On the distance spectral radius of unicyclic graphs with perfect matchings

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1 Electronic Journal of Linear Algebra Volume 27 Article On the distance spectral radius of unicyclic graphs with perfect matchings Xiao Ling Zhang Follow this and additional works at: Recommended Citation Zhang, Xiao Ling. (2014), "On the distance spectral radius of unicyclic graphs with perfect matchings", Electronic Journal of Linear Algebra, Volume 27. DOI: This Article is brought to you for free and open access by Wyoming Scholars Repository. It has been accepted for inclusion in Electronic Journal of Linear Algebra by an authorized editor of Wyoming Scholars Repository. For more information, please contact

2 ON THE DISTANCE SPECTRAL RADIUS OF UNICYCLIC GRAPHS WITH PERFECT MATCHINGS XIAO LING ZHANG Abstract. For a connected graph, the distance spectral radius is the largest eigenvalue of its distance matrix. Let U2k 1 be the graph obtained from C 3 by attaching a path of length n 3 at one vertex. Let U2k 2 be the graph obtained from C 3 by attaching a pendant edge together with k 2 paths of length 2 at the same vertex. In this paper, it is proved that U2k 1 (resp., U2 2k ) is the unique graph with the maximum (resp., minimum) distance spectral radius among all unicyclic graphs with perfect matchings on 2k(k 5) vertices. Key words. Distance spectral radius, Unicyclic graph, Perfect matching. AMS subject classifications. 05C50, 15A Introduction. Let G be a connected graph with vertex set {v 1,v 2,...,v n }. The distance between the vertices v i and v j is the length of a shortest path between them, and is denoted by d G (v i,v j ), or d(v i,v j ). The distance matrix D = D(G) of G is defined so that its (i,j) entry is equal to d G (v i,v j ). The largest eigenvalue of D(G) is called the distance spectral radius, and is denoted by ρ(g). Balaban et al. [1] proposed the use of ρ(g) as a molecular descriptor, while in [4] it was successfully used to infer the extent of branching and model boiling points of alkanes. Therefore, the study concerning the maximum (minimum) distance spectral radius of a given class of graphs is of great interest and significance. Recently, the maximum (minimum) distance spectral radius of a given class of graphs has been studied extensively. For example, Subhi and Powers [8] determined the graph with maximum distance spectral radius among all trees on n vertices; Stevanović and Ilić [9] determined the graph with maximum distance spectral radius among all trees with fixed maximum degree ; Ilić [5] characterized the graph with minimum distance spectral radius among trees with given matching number; Bose et al. [2] studied the graphs with minimum (maximum) distance spectral radius among all graphs of order n with r pendent vertices; Zhang and Godsil [11] determined the graph with minimum distance spectral radius among all graphs of order n with k cut Received by the editors on August 23, Accepted for publication on August 11, Handling Editor: Ravi Bapat. School of Mathematics and Information Science, Yantai University, Yantai, Shandong , P.R. China (zhangxling04@aliyun.com). Supported by NSFC ( ) and NSF of Shandong Province of China (ZR2012AQ022). 569

3 570 X.L. Zhang vertices(resp., k cut edges); Yu et al. [10] determined the unique graph with minimum (maximum) distance spectral radius among unicyclic graphs on n vertices; Milan Nath and Somnath Paul [7] determined the unique graph with minimum distance spectral radius among all connected bipartite graphs of order n with a given matching number (resp., with a given vertex connectivity). A unicyclic graph is a connected graph in which the number of edges equals the number of vertices. A rooted graph has one of its vertex, called the root, distinguished from the others. We use the following notation to represent a unicyclic graph: G = U(C l ;T 1,T 2,...,T l ), where C l is the unique cycle in G with V(C l ) = {v 1,v 2,...,v l } such that v i is adjacent to v i+1 ( mod l) for 1 i l. For each i, let T i be the rooted tree with root v i (see Fig. 1). If V(T i ) = 1, we say T i is a trivial tree. Let U(2k) denote the set of all unicyclic graphs on 2k vertices with perfect matchings. Let U 1 2k be the graph obtained from C 3 by attaching a path of length n 3 at a vertex. Let U 2 2k be the graph obtained from C 3 by attaching a pendant edge together with k 2 paths of length 2 at the same vertex. Fig. 1. Graph U(C l ;T 1,T 2,...,T l ). In this paper, we mainly consider the distance spectral radius of unicyclic graphs on 2k (k 3) vertices with perfect matchings, and prove that U2k 1 (resp., U2 2k ) is the unique graph with the maximum (resp., minimum) distance spectral radius among all unicyclic graphs with perfect matchings on 2k (k 5) vertices. 2. Prelimaries. Wefirstgivesomelemmaswhichwewilluseinthemainresults. Lemma 2.1. [9] Let w be a vertex of the nontrivial connected graph G and for positive integers p and q, let G p,q denote the graph obtained from G by adding pendent paths P = wv 1 v 2 v p and Q = wu 1 u 2 u q of length p and q, respectively, at w. If p q 1, then ρ(g p,q ) < ρ(g p+1,q 1 ).

