DETERMINANT AND PSEUDO-DETERMINANT OF ADJACENCY MATRICES OF DERIVED GRAPHS

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1 Iteratioal Joural Computer Applicatios ( ) Volume * - No*, 016 DETERMINANT AND PSEUDO-DETERMINANT OF ADJACENCY MATRICES OF DERIVED GRAPHS Rajedra P Bharathi College, PG ad Research Cetre, Bharathiagara, Idia prajumaths@gmailcom R Ragaraja DoS i Mathematics Uiversity Mysore, Mysuru, Idia rajra63@gmailcom ABSTRACT The determiat is the product eigevalues adjacecy matrix a graph ad pseudo-determiat is the product o-zero eigevalues adjacecy matrix a graph I this paper, we have computed determiat, pseudo-determiat, characteristic polyomial, spectrum, spread, separator ad eergy derived graph some stadard graphs Keywords Derived graph, Determiat, Pseudo-determiat, Spread ad Separator 1 INTRODUCTION Let G be a simple graph with vertices ad m edges The derived graph G is deoted by G is the graph havig the same vertex set as G i which two vertices are adjacet if ad oly if their distace i G is two [3] Let A(G ) be adjacecy square matrix G order, P G (λ) = φ(g, λ) = λ +a 1 λ 1 + +a 1 λ+a be the characteristic polyomial A(G ) The matrix A(G ) beig symmetric, we have real eigevalues λ 1 λ λ A(G ) So the spectrum G is Spec(G λ1 λ ) = λ 3 λ m 1 m m 3 m r The eergy derived graph G [3] is E(G ) = λ i Details spectrum ad eergy a graph [4, 8] The spread G is S(G ) = λ 1 λ ad separator G is S A (G ) = λ 1 λ, where λ 1 is the largest eigevalue, λ is the secod largest eigevalue ad λ is the least eigevalue G [5] Let K, K, C, P ad S be -vertex complete graph, complemet K, cycle, path, star ad K r,t be complete bipartite graph with r + t vertices respectively S K Ayyaswamy, etc have obtaied the followig graphs[3], (K ) = K, (P ) = P / P /, (S ) = K 1 K 1, (K r,t ) = Kr K t ad C if is odd ad 5, (C ) K = 3 if = 3, K K if = 4, C / C / if is eve ad 6 The derived graph crow graph S 0 is (S) 0 = K K ad derived graph cocktail party graph K is (K ) = K K K }{{} times Here, we recall some results which we eed THEOREM 1 [4] Let P G (λ) = λ +a 1 λ 1 + +a be the characteristic polyomial a arbitrary udirected multigraph G Call a elemetary figure (1) the graph K or () every graph C q (q 1) ( loops beig icludig with q = 1), Call a basic figure U every graph all whose compoets are elemetary figures: Let p(u), c(u) be the umber compoets ad the umber circuits cotaied i U respectively, ad let U i deote the set all basic figures cotaied i G havig exactly i vertices The a i = (i = 1,,, ) ( 1) p(u) c(u) LEMMA [7] Let K be a complete graph with vertices The det(k ) = 1 LEMMA 3 [7] Let K m, be a complete bipartite graph The P det(k m, ) = m LEMMA 4 [7] Let P be a path with vertices The (1) det(p k ) = ( 1) k, k = 1,, 3, () P det(p k+1 ) = ( 1) k (k + 1), k = 1,, 3, LEMMA 5 [7] Let C be a cycle with vertices The (1) det(c k+1 ) =, k = 1,, 3, () det(c 4k+ ) = 4, k = 1,, 3, (3) P det(c 4k ) = 4k, k = 1,, 3, 1

