′ ). By a direct calculation, ρ ( Q ) = ρ ( Q ′ ) = 2 . emma 7.8. For every , the characteristic polynomials of , H , H andsatisfy the following: ) = 2; ) = 1;) = ) + 1; ) = ) + 2
Lemma 7.9.
If is odd, then is an eigenvalue of if and only if is an -eigenvalue ofProof.
Let ) = · · · , P ) = · · ·
As is bipartite, = 0 , j ∈ { , . . . , +1 , k ∈ { , . . . , Hence ) andare odd functions in . By Lemma 7.8, one has) = 0 ) = 2 ) = ) = 0This completes the proof.Combine with Lemmas 7.7 and 7.9, we obtain that the -spectral radius of is 2 for oddThe following result follows directly from Corollary 2.5.
Corollary 7.10.
If is a mixed graph with , then contains no induced positive or oddnegative cycle.
For the other types of mixed cycles, we can also show that thei -spectral radii are strictly lessthan 2, which reads as the following result.
Lemma 7.11.
If is a semi-positive, semi-negative cycle or negative cycle of even order, then the-spectral radius of is strictly less thanProof.
From (5.2)-(5.6), we have ). Similarly, replacing ) by in (5.2)-(5.6) gives | ≤ ), i.e., ) = 2. It is sufficient to show that neither 2 noris an -eigenvalue of if is one of those mixed cycles. This follows directly by substituting2 and 2 into its characteristic polynomial presented in Lemma 7.8. (Note that if is even, then(2) = 2) = 0; if is odd, then (2) = 2) = 0.) Q Q Q Q Q Q Q Q Figure 8: Mixed graphs , . . . Q Figure 8: Mixed graphs Q , . . . Q .We now consider that v ∼ v . In this subcase, it is easy to see that the mixed graph induced on { v , v , v , v } contains either four semi-positive triangles (they are switching equivalent, and call oneof these mixed graphs Q ), or two semi-positive and two semi-negative triangles (they are switchingequivalent, and call one of these mixed graphs Q ′ ), or four semi-negative triangles (they are switchingequivalent, and call one of these mixed graphs Q ′′ ). By a direct calculation, we obtain that ρ ( Q ) = ρ ( Q ′′ ) = 2 . ρ ( Q ′ ) = 2 . Q (resp. Q ′ , Q , Q ′ , Q ′′ ) cannot be an inducedmixed subgraph of M , and so M has no positive quadrangle.Suppose that M contains a negative quadrangle, say F , with consecutive vertices v , v , v , v .Then there is a directed path, say v v v v of length 3 in F , and thus v v is an undirected edge. If F is not an induced mixed subgraph of M , then v ∼ v or/and v ∼ v . If only v ∼ v , as M containsno negative triangle, then v v is either an undirected edge or an arc with direction from v to v . Ineither case, the mixed graph induced on { v , v , v , v } is switching equivalent to Q (see Fig. 8), whoselargest H S -eigenvalue is 2. If only v ∼ v , then by a similar discussion we obtain that the mixed graphinduced on { v , v , v , v } is also switching equivalent to Q (see Fig. 8). If v ∼ v and v ∼ v in F , then v v is either an undirected edge or an arc with direction from v to v and v v is either anundirected edge or an arc with direction from v to v , each of which will deduce that the mixed graphinduced on { v , v , v , v } is switching equivalent to Q ; see Fig. 8. Note that Q contains a positivequadrangle and hence its H S -spectral radius is no less than 2, a contradiction.Suppose that M contains a semi-positive quadrangle, say F . We are to show that F is an inducedmixed subgraph of M . Otherwise, by a similar discussion as above we obtain that the mixed graphinduced on the vertices of this quadrangle is switching equivalent to Q or Q ; see Fig. 8. The largest H S -eigenvalue of Q is 2 . Q contains a positive quadrangle, and hence its H S -spectralradius is no less than 2. Both of them deduce a contradiction. Hence, the semi-positive quadrangle F is an induced mixed subgraph of M .Suppose that M contains a semi-negative quadrangle, say F . We are to show that F is an inducedmixed subgraph of M . Otherwise, by a similar discussion as above we obtain that the mixed graphinduced on the vertices of this quadrangle is switching equivalent to Q , Q , Q or Q ; see Fig. 8.By a direct calculation, we may obtain the largest H S -eigenvalue of Q is 2 . H S -eigenvalue of Q is − . Q and Q contain a positive quadrangle. Hence ρ ( Q i ) ≥ i ∈ { , , , } , a contradiction. Hence, the semi-negative quadrangle F is an induced mixed subgraph26f M .This finishes the proof. Lemma 7.13.
Let M be a connected mixed graph containing a triangle. Then ρ ( M ) < if and only if M is a semi-positive triangle or a semi-negative triangle.Proof. By Corollary 7.10 and Lemma 7.11, the triangle contained in M is semi-positive or semi-negative.If the order of M is 3, the result is clearly true.If the order of M is 4, let C be the triangle contained in the underlying graph of M , and let v bea vertex of M outside C . Then v has only one neighbor in C , otherwise M contains a quadranglewhich is not an induced mixed subgraph of M . By Lemma 7.12, ρ ( M ) ≥
2, a contradiction. Hence, M is switching equivalent to Z or Z ; see Fig. 9. By a direct calculation we obtain that ρ ( Z ) = ρ ( Z ) =2 . M is at least 5, then M contains either a non-induced quadrangle or an inducedmixed subgraph that is switching equivalent to Z or Z . Clearly, this is also impossible. If the order of is 4, let be the triangle contained in the underlying graph of , and let bea vertex of outside . Then has only one neighbor in , otherwise contains a quadranglewhich is not an induced mixed subgraph of . By Lemma 7.12, 2, a contradiction. Hence,is switching equivalent to or ; see Fig. 9. By a direct calculation we obtain that ) = ) =0615, so this cannot happen.If the order of is at least 5, then contains either a non-induced quadrangle or an inducedmixed subgraph that is switching equivalent to or . Clearly, this is also impossible. Z ; 2 . Z ; 2 . Lemma 7.14.
Suppose that is a mixed graph with . Then ∆( Proof.
Suppose to the contrary that there exists a vertex in such that ∆( 4. Let us considerthe mixed graph induced by and four of its neighbors, , v , v , v . If contains a triangle, thenby Lemma 7.13, 2; if contains no triangle, then is switching equivalent to a simple bipartitegraph . By a direct calculation, ) = 2. Hence, ) = 2. By Corollary 2.5, we obtain that2, a contradiction.
Lemma 7.15.
Let be a connected mixed graph with a semi-positive quadrangle . Thenif and only ifProof.