4 On the Distance Spectral Radius of Unicyclic Graphs with Perfect Matchings 571 Lemma 2.2. [11] Let u and v be two adjacent vertices of a connected graph G and for positive integers k and l, let G k,l denote the graph obtained from G by adding paths of length k at u and length l at v. If k > l 1, then ρ(g k,l ) < ρ(g k+1,l 1 ); if k = l 1, then ρ(g k,l ) < ρ(g k+1,l 1 ) or ρ(g k,l ) < ρ(g k 1,l+1 ). Lemma 2.3. [3] ρ(c n ) = n2 4, if n is even; ρ(c n) = n2 1 4, if n is odd. Lemma 2.4. [6] Let A = (a ij ) be an n n nonnegative matrix. Then min a ij ρ(a) max a ij. 1 i n 1 j n 1 i n 1 j n Lemma 2.5. If n 10 and n is even, then ρ(u 2 n) < ρ(c n ). Proof. By Lemma 2.4, we get By Lemma 2.3, we have ρ(un 2 ) max 1 i n 1 j n ρ(c n ) = n2 4. d ij = 7n 2 9. If n = 10, by matlab, we get ρ(u 2 10 ) = < ρ(c 10) = 25. If n 12 and n is even, i.e., ρ(u 2 n ) < ρ(c n). ρ(u 2 n) ρ(c n ) ( 7n ) 2 9 n2 4 = 13 (n 7)2 < 0, 4 So, in either case, we can get ρ(u 2 n ) < ρ(c n), for n 10 and n is even. Let X be the Perron vector of G corresponding to ρ(g). Suppose T i {v i } = αk 2 K 1 and T i+1 {v i+1 } = βk 2 γk 1 for some 1 i l, where α and β are both nonnegative integers, γ = 0 or 1. Using a symmetry, we can denote the coordinates of the Perron vector corresponding to the vertices in V(T i ) and V(T i+1 ) as shown in Fig. 2. Then, we have Lemma 2.6. (i) c + d > b; (ii) h + d > c; (iii) a + b > c; (iv) c + d > h; (v) a+d > b. Proof. We first prove (i). Let S = α(a+b)+c+d and S = v j V(G) D(G)X = ρ(g)x, x j. Since

5 572 X.L. Zhang G e Ti - 1 vi - 1 v i + 1 e h v i g g g d f e a a a c b b b Fig. 2. Graph G. we can easily get (2.1) ρ(g) = 2k j=1 d ij x j. So, we have ρ(g)c+ρ(g)d ρ(g)b 2a+(α+4)b 2c d+(s S 1 h), 2a+(α+4)b 2c d, i.e., (ρ(g)+2)(c+d b) 2a+(α+2)b+d > 0, which implies c+d > b. Similarly, we can prove (ii),(iii), (iv) and (v). Lemma 2.7. Let graphs G 1, G 1, G 2, G 2 p 2 and q 1. Then we have U(2k) be as shown in Fig. 3, where (i) ρ(g 1 ) < ρ(g 1 ); (ii) ρ(g 2) < ρ(g 2 ). Proof. We first prove (i). Let X be the Perron vector of G 1 corresponding to ρ(g 1 ). Using a symmetry, we can denote the coordinates of the Perron vector corresponding to some vertices of G 1 as shown in Fig. 3. Let S = and S = S p(a+b), where p 2. From G 1 to G 1, we have v i V(G 1) ρ(g 1 ) ρ(g 1) X T (D(G 1 ) D(G 1))X = (p 1)(a+b)(S a b).