2 MAIN RESULTS THEOREM 6 Let C be a cycle ad (C ) be the derived graph C The (1) det(c ) = detc if is odd ad 5 () det(c 3 ) = 0 (3) det(c 4 ) = 1 (4) (a) det(c 4k+ ) = 4 if k = 1,, 3, (b) det(c 8k+4 ) = 16 if k = 1,, 3, (c) P det(c 8k ) = 16k 4 if k = 1,, 3, Pro: 1 By defiitio (C ) = C if is odd ad 5 Therefore det(c ) = detc if is odd ad 5 The characteristic polyomial, spectrum, spread, separator ad eergy is same as the results cycles for odd case [7] The derived graph C 3 is K 3 The det(c 3 ) = 0 Let (C 3 ) = K 3, the characteristic polyomial (C 3 ) is φ((c 3 ), λ) = ( φ(k) 3, λ) = λ 3 The spectrum (C 3 ) 0 is Spec(C 3 ) = Spread S(C 3 ) = 0, separator 3 S A (C 3 ) = 0 The eergy (C 3 ) is E(C 3 ) = 0 3 The derived graph C 4 is K by Theorem[1] we have the followig table: No ( 1) p(u) c(u) 1 K 1 [( 1) 0 ] 1 Therefore det(c 4 ) = 1 Let (C 4 ) = K K, the characteristic polyomial (C 4 ) is φ((c 4 ), λ) = φ(k K, λ) = φ(k, λ)φ(k, λ) = (λ 1) 1 1 The spectrum (C 4 ) is Spec(C 4 ) = Spread S(C 4 ) =, separator S A (C 4 ) = 0 The eergy (C 4 ) is E(C 4 ) = 4 4 (a) The basic figures with 4k + vertices i (C 4k+ ), k = 1,, 3, are show i the followig table No ( 1) p(u) c(u) 1 C k+1 1 [( 1) ] 1 Therefore det(c 4k+ ) = 4, k = 1,, 3, (b) The basic figures with 8k + 4 vertices i (C 8k+4 ), k = 1,, 3, are show i the followig table Iteratioal Joural Computer Applicatios ( ) Volume * - No*, 016 No ( 1) p(u) c(u) 1 C 4k+ 1 [( 1) ] 1 = 4 3, 4,, k + 1 k + C 4k+, 4 [( 1) k+ 1 ] 4 = 8 (k + 1)K k + 3,, 4k + 1 4k + (4k + )K 4 [( 1) 4k+ 0 ] 4 = 4 Therefore det(c 8k+4 ) = 16, k = 1,, 3, (c) By defiitio derived graph C, (C 8k ) = C 4k C 4k, we have det(c 8k ) = 0, the pseudo-determiat is P det(c 8k ) = P detc 4k P detc 4k = 16k 4 from Lemma [5] Let (C ) = C C if is eve ad 6, the characteristic polyomial (C ) is φ((c ), λ) = φ(c C, λ) = φ(c, λ) φ(c, λ) ) ) ( λ = T 1 = 4 T ( λ 1) The spectrum (C ) is { Spec(C ) = cos 4π i/i = 1,,, } Therefore spectrum (C ) is ( cos Spec(C ) 4π = cos 8π 1π cos T ( λ 1 cos 4π Largest eigevalue is λ 1 = cos 4π or cos π Least eigevalue is λ = cos 4π 4 or cos π if is eve Spread S(C ) = cos π cos π = 4 if is eve Separator S A (C ) = 0, let (C ) = C C, the eergy (C ) is E(C ) = E(C ) + E(C ) = = 4 cos 4πi + cos 4πi if is eve ad 6 ) cos 4πi THEOREM 7 Let P be a path ad (P ) be derived graph P The (1) det(p 4k ) = 1 if k = 1,, 3, () P det(p 4k+ ) = (k + 1) if k = 1,, 3, (3) P det(p 4k+1 ) = (k + 1) if k = 1,, 3, (4) P det(p 4k+3 ) = (k + 1) if k = 1,, 3, Pro: 1: The basic figures with 4k vertices i (P 4k ), k = 1,, 3, are show i the followig table

