Laplacians of Graphs, Spectra and Laplacian polynomials
Alexander Mednykh
Sobolev Institute of Mathematics Novosibirsk State University
PhD Summer School in Discrete Mathematics Rogla, Slovenia 27 June - 03 July 2015
Mednykh A. D. (Sobolev Institute of Math) Laplacian for Graphs 27 June - 03 July 2015 1 / 30 Laplacian matrix
The Laplacian matrix of a graph and its eigenvalues can be used in several areas of mathematical research and have a physical interpretation in various physical and chemical theories. The related matrix - the adjacency matrix of a graph and its eigenvalues were much more investigated in the past than the Laplacian matrix. In the same time, the Laplacian spectrum is much more natural and more important than the adjacency matrix spectrum because of it numerous application in mathematical physics, chemistry and financial mathematics.
Since the classical paper by Mark Kac “Can one hear the shape of a drum?” (1966), the question of what geometric properties of a manifold are determined by its Laplace operator has inspired many intriguing results. See papers by S. Wolpert (1979), P. Buser (1986), R. Brooks (1987), R. Isangulov (2000) for Riemann surfaces and survey by by E.R.van Dam and W.H.Haemers (2003) for graphs.
Mednykh A. D. (Sobolev Institute of Math) Laplacian for Graphs 27 June - 03 July 2015 2 / 30 Laplacian of Graphs
Laplacian matrix. Laplacian spectrum
The graphs under consideration are supposed to be unoriented and finite. They may have loops, multiple edges and to be disconnected. Let auv be the number of edges between two given vertices u and v of G. The matrix A = A(G) = [auv ]u,v∈V (G), is called the adjacency matrix of the graph G. P Let d(v) denote the degree of v ∈ V (G), d(v) = u auv , and let D = D(G) be the diagonal matrix indexed by V (G) and with dvv = d(v). The matrix L = L(G) = D(G) − A(G) is called the Laplacian matrix of G. It should be noted that loops have no influence on L(G). The matrix L(G) is sometimes called the Kirchhoff matrix of G. It should be mentioned here that the rows and columns of graph matrices are indexed by the vertices of the graph, their order being unimportant.
Mednykh A. D. (Sobolev Institute of Math) Laplacian for Graphs 27 June - 03 July 2015 3 / 30 Laplacian for Graphs
Let G be a given graph. Orient its edges arbitrarily, i.e. for each e ∈ E(G) choose one of its ends as the initial vertex, and name the other end the terminal vertex. The oriented incidence matrix of G with respect to the given orientation is the |V | × |E| matrix C = {cve } with entries +1, if v is the terminal vertex of e, cve = −1, if v is the initial vertex of e, 0, if v and e are not incident. It is well known that L(G) = CC t independently of the orientation given to the edges of G. Since (L(G)x, x) = (CC t x, x) = (C t x, C t x) we have X 2 (L(G)x, x) = avu(xv − xu) . v u∈E(G)
Mednykh A. D. (Sobolev Institute of Math) Laplacian for Graphs 27 June - 03 July 2015 4 / 30 Laplacian for Graphs
Laplacian polynomial and Laplacian spectrum
We denote by µ(G, x) the characteristic polynomial of L(G). We will call it the Laplacian polynomial. Its roots will be called the Laplacian eigenvalues (or sometimes just eigenvalues) of G. They will be denoted by λ1(G) ≤ λ2(G) ≤ ... ≤ λn(G), (n = |V (G)|), always enumerated in increasing order and repeated according to their multiplicity.
We note that λ1 is always equal to 0. Graph G is connected if and only if λ2 > 0. If G consists of k components then
λ1(G) = λ2(G) = ... = λk (G) = 0 and λk+1(G) > 0.
Mednykh A. D. (Sobolev Institute of Math) Laplacian for Graphs 27 June - 03 July 2015 5 / 30 Laplacian for Graphs
We summarise the above results in the following theorem.
Theorem Let G be a graph. Then: (a) L(G) has only real eigenvalues, (b) L(G) is positive semidefinite,
(c) its smallest eigenvalue is λ1 = 0 and a corresponding eigenvector is (1, 1,..., 1)t . The multiplicity of 0 as an eigenvalue of L(G) is equal to the number of components of G.