6 On the Distance Spectral Radius of Unicyclic Graphs with Perfect Matchings 573 H 1 H 2 } a b b d a b p H d a u 1 u a b } p - 1 a b ' a b G 1 G 1 c d a H d b u 2 a b q } u c a b q } a b a b G 2 ' G 2 Fig. 3. Graphs G 1, G 1, G 2, G 2. In the following, we will prove S a b > 0 into two cases. If H 1 = C 3, let C 3 = uvwu. Then S = d+x v +x w. By (2.1), we have ρ(g 1 )d+ρ(g 1 )x v +ρ(g 1 )x w ρ(g 1 )a ρ(g 1 )b = 4a+(p+6)b d 3x v 3x w, i.e., So, we have S a b > 0. (ρ(g 1 )+3)(S a b) = 2d+a+(p+3)b > 0. If H 1 C 3, then V(H 1 ) 4. There must exist some vertex w V(H 1 ) such that d H1 (u,w) = 2. Suppose v N H1 (u) N H1 (w). By (2.1), we have ρ(g 1 )S ρ(g 1 )(a+b) p(a+2b)+4a+6b 4S, i.e., (ρ(g 1 )+4)(S a b) > p(a+2b)+2b > 0, which implies S a b > 0. So, in either case, we can get S a b > 0, which implies ρ(g 1 ) < ρ(g 1). Similarly, we can prove (ii). 3. Main results. Theorem 3.1. U2k 1 is the unique graph with the maximum distance spectral radius among all unicyclic graphs with perfect matchings on 2k (k 3) vertices. Proof. Choose G U(2k) such that ρ(g) is as large as possible. Let G = U(C l ;T 1,T 2,...,T l ) and V(C l ) = {v 1,v 2,...,v l }. Let X = (x 1,x 2,...,x 2k ) T be the

7 574 X.L. Zhang Perron vector of G corresponding to ρ(g), where corresponds to the vertex v i (1 i 2k). Suppose M(G) is any perfect matching of G. If there exists some 1 i l such that v i v (i+1) mod l M(G), we may assume v 1 v 2 M(G). Then v 2 v 3 / M(G) and v 1 v l / M(G). If v i v (i+1) mod l / M(G) for any 1 i l, then v 2 v 3 / M(G) and v 1 v l / M(G). So, in either case, we can always assume v 2 v 3 / M(G) and v 1 v l / M(G). Claim 1. l = 3. Otherwise, we may assume l 4. Case 1. l = 4. If v i V (T 3), let G = G {v 1 v 4 }+{v 1 v 3 }. Then G U(2k). From G to G, the distances between V(T 1 ) and V(T 3 ) are decreased by 1; the distances between V(T 1 ) and V(T 4 ) are increasedby 1; the distances between V(T 2 ) and V(T 1 ) V(T 3 ) V(T 4 ), V(T 3 ) and V(T 4 ) are unchanged. So, we have ρ(g ) ρ(g) X T (D(G ) D(G))X = 2 0. v j V (T 1) In the following, we will prove ρ(g) ρ(g ). x j v i V (T 4) v i V(T 3) If not, then X is also the Perron vector of G corresponding to ρ(g ). According to (2.1), we have ρ(g)x 4 = d 4jx j + (d 1j +1)x j + (d 2j +2)x j + (d 3j +1)x j, v j V(T 4 ) ρ(g )x 4 = v j V (T 4 ) v j V (T 1 ) d 4jx j + v j V (T 1 ) v j V (T 2 ) (d 1j +2)x j + v j V (T 2 ) Since ρ(g) = ρ(g ), from the above two equations, we get x j = 0, v j V(T 1) which contradicts to the fact that X is a Perron eigenvector. v j V(T 3 ) (d 2j +2)x j + v j V (T 3 ) (d 3j +1)x j.

