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3/1/2015 Ma/CS 6b Class 23: Eigenvalues in Regular Graphs By Adam Sheffer Recall: The Spectrum of a Graph ο Consider a graph πΊ = π, πΈ and let π΄ be the adjacency matrix of πΊ. β¦ The eigenvalues of πΊ are the eigenvalues of π΄. β¦ The characteristic polynomial π πΊ; π is the characteristic polynomial of π΄. β¦ The spectrum of πΊ is π1 , β¦ , ππ‘ π πππ πΊ = , π1 , β¦ , ππ‘ where π1 , β¦ , ππ‘ are the eigenvalues of π΄ and ππ is the multiplicity of ππ . 1 3/1/2015 Example: Spectrum π΄= 0 1 0 1 1 0 1 0 0 1 0 1 1 0 1 0 π£1 π£2 π£4 π£3 π β1 0 β1 β1 0 det ππΌ β π΄ = det β1 π 0 β1 π β1 β1 0 β1 π = π2 π β 2 π + 2 . π πππ πΆ4 = 0 2 β2 2 1 1 Slight Change of Notation Instead of multiplicities, let π1 , β¦ , ππ be the not necessarily distinct eigenvalues of π. 2 β1 ο For example, if the spectrum is , 2 2 we write π1 = π2 = 2 and π3 = π4 = β1 (instead of π1 = 2, π1 = 2, π2 = β1, π2 = 2). ο 2 3/1/2015 Recall: The Spectral Theorem ο Theorem. Any real symmetric π × π matrix has real eigenvalues and π orthonormal eigenvectors. β¦ By definition, any adjacency matrix π΄ is symmetric and real. β¦ The algebraic and geometric multiplicities are the same in this case. β¦ We have π π΄; π = ππ=1 π β ππ . β¦ The multiplicity of an eigenvalue π is π β ππππ ππΌ β π΄ . More Examples ο We already derived the following: β¦ π πππ πΎπ = β¦ π πππ πΎπ,π ο β1 πβ1 πβ1 1 0 = π+πβ2 ππ 1 β ππ 1 Our next goal is to study regular graphs. β¦ Can you come up with an eigenvector of any regular graph with π vertices? 3 3/1/2015 Eigenvalues of Regular graphs ο If π΄ is the adjacency matrix of a π-regular graph, then any row of π΄ contains exactly π 1βs. β¦ Thus, the vector 1π = 1,1, β¦ , 1 is an eigenvector of π΄ with eigenvalue π. ο Theorem. Let πΊ be a connected graph. The eigenvalue of πΊ of largest absolute value is the maximum degree if and only if πΊ is regular. Proof π΄ β π × π adjacency matrix of a graph πΊ. ο Ξ πΊ β the maximum degree of πΊ. ο π₯ = π₯1 , β¦ , π₯π β eigenvector of eigenvalue π of largest absolute value. ο π₯π = max |π₯π | . ο π π π₯π = π΄π₯ π = π₯π π£π βπ π£π β€ deg π£π π₯π β€ Ξ πΊ π₯π , ο So π β€ Ξ πΊ . 4 3/1/2015 ο ο Ξ πΊ β the maximum degree of πΊ. We proved that the absolute value of any eigenvalue of π΄ is at most Ξ πΊ , using π π₯π = π΄π₯ π = π₯π β€ deg π£π π₯π β€ Ξ πΊ π₯π . π£π βπ π£π ο For equality to hold, we need β¦ deg π£π = Ξ πΊ . β¦ π₯π = π₯π for each π£π β π π£π . ο That is, π£π and all of its neighbors are of degree Ξ πΊ . Repeating the same argument for a neighbor π£π implies that π£π βs neighbors are also of degree Ξ πΊ . We continue to repeat the argument to obtain that the graph is regular. Multiplicity The previous proof also shows that any eigenvector π₯1 , β¦ , π₯π of the eigenvalue π satisfies π₯1 = π₯2 = β― = π₯π . ο Thus, the space of eigenvectors of the eigenvalue π is of dimension 1 (that is, π has multiplicity 1). ο 5 3/1/2015 The Spectrum of the Petersen Graph ο The Petersen graph πΊ = π, πΈ is a 3regular graph with 10 vertices. β¦ We know that it has eigenvalue 3 with eigenvector 110 . ο To find the other eigenvalues, we notice some useful properties: β¦ If π’, π£ β πΈ then π’ and π£ have no common neighbors. β¦ If π’, π£ β πΈ then π’ and π£ have exactly one common neighbor. The Adjacency Matrix ο The number of neighbors shared by π£π and π£π is π΄2 ππ . That is π΄2 ο ππ 3, = 0, 1, if π = π, if π β π and π£π , π£π β πΈ, if π β π and π£π , π£π β πΈ. That is, π΄2 + π΄ β 2πΌ = 1π×π . 