3 Iteratioal Joural Computer Applicatios ( ) Volume * - No*, 016 No ( 1) p(u) c(u) 1,,, k 1 k k K 1 [( 1) k 0 ] 1 Therefore det(p 4k ) = 1, k = 1,, 3, : There is o basic figure with 4k + ad 4k + 1 vertices i (P 4k+ ), k = 1,, 3, The the basic figures with 4k vertices i (P 4k+ ), k = 1,, 3, are show i the followig table No ( 1) p(u) c(u) 1,,, k 1 k k K (k+1) [( 1) k 0 ] (k + 1) Therefore P det(p 4k+ ) = (k + 1), k = 1,, 3, 3: There is o basic figure with 4k + 1 vertices i (P 4k+1 ), k = 1,, 3, The the basic figures with 4k vertices i (P 4k+1 ), k = 1,, 3, are show i the followig table No ( 1) p(u) c(u) 1,,, k 1 k k K (k + 1) [( 1) k 0 ] (k + 1) Therefore P det(p 4k+1 ) = (k + 1), k = 1,, 3, 4: There is o basic figure with 4k + 3 vertices i (P 4k+3 ), k = 1,, 3, The the basic figures with 4k + vertices i (P 4k+3 ), k = 1,, 3, are show i the followig table No ( 1) p(u) c(u) 1,,, k k + 1 (k+1) K (k + 1) [( 1) k+1 0 ] (k + 1) Therefore P det(p 4k+3 ) = (k + 1), k = 1,, 3, Let (P ) = P P, the characteristic polyomial (P ) is φ((p ), λ) = φ(p P, λ) = φ(p, λ) φ(p, λ) λ λ = U U, where U (x) is the chebyshev polyomial the secod kid The spectrum P is { / πi Spec(P ) = cos + 1 i = 1,, 3,, } The spectrum P is { Spec(P ) = cos / πi + 1 i = 1,, 3,, } The eergy (P ) is E(P ) = E(P ) + E(P ) = cos πi cos πi + 1 THEOREM 8 Let K r,t be a complete bipartite graph with r + t vertices ad (K r,t ) be the derived graph K r,t The det(k r,t ) = (r 1)(t 1) Pro Let v 1, v,, v r, u 1, u,, u t be vertices (K r,t ) The the adjacecy matrix (K r,t ) is The determiat is det(k r,t ) = A(K r) 0 r t 0 t r A(K t ) From Lemma [], we have det(k r,t ) = det(k r ) det(k t ) det(k r,t ) = (r 1)(t 1) (r+t) (r+t) We also compute some relatios o spectrum (K r,t ) We kow that (K r,t ) = K r K t, the characteristic polyomial (K r,t ) is φ((k r,t ), λ) = φ(k r K t, λ) = φ(k r, λ) φ(k t, λ) = (λ r + 1)(λ + 1) r 1 (λ t + 1)(λ + 1) t 1 = [λ (r + t )λ + (r 1)(t 1)](λ + 1) r+t Spectrum (K r,t ) is ( ) (r+t )± Spec(K r,t ) (r+t ) 4(r 1)(t 1) = 1 1 r + t Spread S(K r,t ) = (r+t )+ (r+t ) 4(r 1)(t 1) + 1 Separator S A (K r,t ) = (r + t ) 4(r 1)(t 1) The eergy (K r,t ) is E(K r,t ) = (r + t ) THEOREM 9 Let S be a star with vertices ad (S ) be derived graph S The P det(s ) = ( ) Pro: Let v 1, v,, v 1, v be vertices (S ), v 1, v,, v 1 form complete graph with 1 vertices ad v is a isolated vertex The the adjacecy matrix (S ) is 3