8 On the Distance Spectral Radius of Unicyclic Graphs with Perfect Matchings 575 So, we have ρ(g ) > ρ(g), which is a contradiction. If <, let v i V (T 3) G = G {v 2 v 3 }+{v 2 v 4 }. Then G U(2k). Similar to the above, we can also get a contradiction. Case 2. l 5. If v i V(T 3) V(T l ) +1 2 v i V(T l 2 +2 ) V(T l) G = G {v 2 v 3 }+{v 2 v l }., let Then G U(2k). From G to G, the distances between V(T 1 ) and V(T 3 ) V(T l ) are increased by at least 1; the distances between V(T 2) and V(T 3 ) 2 V(T l +1) are increased by at least 1; the distances between V(T 2) and V(T 2 l +2) 2 V(T l ) are decreased by 1; the distances between V(T i ) (3 i l 1) and V(T j ) (i < j l) are unchanged or increased by at least 1. So, we have ρ(g ) ρ(g) X T (D(G ) D(G))X > 2 x j v i V(T 3) V (T l ) , v j V(T 2) i.e., ρ(g ) > ρ(g), which is a contradiction. If < v i V(T 3) V(T l ) +1 2 v i V(T l 2 +2 ) V(T l) G = G {v 1 v l }+{v 1 v 3 }., let v i V(T l 2 +2 ) V(T l) Then G U(2k). Similar to the above, we can also get a contradiction. Claim 2. G = U 1 2k. Since G = U(C 3 ;T 1,T 2,T 3 ), using Lemma 2.1 frequently, we can first get each T i (1 i 3) is a path. Then using Lemma 2.2 at most twice, we can get G = U 1 2k. Theorem 3.2. H 2 (see Fig. 4) is the unique graph with the minimum distance spectral radius among all unicyclic graphs with perfect matchings on 6 vertices. Proof. There are 8 graphs in U(6) (see Fig. 4). By Lemma 2.1, we have (3.1) ρ(h 8 ) > ρ(h 5 ).

9 576 X.L. Zhang H1 H 2 H3 H 4 H5 H6 H7 H8 Fig. 4. Graphs H 1 H 8. By Lemma 2.2, we have (3.2) ρ(h 3 ) > ρ(h 4 ), ρ(h 7 ) > ρ(h 6 ). Combining (3.1), (3.2) and Table 3.1, we get G = H 2. Table 3.1 G H 1 H 2 H 4 H 5 H 6 ρ(g) Theorem 3.3. G 9 (see Fig. 5) is the unique graph with the minimum distance spectral radius among all unicyclic graphs with perfect matchings on 8 vertices. Proof. Choose G U(8) such that ρ(g) is as small as possible. Let G = U(C l ;T 1,T 2,...,T l ). Then, we get l 8. By Lemma 2.1 and Lemma 2.7, we get T i {v i } = ak 1 bk 2 for 1 i l, where a = 0 or 1. Since V(G) = 8, we have G {G i 1 i 18 and i is an integer} (see Fig. 5). (3.3) By Lemma 2.2, we have ρ(g 11 ) > ρ(g 10 ), ρ(g 15 ) > ρ(g 17 ). Combining (3.3) and Table 3.2, we get G = G 9. Table 3.2 G G 1 G 2 G 3 G 4 G 5 G 6 G 7 G 8 ρ(g) G G 9 G 10 G 12 G 13 G 14 G 16 G 17 G 18 ρ(g) Theorem 3.4. U2k 2 is the unique graph with the minimum distance spectral radius among all unicyclic graphs with perfect matchings on 2k (k 5) vertices.

10 On the Distance Spectral Radius of Unicyclic Graphs with Perfect Matchings 577 G1 G 2 G3 G4 G5 G6 G7 G8 G9 G10 G11 G12 G13 G14 G15 G16 G17 G18 Fig. 5. Graphs G 1 G 18. Proof. Choose G U(2k) such that ρ(g) is as small as possible. Let G = U(C l ;T 1,T 2,...,T l ) and C l = v 1 v 2 v l v 1. By Lemma 2.1 and Lemma 2.7, we get T i {v i } = ak 1 bk 2 for 1 i l, where a = 0 if V(T i ) is odd, and a = 1 if V(T i ) is even. In the following, we always assume N Ti (v i ) = {v j v j V(T i ) and v j is adjacent to v i }, N T i (v i ) = {v j v j N Ti (v i ),d(v j ) = 2} and R Ti = {v j v j N Ti (v i ),d(v j ) = 1} {v i }. Then R Ti = 1 or 2. Suppose G is the graph obtained from G by grafting some edges and G U(2k). For some i, if T i is still a rooted tree of G, we still use T i to denote the rooted tree with root v i in G ; if T i is not a rooted tree of G any more, but v i is still a root, we always use T i to denote the rooted tree with root v i in G. Let M(G) be any perfect matching of G. Claim 1. l 4. Otherwise, we may assume l 5. Case 1. There exists some 1 i l such that V(T i ) 3. Without loss of generality, we may assume V(T 1 ) 3. Subcase 1.1. v l 1 v l / M(G) and v 2 v 3 / M(G). Suppose N T l (v l ) = {v l1,...,v lr } and N T 2 (v 2 ) = {v 21,...,v 2s }. If l is even, let G = G {v l v l 1,v l v l1,...,v l v lr }+{v 1 v l 1,v 1 v l1,...,v 1 v lr }. Then G U(2k) and G = U(C l 1 ;T 1,T 2,...,T l 1 ). Let X = (x 1,x 2,...,x 2k ) T be the