By Lemmas 7.12 and 7.13, cannot contain a triangle, and the quadrangle is an inducedmixed subgraph of . Suppose that Then choose ) such that isadjacent to some vertices ofIf has just one neighbor in , then the mixed graph induced on ∪ { is switching equivalentto as depicted in Fig. 10. By a direct calculation, ) = 2 074 This can not happen. Ascontains no triangle, cannot have three neighbours in . Next we consider that has two neighborsin . The vertex is adjacent to two non-adjacent vertices of . Thus, )] contains threequadrangles, say
Q, Q , Q . Since there are no positive quadrangles (by Corollary 7.10), (resp.is semi-positive, negative or semi-negative.It is routine to check that if one quadrangle in , Q is semi-positive, then the other is semi-negative (based on Lemma 7.12). Thus the mixed graph induced by ∪ { is switching equivalentto as depicted in Fig. 10, whose -spectral radius is 2 236; if one quadrangle in , Q is negative,then the other is semi-negative. Thus the mixed graph induced by ∪ { is switching equivalent toas depicted in Fig. 10, whose -spectral radius is 2; if one quadrangle in , Q is semi-negative,then the other is semi-positive or negative, which have been discussed as above. All of these cases cannot happen.Hence, is just the , as desired.
Lemma 7.16.
Let be a connected mixed graph containing a pentagon. Then if and onlyif is a semi-positive pentagon or a semi-negative pentagon. Figure 9: Mixed graphs Z and Z together with their H S -spectral radii. Lemma 7.14.
Suppose that M = M G is a mixed graph with ρ ( M ) < . Then ∆( G ) ≤ .Proof. Suppose to the contrary that there exists a vertex v in G such that ∆( G ) ≥
4. Let us considerthe mixed graph Z induced by v and four of its neighbors, v , v , v , v . If Z contains a triangle, thenby Lemma 7.13, ρ ( Z ) ≥
2; if Z contains no triangle, then Z is switching equivalent to a simple bipartitegraph K , . By a direct calculation, ρ ( K , ) = 2. Hence, ρ ( Z ) = 2. By Corollary 2.5, we obtain that ρ ( M ) ≥ ρ ( Z ) ≥
2, a contradiction.
Lemma 7.15.
Let M be a connected mixed graph with a semi-positive quadrangle Q . Then ρ ( M ) < if and only if M = Q Proof.
By Lemmas 7.12 and 7.13, M cannot contain a triangle, and the quadrangle Q is an inducedmixed subgraph of M . Suppose that V ( M ) \ V ( Q ) = ∅ . Then choose v ∈ V ( M ) \ V ( Q ) such that v isadjacent to some vertices of Q .If v has just one neighbor in Q , then the mixed graph induced on V ( Q ) ∪ { v } is switching equivalentto Z as depicted in Fig. 10. By a direct calculation, ρ ( Z ) = 2 . . This can not happen. As M contains no triangle, v cannot have three neighbours in Q . Next we consider that v has two neighborsin Q . The vertex v is adjacent to two non-adjacent vertices of Q . Thus, M [ v ∪ V ( Q )] contains threequadrangles, say Q, Q , Q . Since there are no positive quadrangles (by Corollary 7.10), Q (resp. Q )is semi-positive, negative or semi-negative.It is routine to check that if one quadrangle in { Q , Q } is semi-positive, then the other is semi-negative (based on Lemma 7.12). Thus the mixed graph induced by V ( Q ) ∪ { v } is switching equivalent27 ; 2 . Z ; 2 . Z ; 2Figure 10: Mixed graphs , Z and together with their -spectral radii. Proof.
First of all, by Lemma 7.13, can not contain a triangle, the pentagon is an induced mixedsubgraph of . By Corollary 7.10 and Lemma 7.11, the pentagon contained in is semi-positive orsemi-negative. If the order of is 5, the result is clear true.If the order of is 6, let be the pentagon contained in the underlying graph of , and let bea vertex of outside . As contains no triangle, can not have three or more neighbours inIf has just one neighbour in , then is switching equivalent to or ; see Fig. 11. By a directcalculation we obtain that ) = ) = 2 076, this can not happen. If has two neighbours init is adjacent to two non-adjacent vertices of . Thus contains two pentagons and one quadrangle.By Corollary 7.10 and Lemma 7.15, contains no induced positive quadrangle, no induced positive ornegative pentagon, no semi-positive quadrangle. is switching equivalent to , Y or ; see Fig. 11.By a direct calculation we obtain that ) = ) = 2 199 and ) = 2, so this can not happen.If the order of is at least 7, let be the pentagon contained in the underlying graph ofChoose ) such that is adjacent to some vertices of . Then by the discussion above,the -spectral radius of ∪ { ] is at least 2. Clearly, this is also impossible. ; 2 076 ; 2 199 ; 2 ; 2 076 ; 2 199Figure 11: Mixed graphs , . . . , Y together with their -spectral radii.
Lemma 7.17.
Let be a connected mixed graph containing a semi-negative quadrangle . Ifcontains no induced subgraph obtained from two semi-negative quadrangles sharing with two consecutiveedges. Then if and only if is switching equivalent to , Z , Z or , where , Z and are depicted in Fig. 12.Proof. By Lemma 7.12, the quadrangle is an induced mixed subgraph ofSuppose that Then choose ) such that is adjacent to somevertices of . Let ) = , v , v , v . By Lemmas 7.12, 7.13 and 7.15, contains no triangle, nopositive or semi-positive quadrangle. Also, contains no other semi-negative quadrangle sharing twoconsecutive edges with . Then has just one neighbor, say , in . Thus the mixed graph inducedon ∪ { is switching equivalent to . By a direct calculation, ) = 1 902. Suppose that ∪ {
Then choose ∪ { ) such that is adjacent to somevertices in ∪ { . Also, has just one neighbor in . Since the maximum degree of is at most3 (by Lemma 7.14), . If is adjacent to one vertex in , v , but , then the mixed graph27 Figure 10: Mixed graphs Z , Z and Z together with their H S -spectral radii.to Z as depicted in Fig. 10, whose H S -spectral radius is 2 . { Q , Q } is negative,then the other is semi-negative. Thus the mixed graph induced by V ( Q ) ∪ { v } is switching equivalent to Z as depicted in Fig. 10, whose H S -spectral radius is 2; if one quadrangle in { Q , Q } is semi-negative,then the other is semi-positive or negative, which have been discussed as above. All of these cases cannot happen.Hence, M is just the Q , as desired. Lemma 7.16.
Let M be a connected mixed graph containing a pentagon. Then ρ ( M ) < if and onlyif M is a semi-positive pentagon or a semi-negative pentagon.Proof. First of all, by Lemma 7.13, M can not contain a triangle, the pentagon is an induced mixedsubgraph of M . By Corollary 7.10 and Lemma 7.11, the pentagon contained in M is semi-positive orsemi-negative. If the order of M is 5, the result is clear true.If the order of M is 6, let C be the pentagon contained in the underlying graph of M , and let v bea vertex of M outside C . As M contains no triangle, v can not have three or more neighbours in C .If v has just one neighbour in C , then M is switching equivalent to Y or Y ; see Fig. 11. By a directcalculation we obtain that ρ ( Y ) = ρ ( Y ) = 2 . v has two neighbours in C ,it is adjacent to two non-adjacent vertices of C . Thus M contains two pentagons and one quadrangle.By Corollary 7.10 and Lemma 7.15, M contains no induced positive quadrangle, no induced positive ornegative pentagon, no semi-positive quadrangle. M is switching equivalent to Y , Y or Y ; see Fig. 11.By a direct calculation we obtain that ρ ( Y ) = ρ ( Y ) = 2 .