4 the determiat is det(s ) = A(K 1) so det(s ) = 0, we ca compute pseudo-determiat (S ), From Lemma [], we have that is P det(s ) = det(k 1 ) P det(s ) = ( ) We compute spectrum related to (S ), from defiitio (S ) = K 1 K 1, the characteristic polyomial is φ((s ), λ) = φ(k 1 K 1, λ) = φ(k 1, λ) φ(k 1, λ) = (λ + )(λ + 1) λ 0 1 Spectrum (S ) is Spec(S ) = 1 1 Spread S(S ) = 1 ad separator S A (S ) = The eergy (S ) is E(S ) = ( ) Defiitio: [1] Crow graph S 0 for a iteger 3 is the graph with vertex set V = {v 1, v,, v, u 1, u,, u } ad edge set {v i u j / 1 i, j, i j} Therefore S 0 coicides with the complete bipartite graph K, with the horizotal edges removed THEOREM 10 Let S 0 be crow graph ad (S) 0 be derived graph S 0 The det(s) 0 = ( 1) Pro: We kow that the derived graph crow graph S 0 is (S) 0 = K K The det(s) 0 = detk detk = ( 1) from [] Spectrum related to crow graph S, 0 we kow that (S) 0 = K K, the characteristic polyomial (S) 0 is φ((s) 0, λ) = φ(k K, λ) = φ(k, λ)φ(k, λ) = (λ + 1) (λ + 1) 1 1 Spectrum (S) 0 is Spec(S) 0 = Spread S(S 0 ) =, separator S A (S 0 ) = 0 The eergy (S 0 ) is E(S 0 ) = 4 4 Defiitio: [1] The cocktail party graph, deoted by K, is a graph havig vertex set V = {u i, v i } ad edge set E = {u i u j, v i v j, u i v j, v i u j : 1 i < j } This graph is also called as complete -partite graph THEOREM 11 Let K be cocktail party graph ad (K ) be derived graph K The det(k ) = ( 1) Pro: Let K has vertices, the the basic figures with vertices i (K ), are show i the followig table Iteratioal Joural Computer Applicatios ( ) Volume * - No*, 016 No ( 1) p(u) c(u) 1,,, 1 K 1 [( 1) 0 ] 1 Therefore det(k ) = ( 1) Some studies spectrum related to derived graph cocktail party graph K We kow that (K ) = K K K }{{}, characteristic times polyomial (K ) is φ((k ), λ) = φ(k K K }{{}, λ) times = φ(k, λ)φ(k, λ) φ(k, λ) = (λ 1)(λ 1) (λ 1) = (λ 1) }{{} times 1 1 Spectrum (K ) is Spec(K ) =, Spread S(K ) = ad separator S A (K ) = 0 The eergy (K ) is E(K ) = 3 DERIVED GRAPH OF TADPOLE GRAPH The T m, -tadpole graph, is the graph obtaied by joiig a cycle graph C m to a path graph P with a bridge [6] 31 Derived graph tadpole graph T 3, THEOREM 1 The derived graph tadpole graph T 3, is (T 3, ) ad Z + The (1) P det(t 3,k 1 ) = ( 1) k (k + 1) if k = 1,, 3, () P det(t 3,4k ) = if k = 1,, 3, (3) P det(t 3,4k ) = k + 3k + 1 if k = 1,, 3, Pro: 1 There is o basic figure with 3+(k 1) ad 3+(k ) vertices i (T 3,k 1 ), k = 1,, 3, The the basic figures with 3 + (k 3) or k vertices i (T 3,k 1 ), k = 1,, 3, are show i the followig table No ( 1) p(u) c(u) 1,,, k 1 k k K (k + 1) [( 1) k 0 ] (k + 1) Therefore P det(t 3,k 1 ) = ( 1) k (k + 1), k = 1,, 3, There is o basic figure with 3 + (4k ) vertices i (T 3,4k ), k = 1,, 3, The the basic figures with 4k vertices i (T 3,4k ), k = 1,, 3, are show i the followig table No ( 1) p(u) c(u) 1,,, k 1 k kk [( 1) k 0 ] 4