Mednykh A. D. (Sobolev Institute of Math) Laplacian for Graphs 27 June - 03 July 2015 6 / 30 Laplacian for Graphs
Many published works relate the Laplacian eigenvalues of graphs with the eigenvalues of graphs obtained by means of some operations on the graphs we start with. The first result is obvious but very useful.
Theorem
Let G be the disjoint union of graphs G1, G2,..., Gk . Then
k Y µ(G, x) = µ(Gi , x). i=1
Mednykh A. D. (Sobolev Institute of Math) Laplacian for Graphs 27 June - 03 July 2015 7 / 30 Laplacian for Graphs.
The complement of a graph G is a graph G on the same vertices such that two distinct vertices of G are adjacent if and only if they are not adjacent in G. The next two results were first observed by A. K. Kelmans.
Theorem (Kelmans, 1966) If G denotes the complement of the graph G then x µ(G, x) = (−1)n−1 µ(G, n − x) n − x
and so the eigenvalues of G are λ1(G) = 0, and
λi+1(G) = n − λn−i+1(G), i = 1, 2,..., n − 1.
Mednykh A. D. (Sobolev Institute of Math) Laplacian for Graphs 27 June - 03 July 2015 8 / 30 Laplacian for Graphs.
As a corollary from the previous result one can get the following beautiful theorem.
Theorem (Kel’mans, 1965)
Let X1 ∗ X2 denote the join of X1 and X2, i.e. the graph obtained from the disjoint union of X1 and X2 by adding all possible edges uv, u ∈ V (X1), v ∈ V (X2). Then
x(x − n1 − n2) µ(X1 ∗ X2, x) = µ(X1, x − n2)µ(X2, x − n1). (x − n1)(x − n2)
where n1 and n2 are orders of X1 and X2, respectively and µ(X , x) is the characteristic polynomial of the Laplacian matrix of X .
Mednykh A. D. (Sobolev Institute of Math) Laplacian for Graphs 27 June - 03 July 2015 9 / 30 Laplacian for Graphs
Let G be a simple graph (without multiple edges). The line graphL (G) of G is the graph whose vertices correspond to the edges of G with two vertices of L(G) being adjacent if and only if the corresponding edges in G have a vertex in common. The subdivision graphS (G) of G is obtained from G by inserting, into each edge of G, a new vertex of degree 2. The total graphT (G) of G has its vertex set equal to the union of vertices and edges of G, and two of them being adjacent if and only if they are incident or adjacent in G.
Mednykh A. D. (Sobolev Institute of Math) Laplacian for Graphs 27 June - 03 July 2015 10 / 30 Laplacian for Graphs.
Theorem (Kel’mams, 1967) Let G be a d-regular simple graph with m edges and n vertices. Then (a) µ(L(G), x) = (x − 2d)m−nµ(G, x), (b) µ(S(G), x) = (−1)m(2 − x)m−nµ(G, x(d + 2 − x)), m n m−n x(d+2−x) (c) µ(T (G), x) = (−1) (d + 1 − x) (2d + 2 − x) µ(G, d+1−x ).
Mednykh A. D. (Sobolev Institute of Math) Laplacian for Graphs 27 June - 03 July 2015 11 / 30 Laplacian for Graphs
The Cartesian productG × H (simetimes GH) of graphs G and H is a graph such that the vertex set of G × H is the Cartesian product V (G) × V (H); and any two vertices (u, u0) and (v, v 0) are adjacent in G × H if and only if either u = v and u0 is adjacent with v 0 in H, or u0 = v 0 and u is adjacent with v in G. Examples : (a) The Cartesian product of two edges is a cycle on four vertices: K2 × K2 = C4.
(b) The Cartesian product of K2 and a path graph is a ladder graph. (c) The Cartesian product of two path graphs is a grid graph. (d) The Cartesian product of two hypercube graphs is another hypercube: Qi × Qj = Qi+j . (e) The graph of vertices and edges of an n-prism is the Cartesian product graph K2 × Cn.
Mednykh A. D. (Sobolev Institute of Math) Laplacian for Graphs 27 June - 03 July 2015 12 / 30 Laplacian for Graphs
Theorem (M. Fiedler (1973))
The Laplacian eigenvalues of the Cartesian product X1 × X2 of graphs X1 and X2 are equal to all the possible sums of eigenvalues of the two factors:
λi (X1) + λj (X2), i = 1,..., |V (X1)|, j = 1,..., |V (X2)|.