199 and ρ ( Y ) = 2, so this can not happen.If the order of M is at least 7, let C be the pentagon contained in the underlying graph of M .Choose v ∈ V ( M ) \ V ( C ) such that v is adjacent to some vertices of C . Then by the discussion above,the H S -spectral radius of M [ V ( C ) ∪ { v } ] is at least 2. Clearly, this is also impossible. ; 2 074 ; 2 236 ; 2Figure 10: Mixed graphs , Z and together with their -spectral radii. Proof.
First of all, by Lemma 7.13, can not contain a triangle, the pentagon is an induced mixedsubgraph of . By Corollary 7.10 and Lemma 7.11, the pentagon contained in is semi-positive orsemi-negative. If the order of is 5, the result is clear true.If the order of is 6, let be the pentagon contained in the underlying graph of , and let bea vertex of outside . As contains no triangle, can not have three or more neighbours inIf has just one neighbour in , then is switching equivalent to or ; see Fig. 11. By a directcalculation we obtain that ) = ) = 2 076, this can not happen. If has two neighbours init is adjacent to two non-adjacent vertices of . Thus contains two pentagons and one quadrangle.By Corollary 7.10 and Lemma 7.15, contains no induced positive quadrangle, no induced positive ornegative pentagon, no semi-positive quadrangle. is switching equivalent to , Y or ; see Fig. 11.By a direct calculation we obtain that ) = ) = 2 199 and ) = 2, so this can not happen.If the order of is at least 7, let be the pentagon contained in the underlying graph ofChoose ) such that is adjacent to some vertices of . Then by the discussion above,the -spectral radius of ∪ { ] is at least 2. Clearly, this is also impossible. Y ; 2 . Y ; 2 . Y ; 2 Y ; 2 . Y ; 2 . , . . . , Y together with their -spectral radii. Lemma 7.17.
Let be a connected mixed graph containing a semi-negative quadrangle . Ifcontains no induced subgraph obtained from two semi-negative quadrangles sharing with two consecutiveedges. Then if and only if is switching equivalent to , Z , Z or , where , Z and are depicted in Fig. 12.Proof. By Lemma 7.12, the quadrangle is an induced mixed subgraph ofSuppose that Then choose ) such that is adjacent to somevertices of . Let ) = , v , v , v . By Lemmas 7.12, 7.13 and 7.15, contains no triangle, nopositive or semi-positive quadrangle. Also, contains no other semi-negative quadrangle sharing twoconsecutive edges with . Then has just one neighbor, say , in . Thus the mixed graph inducedon ∪ { is switching equivalent to . By a direct calculation, ) = 1 902. Suppose that ∪ {
Then choose ∪ { ) such that is adjacent to somevertices in ∪ { . Also, has just one neighbor in . Since the maximum degree of is at most3 (by Lemma 7.14), . If is adjacent to one vertex in , v , but , then the mixed graph27 Figure 11: Mixed graphs Y , . . . , Y together with their H S -spectral radii. Lemma 7.17.
Let M be a connected mixed graph containing a semi-negative quadrangle Q . If M contains no induced subgraph obtained from two semi-negative quadrangles sharing with two consecutive dges. Then ρ ( M ) < if and only if M is switching equivalent to M C , Z , Z or Z , where Z , Z and Z are depicted in Fig. 12.Proof. By Lemma 7.12, the quadrangle Q is an induced mixed subgraph of M .Suppose that V ( M ) \ V ( Q ) = ∅ . Then choose v ∈ V ( M ) \ V ( Q ) such that v is adjacent to somevertices of Q . Let V ( Q ) = { v , v , v , v } . By Lemmas 7.12, 7.13 and 7.15, M contains no triangle, nopositive or semi-positive quadrangle. Also, M contains no other semi-negative quadrangle sharing twoconsecutive edges with Q . Then v has just one neighbor, say v , in Q . Thus the mixed graph inducedon V ( Q ) ∪ { v } is switching equivalent to Z . By a direct calculation, ρ ( Z ) = 1 . V ( M ) \ ( V ( Q ) ∪ { v } ) = ∅ . Then choose v ′ ∈ V ( M ) \ ( V ( Q ) ∪ { v } ) such that v ′ is adjacent to somevertices in V ( Q ) ∪ { v } . Also, v ′ has just one neighbor in Q . Since the maximum degree of M is at most3 (by Lemma 7.14), v ′ v . If v ′ is adjacent to one vertex in { v , v } , but v ′ v , then the mixed graphinduced on V ( Q ) ∪ { v, v ′ } is switching equivalent to Z as depicted in Fig. 12. By a direct calculation, ρ ( Z ) = 2 . induced on ∪ { v, v is switching equivalent to as depicted in Fig. 12. By a direct calculation,) = 2 029. This can not happen. Z ; 1 . Z ; 2 . Z ; 2 Z ; 2 . Z ; 1 . Z ; 1 . Z ; 2 . Z ; 2 Z ; 2 . Z ; 2 Z ; 2 Z ; 2Figure 12: Mixed graphs , . . . , Z together with their -spectral radii.If v, v or v, v , then ∪ { v, v ] contains either two semi-negativequadrangles (in this case, ∪ { v, v ] is switching equivalent to or ), or a semi-negative anda negative quadrangle (in this case, ∪ { v, v ] is switching equivalent to ), where , Z and are depicted in Fig. 12. By a direct calculation, ) = 2 101 , ρ ) = 2 and ) = 1 950.If , v , then ∪ { v, v ] is switching equivalent to as depicted in Fig.12. By adirect calculation, ) = 2. If , and , then ∪ { v, v ] contains a pentagon. ByLemma 7.16, this can not happen.If is adjacent to , but it is adjacent to no vertex of , then ∪ { v, v ] is switchingequivalent to ; see Fig. 12. By a direct calculation, ) = 1 970.From the discussion above, we know that if ∪ { and is adjacent to somevertices in ∪ { , then ∪ { v, v ] must be switching equivalent to or . Supposethat ∪ { v, v Then choose ′′ ∪ { v, v ) such that ′′ is adjacentto some vertices in ∪ { v, v First we consider that ∪ { v, v ] is switching equivalent to . If ′′ has only one neighbourin ∪{ v, v , then as the maximal degree of is at most 3 (by Lemma 7.14), ∪{ v, v , v ′′ ] isswitching equivalent to or ; see Fig.12. By a direct calculation, ) = 2 061 and ) = 2.If ′′ has two neighbours in ∪ { v, v , then by Lemmas 7.12, 7.13, 7.15 and 7.16, containsno triangle, no positive or semi-positive quadrangle, no pentagon, ∪ { v, v , v ′′ ] is switchingequivalent to or ; see Fig. 12. By a direct calculation, ) = 2 101 and ) = 2. Ascontains no triangle, ′′ can not have three or more neighbours in ∪ { v, v . This is impossible.Now we consider that ∪ { v, v ] is switching equivalent to . According to the discussionabove, ′′ can be only adjacent to one of the vertices and . If ′′ , then , v , v , v, v , v ′′ is a mixed tree with two vertices of degree 3. By Corollary 7.6, its -spectral radius is no less than 2.If ′′ , then ∪ { v, v , v ′′ ] is switching equivalent to as depicted in Fig.12. By a directcalculation, ) = 2. This is also impossible. 