5 Therefore P det(t 3,4k ) =, k = 1,, 3, 3 There is o basic figure with 3 + 4k, 3 + (4k 1) ad 3 + (4k ) vertices i (T 3,4k ), k = 1,, 3, The the basic figures with 4k vertices i (T 3,4k ), k = 1,, 3, are show i the followig table No 1,,, k 1 k kk k + 3k+1 ( 1) p(u) c(u) [( 1) k 0 ] (k + 3k + 1) Therefore P det(t 3,4k ) = k + 3k + 1, k = 1,, 3, 3 Derived graph T m, where m is odd ad Z + THEOREM 13 The derived graph tadpole graph T m, is (T m, ) Let m = 5, 9, 13, 17, ad Z + The (1) det(t m,4k 3 ) = 4 if k = 1,, 3, () det(t m,4k 1 ) = 1 if k = 1,, 3, ( (3) P det(t m,4k ) = 5k + m 1 ) if k = 1,, 3, (4) det(t m,4k ) = if k = 1,, 3, Pro: 1 The basic figures with m + (4k 3) vertices i (T m,4k 3 ), k = 1,, 3, are show i the followig table No ( 1) p(u) c(u) k 1 1 ] 1 1,,, (k ) k 1 C m+1, 1 [( 1) (k ) K = k,, (k + m 5 k + (k + k+ [( 1) 0 ] = Therefore det(t m,4k 3 ) = 4, k = 1,, 3, The basic figures with m + (4k 1) vertices i (T m,4k 1 ), k = 1,, 3, are show i the followig table No ( 1) p(u) c(u) 1,,, k + k + (k+ 1 [( 1) = 1 k+ Therefore det(t m,4k 1 ) = 1, k = 1,, 3, 0 ] 1 3 There is o basic figure with m+(4k ) vertices i (T m,4k ), k = 1,, 3, The the basic figures with m + (4k 3) vertices i (T m,4k ), k = 1,, 3, are show i the followig table Iteratioal Joural Computer Applicatios ( ) Volume * - No*, 016 k 1 1 ] k No ( 1) p(u) c(u) 1,,, k k 1 C m+1, k [( 1) (k ) K = k k,, k + m 5 k + ) 0 ] (k + ) K (3k + ) (3k + [( 1) (k+ ) Therefore P det(t m,4k ) = k (3k + ) = 5k +, k = 1,, 3, 4 The basic figures with m + 4k vertices i (T m,4k ), k = 1,, 3, are show i the followig table No ( 1) p(u) c(u) 1,,, k k + 1 C m, k K 1 [( 1) k+1 1 ] 1 = Therefore det(t m,4k ) =, k = 1,, 3, THEOREM 14 The derived graph tadpole graph T m, is (T m, ) Let m = 7, 11, 15, 19, ad Z + The (1) P det(t m,4k 3 ) = (m + 4k 3) if k = 1,, 3, () det(t m,4k 1 ) = 1 if k = 1,, 3, (3) det(t m,4k ) = if k = 1,, 3, (4) det(t m,4k ) = if k = 1,, 3, k 1 1 ] k Pro: 1 There is o basic figure with m + (4k 3) vertices i (T m,4k 3 ), k = 1,, 3, The the basic figures with m + (4k 4) vertices i (T m,4k 3 ), k = 1,, 3, are show i the followig table No ( 1) p(u) c(u) 1,,, k k 1 C m, k [( 1) (k ) K = k k,, (k + m 7 k + m 5 C 3, (k + m 7 ) K k + m 5 (k+ [( 1) ) 1 ] (k + Therefore P det(t m,4k 3 ) = (m + 4k 3), k = 1,, 3, The basic figures with m + (4k 1) vertices i (T m,4k 1 ), k = 1,, 3, are show i the followig table 5

