Solution to a conjecture on the maximum skew-spectral radius of odd-cycle graphs

Let $G$ be a simple graph with no even cycle, called an odd-cycle graph. Cavers et al. [Cavers et al. Skew-adjacency matrices of graphs, Linear Algebra Appl. 436(2012), 4512--1829] showed that the spectral radius of $G^\sigma$ is the same for every orientation $\sigma$ of $G$, and equals the maximum matching root of $G$. They proposed a conjecture that the graphs which attain the maximum skew spectral radius among the odd-cycle graphs $G$ of order $n$ are isomorphic to the odd-cycle graph with one vertex degree $n-1$ and size $m=\lfloor 3(n-1)/2\rfloor$. This paper, by using the Kelmans transformation, gives a proof of the conjecture. Moreover, sharp upper bounds of the maximum matching roots of the odd-cycle graphs with given order $n$ and size $m$ are given and extremal graphs are characterized.


Introduction
Let G be a finite simple graph with vertex set V = {v 1 , . . ., v n } and edge set E(G).For more notation and terminology that will be used in the sequel, we refer to [2,3].Given an orientation σ of G which makes any edge uv to be an arc u → v or v → u, we get an oriented graph G σ .The skew-adjacency matrix is the square matrix S(G σ ) = [s ij ] of order n, where s ij = 1 and s ji = −1 if u → v, s ij = s ji = 0 otherwise.By the definition, S(G σ ) is a skew symmetric matrix.Let the skew-characteristic polynomial be φ s (G σ , x) = det(xI n − S(G σ )), where I n is the unit matrix with order n.Thus the eigenvalues of S(G σ ) are all pure imaginary numbers, which form the skew-spectrum of G σ .The spectral radius ρ(G σ ) of an oriented graph G σ is the maximum norm of its all eigenvalues.In recent years, there are many results about the skew-spectrum and the skew-spectral radius, see [1,5,6,18].
For a given graph G, different orientations may lead to different skew-spectra, and thus different skew-spectral radius.In [5], Cavers et al. introduced the maximum skewspectral radius ρ s (G) which is defined as ρ s (G) = max{ρ(G σ ) : σ is an orientation of G}.They studied a special class of graphs, called the odd-cycle graphs.A graph is called an odd-cycle graph if it have no even cycles.In [5], Cavers et al. showed that when the graph G is an odd-cycle graph, the skew-spectra are independent of the orientation σ.In fact, the skew-characteristic polynomial φ s (G σ , x) of G σ can be determined by the matching polynomial m(G, x) of the graph G. Anuradha and Balakrishnan also got some skew-spectral properties of the odd-cycle graphs in [1].Now, we recall the definition of the matching polynomial [12].Definition 1.1 Let m r (G) denote the number of r independent edges in a graph G. Define the matching polynomial of G as Note that many results about the roots of the matching polynomial have been obtained; see [9,10,12,15], such as, the matching roots are real-valued, and symmetric.Denote by t(G) the maximum matching root of G.For the applications of the matching polynomial in chemistry and statistical physics, we refer the reader to [14,15].
Given an odd-cycle graph G, the following lemma in [5] implies the relationship between the skew-characteristic polynomial of an oriented graph G σ and the matching polynomial of the graph G.

Lemma 1.2 ( [5]
) Let G be a graph of order n.Then G is an odd-cycle graph if and only if φ s (G σ , x) = (−i) n m(G, ix) for all orientations σ of G.
From the above lemma, we note that when G is an odd-cycle graph, the skew-spectral radius ρ(G σ ) under any orientation σ equals the maximum skew-spectral radius ρ s (G).Moreover, the maximum skew-spectral radius of the graph G equals the maximum matching root of the graph G, i.e. ρ s (G) = t(G).
In [5], Cavers et al. studied the upper bound of the maximum skew-spectral radius among the odd-cycle graphs with order n, and proposed a conjecture.Let H n be the odd-cycle graph of order n with one vertex degree n − 1 and size m = ⌊3(n − 1)/2⌋.It is easy to check that H n is unique (up to isomorphism).Moreover, we can find that H n is the maximal odd-cycle graph of order n, since the size m of an odd-cycle graph of order n is no more than ⌊3(n − 1)/2⌋ ( [1], Theorem 3.2).For convenience, in the remainder of this paper, we just study the maximum matching root t(G) of an odd-cycle graph.We will prove Conjecture 1.4.In fact, we get an even stronger result.We characterize the extremal graphs with maximum t(G) among the odd-cycle graphs with order n and size m.(
We organize the rest of this paper as follows.In Section 2, we study the Kelmans transformation acting on odd-cycle graphs.In Section 3, we find the monotone property of the maximum matching roots after an acting of the Kelmans transformation.In Section 4, we give our proofs of Conjecture 1.4 and Theorem 1.5.