28 Figure 12: Mixed graphs Z , . . . , Z together with their H S -spectral radii.If v ′ ∼ v, v ′ ∼ v or v ′ ∼ v, v ′ ∼ v , then M [ V ( Q ) ∪ { v, v ′ } ] contains either two semi-negativequadrangles (in this case, M [ V ( Q ) ∪ { v, v ′ } ] is switching equivalent to Z or Z ), or a semi-negative anda negative quadrangle (in this case, M [ V ( Q ) ∪ { v, v ′ } ] is switching equivalent to Z ), where Z , Z and Z are depicted in Fig. 12. By a direct calculation, ρ ( Z ) = 2 . , ρ ( Z ) = 2 and ρ ( Z ) = 1 . v ′ ∼ v , v ′ v , then M [ V ( Q ) ∪ { v, v ′ } ] is switching equivalent to Z as depicted in Fig.12. By adirect calculation, ρ ( Z ) = 2. If v ′ ∼ v , and v ′ ∼ v , then M [ V ( Q ) ∪ { v, v ′ } ] contains a pentagon. ByLemma 7.16, this can not happen.If v ′ is adjacent to v , but it is adjacent to no vertex of Q , then M [ V ( Q ) ∪ { v, v ′ } ] is switchingequivalent to Z ; see Fig. 12. By a direct calculation, ρ ( Z ) = 1 . V ( M ) \ ( V ( Q ) ∪ { v } ) = ∅ and v ′ is adjacent to somevertices in V ( Q ) ∪ { v } , then M [ V ( Q ) ∪ { v, v ′ } ] must be switching equivalent to Z or Z . Suppose29hat V ( M ) \ ( V ( Q ) ∪ { v, v ′ } ) = ∅ . Then choose v ′′ ∈ V ( M ) \ ( V ( Q ) ∪ { v, v ′ } ) such that v ′′ is adjacentto some vertices in V ( Q ) ∪ { v, v ′ } .First we consider that M [ V ( Q ) ∪ { v, v ′ } ] is switching equivalent to Z . If v ′′ has only one neighbourin V ( Q ) ∪{ v, v ′ } , then as the maximal degree of M is at most 3 (by Lemma 7.14), M [ V ( Q ) ∪{ v, v ′ , v ′′ } ] isswitching equivalent to Z or Z ; see Fig.12. By a direct calculation, ρ ( Z ) = 2 .
061 and ρ ( Z ) = 2.If v ′′ has two neighbours in V ( Q ) ∪ { v, v ′ } , then by Lemmas 7.12, 7.13, 7.15 and 7.16, M containsno triangle, no positive or semi-positive quadrangle, no pentagon, M [ V ( Z ) ∪ { v, v ′ , v ′′ } ] is switchingequivalent to Z or Z ; see Fig. 12. By a direct calculation, ρ ( Z ) = 2 .
101 and ρ ( Z ) = 2. As M contains no triangle, v ′′ can not have three or more neighbours in V ( Q ) ∪ { v, v ′ } . This is impossible.Now we consider that M [ V ( Q ) ∪ { v, v ′ } ] is switching equivalent to Z . According to the discussionabove, v ′′ can be only adjacent to one of the vertices v and v ′ . If v ′′ ∼ v , then M [ { v , v , v , v, v ′ , v ′′ } ]is a mixed tree with two vertices of degree 3. By Corollary 7.6, its H S -spectral radius is no less than 2.If v ′′ ∼ v ′ , then M [ V ( Q ) ∪ { v, v ′ , v ′′ } ] is switching equivalent to Z as depicted in Fig.12. By a directcalculation, ρ ( Z ) = 2. This is also impossible.This finishes the proof. Lemma 7.18.
Let M be a connected mixed graph containing a subgraph obtained from two semi-negativequadrangles sharing with two consecutive edges. Then ρ ( M ) < if and only if M is switching equivalentto Θ or Θ , where Θ and Θ are depicted in Fig. 13.Proof. It is straightforward to check that all mixed graphs obtained from two semi-negative quadranglessharing with two consecutive edges are switching equivalent to Θ or contain a positive quadrangle.Without loss of generality, we assume that Θ is a mixed subgraph of M, otherwise ρ ( M ) ≥ M contains no triangle, Θ is an induced mixed subgraph of M .By a direct calculation, ρ (Θ) = √ V ( M ) \ V (Θ) = ∅ . Then choose v ∈ V ( M ) \ V (Θ) such that v is adjacent to somevertices of Θ. Thus, v has at most three neighbors in Θ. If v has two or three neighbors in Θ, thenby Lemmas 7.12 and 7.15, M contains no positive or semi-positive quadrangle, and so M [ V (Θ) ∪ { v } ]is switching equivalent to Θ as depicted in Fig. 13. By a direct calculation, ρ (Θ ) = 2. This can nothappen.If v has only one neighbor in Θ, then by Lemma 7.14, this neighbor has degree 2 in Θ. M [ V (Θ) ∪{ v } ]is switching equivalent to Θ (see Fig. 13). By a direct calculation, ρ (Θ ) = 1 . V ( M ) \ ( V (Θ) ∪{ v } ) = ∅ , then choose v ′ ∈ V ( M ) \ ( V (Θ) ∪ { v } ) such that v ′ is adjacent to some vertices in V (Θ) ∪ { v } .By Lemmas 7.16 and 7.17, M contains no pentagon, no induced Z and v ′ can be only adjacent to v .Then M [ V (Θ) ∪ { v, v ′ } ] is switching equivalent to Θ (see Fig. 13). By a direct calculation, ρ (Θ ) = 2,this deduces a contradiction.This completes the proof.Combine with Lemmas 7.17 and 7.18, the following result is clear. Lemma 7.19.
Let M be a connected mixed graph with a semi-negative quadrangle, then ρ ( M ) < ifand only if M is switching equivalent to M C , Z , Z , Z , Θ or Θ . Lemma 7.20.
Let M = M G be a connected mixed graph with ρ ( M ) < . his finishes the proof. Lemma 7.18.
Let be a connected mixed graph containing a subgraph obtained from two semi-negativequadrangles sharing with two consecutive edges. Then if and only if is switching equivalentto or , where and are depicted in Fig. 13.Proof.
It is straightforward to check that all mixed graphs obtained from two semi-negative quadranglessharing with two consecutive edges are switching equivalent to Θ. Without loss of generality, we assumethat Θ is a mixed subgraph of . By Lemma 7.13, contains no triangle, Θ is an induced mixedsubgraph of . By a direct calculation, (Θ) = 3.Suppose that (Θ) Then choose (Θ) such that is adjacent to somevertices of Θ. Thus, has at most three neighbors in Θ. If has two or three neighbors in Θ, thenby Lemmas 7.12 and 7.15, contains no positive or semi-positive quadrangle, and so (Θ) ∪ { is switching equivalent to Θ as depicted in Fig. 13. By a direct calculation, (Θ ) = 2. This can nothappen.If has only one neighbor in Θ, then by Lemma 7.14, this neighbor has degree 2 in Θ. (Θ) ∪{ is switching equivalent to Θ (see Fig. 13). By a direct calculation, (Θ ) = 1 932. If (Θ)then choose (Θ) ∪ { ) such that is adjacent to some vertices in (Θ) ∪ { By Lemmas 7.16 and 7.17, contains no pentagon, no induced and can be only adjacent toThen (Θ) ∪ { v, v ] is switching equivalent to Θ (see Fig. 13). By a direct calculation, (Θ ) = 2,this deduces a contradiction.This completes the proof.Θ; √ ; 2 Θ ; 1 .