11 578 X.L. Zhang Perron vector of G corresponding to ρ(g ), where corresponds to the vertex v i (1 i 2k). For convenience, we can denote the coordinates of Perron vector X corresponding to vertices in V(T 1) the same as V(T i+1 ) in Fig. 2. Since V(T 1 ) 3 and v i R Tl = e+g or f, from G to G, we have = 2 x j v i V(T 2) v i R Tl > 2 v i V (T 1) v j V (T l2 +1 ) V(T l 1) v j V(T l2 +2 ) V(T l 1) v i V(T l2 1 ) v j V(T l 1 ) v j V (T l2 +1 ) V(T l 1) v i R Tl > 2[e+g +h (e+g)] > 0, which is a contradiction. If l is odd, let x j x j +2 x j 2 V(T l2 ) v i V (T 3) v i R Tl v j V (T l )\R Tl x j v j V(T l2 +3 ) V(T l 1) v j V(T l )\R Tl x j v j V(T l2 +1 ) V (T l 1) V(T l )\R Tl x j v j V(T l2 +1 ) V (T l 1) V(T l )\R Tl x j G = G {v 2 v 3,v 2 v 21,...,v 2 v 2s } {v l v l 1,v l v l1,...,v l v lr } +{v 1 v 3,v 1 v 21,...,v 1 v 2s }+{v 1 v l 1,v 1 v l1,...,v 1 v lr }. Then G U(2k) and G = U(C l 2 ;T 1,T 3,...,T l 1 ). From G to G, we have > 2 +2 v i R T2 v i R Tl v i V(T 3) V (T l+1 ) 2 + v i V(T l ) V(T l 1) v i V(T 2)\R T2 + v i V (T l )\R Tl x j.

12 On the Distance Spectral Radius of Unicyclic Graphs with Perfect Matchings 579 Similar to the case that l is even, we can also get a contradiction. Subcase 1.2. v l 1 v l / M(G) and v 2 v 3 M(G). Suppose N T l (v l ) = {v l1,...,v lr } and N T 3 (v 3 ) = {v 31,...,v 3t }. Let G = G {v l v l 1,v l v l1,...,v l v lr } {v 3 v 4,v 3 v 31,...,v 3 v 3t } +{v 1 v l 1,v 1 v l1,...,v 1 v lr }+{v 2 v 4,v 2 v 31,...,v 2 v 3t }. Then G U(2k) and G = U(C l 2 ;T 1,T 2,T 4,...,T l 1 ). From G to G, we have ρ(g) ρ(g ) X T ( D(G) D(G ) ) X > 2 + v i V (T 1 ) (3.4) 2 x 3 + v i R Tl v i V (T 2 ) v i V(T 4 ) V(T l 1 ) v i V (T 4 ) V(T l 1 ) + + v i (V (T 3 )\R T3 ) (V (T l )\R Tl ) v i (V (T 3 )\R T3 ) (V (T l )\R Tl ). We can denote the coordinates of Perron vector X correspondingto vertices in V(T 1) and V(T 2 ) the same as V(T i+1) and V(T i ) in Fig. 2. Then, by Lemma 2.3 (ii), we have (3.5) v i R Tl + v i V(T 2) x 3 (e+g +h) (e+g)+d c = h+d c > 0. Combining (3.4) and (3.5), we get ρ(g) > ρ(g ), which is a contradiction. Subcase 1.3. v l 1 v l M(G) and v 2 v 3 M(G). If there exists some i = 2,3,l 1,l such that V(T i ) 3, then dealing with this case the same as Subcase 1.1 and Subcase 1.2, respectively, we can get a contradiction. Otherwise, V(T 2 ) = V(T 3 ) = V(T l 1 ) = V(T l ) = 1. If l = 5, let G = G {v 3 v 4 }+{v 1 v 3 }. Then G U(2k) and G = U(C 3 ;T 1,T 2,T 3 ). We can denote the coordinates of Perron vector X corresponding to vertices in V(T 1 ) the same as V(T i) in Fig. 2. Using a symmetry, we can get x 2 = x 3. Since V(G) 10, we have V(T 1 ) 6. By