The Kelmans transformation on odd-cycle graphs
In [5], Cavers et al. suggested that the standard techniques called edge-switching will be useful to prove Conjecture 1.3.In this paper, we mainly use a graph transformation called the Kelmans transformation.
We first recall some results about the Kelmans transformation.In [16], Kelmans introduced a graph transformation for the extremal problems related with the synthesis of reliable networks.In [17], Satyanarayana et al. also introduced a reliability improving graph transformation, named the "swing surgery", which can be thought as the inverse of the Kelmans transformation.Brown et al. [4] called the Kelmans transformation as shift operation, and studied many applications in network reliability.Csikvári in [7] made a breakthrough in solving a conjecture of Nikiforov by the Kelmans transformation.Moreover, Csikvári in [8] found several applications of the Kelmans transformation in the extremal problems on graph polynomials, such as, matching polynomial, independent polynomial, chromatic polynomial, Laplacian polynomial.For more interesting details, see [8].Next, we give the definition of the Kelmans transformation.Definition 2.1 Let u and v be two vertices of a graph G. Let N(v) be the neighbor set of v in G.For any w ∈ N(v), if w is not u and u and w are nonadjacent, then we delete the edge vw and add the edge uw.After the above operations, we get a new graph G ′ .Then G ′ is a graph obtained from G by the Kelmans transformation between u and v; see Fig.

2.1.
For convenience, Csikvári [8] called u and v the beneficiary and the co-beneficiary of the transformation, respectively.For brevity, we denote by KT (G, u, v) the Kelmans transformation acting on a graph G with the beneficiary u and the co-beneficiary v.It is easy to check that the Kelmans transformation does not change the number of edges.Let G be a connected odd-cycle graph with order n and size m.
Theorem 2.2 Let G be a connected odd-cycle graph with order n and size m.Then F (n, m) can be obtained from G by a number of Kelmans transformations.
Proof.Let G be a connected odd-cycle graph with order n and size m, which implies that n − 1 ≤ m ≤ n + ⌊(n − 1)/2⌋ − 1.Then G must be a cactus, that is, every block of G is either a cycle or an edge ( [2], EX. 3.2.3).It follows that any two cycles of G are edge-disjoint.Let C(G) be the set of odd cycles with order larger than 3. Next, we will prove the theorem by three steps. Step and u, v be two vertices of C whose distance in C is 2. Then there exists a path uwvw ′ in C. By the Kelmans transformation KT (G, u, v), we get a graph G ′ .It is easy to find that the degree of the vertex v in G ′ is 1; see Fig. 2.2.
We first show that G ′ is connected.For every vertex u ′ in G, there is a path u ′ P u in G connected u ′ and u.If the vertex v is in the path P , then the path P can be decomposed as P 1 vP 2 .By the definition of the Kelmans transformation, we can check that u ′ P 1 u is in the graph G ′ and u ′ is connected to u in G ′ .If v is not in the path P , then u ′ and u are connected by the path P in G ′ .
Secondly, we claim that G ′ is still an odd-cycle graph.Suppose that there exists an even cycle C ′ in G ′ .Then u must be in C ′ .Let C ′ = uh 1 . . .h s u.If uh 1 and uh s both belong to G, then C ′ is also an even cycle of G, a contradiction.If neither uh 1 nor uh s belongs to G, then vh 1 . . .h s v is an even cycle of G, again a contradiction.Therefore, without loss of generality, suppose that uh 1 is in G and uh s is not in G, which implies that h 1 = h s and vh s is an edge of G.We then obtain a path P = uh 1 • • • h s v with even length in G. Since G is a cactus and u, v are both in the cycle C, we can deduce that P is in the cycle C.It follows that P = uwv, contradicting with h 1 = h s .
Finally, we consider the cycle set C(G ′ ).By analyzing the corresponding relationship between C(G) and C(G ′ ), it can be verified that Thus, by less than the number m 2 of Kelmans transformations as above, we can get a connected odd-cycle graph G ′ whose cycles are all triangles.Then, we turn to Step 3.
Step 2. If C(G) is empty, then G contains no cycles or each cycle of G is a triangle.Let G ′ := G.
Step 3. Assume that a vertex u of G ′ has the maximum degree in G ′ and its degree is denoted by d G ′ (u).If d G ′ (u) is less than n − 1, then there exists a vertex v such that the distance of u, v in G ′ is 2. Suppose that uwv is a 2-path in G ′ .We apply the Kelmans transformation KT (G ′ , u, w) and get a graph G ′′ .By the similar analysis in the first step, we find that G ′′ is a connected odd-cycle graph whose cycles are all triangles.Moreover, we get d G ′ (u) ≥ d G (u) + 1.Then by at most n − 1 kelmans transformations as above, we get a graph G ′′ with the maximum degree n − 1, order n and size m.Moreover, G ′′ satisfies that all cycles are triangles.It is easy to see that the graph G ′′ is isomorphic to the graph F (n, m); see Fig. 2

.3.
The proof is this complete.