932 Θ ; 2Figure 13: Mixed graphs Θ , Θ together with their -spectral radiiCombine with Lemmas 7.17 and 7.18, the following result is clear. Lemma 7.19.
Let be a connected mixed graph with a semi-negative quadrangle, then ifand only if is switching equivalent to , Z , Z , Z or Lemma 7.20.
Let be a connected mixed graph with (a)
If contains an induced mixed hexagon and at most one induced mixed quadrangle, then isswitching equivalent to , M , M , X , X or , where and are depicted inFig. 14. (b)
If contains a non-induced mixed hexagon, then the mixed graph induced on the vertices of thismixed hexagon is switching equivalent to (see Fig. 12) or (see Fig. 14). (c) If contains an induced mixed -cycle for , then Figure 13: Mixed graphs Θ , Θ , Θ , Θ together with their H S -spectral radii v v v v v v v v v v v v v v v v v v Θ Θ ′ Θ ′′ X ; 1 . X ; 1 . X ; 1 . X ; 2Figure 14: Mixed graphs , X , X , X and Θ ′′ together with some of their -spectral radii. Proof. (a) Suppose that contains an induced mixed hexagon . By Corollary 7.10 and Lemma 7.11,is switching equivalent to , M or . Suppose that Then choose) such that is adjacent to some vertices of . As contains no triangle (by Lemma 7.13)and at most one quadrangle, has at most two neighbors inIf is a semi-positive (resp. semi-negative) hexagon, and has only one neighbor in , then by adirect calculation, the -spectral radius of ∪ { ] is 2 074 (resp. 2). If is a semi-positive orsemi-negative hexagon, and has two neighbors in , then by Lemmas 7.12, 7.13, 7.15, 7.16 and 7.19,contains no triangle or pentagon, and quadrangles in it are negative. ∪ { ] is switching equivalent to as depicted in Fig. 14. By a direct calculation, ) = 2. This cannot happen.Suppose now that is a negative hexagon with consecutive vertices , v , v , v , v , v . Let us firstconsider that has two neighbors in . Since contains no triangle (by Lemma 7.13), no positive orsemi-positive quadrangle (by Lemmas 7.12 and 7.15), no pentagon (by Lemma 7.16), no induced positivehexagon (by Corollary 7.10), and the case that contains semi-negative quadrangles is discussed inLemma 7.19, this case cannot happen.If has only one neighbor, say in X, then ∪ { ] is switching equivalent to asdepicted in Fig. 14. By a direct calculation, ) = 1 932. Suppose that ∪ { Then choose ∪ { ) such that is adjacent to some vertices in ∪ {
Similar to v, v has only one neighbor inFirst we consider that . In this subcase, , otherwise v, v , v , v , v ] containspentagon, a contradiction to Lemma 7.16. Thus, [( \ { ∪ { v, v ] is a mixed tree, whichcontains two vertices of degree 3. Hence, by Corollary 7.6, the -spectral radius of this mixed tree isno less than 2. Hence, this case is impossible. By a similar discussion, we may show that andis also impossible for ∈ { Next we consider and . By Lemmas 7.12, 7.15 and 7.19, the mixed graph induced on , v , v, v is a negative quadrangle. The mixed graph induced on ∪ { v, v is switchingequivalent to . By a direct calculation, ) = 1 956. In this case, if ∪ { v, v then choose ′′ ∪ { v, v ) such that ′′ is adjacent to some vertices in ∪ { v, v According to the discussion above, ′′ can only be adjacent to one of and . Then by a directcalculation, the -spectral radius of ∪ { v, v , v ′′ ] is 2, this cannot happen. Similarly, we canshow that if and , then is switching equivalent toNow we consider and has no neighbor in . In this case, we obtain an induced mixed subgraph . By Corollary 7.6 its -spectral radius is no less than 2. At last we consider . Up to now we have shown that (resp. ) is vertex of degree3; (resp. , v , v ) is of degree 2; (resp. ) has no neighbor in \ { , v , . . . , v (resp. Figure 14: Mixed graphs X , X , X , X , Θ , Θ ′ and Θ ′′ together with some of their H S -spectral radii.(a) If M contains an induced mixed hexagon and at most one induced mixed quadrangle, then M isswitching equivalent to M C , M C , M C , X , X or X , where X , X and X are depicted inFig. 14. (b) If M contains a non-induced mixed hexagon, then the mixed graph induced on the vertices of thismixed hexagon is switching equivalent to Z (see Fig. 12) or Θ (see Fig. 14). (c) If M contains an induced mixed k -cycle for k ≥ , then G ∼ = C k .Proof. (a) Suppose that M contains an induced mixed hexagon X . By Corollary 7.10 and Lemma 7.11, X is switching equivalent to M C , M C or M C . Suppose that V ( M ) \ V ( X ) = ∅ . Then choose v ∈ V ( M ) \ V ( X ) such that v is adjacent to some vertices of X . As M contains no triangle (by Lemma 7.13)and at most one quadrangle, v has at most two neighbors in X .If X is a semi-positive (resp. semi-negative) hexagon, and v has only one neighbor in X , then by adirect calculation, the H S -spectral radius of M [ V ( X ) ∪ { v } ] is 2 .