6 3/1/2015 The Additional Eigenvalues ο ο ο ο ο ο We have π΄2 + π΄ β 2πΌ = 1π×π . Since 3 is an eigenvalue with eigenvector 1π , the other eigenvectors are orthogonal to 1π . Thus, for an eigenvector π£ of eigenvalue π β 3: 1π×π π£ = 0π . That is, π΄2 + π΄ β 2πΌ π£ = 0π . If π£ is an eigenvector of π then we have π2 π£ + ππ£ β 2π£ = 0π . Thus, the additional eigenvalues of π΄ are 1, β2. 3 1 β2 π πππ πΊ = . 1 π2 π3 The Multiplicities 1 β2 . π 2 π3 10 ο Recall that π=1 ππ = π‘ππππ π΄ = 0. ο That is, 3 + π2 β 2π3 = 0. ο Combining this with π2 + π3 = 9 and π2 , π3 β₯ 0, we obtain the unique solution π2 = 5, π3 = 4. 3 1 β2 ο π πππ πΊ = . 1 5 4 ο π πππ πΊ = 3 1 7 3/1/2015 Petersen Graph and πΎ10 ο Problem. Can we partition the edges of πΎ10 into three disjoint sets, such that each set forms a Petersen graph? 10 = 45 edges, and the Petersen 2 graph has 15. β¦ In πΎ10 every vertex is of degree 9, and in the Petersen graph 3. β¦ πΎ10 has Disproof ο ο ο ο ο ο Assume, for contradiction, that the partition exists, and let π΄, π΅, πΆ be the adjacency matrices of the three copies of the Petersen graph. The adjacency matrix of πΎ10 is 110×10 β πΌ. That is, π΄ + π΅ + πΆ = 110×10 β πΌ. ππ΄ , ππ΅ β the vector subspaces of eigenvectors corresponding to the eigenvalue 1 in π΄ and π΅. We know that dim ππ΄ = dim ππ΅ = 5. Since both ππ΄ and ππ΅ are orthogonal to 110 , they are not disjoint (otherwise we would have a set of 11 orthogonal vectors in β10 ). 8 3/1/2015 Disproof (cont.) π΄, π΅, πΆ β the adjacency matrices of the three copies of the Petersen graph in πΎ10 . ο π΄ + π΅ + πΆ = 1π×π β πΌ. ο ππ΄ , ππ΅ β the vector subspaces of eigenvectors corresponding to the eigenvalue 1 in π΄ and π΅. ο ππ΄ and ππ΅ are not disjoint. ο Let π§ β ππ΄ β© ππ΅ . Since every vector in ππ΄ and ππ΅ is orthogonal to 1π , so is π§. ο We have πΆπ§ = 1π×π β πΌ β π΄ β π΅ π§ = 0 β π§ β π΄π§ β π΅π§ = β3π§. ο Contradiction since -3 is not an eigenvalue of πΆ. ο Four Things You Did Not Know About The Petersen Graph It has 15 edges and 2000 spanning trees. ο It is the smallest 3-regular graph of girth 5 (this is called a 3,5 -cage graph). ο It likes gardening, ballet, and building airplane models. ο It has gotten divorced three times. ο 9 3/1/2015 Moore Graphs ο A Moore graph is a graph that is πregular, of diameter π, and whose number of vertices is πβ1 πβ1 π. 1+π π=0 ο As can be easily checked, this is the minimum possible number of vertices of any graph of diameter π and minimum degree π. Examples of Moore Graphs ο A Moore graph is a graph that is π-regular, of diameter π, and whose number of vertices is πβ1 πβ1 π. 1+π π=0 ο What Moore graphs do we know? β¦ The Petersen graph is 3-regular, of diameter 2, and contains 1 + 3 1π=0 3 β 1 π = 10 vertices. 10 3/1/2015 Examples of Moore Graphs ο A Moore graph is a graph that is π-regular, of diameter π, and whose number of vertices is πβ1 πβ1 π. 1+π π=0 ο ο The Petersen graph is 3-regular, of diameter 2, and contains 1 + 3 1π=0 3 β 1 π = 10 vertices. Is there a Moore graph that is 2-regular and of diameter 2? πΆ5 Moore Graphs of Diameter 2 and Girth 5 Recall that the girth of a graph is the length of the shortest cycle in it. ο Theorem. There exist π-regular Moore graphs with diameter 2 and girth 5 only for π = 2,3,7, and possibly 57. ο β¦ The case of π = 57 is an open problem. β¦ If it exists, it has 3250 vertices, and 92,625 edges. The case of π = 7 11 3/1/2015 ο πΊ = π, πΈ β a π-regular