13 580 X.L. Zhang Lemma 2.3 (i), from G to G, we have which is a contradiction. If l = 6, let = 2x 3 2x 2 x 4 4x 3 x 4 = 2x 3 3x 4 2x 3 [2(a+b)+c+d 3b] = 2x 3 (2a+c+d b) > 0, Then G U(2k). From G to G, we have G = G {v 3 v 4,v 4 v 5 }+{v 1 v 3,v 1 v 4 }. = 2 x x 3 x 6 v i V (T 1) = 2x 3 > 0. x 5 +x x 5 4x 5 x 3 2x 5 If l = 7, V(T 4 ) 3 or V(T 5 ) 3, deal with the case the same as Subcase 1.2. If l = 7, V(T 4 ) = V(T 5 ) = 1 and V(T 1 ) = 4, by matlab, we get ρ(g) = > ρ(u 2 10) = , which is a contradiction. If l = 7, V(T 4 ) = V(T 5 ) = 1 and V(T 1 ) > 4, let Then G U(2k). From G to G, we have G = G {v 5 v 6 }+{v 1 v 5 }. (3.6) = 2 x 4 +4 x 5 +2x 2 x 5 2x 3 x 6 4x 4 x 6 4x 5 x 6 v i V (T 1) > 2 x 4 x 3 x 6 2x 4 x 6 +4x 5 x 6.

14 On the Distance Spectral Radius of Unicyclic Graphs with Perfect Matchings 581 Since v 3 and v 4 are symmetric in G, we can assume x 3 = x 4. We can denote the coordinates of Perron vector X corresponding to vertices in V(T 1 ) the same as V(T i ) in Fig. 2. By Lemma 2.3 (i), we have x 4 x 3 x 6 2x 4 x 6 = 3x 6 x 4 v i V (T 1) (2(a+b)+c+d 3b)x 4 (3.7) and (3.8) = (2a+c+d b)x 4 > 0, x 6 > 3x 6 > 0. Combining (3.6), (3.7) and (3.8), we get ρ(g) > ρ(g ), which is a contradiction. If l = 7 and V(T 4 ) = V(T 5 ) = 2, let G = G {v 3 v 4,v 4 v 5,v 5 v 6 }+{v 1 v 4,v 1 v 5,v 1 v 6 }. Then G U(2k) and G = U(C 3 ;T 1,T 6,T 7 ). From G to G, we have (3.9) = x > 4 v i V (T 1) +2 2 v i V(T 5) x 7 v i V(T 5) x 3 v i V (T 4) v i V(T 5) x 3 2 +x 6 +2x 2 v i V(T 5) +2x 6 v i V(T 5) x 6. v i V(T 5) v i V (T 5) +x 6 We can denote the coordinates of Perron vector X corresponding to vertices in V(T 1) the same as V(T i ) in Fig. 2. Since V(T 1 ) 3, we have (3.10) x 3 > a+b+d b = a+d > 0.

15 582 X.L. Zhang Similarly, we can have (3.11) 2 > 0, v i V (T 1) v i V(T 5) x 3 > 0. Combining (3.9), (3.10) and (3.11), we get ρ(g) > ρ(g ), which is a contradiction. If l 8, let G = G {v 3 v 4,v l 2 v l 1 }+{v 2 v 4,v l 2 v l }. Then G U(2k) and G = U(C l 2 ;T 1,T 2,T 4,...,T l 2,T l ). From G to G, we have > 2 +x 2 +x l x 3 x l 1 v i V (T 4) V(T l 2 ) Similarly to the case l = 7, we can also get a contradiction. Case 2. For any 1 i l, V(T i ) = 2. If l = 5, by matlab, we get ρ(g) = > ρ(u10 2 ) = , which is a contradiction. If l = 6, by matlab, we get ρ(g) = > ρ(u12 2 ) = , which is a contradiction. If l = 7, by matlab, we get ρ(g) = > ρ(u14 2 ) = , which is a contradiction. If l = 8, by matlab, we get ρ(g) = > ρ(u 2 16) = , which is a contradiction. If l 9, let G = G {v 2 v 3,v 3 v 4,...,v l 2 v l 1 }+{v 1 v 3,v 1 v 4,...,v 1 v l 1 }. Then G U(2k) andg = U(C 3 ;T 1,T l 1,T l,). Since = = v i V(T 2) v i V (T l 2 )