074 (resp. 2). If X is a semi-positive orsemi-negative hexagon, and v has two neighbors in X , then by Lemmas 7.12, 7.13, 7.15, 7.16 and 7.19, M contains no triangle or pentagon, and quadrangles in it are negative. M [ V ( X ) ∪ { v } ] is switchingequivalent to X as depicted in Fig. 14. By a direct calculation, ρ ( X ) = 2. This cannot happen.Suppose now that X is a negative hexagon with consecutive vertices v , v , v , v , v , v . Let us firstconsider that v has two neighbors in X . Since M contains no triangle (by Lemma 7.13), no positive orsemi-positive quadrangle (by Lemmas 7.12 and 7.15), no pentagon (by Lemma 7.16), no induced positivehexagon (by Corollary 7.10), and the case that M contains semi-negative quadrangles is discussed inLemma 7.19, this case cannot happen.If v has only one neighbor, say v , in X, then M [ V ( X ) ∪ { v } ] is switching equivalent to X asdepicted in Fig. 14. By a direct calculation, ρ ( X ) = 1 . V ( M ) \ ( V ( X ) ∪ { v } ) = ∅ . v ′ ∈ V ( M ) \ ( V ( X ) ∪ { v } ) such that v ′ is adjacent to some vertices in V ( X ) ∪ { v } . Similarto v, v ′ has only one neighbor in X .First we consider that v ′ ∼ v . In this subcase, v v ′ , otherwise M [ { v, v , v , v , v ′ } ] containspentagon, a contradiction to Lemma 7.16. Thus, M [( V ( X ) \ { v } ) ∪ { v, v ′ } ] is a mixed tree, whichcontains two vertices of degree 3. Hence, by Corollary 7.6, the H S -spectral radius of this mixed tree isno less than 2. Hence, this case is impossible. By a similar discussion, we may show that v ′ v and v ′ ∼ v i is also impossible for i ∈ { , , } .Next we consider v ′ ∼ v and v ′ ∼ v . By Lemmas 7.12, 7.15 and 7.19, the mixed graph inducedon { v , v , v, v ′ } is a negative quadrangle. The mixed graph induced on V ( X ) ∪ { v, v ′ } is switchingequivalent to X . By a direct calculation, ρ ( X ) = 1 . V ( M ) \ ( V ( X ) ∪ { v, v ′ } ) = ∅ , then choose v ′′ ∈ V ( M ) \ ( V ( X ) ∪ { v, v ′ } ) such that v ′′ is adjacent to some vertices in V ( X ) ∪ { v, v ′ } . According to the discussion above, v ′′ can only be adjacent to one of v and v ′ . Then by a directcalculation, the H S -spectral radius of M [ V ( X ) ∪ { v, v ′ , v ′′ } ] is 2, this cannot happen. Similarly, we canshow that if v ′ ∼ v and v ′ ∼ v , then M is switching equivalent to X .Now we consider v ′ ∼ v and v ′ has no neighbor in X . In this case, we obtain an induced mixedsubgraph Y , , . By Corollary 7.6 its H S -spectral radius is no less than 2.At last we consider v ′ ∼ v . Up to now we have shown that v (resp. v ) is vertex of degree3; v (resp. v , v , v ) is of degree 2; v (resp. v ′ ) has no neighbor in V ( M ) \ { v ′ , v , . . . , v } (resp. V ( M ) \ { v, v , . . . , v } ). Thus, in order to complete the proof of (a), it suffices to consider whether v is adjacent to v ′ or not. If v v ′ , then the graph induced by { v, v ′ , v , . . . , v } is just the mixed graph M , i.e., M is switching equivalent to X ; see Fig. 14. By a direct calculation, ρ ( X ) = 1 . v ∼ v ′ ,then we obtain three induced hexagons in M , and not all of them can be negative. Thus, this can nothappen.(b) If M contains a non-induced hexagon X , then X has chords (edges between non-consecutivevertices in a cycle). By Lemma 7.13, M contains no triangle, and so the chords join opposite verticeson the X . Thus there could be one, two or three of them. Let M ′ be the mixed subgraph induced onthe vertices of X . By Lemmas 7.12 and 7.15, M ′ contains no positive or semi-positive quadrangle. Thecase that M ′ contains semi-negative quadrangles is discussed in Lemma 7.19. We consider the case thatall the quadrangles in M ′ are negative here.As each quadrangle is negative, it is easy to check that if X contains two or three chords. Then X is a directed hexagon (all the edges of it are arcs with the same direction) with two or three undirectedchords in it. By a direct calculation, in these two cases, ρ ( M ′ ) = 2. It remains to consider the casethere is only one chord in X .If the only chord is undirected, then M ′ is switching equivalent to Θ or Θ ′ , whereas if the onlychord is directed, then M ′ is switching equivalent to Θ ′′ , where Θ , Θ ′ and Θ ′′ are depicted in Fig. 14.We will see that Θ ′ and Θ ′′ are all switching equivalent to Θ .In fact, for Θ ′′ , we can take V = { v } , V i √ = V ( M ′ ) \{ v } and remains being null. It is easy tocheck that this partition is admissible and by a three-way switching with respect to this partition, wecan obtain Θ ′ .For Θ ′ , we can take V = { v , v , v , v } , V − − i √ = { v } , V − i √ = { v } and remains being null.It is easy to check that this partition is admissible and by a three-way switching with respect to thispartition, we can obtain Θ .(c) Suppose that M contains an induced mixed k -cycle M C k with consecutive vertices v , . . . , v k − ( k ≥ v v v v v v n − v n − i √ a, b, c, d be nonnegative integers. Let a,b,c,d be a mixed graph obtained from a negativequadrangle with consecutive vertices , v , v , v by attaching undirected paths of lengths a, b, c, d to , v , v and , respectively. This graph has + 4 vertices. It is easy to see that theresulting mixed graph is unique up to switching equivalenceIn the discussion of Hermitian adjacency matrix (of the first kind) for mixed graphs, a mixed cycleis negative if and only if its weight is 1 (see Liu and Li [20], Guo and Mohar [14]). This definitioncoincides with ours. By comparing the characteristic polynomials of the Hermitian adjacency matrix (ofthe first kind) [20, Theorem 2.8] and the Hermitian adjacency matrix of the second kind (Theorem 3.2)for a mixed graph, we know that if is a unicyclic mixed graph (i.e, the underlying graph of is aunicyclic graph) with the unique mixed cycle negative, then the characteristic polynomials of the twokind of Hermitian adjacency matrices for are the same. Hence [14, Lemma 4.11] gives Lemma 7.21.
Let be a unicyclic mixed graph with a negative quadrangle, then if and onlyif is switching equivalent to one of the following mixed graphs (1) a, ,c, , where (2) or By a similar discussion as the proof of [14, Lemma 4.13], we obtain the following lemma.
Lemma 7.22.
Suppose that is a connected mixed graph with , then any two verticesand of degree are at distance at most in
Lemma 7.23.
Let be a connected mixed graph with at least two induced quadrangles. If, then is switching equivalent to one of the mixed graphs and , . . . , where , . . . , are depicted in Fig. 16.Proof. By Lemmas 7.12, 7.15 and 7.19, it is sufficient to consider the case that all the quadrangles inare negative. Let , Q be two induced quadrangles inSuppose that and have no vertex in common. Let be a shortest path in from to. Take together with the neighbors of the ends of . This subgraph is a mixed tree and has twovertices of degree 3. By Corollary 7.6, its -spectral radius is at least 2. So it cannot be an inducedsubgraph. By Lemmas 7.13 and 7.22, has no triangle and the length of is at most 3, and sothe only possibility is that a vertex in is adjacent to a vertex in . Since is a shortest path, itmeans that is a single edge and thus we have two edges joining adjacent vertices in with adjacentpair of vertices in . This forms a new quadrangle having common vertices with . So it is enoughto consider that and have at least one vertex in common. By Lemmas 7.13 and 7.14, has32
Figure 15: Long cycle plus a vertex with two neighbors.7). If G = C k , then V ( M ) \ V ( M C k ) = ∅ . Choose v ∈ V ( M ) \ V ( M C k ) such that v is adjacent to somevertices of M C k . By Lemma 7.14, v has at most three neighbours in V ( C k ).If v has only one neighbour in M C k , say v , then M [( V ( M C k ) \{ v } ) ∪ { v } ] is switching equivalentto Y k − , , . By Corollary 7.6, k − ≤
4, i.e., k = 7 or 8. If k = 7, by Corollary 7.10 and Lemma 7.11, M C is semi-positive or semi-negative. By a direct calculation, in both cases, the H S -spectral radius of M [ V ( M C ) ∪ { v } ] is 2 . k = 8, then M [( V ( M C ) \{ v } ) ∪ { v } ] is switching equivalent to Y , , . ByCorollary 7.6, it can not happen.Now we consider that v has two or three neighbors in M C k . By Lemma 7.16, M does not containinduced pentagon; By Lemma 7.20(a), M does not contain induced hexagon. Similarly, we may showthat G does not contain induced cycle C t for 7 ≤ t ≤ n −
1. Thus, any two neighbors of v must be atdistance precisely 2 on the cycle C k . This in particular means that v has just two neighbors, say v and v , on the cycle C k . By Lemmas 7.12, 7.15 and 7.19, the subgraph induced on { v , v , v , v } forms anegative quadrangle.We claim that the H S -spectral radius of M [ V ( M C k ) ∪ { v } ] is at least 2. In fact, M [ V ( M C k ) ∪ { v } ]contains two induced mixed k -cycles and one induced negative quadrangle. Furthermore, one mixed k -cycle is positive if and only if the other is negative. In this case, by Corollary 7.10, the H S -spectralradius of M [ V ( M C k ) ∪ { v } ] is at least 2. On the other hand, one mixed k -cycle is semi-positive if andonly if the other is semi-negative. In this case, M [ V ( M C k ) ∪ { v } ] is switching equivalent to the mixedgraph on the left in Fig. 15, and the labels at vertices on the right in Fig. 15 show an eigenvector forthe H S -eigenvalue 2. This implies that the H S -spectral radius of M [ V ( M C k ) ∪ { v } ] is at least 2. So thiscase cannot happen.This completes the proof.Let a, b, c, d be nonnegative integers. Let (cid:3) a,b,c,d be a mixed graph obtained from a negativequadrangle with consecutive vertices v , v , v , v by attaching undirected paths of lengths a, b, c, d to v , v , v and v , respectively. This graph has a + b + c + d + 4 vertices. It is easy to see that theresulting mixed graph is unique up to switching equivalence.In the discussion of Hermitian adjacency matrix (of the first kind) for mixed graphs, a mixed cycleis negative if and only if its weight is − M is a unicyclic mixed graph (i.e, the underlying graph of M is aunicyclic graph) with the unique mixed cycle negative, then the characteristic polynomials of the two33 ; 1 .
