Spring 2017, MATH-595 http://orion.math.iastate.edu/butler/2017/spring/x95/ Laplacian Matrix - Introduction Let G be a graph of order n. Recall adjacency matrix A of G is n × n matrix indexed by vertices of G where ( 1 if uv 2 E(G); Au;v = 0 otherwise. The Laplacian Matrix of G, denoted by L, is defined as 8 deg(u) if u = v <> Lu;v = −1 if uv 2 E(G); :>0 otherwise. Let D be a diagonal degree matrix. Then L = D − A. − 1: Find the Laplacian matrix of the following graph K4 : 3 0 3 −1 −1 −11 B−1 3 −1 −1C 1 2 L = B C @−1 −1 2 0 A −1 −1 0 2 4 2: Find some (obvious) eigenvector and corresponding eigenvalue of L. Hint: Look at row/column sum. Solution: Notice the sum in every row is 0. Hence 1 is an eigenvector and corresponds to eigenvalue 0. This is nice that we always have 1 and an eigenvector. Unlike for A, it is the SMALLEST eigenvalue 0. 3: Let G be a k-regular graph with AG having eigenvalues λ1; : : : ; λn. Show that LG has eigenvalues k − λ1; : : : ; k − λn. Hint: L = D − A and D is special. Solution: L = D − A = kI − A. If vi is an eigenvector for λi, we get Lvi = (kI − A)vi = kvi − λivi = (k − λi)vi: 4: Let G = Kn. Find spec(AG) and spec(LG). Hint for LG: Kn is n − 1 regular. Solution: AG : : : n − 1; −1; −1;:::; −1 LG ::: 0; n; n; : : : ; n 5: Let G = Cn. Find spec(LG). Hint: spec(AG) contains cos. 2πk Solution: spec(AG) = f2 cos n for k = 0; : : : ; n − 1g. 2πk spec(LG) = f2 − 2 cos n for k = 0; : : : ; n − 1g. 6: Let G = Pn. Find spec(LG). Hint1: Write L as 2I − X for some matrix X. πk Hint2: spectrum of path of order n with loops at the end is f2 cos n for k = 0; : : : ; n − 1g. Solution: 0 1 −1 1 02 1 01 1 1 B−1 2 −1 C B 2 C B1 0 1 C L = B C = B C − B C @ −1 2 −1A @ 2 A @ 1 0 1A −1 1 2 1 1 Notice the second matrix is a adjacency matrix of a path with loops. πk Hence spec(LG) = f2 − 2 cos n for k = 0; : : : ; n − 1g. from S. Butler https://www.youtube.com/watch?v=Dm_FuQDjGHI MATH X95 - 1 - page 1/4 P 2 Recall λi = 2jEj where λi 2 spec(A). P 7: Show µi = 2jEj where µi 2 spec(L). Solution: Sum of the entries on the diagonal is equal to the sum of the eigenvalues. Rest of the lecture is proving the following claim in different ways. Claim 1 All eigenvalues of L are ≥ 0. 8: Prove the Claim 1 using using the following Lemma (Gershgorin): n×n If M 2 C then its eigenvalues lie in 8 9 [ < X = z 2 C : jz − mi;ij ≤ jmi;jj : i : j6=i ; − 2 Hint: Try to evaluate the formula on K4 , draw the solutions in the plane (see C as R ). Solution: Notice that the value on the diagonal is equal to the sum of all the other values in a row. So the formula is saying that the distance of z from (mi;i; 0) is at most mi;i, which creates a circle with (0; 0) being − the most left point. For the case K4 , there are two circles, one centered at (3; 0) of radius and one centered at (2; 0) of radius 2. Notice we also showed that all eigenvalues are ≤ 2∆. x?Mx Define Raleigh quotient as x?x . Main use is x?Mx λ ≤ ≤ λ min x?x max So we try to show that Raleigh quotient with L is ≥ 0. Notice that it is enough to show x?Lx ≥ 0. Also, x?Mx λmin = min x6=0 x?x x?Mx λmax = max x6=0 x?x If λ1 ≤ λ2 ≤ · · · ≤ λn then y?My λi = max min i−1 dim subspace S y?S y?y y?My = min max n−i−1 dim subspace S y?S y?y from S. Butler https://www.youtube.com/watch?v=Dm_FuQDjGHI MATH X95 - 1 - page 2/4 T − 9: Compute x Lx for K4 and show it is ≥ 0. (hint: complete to square) 0 1 0 1 3 −1 −1 −1 x1 B−1 3 −1 −1C Bx2C 2 2 2 2 (x1; x2; x3; x4) B C B C =3 x1 + 3x2 + 2x3 + 2x4 − 2x1x2 − 2x1x3 − 2x1x4 − 2x2x3 − 2x2x4 @−1 −1 2 0 A @x3A −1 −1 0 2 x4 2 2 2 2 2 =( x1 − x2) + (x1 − x3) + (x1 − x4) + (x2 − x3) + (x3 − x4) ≥ 0 Notice that every term comes from one edge. 10: Find a general formula for xT Lx. Solution: T X 2 x Lx = (xi − xj) ij2E(G) Recall basic property of Laplacian matrix: it has eigenvalue 0. 11: What are eigenvectors corresponding to the eigenvalue 0? What is the multiplicity of 0? Hint: Use the previous formula. Solution: If x is an eigenvector for 0, it must be true that T X 2 x Lx = (xi − xj) = 0: ij2E(G) Hence for every edge ij, xi = xj. So values on one connected component must be all equal. Hence the multiplicity of 0 in spec(L) is equal to the number of components. Once could take base that assigns one to xi' from one component and 0 elsewhere. Exists notion of algebraic connectivity which is λ2, where 0 = λ1 ≤ λ2 ≤ · · · . Note: Laplacian is not always better than adjacency matrix. Pair distinguished by L but not by A: Pair distinguished by A but not by L: from S. Butler https://www.youtube.com/watch?v=Dm_FuQDjGHI MATH X95 - 1 - page 3/4 Suppose Lx = λx. Then x?Lx = λx?x = λjjxjj2. Hence x?Lx ≥ 0 implies Claim 1. 12: How could one show x?Mx ≥ 0 for all x for some matrix M? Solution: - show eigenvalues are all ≥ 0 - not good here - for real matrix are some sufficient conditions based on determinants - we use M = CC? for some other matrix and then x?Mx = x?CC?x = (C?x)?(C?x) ≥ 0 Recall incidence matrix of G is jV j × jEj and every columns has exactly to entries 1 that correspond to the two vertices incident with the edge corresponding to the column. 3 0 1 1 1 1 0 0 b d B C B C a B1 0 0 1 1C 1 2 Incidence Matrix = B C B C B0 1 0 1 0C c e @ A 0 0 1 0 1 4 The incidence matrix can encode orientation of edges. Then it is called signed incidence matrix. Entry is −1 if edges is going out from v and 1 if edge is going in v. 13: Finish the signed adjacent matrix 3 0 1 −1 1 1 0 0 b d B C B C a B 1 0 0 1 1C 1 2 Signed Incidence Matrix = B C = C B C B 0 1 0 1 0C c e @ A 0 0 1 0 1 4 14: Compute CCT for C from the previous question. Don't forget to fill the signs! 0 1 0 1 −1 1 0 0 0 1 −1 1 1 0 0 B C 3 −1 −1 −1 B C B C B 1 0 1 0C B C B 1 0 0 1 1C B C B−1 3 −1 −1C B C B C B C B C · B 1 0 0 1C = B C B 0 1 0 1 0C B C B−1 −1 2 0 C B C B C B C @ A B 0 1 1 0C @ A 0 0 1 0 1 @ A −1 −1 0 2 0 1 0 1 15: Compute CCT for any graph, where C is a signed incidence matrix. Notice that for a graph without any orientation, we first pick some orientation and then do the product. Interestingly, the result does NOT depend on the orientation you pick. T T T (CC )v;v = deg(v); (CC )u;v = −1 if uv 2 E and 0 otherwise. So CC = L. from S. Butler https://www.youtube.com/watch?v=Dm_FuQDjGHI MATH X95 - 1 - page 4/4.