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 ∈ E(G), Au,v = 0 otherwise.
The Laplacian Matrix of G, denoted by L, is defined as deg(u) if u = v Lu,v = −1 if uv ∈ 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 3 −1 −1 −1 −1 3 −1 −1 1 2 L = −1 −1 2 0 −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) = {2 cos n for k = 0, . . . , n − 1}. 2πk spec(LG) = {2 − 2 cos n for k = 0, . . . , n − 1}.
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 {2 cos n for k = 0, . . . , n − 1}. Solution: 1 −1 2 1 1 −1 2 −1 2 1 0 1 L = = − −1 2 −1 2 1 0 1 −1 1 2 1 1 Notice the second matrix is a adjacency matrix of a path with loops. πk Hence spec(LG) = {2 − 2 cos n for k = 0, . . . , n − 1}. from S. Butler https://www.youtube.com/watch?v=Dm_FuQDjGHI MATH X95 - 1 - page 1/4 P 2 Recall λi = 2|E| where λi ∈ spec(A). P 7: Show µi = 2|E| where µi ∈ 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 ∈ C then its eigenvalues lie in [ X z ∈ C : |z − mi,i| ≤ |mi,j| . 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) 3 −1 −1 −1 x1 −1 3 −1 −1 x2 2 2 2 2 (x1, x2, x3, x4) =3 x1 + 3x2 + 2x3 + 2x4 − 2x1x2 − 2x1x3 − 2x1x4 − 2x2x3 − 2x2x4 −1 −1 2 0 x3 −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) ij∈E(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. ij∈E(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 = λ||x||2. 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 |V | × |E| and every columns has exactly to entries 1 that correspond to the two vertices incident with the edge corresponding to the column.
3 1 1 1 0 0 b d a 1 0 0 1 1 1 2 Incidence Matrix = 0 1 0 1 0 c e 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 −1 1 1 0 0 b d a 1 0 0 1 1 1 2 Signed Incidence Matrix = = C 0 1 0 1 0 c e 0 0 1 0 1 4
14: Compute CCT for C from the previous question. Don’t forget to fill the signs! −1 1 0 0 −1 1 1 0 0 3 −1 −1 −1 1 0 1 0 1 0 0 1 1 −1 3 −1 −1 · 1 0 0 1 = 0 1 0 1 0 −1 −1 2 0 0 1 1 0 0 0 1 0 1 −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 ∈ E and 0 otherwise. So CC = L. from S. Butler https://www.youtube.com/watch?v=Dm_FuQDjGHI MATH X95 - 1 - page 4/4