902 Θ ; √ ; 1 . ; 1 .
956 Θ ; 1 .
970 Θ ; 1 . · · · together with their -spectral radii.maximum degree no more than 3, and has no triangle. and either share one edge or share twoconsecutive edges.If and share two consecutive edges, then )] contains three quadrangles. Itis easy to see that not all of them are negative, which is impossible (based on Lemmas 7.12, 7.15 and7.19).If and share one edge, then by Lemma 7.20(b) )] is switching equivalent to(see Fig 14). Without loss of generality, let )] be Θ with labelled vertices as shownin Fig 14. Suppose that )) . Choose )) such thatis adjacent to some vertices in ).By Lemma 7.14, the maximum degree of is at most 3. If has only one neighbor in ),then is adjacent to some for ∈ { . In each case, ∪ { ] is switchingequivalent to Θ (see Fig. 16). By a direct calculation, (Θ ) = 1 902By Lemmas 7.13 and 7.16, contains no triangle or pentagon. If has two neighbors in), then is adjacent to either and , or and . In either case, ∪ { ] isswitching equivalent to Θ . By a direct calculation, (Θ ) =Hence, in order to characterize the structure of with 2, we proceed by considering thefollowing two possible cases. Case 1. has two neighbors in ), say and . Suppose that ∪ { . Choose ∪ { ) such that is adjacent to some verticesin ∪ { . If has only one neighbor in ∪ { then is adjacent to oneof , v and . In either case, ∪ { v, v ] is switching equivalent to Θ (see Fig. 16).By a direct calculation, (Θ ) = 1 932 If has two neighbors in ∪ { then isadjacent to two of , v and . In either case, by a directed calculation, the -spectral radius of ∪ { v, v ] is 2. If has three neighbors in ∪ { then is adjacentto , v and . As every quadrangle is negative, it is easy to check that this cannot happen.
Case 2. has only one neighbor in ), say . Suppose that. Choose ∪ { ) such that is adjacent to some vertices in ∪ {
If has only one neighbor in ∪ { then is adjacent to one of , v , v andIf , then ∪ { v, v ] is switching equivalent to Θ (see Fig. 16). By a directcalculation, (Θ ) = 1 956. If , then ∪ { v, v ] is switching equivalent to Θ (seeFig. 16). By a direct calculation, (Θ ) = 1 970. If , then by a direct calculation, the -spectralradius of ∪ { v, v ] is 2. If , then contains an induced mixed tree with two33
Figure 16: Mixed graphs Θ , · · · , Θ together with their H S -spectral radii.kind of Hermitian adjacency matrices for M are the same. Hence [14, Lemma 4.11] gives Lemma 7.21.
Let M be a unicyclic mixed graph with a negative quadrangle, then ρ ( M ) < if and onlyif M is switching equivalent to one of the following mixed graphs: (1) (cid:3) a, ,c, , where a ≥ c ≥ ; (2) (cid:3) , , , , (cid:3) , , , , (cid:3) , , , , (cid:3) , , , , (cid:3) , , , or (cid:3) , , , . By a similar discussion as the proof of [14, Lemma 4.13], we obtain the following lemma.
Lemma 7.22.
Suppose that M = M G is a connected mixed graph with ρ ( M ) < , then any two vertices u and v of degree are at distance at most in G . Lemma 7.23.
Let M = M G be a connected mixed graph with at least two induced quadrangles. If ρ ( M ) < , then M is switching equivalent to one of the mixed graphs Z , Θ , Θ , Θ and Θ , . . . , Θ , where Θ , . . . , Θ are depicted in Fig. 16.Proof. By Lemmas 7.12, 7.15 and 7.19, it is sufficient to consider the case that all the quadrangles in M are negative. Let Q , Q be two induced quadrangles in M .Suppose that Q and Q have no vertex in common. Let P be a shortest path in G from Q to Q . Take P together with the neighbors of the ends of P . This subgraph is a mixed tree and has twovertices of degree 3. By Corollary 7.6, its H S -spectral radius is at least 2. So it cannot be an inducedsubgraph. By Lemmas 7.13 and 7.22, M has no triangle and the length of P is at most 3, and sothe only possibility is that a vertex in Q is adjacent to a vertex in Q . Since P is a shortest path, itmeans that P is a single edge and thus we have two edges joining adjacent vertices in Q with adjacentpair of vertices in Q . This forms a new quadrangle having common vertices with Q . So it is enoughto consider that Q and Q have at least one vertex in common. By Lemmas 7.13 and 7.14, G hasmaximum degree no more than 3, and has no triangle. Q and Q either share one edge or share twoconsecutive edges.If Q and Q share two consecutive edges, then M [ V ( Q ) ∪ V ( Q )] contains three quadrangles. Itis easy to see that not all of them are negative, which is impossible (based on Lemmas 7.12, 7.15 and7.19). 34f Q and Q share one edge, then by Lemma 7.20(b) M [ V ( Q ) ∪ V ( Q )] is switching equivalent toΘ (see Fig 14). Without loss of generality, let M [ V ( Q ) ∪ V ( Q )] be Θ with labelled vertices as shownin Fig 14. Suppose that V ( M ) \ ( V ( Q ) ∪ V ( Q )) = ∅ . Choose v ∈ V ( M ) \ ( V ( Q ) ∪ V ( Q )) such that v is adjacent to some vertices in V ( Q ) ∪ V ( Q ).By Lemma 7.14, the maximum degree of G is at most 3. If v has only one neighbor in V ( Q ) ∪ V ( Q ),then v is adjacent to some v i for i ∈ { , , , } . In each case, M [ V ( Q ) ∪ V ( Q ) ∪ { v } ] is switchingequivalent to Θ (see Fig. 16). By a direct calculation, ρ (Θ ) = 1 . . By Lemmas 7.13 and 7.16, M contains no triangle or pentagon. If v has two neighbors in V ( Q ) ∪ V ( Q ), then v is adjacent to either v and v , or v and v . In either case, M [ V ( Q ) ∪ V ( Q ) ∪ { v } ] isswitching equivalent to Θ . By a direct calculation, ρ (Θ ) = √ . Hence, in order to characterize the structure of M with ρ ( M ) <
2, we proceed by considering thefollowing two possible cases.