16 and v i V(T l 1 ) On the Distance Spectral Radius of Unicyclic Graphs with Perfect Matchings 583 = v i V(T l ) in G, from G to G, we have > v i V(T 2) v i V(T 3) > 0, which is a contradiction. v i V (T l 2 ) v i V (T 3) v i V(T l 3 ) v i V(T 3) v i V (T l 1 ) + v i V (T l 1 ) v i V(T 5) + v i V (T l 2 ) v i V(T 6) + + v i V(T l 1 ) v i V(T l ) v i V(T l ) Case 3. For any 1 i l, V(T i ) 2, and there exists some 1 j l such that V(T j ) = 1. For convenience, we may assume that V(T 1 ) = 2 and V(T l ) = 1. Subcase 3.1. V(T 2 ) = 1. Then v 2 v 3,v l 1 v l M(G). Dealing with this case the same as Subcase 1.3. Subcase 3.2. V(T 2 ) = V(T 3 ) = 2. Then v l 1 v l M(G). Since V(G) 10, we can get l 6. If l = 6, we get ρ(g) = > ρ(u10 2 ) = , which is a contradiction. If l = 7 and V(T 4 ) = 1, we get ρ(g) = > ρ(u10 2 ) = , which is a contradiction. If l = 7 and V(T 4 ) = 2, we get ρ(g) = > ρ(u12 2 ) = , which is a contradiction. If l 8 is odd, let G = G {v 2 v 3,v 3 v 4,v l 2 v l 1 }+{v 1 v 3,v 1 v 4,v 1 v l 2 }.

17 584 X.L. Zhang Then G U(2k) and G = U(C l 4 ;T 1,T 4,...,T l 2 ). We can denote the coordinates of Perron vector X corresponding to vertices in V(T 1 ) the same as V(T i) in Fig. 2. From G to G, we have > 2 x l 1 v i V (T 1) = (c+d b) > 0. v i V (T 4) V(T l 2 ) V(T l 2 ) If l 8 is even, let G = G {v 1 v l,v 3 v 4,v l 2 v l 1 }+{v 2 v l,v 2 v 4,v l 2 v l }. Then G U(2k) and G = U(C l 3 ;T 2,T 4,...,T l 2,T l ). We can denote the coordinates of Perron vector X corresponding to vertices in V(T 2 ) and V(T l ) the same as V(T i+1 ) and V(T i ) in Fig. 2, respectively. From G to G, we have > 2 +x l x l 1 v i V(T 2) = (h+f +d c) > 0, which is a contradiction. Subcase 3.3. V(T 2 ) = 2, V(T 3 ) = 1. v i V(T l2 +1 ) V (T l 2) v i V(T l2 +1 ) V (T l 2) Then v 3 v 4,v l 1 v l M(G). Since V(G) 10, V(T 1 ) = 2 and V(T l ) = 1, we get l 7. If l = 7, we get ρ(g) = > ρ(u10 2 ) = , which is a contradiction. If l = 8 and V(T 5 ) = 1, we get ρ(g) = > ρ(u 2 10) = , which is a contradiction. If l = 8 and V(T 5 ) = 2, we get ρ(g) = > ρ(u12 2 ) = , which is a contradiction.

18 On the Distance Spectral Radius of Unicyclic Graphs with Perfect Matchings 585 If l = 9 and V(T 5 ) = 1 or V(T 7 ) = 1, we can deal with the case similarly to Subcase 1.3. If l = 9 and V(T 5 ) = 2, V(T 7 ) = 2, we can deal with the case similarly to Subcase 3.2. If l 10, Let G = G {v 4 v 5,v l 2 v l 1 }+{v 2 v 5,v 1 v l 2 }. Then G U(2k) and G = U(C l 4 ;T 1,T 2,T 5,...,T l 2 ). We can denote the coordinates of Perron vector X corresponding to vertices in V(T 1 ) and V(T 2 ) the same as V(T i ) and V(T i+1 ) in Fig. 2, respectively. By Lemma 2.3 (i), from G to G, we have > 4 + v i V(T 2) = 4(c+d+h+f b g) > 0, x 4 x l 1 v i V (T 5) V(T l 2 ) v i V(T 5) V(T l 2 ) which is a contradiction. Case 4. For any 1 i l, V(T i ) = 1. Then G = C n. By Lemma 2.5, we have ρ(g) > ρ(u 2 n), which is a contradiction. Claim 2. l = 3. Otherwise, we have l = 4. Let G = U(C 4 ;T 1,T 2,T 3,T 4 ) and C 4 = v 1 v 2 v 3 v 4 v 1. Since V(G) 10, there must exist some 1 i 4 such that V(T i ) 3. We may assume that V(T 1 ) 3 and V(T 1 ) has the same parity as V(T 2 ). Suppose N T 2 (v 2 ) = {v 21,...,v 2s }. Let G = G {v 2 v 21,...,v 2 v 2s }+{v 1 v 21,...,v 1 v 2s }.