Using this theorem we can easily determine the spectrum of “lattice” graphs. The m × n lattice graph is just the Cartesian product of paths, Pm × Pn. Below we will show that the spectrum of path-graph Pk is πi `(k) = 4 sin2 , i = 0, 1,..., k − 1. i 2k
So Pm × Pn has eigenvalues πi πj λ = `(m)+ `(n) = 4 sin2 +4 sin2 , i = 0, 1,..., m−1, j = 0, 1,..., n−1. i, j i j 2m 2n
Mednykh A. D. (Sobolev Institute of Math) Laplacian for Graphs 27 June - 03 July 2015 13 / 30 Laplacian for Graphs
Circulant matrices Fix a positive integer n ≥ 2 and let v = (v0, v1,..., vn−1) be a row vector n n n in C . Define the shift operator T : C → C by
T (v0, v1,..., vn−1) = (vn−1, v0,..., vn−2).
The circulant matrix associated to v is the n × n matrix whose rows are given by iteration of the shift operator acting on v, that is to say the k-th row is given by T k−1v, k = 1,..., n. Such a matrix will be denoted by
V = circ{v} = circ{v0, v1,..., vn−1}. The following theorem shows how one can calculate eigenvalues and eigenvectors of V .
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Theorem (Eigenvalues of circulant matrix) n Let v = (v0, v1,..., vn−1) be a row vector in C , and V = circ{v}. If ε is primitive n−th root of unity, then v0 v1 . . . vn−2 vn−1 vn−1 v0 . . . vn−3 vn−2 n−1 n−1 ...... Y X j` det V = det . . . . . = ( ε vj ). l=0 j=0 v2 v3 . . . v0 v1 v1 v2 . . . vn−1 v0
Corollary Eigenvalues of circulant matrix V is given by the formulae
n−1 X j` λ` = ε vj , ` = 0,..., n − 1. j=0
Mednykh A. D. (Sobolev Institute of Math) Laplacian for Graphs 27 June - 03 July 2015 15 / 30 Laplacian for Graphs
n Proof. We view the matrix V as a a self map (linear operator) of C . For n each integer `, 0 ≤ ` ≤ n − 1, let x` ∈ C be a transpose of the row vector (1, ε`, ε2`, . . . , ε(n−1)`) and ` (n−1)` λ` = v0 + ε v1 + ... + ε vn−1. A quite simple calculation shows that
v0 v1 ... vn−2 vn−1 1 1 ` ` vn−1 v0 ... vn−3 vn−2 ε ε ...... . . . . . . . . = λ` . . (n−2)` (n−2)` v2 v3 ... v0 v1 ε ε (n−1)` (n−1)` v1 v2 ... vn−1 v0 ε ε
Thus λ` is an eigenvalue of V with eigenvector x`. Since {x0, x1,..., xn−1} is linearly independent set, we conclude that n−1 Q det V = λ`. `=0 Mednykh A. D. (Sobolev Institute of Math) Laplacian for Graphs 27 June - 03 July 2015 16 / 30 Laplacian for Graphs
Circulant graphs
Circulant graphs can be described in several equivalent ways:
(a) The graph has an adjacency matrix that is a circulant matrix. (b) The automorphism group of the graph includes a cyclic subgroup that acts transitively on the graph’s vertices. (c) The n vertices of the graph can be numbered from 0 to n − 1 in such a way that, if some two vertices numbered x and y are adjacent, then every two vertices numbered z and (z − x + y) mod n are adjacent. (d) The graph can be drawn (possibly with crossings) so that its vertices lie on the corners of a regular polygon, and every rotational symmetry of the polygon is also a symmetry of the drawing. (e) The graph is a Cayley graph of a cyclic group.
Mednykh A. D. (Sobolev Institute of Math) Laplacian for Graphs 27 June - 03 July 2015 17 / 30 Laplacian for Graphs
Examples
(a) The circulant graph Cn(s1,..., sk ) with jumps s1,..., sk is defined as the graph with n vertices labeled 0, 1,..., n − 1 where each vertex i is adjacent to 2k vertices i ± s1,..., i ± sk mod n.
(b) n-cycle graph Cn = Cn(1).
(c) n-antiprism graph C2n(1, 2).
(d) n-prism graph Yn = C2n(2, n), n odd.
(e) The Moebius ladder graph Mn = C2n(1, n). n (f) The complete graph Kn = Cn(1, 2, ··· , [ 2 ]). n (g) The complete bipartite graph Kn,n = Cn(1, 3, ··· , 2[ 2 ] + 1).