Case 1. v has two neighbors in V ( Q ) ∪ V ( Q ), say v and v . Suppose that V ( M ) \ ( V ( Q ) ∪ V ( Q ) ∪ { v } ) = ∅ . Choose v ′ ∈ V ( M ) \ ( V ( Q ) ∪ V ( Q ) ∪ { v } ) such that v ′ is adjacent to some verticesin V ( Q ) ∪ V ( Q ) ∪ { v } . If v ′ has only one neighbor in V ( Q ) ∪ V ( Q ) ∪ { v } , then v ′ is adjacentto one of v , v and v . In either case, M [ V ( Q ) ∪ V ( Q ) ∪ { v, v ′ } ] is switching equivalent to Θ (seeFig. 16). By a direct calculation, ρ (Θ ) = 1 . . If v ′ has two neighbors in V ( Q ) ∪ V ( Q ) ∪ { v } , then v ′ is adjacent to two of v , v and v . In either case, by a direct calculation, the H S -spectral radius of M [ V ( Q ) ∪ V ( Q ) ∪ { v, v ′ } ] is 2. If v ′ has three neighbors in V ( Q ) ∪ V ( Q ) ∪ { v } , then v ′ is adjacentto v , v and v . As every quadrangle is negative, it is easy to check that this cannot happen. Case 2. v has only one neighbor in V ( Q ) ∪ V ( Q ), say v . Suppose that V ( M ) \ ( V ( Q ) ∪ V ( Q ) ∪{ v } ) = ∅ . Choose v ′ ∈ V ( M ) \ ( V ( Q ) ∪ V ( Q ) ∪ { v } ) such that v ′ is adjacent to some vertices in V ( Q ) ∪ V ( Q ) ∪ { v } .If v ′ has only one neighbor in V ( Q ) ∪ V ( Q ) ∪ { v } , then v ′ is adjacent to one of v , v , v and v .If v ′ ∼ v , then M [ V ( Q ) ∪ V ( Q ) ∪ { v, v ′ } ] is switching equivalent to Θ (see Fig. 16). By a directcalculation, ρ (Θ ) = 1 . v ′ ∼ v , then M [ V ( Q ) ∪ V ( Q ) ∪ { v, v ′ } ] is switching equivalent to Θ (seeFig. 16). By a direct calculation, ρ (Θ ) = 1 . v ′ ∼ v , then by a direct calculation, the H S -spectralradius of M [ V ( Q ) ∪ V ( Q ) ∪ { v, v ′ } ] is 2. If v ′ ∼ v , then M contains an induced mixed tree with twovertices of degree 3, by Corollary 7.6, this cannot happen.The case v ′ has two neighbors in V ( Q ) ∪ V ( Q ) is the same as the case v has two neighbors in V ( Q ) ∪ V ( Q ). This case has been discussed.It remains to consider the case v ′ ∼ v and v ′ has only one neighbor in V ( Q ) ∪ V ( Q ). Then v ′ ∼ v and v ′ is adjacent to one vertex in { v , v , v } . If v ′ ∼ v , v ′ ∼ v , then M [ V ( Q ) ∪ V ( Q ) ∪ { v, v ′ } ]contains either an induced positive hexagon, or an induced subgraph which is switching equivalent to X (see Fig. 14). By Lemma 7.20, this cannot happen. If v ′ ∼ v , v ′ ∼ v , then M [ V ( Q ) ∪ V ( Q ) ∪ { v, v ′ } ]is switching equivalent to Θ (see Fig. 16). By a direct calculation, ρ (Θ ) = 1 . v ′ ∼ v , v ′ ∼ v ,then M [ V ( Q ) ∪ V ( Q ) ∪ { v, v ′ } ] contains a pentagon. By Lemma 7.16, this cannot happen.By a direct calculation, adding any new vertex out of Θ , Θ , Θ and Θ raises the H S -spectralradius to at least 2. This completes the proof. Theorem 7.24.
Let M = M G be a connected mixed graph. Then ρ ( M ) < if and only if M is switchingequivalent to one of the following: (a) M C n , M C n ; 35b) M C n with n ≡ P n ;(d) Y a,b, where a ≥ b ≥ and either b = 1 and a ≥ , or b = 2 and ≤ a ≤ Z , Z , Z ; see Fig. and Θ ; see Fig. ; see Fig. X , X and X ; see Fig. (cid:3) a, ,c, , where a ≥ c ≥ (cid:3) , , , , (cid:3) , , , , (cid:3) , , , , (cid:3) , , , , (cid:3) , , , or (cid:3) , , , ;(k) Θ , . . . , Θ ; see Fig. . Proof.
First we note that the H S -spectral radius of every mixed graph in items (a)-(k) is strictly lessthan 2.Let M = M G be a connected mixed graph with ρ ( M ) <
2. Suppose G is a tree. Then by Corol-lary 7.6, M is switching equivalent to one of the mixed graphs in items (c) and (d).If G is a cycle, then by Corollary 7.10 and Lemma 7.11, M is switching equivalent to one of themixed graphs in items (a) and (b).Now we assume that G contains a cycle and some other edges if possible. If G contains a triangle,then by Lemma 7.13, G is a triangle. Thus assume from now on that G has no triangle.Suppose that M has a mixed quadrangle Q , then by Lemma 7.12, Q cannot be positive. If Q issemi-positive, then by Lemma 7.15, M is switching equivalent to M C . If Q is semi-negative, then byLemma 7.19, M is switching equivalent to M C , or one of the mixed graphs in items (e) and (f).Suppose that Q is negative. If M contains no other cycle, then M is switching equivalent to (cid:3) a,b,c,d for some a, b, c, d . By Lemma 7.21, M is switching equivalent to one of the mixed graphs in items (i)and (j). If G has at least two cycles, then by Lemma 7.23, M is switching equivalent to Z or one ofthe mixed graphs in items (g) and (k); by Lemmas 7.16 and 7.20, M is switching equivalent to X .Now we may suppose the shortest cycle of G is of length k where k ≥
5. Let C be an induced k -cycle in G . If k = 5 or k ≥
7, then by Lemmas 7.16 and 7.20, G = C. Suppose now that k = 6. ByLemma 7.20, G = C, or M is switching equivalent to X or X .This completes the proof. Remark 2.
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