Maximum matching root and Kelmans transformation
In this section, we will show that the maximum matching roots of the graph strictly increase after the Kelmans transformation.Before proceeding, we first review some useful lemmas of the matching polynomial.Let G and H be two disjoint graphs.Then the graph G H denotes the union of G and H.
In [13], Gutman showed some parallel results for the roots of the matching polynomial and the spectra of the characteristic polynomial.
Now, we give some propositions of the maximum matching root under the Kelmans transformation.
It is easy to check that G 1 ≻ G 2 can deduce G 1 G 2 .Note that Csikvári [8] introduced the definition ≻ which has the same meaning as our definition , and they obtained many good results which inspire us.In the following, we obtain some results about ≻ and .
Before proceeding, we list some useful results about which were proved in [8].
Next, we give our results.
is the maximum root of the matching polynomial, it implies that for x ≥ t(G 1 ), we have m(G 2 , x) > m(G 1 , x) ≥ 0. Due to the fact that the leading coefficient of the matching polynomial is 1, we get t(G Proof.We only prove the first part, the second part can be proved similarly.For Proof.Since H is a proper spanning subgraph of G, suppose H is obtained by deleting l edges {e 1 , . . ., e l } from G.Then, m(H, x) = m(G − e 1 − • • • − e l , x).Suppose e 1 is uv.By Lemma 3.1, we have Since G is connected, by Lemma 3.3, we have t(G − u − v) < t(G) and t(G − e 1 ) < t(G).
Since the leading coefficient of the matching polynomial is 1, it follows that for x ≥ t(G), we get Since H is the proper spanning subgraph of G, we can find that for some i, H i is the proper spanning subgraph of G i .Without loss of generality, assume that H 1 is the proper spanning subgraph of G 1 .By Proposition 3.10, we have For other subgraphs H i where i ≥ 2, by Proposition 3.6, we have Using Equations 3.4 and 3.5, we deduce that for The proof is now complete.
Theorem 3.12 Let G be a connected graph.Assume that G ′ is obtained from G by a number of Kelmans transformations, and Proof.Let G be a connected graph.Let G 1 be a graph obtained from G by the Kelmans transformation between u and v, where u is the beneficiary, and G 1 is not isomorphic to G. We just need to prove G 1 ≻ G.
Firstly, we find that u has an adjacent vertex w which is not adjacent to v, and v has an adjacent vertex w ′ which is not in the set of neighbors of the vertex u.Otherwise, G 1 is isomorphic to G.This is a contradiction.
Secondly, we use Lemma 3.1 and get Then we have the following equation (3.6) Next, we consider the right part of the above equation.If we apply the Kelmans transformation between u and v in G − uw, then we can get the graph G 1 − uw.Thus, by Theorem 3.7, we get Otherwise, suppose that G is connected and G 1 is not connected.Then it follows that the set of neighbors of u and the set of neighbors of v are disjoint.Moreover, G 1 is the disjoint union of a isolated vertex v and a connected component . By Proposition 3.9, we get G ′ ≻ G.The proof is thus complete.