Mednykh A. D. (Sobolev Institute of Math) Laplacian for Graphs 27 June - 03 July 2015 18 / 30 Exercises
Exercise 1.1. Find Laplacian spectrum of the complete graph on n vertices Kn.
n−1 Solution: We want to show that µ(Kn, x) = x(x − n) . To solve this problem we use induction by number of vertices n. For n = 1, K1 is a singular vertex. Its Laplacian matrix L(Kn) = {0}. Hence µ(K1, x) = x. Hence the statement is true for n = 1. Suppose that for given n the n−1 equality µ(Kn, x) = x(x − n) is already proved. It is easy to see that Kn+1 is a join of Kn and K1. By Kel’mans theorem we get x(x − n − 1) µ(K , x) = µ(K , x − n)µ(K , x − 1) = n+1 (x − 1)(x − n) 1 n
x(x − n − 1) = (x − n)(x − 1)(x − n − 1)n−1 = x(x − n − 1)n. (x − 1)(x − n) 1 n−1 Hence, the Laplacian spectrum of Kn is {0 , n }.
Mednykh A. D. (Sobolev Institute of Math) Laplacian for Graphs 27 June - 03 July 2015 19 / 30 Exercises
Exercise 1.2. Find Laplacian spectrum of the complete bipartite graph Kn,m.
Solution: Let us note that Kn,m is a join of Xm and Xn, where Xk is a disjoint union of k vertices. We have L(Xk ) = D(Xk ) − A(Xk ) = Ok − Ok = Ok , where Ok is k × k zero matrix. k Hence, µ(Xk , x) = x . By Kel’mans theorem we obtain x(x − m − n) µ(K , x) = µ(X , x − m)µ(X , x − n) = n,m (x − n)(x − m) n m
x(x − m − n) = (x − m)n(x − n)m = x(x − m − n)(x − n)m−1(x − m)n−1. (x − n)(x − m) 1 m−1 n−1 1 Hence, the Laplacian spectrum of Kn,m is {0 , n , m , (m + n) }.
Mednykh A. D. (Sobolev Institute of Math) Laplacian for Graphs 27 June - 03 July 2015 20 / 30 Exercises
Exercise 1.3. Find Laplacian spectrum of the cycle graph Cn.
Solution: The Laplacian matrix L(Cn) is the circulant matrix with entities
v0 = 2, v1 = −1, v2 = ... = vn−2 = 0, vn−1 = −1. Then by properties of circulant matrices its eigenvalues are
k (n−1)k λk = v0 + v1ε + ... + vn−1ε , k = 0,..., n − 1,
2πi where ε = e n is the n-th primitive root of the unity. Hence,
k (n−1)k λk = 2 − ε − ε . Since 2πi k − 2πi k 2πi k 2πi (n−1)k e n + e n 2πk e n + e n = 2( ) = 2 cos , 2 n 2πk we have λk = 2 − 2 cos n , k = 0,..., n − 1. Mednykh A. D. (Sobolev Institute of Math) Laplacian for Graphs 27 June - 03 July 2015 21 / 30 Exercises
Exercise 1.4. Find Laplacian spectrum of the path graph Pn.
Solution: The Laplacian matrix for path graph Pn has the form
1 −1 0 . . . 0 0 −1 2 −1 . . . 0 0 0 −1 2 . . . 0 0 0 0 −1 . . . 0 0 Ln = L(Pn) = . ...... ...... 0 0 0 . . . 2 −1 0 0 0 . . . −1 1
Mednykh A. D. (Sobolev Institute of Math) Laplacian for Graphs 27 June - 03 July 2015 22 / 30 Then its characteristic matrix is given by
1 − λ −1 0 . . . 0 0 −1 2 − λ 1 . . . 0 0 0 −1 λ − 2 . . . 0 0 0 0 1 . . . 0 0 Ln − λIn = . ...... ...... 0 0 0 . . . 2 − λ −1 0 0 0 . . . −1 1 − λ
Mednykh A. D. (Sobolev Institute of Math) Laplacian for Graphs 27 June - 03 July 2015 23 / 30 Let Vn = det(Ln − λIn). Then det Vn is equal
1 − λ −1 . . . 0
−1 2 − λ . . . 0 = ......