6 Iteratioal Joural Computer Applicatios ( ) Volume * - No*, 016 No 1,,, k + k + (k + ( 1) p(u) c(u) 1 k+ [( 1) 0 ] 1 = 1 Therefore det(t m,4k 1 ) = 1, k = 1,, 3, 3 The basic figures with m + (4k ) vertices i (T m,4k ), k = 1,, 3, are show i the followig table No ( 1) p(u) c(u) 1,,, k + m 5 k + k+ C 3, 1 [( 1) 1 ] 1 (k+ m 5 = Therefore det(t m,4k ) =, k = 1,, 3, 4 The basic figures with m + 4k vertices i (T m,4k ), k = 1,, 3, ad table as Pro is similar to the pro Theorem 13(4) 33 Derived graph T 4,, Z + THEOREM 15 The derived graph tadpole graph T 4, is (T 4, ) ad Z + The (1) det(t 4,4k 3 ) = if k = 1,, 3, () P det(t 4,4k ) = (k + 1) if k = 1,, 3, (3) P det(t 4,4k 1 ) = (k + 1) if k = 1,, 3, (4) det(t 4,4k ) = 1 if k = 1,, 3, Pro: 1 The basic figure with 4 + (4k 3) vertices i (T 4,4k 3 ), k = 1,, 3, are show i the followig table No ( 1) p(u) c(u) 1,,, k 1 k C 3, 1 [( 1) k 1 ] 1 = (k 1)K Therefore det(t 4,4k 3 ) =, k = 1,, 3, There is o basic figure with [4 + (4k )] vertices i (T 4,4k ), k = 1,, 3, The basic figure with 4 + (4k 3) vertices i (T 4,4k ), k = 1,, 3, are show i the followig table No ( 1) p(u) c(u) 1,,, k 1 k C 3, (k+1) [( 1) k 1 ] (k + 1) (k 1)K = (k + 1) 3 There is o basic figure with [4+(4k 1)] vertices i (T 4,4k 1 ), k = 1,, 3, The basic figure with [4 + (4k )] vertices i (T 4,4k 1 ), k = 1,, 3, are show i the followig table No ( 1) p(u) c(u) 1,,, k k + 1 (k +1)K k+1 [( 1) k+1 0 ] (k + 1) = ( 1)(k + 1) Therefore P det(t 4,4k 1 ) = (k + 1), k = 1,, 3, 4 The basic figure with 4 + 4k vertices i (T 4,4k ), k = 1,, 3, are show i the followig table No ( 1) p(u) c(u) 1,,, k + 1 k + (k +)K 1 [( 1) k+ 0 ] 1 = 1 Therefore det(t 4,4k ) = 1, k = 1,, 3, Ackowledgemet: The secod author is thakful to UGC-SAP, DRS-I, No F510//DRS/011(SAP-I) 4 REFERENCES [1] C Adiga, A Bayad, I Gutma ad Shrikath A S, The miimum coverig eergy a graph, Kragujevac J Sci, 34 (01), [] Alireza Abdollahi, Determiats adjacecy matrices graphs, Trasactios o Combiatorics, 1(4) (01), 9-16 [3] S K Ayyaswamy, S Balachadra ad I Gutma, O secodstage spectrum ad eergy a graph, Kragujevac J Sci, 34 (010), [4] D M Cvetković, M Doob ad H Sachs, Spectra Graphs, Theory ad Applicatio, Academic Press, New York, USA(1980) [5] W X Hog ad L H You, O the eigevalues firefly graph, Trasactios o Combiatorics, 3 (3)(014), 1-9 [6] [7] R Ragaraja ad P Rajedra, Computatio spectral parameters graphs, ifrsa Iteratioal Joural Computig, 4(4)(014), [8] Xueliag Li, Yogtag Shi ad I Gutma, Graph Eergy, Spriger New York, 01 Therefore P det(t 4,4k ) = (k + 1), k = 1,, 3, 6