19 586 X.L. Zhang Then G U(2k) and G = U(C 3 ;T 1,T 3,T 4 ). From G to G, we have = 2 x j +2 v j V(T 2)\R T2 2 x j 2 v i V (T 2)\R T2 v j R T2 v i R T2 v j V (T 3) = 2( x j )( x j + v j R T2 v j V(T 2)\R T2 > 0, v i V (T 1) v j V(T 3) x j x j v j V(T 3) x j ) which is a contradiction. Claim 3. G = U 2 2k. Otherwise, let G = U(C 3 ;T 1,T 2,T 3 ). There must exist some 1 i,j 3 such that V(T i ) is even and V(T j ) > 1. We may assume V(T 1 ) is even, V(T 2 ) > 1 and V(T 2 ) V(T 3 ). If V(T 2 ) = 2, then we have V(T 3 ) = 2. Suppose R T3 \{v 3 } = {v 3 }. Let G = G {v 2 v 3 }+{v 1 v 3}. Then G = U 2 2k. Using a symmetry, we get x 3 = x v 3 in G. From G to G, we have = 2x v 3 2x 3 = 2x 3 v i V(T 2) v i V(T 2) Since V(G) 10 and V(T 2 ) = V(T 3 ) = 2, we have V(T 1 ) 6. So, we have > 0. This implies ρ(g) > ρ(g ), which is a contradiction. v i V(T 2) If V(T 2 ) > 2, we may assume N T 1 (v 1 ) = {v 11,...,v 1r }, R T1 \{v 1 } = {v 1} and N T 3 (v 3 ) = {v 31,...,v 3t }. Let. G = G {v 1 v 3,v 1 v 11,...,v 1 v 1r } {v 3 v 31,...,v 3 v 3t } +{v 2 v 1,v 2 v 11,...,v 2 v 1r }+{v 2 v 31,...,v 2 v 3t }.

20 On the Distance Spectral Radius of Unicyclic Graphs with Perfect Matchings 587 Then G = U 2 2k. Using a symmetry, we get x 1 = x v 1 in G. From G to G, we have = 2x x v 1 + \R T1 v i R T3 v i V (T 2) +2 + v i V(T 2) \R T1 v i V(T 3)\R T3 +2 v i V(T 3)\R T3 v i V (T 1)\R T1 v i R T3 > 2 x 1 x 1 + > 0, v i V(T 2) which is a contradiction. \R T1 +2 v i V(T 3)\R T3 v i V(T 3)\R T3 ) v i V(T 2) Acknowledgment. The author would like to thank the anonymous referees for their valuable comments and suggestions. v i R T3 REFERENCES [1] A.T. Balaban, D. Ciubotariu, and M. Medeleanu. Topological indices and real number vertex invariants based on graph eigenvalues or eigenvectors. Journal of Chemical Information and Computer Sciences, 31: , [2] S.S. Bose, M. Nath, and S. Paul. Distance spectral radius of graphs with r pendent vertices. Linear Algebra and its Applications, 435: , [3] P.W. Fowler, G. Caporossi, and P. Hansen. Distance matrices, Wiener indices, and related invariants of fullerenes. Journal of Physical Chemistry A, 105: , [4] I. Gutman and M. Medeleanu. On the structure-dependence of the largest eigenvalue of the distance matrix of an alkane. Indian Journal of Chemistry Section A, 37: , [5] A. Ilić. Distance spectral radius of trees with given matching number. Discrete Applied Mathematics, 158: , [6] H. Minc. Nonnegative Matrices. John Wiley and Sons Inc., New York, [7] M. Nath and S. Paul. On the distance spectral radius of bipartite graphs. Linear Algebra and its Applications, 436: , [8] S. Ruzeih and D. Powers. The distance spectrum of the path P n and the first distance eigenvector of connected graphs. Linear and Multilinear Algebra, 28:75 81, [9] D. Stevanović and A. Ilić. Distance spectral radius of trees with fixed maximum degree. Electronic Journal of Linear Algebra, 20: , [10] G.L. Yu, Y.R. Wu, and J.L. Shu. Some graft transformations and its application on a distance spectrum. Discrete Mathematics, 311: , [11] X.L. Zhang and C. Godsil. Connectivity and minimal distance spectral radius. Linear and Multilinear Algebra, 59: , 2011.