0 0 . . . 1 − λ
2 − λ −1 . . . 0 1 0 . . . 0
−1 2 − λ . . . 0 −1 2 − λ . . . 0 − = ......
0 0 . . . 1 − λ 0 0 . . . 1 − λ
Mednykh A. D. (Sobolev Institute of Math) Laplacian for Graphs 27 June - 03 July 2015 24 / 30
2 − λ −1 . . . 0 2 − λ −1 . . . 0
−1 2 − λ . . . 0 −1 2 − λ . . . 0 − = ......
0 0 . . . 1 − λ 0 0 . . . 1 − λ n×n n−1×n−1
Dn − Dn−1.
In a similar way Dn = Un − Un−1, where
2 − λ −1 . . . 0
−1 2 − λ . . . 0 2 − λ U = = U ( ). n . . .. . n . . . . 2
0 0 . . . 2 − λ n×n
Mednykh A. D. (Sobolev Institute of Math) Laplacian for Graphs 27 June - 03 July 2015 25 / 30 Exercises
sin(n + 1) arccos x Here U (x) = is a Chebyshev polynomial of the second n sin arccos x
kind. Since Un(x) − 2xUn−1(x) + Un−2(x) = 0, we obtain
Vn = Dn − Dn−1 = Un(x) − 2Un−1(x) + Un−2(x) λ − 2 = (2x − 2)U (x) = −λU ( ), n−1 n−1 2 λ−2 where x = 2 . The equation λ − 2 λU ( ) = 0 n−1 2
πk has the following solutions λk = 2 − 2 cos( n ), k = 0,..., n − 1.
Mednykh A. D. (Sobolev Institute of Math) Laplacian for Graphs 27 June - 03 July 2015 26 / 30 Exercises
Exercise 1.5. Show that Laplacian polynomial of the path graph Pn has the following form x − 2 µ(P , x) = x U ( ), n n−1 2 sin(n arccos x) where U (x) = is the Chebyshev polynomial of the n−1 sin(arccos x) second kind.
Solution: Follows from the previous exercise.
Mednykh A. D. (Sobolev Institute of Math) Laplacian for Graphs 27 June - 03 July 2015 27 / 30 Exercises
Exercise 1.6. Find Laplacian spectrum of the wheel graph Wn = K1 ∗ Cn.
2πk Answer: {0, n + 1, 3 − 2 cos n , k = 1,..., n − 1} Exercise 1.7.
Find Laplacian spectrum of the fan graph Fn = K1 ∗ Pn. πk Answer: {0, n + 1, 3 − 2 cos n , k = 1,..., n − 1} Exercise 1.8. Show that the Laplacian polynomial of the fan graph Fn = K1 ∗ Pn is given by the formula x − 3 µ(F , x) = x(x − n − 1)U ( ) n n−1 2
where Un−1(x) is the Chebyshev polynomial of the second kind. Mednykh A. D. (Sobolev Institute of Math) Laplacian for Graphs 27 June - 03 July 2015 28 / 30 Exercises
Exercise 1.9. Find Laplacian spectrum of the cylinder graph Pm × Cn.
Solution: 2 πj The spectrum of Pm is λj = 4 sin ( 2m ), j = 0,..., m − 1 and the 2 2πk spectrum of Cn is µk = 4 sin ( 2n ), k = 0,..., n − 1. As a result we have the following spectrum for Pm × Cn.
πj 2πk ` = 4 sin2( ) + 4 sin2( ), j = 0,..., m − 1, k = 0,..., n − 1. j,k 2m 2n
Mednykh A. D. (Sobolev Institute of Math) Laplacian for Graphs 27 June - 03 July 2015 29 / 30 Exercises
Exercise 1.10. Find Laplacian spectrum of the Moebius ladder graph Mn. Moebius ladder graph is a cycle graph C2n with additional edges, connecting opposite vertices in cycle.
Solution: We note that the Laplacian matrix for Mn is circulant circ{v0 ..., v2n−1}, where v0 = 3, v1 = −1, v2 = ... = vn−1 = 0, 2πi vn = −1, vn+1 = ... = v2n−2 = 0, v2n−1 = −1. Let ε = e 2n be the 2n-th primitive root of unity. Then L(Mn) has the following spectrum
2n−1 X πk λ = εk j v = 3 + (−1)k+1 − 2 cos , k = 0,..., 2n − 1. k j n j=0
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