The maximum matching roots of odd-cycle graphs
In this section, we first prove a theorem about the maximum matching root of F (n, m).Given a positive integer m with m > 4, when ⌈ 2m+1 3 ⌉ ≤ n ≤ m, suppose that G ′ is the disjoint union of F (n, m) and an isolate vertex named w.Since n ≤ m, it follows that there exists a cycle in F (n, m).Due to the definition of F (n, m), the cycle in F (n, m) is a triangle.Then there exists a triangle {v 1 , v 2 , v 3 } where v 1 has the maximum degree in F (n, m).Up to isomorphism, the graph F (n + 1, m) can be obtained from G ′ by deleting the edge v 2 v 3 and adding the edge v 1 w.By Lemma 3.1, we have Note that The graph Above all, by Lemma 3.11, for Proof.If G is disconnected, then there is a connected odd-cycle graph G ′ with order n which contains G as a proper subgraph.By Lemma 3.3, t(G) is less than t(G ′ ).Then, suppose that G is a connected odd-cycle graph with order n and size m.
By Kelmans transformations, we transfer the graph G into F (n, m).By Theorem 3.12, we deduce that t(G) ≤ F (n, m) and equality holds if and only Then, we conclude that t(G) ≤ t(H n ) and equality holds if and only if G ∼ = H n .The proof is thus complete.Now, it is time to prove Theorem 1.5.
Proof of Theorem 1.5.For m ≤ 2, it is easy to check that the first two parts of the theorem is true.
For m = 3, G is one of the following graphs: the disjoint union of three edges and n − 6 isolated vertices, the disjoint union of the star K 1,2 , an edge and n − 5 isolated vertices, the disjoint union of the star K 1,3 and n − 4 isolated vertices, the disjoint union of a triangle and n − 4 isolated vertices.By some computations, it is easy to check that the third part of the theorem is true.In every nontrivial connected component F (n i , m i ), one of the vertices which have the maximum degree in F (n i , m i ) is named the root vertex of F (n i , m i ).By identifying all root vertices, we obtain a graph F (n − s + 1, m).Let G ′′ be the graph obtained from the disjoint union of the graph F (n − s + 1, m) and s − 1 isolated vertices.
If there are at least two nontrivial connected components, then every nontrivial connected component F (n i , m i ) is a proper subgraph of F (n − s + 1, m).Then by Lemma 3.2 and Lemma 3. Above all, together with Theorem 3.12, we conclude that t(G) ≤ t(F (m + 1, m)) and the equality holds if and only if G has only one nontrivial connected component which is isomorphic to F (m + 1, m).Thus, the fourth part of the theorem is true.
When n − 1 ≤ m ≤ ⌊ 3(n−1) 2 ⌋ and m ≥ 4, suppose that G is disconnected and has s connected components.Then by the similar method to the fourth part, we can deduce that the maximum matching root of G is smaller than the maximum matching root of F (n − s + 1, m).By Theorem 4.1, for s > 1 we have t(F (n − s + 1, m)) < t(F (n, m)).
Then, we conclude that t(G) ≤ t(F (n, m)), and the equality holds if and only if G is connected and isomorphic to F (n, m) from Theorem 3.12.Thus, the fifth part of the theorem is true.
The proof is thus complete.

Theorem 1 . 5 2 ⌋.( 1 )
Let G be an odd-cycle graph with order n and size m, where 1 ≤ m ≤ ⌊ 3(n−1) Let F (n, m) be an odd-cycle graph with one vertex degree n − 1 and size m, which is unique up to isomorphism.If m = 1, then t(G) = 1.

( 2 ) 2 ,
If m = 2, then t(G) ≤ √ and equality holds if and only if G is the disjoint union of a star K 1,2 and n − 2 isolated vertices.(3) If m = 3, then t(G) ≤ √ 3, and equality holds if and only if G is the disjoint union of a triangle and n − 3 isolated vertices, or the disjoint union of a star graph K 1,3 and n − 4 isolated vertices.

( 4 )
If 4 ≤ m ≤ n − 2, then t(G) ≤ t(F (m + 1, m)), and equality holds if and only if G is the disjoint union of the graph F (m + 1, m) and n − m − 1 isolated vertices.

(4. 3 )
Thanks to the structure of F (n + 1, m), the graph F (n + 1, m) − v 1 − w is the disjoint union of m − n edges and 3n − 2m − 1 isolated vertices.The graph G ′ − v 2 − v 3 is the disjoint union of the graph F (n − 2, m − 3) and an isolated vertex w.By the definition of F (n − 2, m − 3), we know that the graph F
and equality holds if and only ifG ∼ = H n .Based on Lemma 1.2, we can find that Conjecture 1.3 can be equivalently rewritten as follows.
Conjecture 1.4 If G is an odd-cycle graph of order n, then t(G) ≤ t(H n ), and equality holds if and only if G ∼ = H n .
.3) Thus we get G ≻ G − e 1 .Since H is a spanning subgraph of G − e 1 , by Proposition 3.6, we have G − e 1 H.By Proposition 3.9, we deduce that G ≻ H. Proof.If G is a connected graph, then by Proposition 3.10, we get G ≻ H and m(G, x) > m(H, x) for x > t(G).Next, we suppose that G has k connected components G 1 , . . ., G k where k ≥ 2. Since H is an edge-deleted subgraph of G, let H be a union of k disjoint graphs {H 1 , . . ., H k }, where H i is the spanning subgraph of G i .By Lemma 3.2, we know The proof is now complete.Next, we give a proof of Conjecture 1.4 by proving the following theorem.Theorem 4.2 If G is an odd-cycle graph of order n, then t(G) ≤ t(H n ), and equality holds if and only if G ∼ = H n .