Laplacian for Graph Exercises 1. Solutions

Transcription

1 Laplacia for Graph Exercises 1 Solutios Istructor: Medykh I A Sobolev Istitute of Mathematics Novosibirsk State Uiversity Witer School i Harmoic Fuctios o Graphs ad Combiatorial Desigs Jauary, 2014 Medykh I A (Sobolev Istitute of Math) Laplacia for Graph Jauary / 13

2 Exercise 11 Fid Laplacia spectrum of the complete graph o vertices K Solutio: We wat to show that µ(k, x) = x(x ) 1 To solve this problem we use iductio by umber of vertices For = 1, K 1 is a sigular vertex Its Laplacia matrix L(K ) = {0} Hece µ(k 1, x) = x Hece the statemet is true for = 1 Suppose that for give the equality µ(k, x) = x(x ) 1 is already proved It is easy to see that K +1 is a joi of K ad K 1 By Kel mas theorem we get = µ(k +1, x) = x(x 1) (x 1)(x ) µ(k 1, x )µ(k, x 1) = x(x 1) (x 1)(x ) (x )(x 1)(x 1) 1 = x(x 1) Hece, the Laplacia spectrum of K is {0 1, 1 } Medykh I A (Sobolev Istitute of Math) Laplacia for Graph Jauary / 13

3 Exercise 12 Fid Laplacia spectrum of the complete bipartite graph K,m Solutio: Let us ote that K,m is a joi of X m ad X, where X k is a disjoit uio of k vertices We have L(X k ) = D(X k ) A(X k ) = O k O k = O k, where O k is k k zero matrix Hece, µ(x k, x) = x k By Kel mas theorem we obtai = µ(k,m, x) = x(x m ) (x )(x m) µ(x, x m)µ(x m, x ) = x(x m ) (x )(x m) (x m) (x ) m = x(x m )(x ) m 1 (x m) 1 Hece, the Laplacia spectrum of K,m is {0 1, m 1, m 1, (m + ) 1 } Medykh I A (Sobolev Istitute of Math) Laplacia for Graph Jauary / 13

4 Exercise 13 Fid Laplacia spectrum of the cycle graph C Solutio: The Laplacia matrix L(C ) is the circulat matrix with etities v 0 = 2, v 1 = 1, v 2 = = v 2 = 0, v 1 = 1 The by properties of circulat matrices its eigevalues are λ k = v 0 + v 1 ε k + + v 1 ε ( 1)k, k = 0,, 1, where ε = e 2πi is the -th primitive root of the uity Hece, Sice e 2πi k + e 2πi λ k = 2 ε k ε ( 1)k ( 1)k = 2( e 2πi k + e 2πi k we have λ k = 2 2 cos 2πk, k = 0,, 1 2 ) = 2 cos 2πk, Medykh I A (Sobolev Istitute of Math) Laplacia for Graph Jauary / 13

5 Exercise 14 Fid Laplacia spectrum of the path graph P Solutio: The Laplacia matrix for path graph P has the form L = L(P ) = Medykh I A (Sobolev Istitute of Math) Laplacia for Graph Jauary / 13

6 The its characteristic matrix is give by 1 λ λ λ L λi = λ λ Medykh I A (Sobolev Istitute of Math) Laplacia for Graph Jauary / 13

7 Let V = det(l λi ) The det V is equal 1 λ λ λ = 2 λ λ λ λ λ = Medykh I A (Sobolev Istitute of Math) Laplacia for Graph Jauary / 13

8 2 λ λ λ 2 λ λ λ 1 1 = D D 1 I a similar way D = U U 1, where 2 λ λ 0 U = λ = U ( 2 λ ) 2 Medykh I A (Sobolev Istitute of Math) Laplacia for Graph Jauary / 13

9 si( + 1) arccos x Here U (x) = is a Chebyshev polyomial of the secod si arccos x kid Sice U (x) 2xU 1 (x) + U 2 (x) = 0, we obtai V = D D 1 = U (x) 2U 1 (x) + U 2 (x) where x = λ 2 2 The equatio = (2x 2)U 1 (x) = λu 1 ( λ 2 ), 2 λu 1 ( λ 2 ) = 0 2 has the followig solutios λ k = 2 2 cos( πk ), k = 0,, 1 Medykh I A (Sobolev Istitute of Math) Laplacia for Graph Jauary / 13

10 Laplacia for Graphs Exercise 15 Show that Laplacia polyomial of the path graph P has the followig form µ(p, x) = x U 1 ( x 2 2 ), si( arccos x) where U 1 (x) = is the Chebyshev polyomial of the si(arccos x) secod kid Solutio: Follows from the previous exercise Medykh I A (Sobolev Istitute of Math) Laplacia for Graph Jauary / 13

11 Laplacia for Graphs Exercise 16 Fid Laplacia spectrum of the wheel graph W = K 1 C Aswer: {0, + 1, 3 2 cos 2πk, k = 1,, 1} Exercise 17 Fid Laplacia spectrum of the fa graph F = K 1 P Aswer: {0, + 1, 3 2 cos πk, k = 1,, 1} Exercise 18 Show that the Laplacia polyomial of the fa graph F = K 1 P is give by the formula µ(f, x) = x(x 1)U 1 ( x 3 2 ) where U 1 (x) is the Chebyshev polyomial of the secod kid Medykh I A (Sobolev Istitute of Math) Laplacia for Graph Jauary / 13

12 Laplacia for Graphs Exercise 19 Fid Laplacia spectrum of the cylider graph P m C Solutio: The spectrum of P m is λ j = 4 si 2 ( πj 2m ), j = 0,, m 1 ad the spectrum of C is µ k = 4 si 2 ( 2πk 2 ), k = 0,, 1 As a result we have the followig spectrum for P m C l j,k = 4 si 2 ( πj 2m ) + 4 si2 ( 2πk ), j = 0,, m 1, k = 0,, 1 2 Medykh I A (Sobolev Istitute of Math) Laplacia for Graph Jauary / 13

13 Laplacia for Graphs Exercise 110 Fid Laplacia spectrum of the Moebius ladder graph M Moebius ladder graph is a cycle graph C 2 with additioal edges, coectig opposite vertices i cycle Solutio: We ote that the Laplacia matrix for M is circulat circ{v 0, v 2 1 }, where v 0 = 3, v 1 = 1, v 2 = = v 1 = 0, v = 1, v +1 = = v 2 2 = 0, v 2 1 = 1 Let ε = e 2πi 2 primitive root of uity The L(M ) has the followig spectrum λ k = 2 1 j=0 be the 2-th ε k j v j = 3 + ( 1) k+1 2 cos πk, k = 0,, 2 1 Medykh I A (Sobolev Istitute of Math) Laplacia for Graph Jauary / 13