graph of diameter 2, girth 5, and with 1 π =1+π πβ1 π = 1 + π2 . π=0 ο Since the girth is five, if π£π , π£π β πΈ then π£π and π£π have no common neighbors. ο Since the diameter is two, if π£π , π£π β πΈ then π£π and π£π have exactly one common neighbor. Thus, the adjacency matrix π΄ satisfies ο π΄2 ο ππ π, = 0, 1, if π = π, if π β π and π£π , π£π β πΈ, if π β π and π£π , π£π β πΈ. That is, we have π΄2 + π΄ β π β 1 πΌ = 1π×π . Proof (cont.) πΊ = π, πΈ β a π-regular graph of diameter 2, girth 5, and with π = 1 + π 2 . ο π΄2 + π΄ β π β 1 πΌ = 1π×π . ο π β π β an eigenvalue of π΄ with eigenvector π£. Since π β π, π£ is orthogonal to 1π . Thus π΄2 + π΄ β π β 1 πΌ π£ = 0π , or 2 π π£ + ππ£ β π β 1 π£ = 0. ο This implies that π2 + π β π + 1 = 0, so 1± 1+4 πβ1 1 ± 4π β 3 π2,3 = β =β . 2 2 π π2 π3 ο We thus have π πππ πΊ = . 1 π2 π3 ο 12 3/1/2015 Finding the Multiplicities ο We have π2 = β 1+ 4πβ3 π πππ πΊ = ο ο ο 2 π 1 1β 4πβ3 , π3 = β π2 π3 . π2 π3 2 , 0 = π‘ππππ π΄ = π + π2 π2 + π3 π3 π2 + π3 π3 β π2 =πβ + 4π β 3. 2 2 Since π2 + π3 = π β 1 = π 2 , we have π 2 β 2π = π3 β π2 4π β 3. This can happen if either π2 = π3 or 4π β 3 = π 2 for some integer π . If π2 = π3 than π 2 β 2π = 0, implying π = 2. The Case of 4π β 3 = π 2 ο ο ο π 2 β 2π = π3 β π2 4π β 3. Assume that 4π β 3 = π 2 for some integer π . That is, π = π 2 + 3 /4. Substituting into the above equation: π 2 + 3 2 2π 2 + 6 β = π3 β π2 π . 16 4 Setting π3 β π2 = 2π3 β π 2 , we have π 4 + 6π 2 + 9 β 8π 2 β 24 = 32π3 π β π 5 β 6π 3 β 9π . ο ο π 5 + π 4 + 6π 3 β 2π 2 + 9 β 32π3 π = 15. So π must divide 15, and we get π β ± 1,3,5,15 , which implies π β 1,3,7,57 . The case π = 1 leads to πΎ2 . Not a Moore graph. 13 3/1/2015 Recall: Independent Sets Consider a graph πΊ = π, πΈ . An independent set in πΊ is a subset π β² β π such that there is no edge between any two vertices of πβ². ο Finding a maximum independent set in a graph is a major problem in theoretical computer science. ο β¦ No polynomial-time algorithm is known. Past Bounds Let πΊ = (π, πΈ) be a graph. ο Already proved: ο β¦ πΊ has an independent set of size at least π£βπ 1 . 1 + deg π£ π β¦ If πΈ = π β , then πΊ has an independent 2 set of size at least π /2π. 14 3/1/2015 An Upper Bound ο Theorem. For a π-regular graph πΊ = π, πΈ with smallest (most negative) eigenvalue ππ , the size of the largest π independent set of πΊ is at most . 1βπ/ππ ο Example. 0 2 β2 . 2 1 1 4 β¦ So at most 2 = 2. β¦ π πππ πΆ4 = 1β π£1 π£2 π£4 π£3 β2 Recall: The Rayleigh Quotient ο The Rayleigh quotient is π π΄, π₯ = π (π₯) = ο π₯ π π΄π₯ π₯ππ₯ for π × π matrix π΄ and π₯ β βπ . Lemma. Let π΄ be a real symmetric π × π matrix. Then π π₯ attains its maximum and minimum at eigenvectors of π΄. (We do not prove the lemma.) ο Question. What is π π₯ when π₯ is an eigenvector of eigenvalue π? β¦ π₯ π π΄π₯ π₯ππ₯ = π₯ π ππ₯ π₯ππ₯ = π. β¦ Thus, the min and max values of π π₯ are the min and max eigenvalues of π΄. 15 3/1/2015 ο ο ο ο ππ β the most negative eigenvalue of πΊ. π β a largest independent set of πΊ. 1π = π₯1 , β¦ , π₯π β a vector with π₯π = 1 if π£π β π (otherwise π₯π = 0). π¦ = π1π β 1π β π . π¦ π π΄π¦ = π2 β 1ππ π΄1π β 2 π π β 1ππ π΄1π + π ο 2 β 1ππ π΄1π . Since π is an independent set, we have 1ππ π΄1π = π,πβS π΄ππ = 0. Since πΊ is π-regular, 1ππ π΄1π = 1ππ β π1π = π π , and also 1ππ π΄1π = 1ππ β π1π = ππ. ο Combining the above, we have π¦ π π΄π¦ = 0 β 2 π π β π π + π 2 β ππ = β π 2 ππ. π¦ π π¦ = π2 1ππ 1π β 2 π π1ππ 1π + π 2 1ππ 1π = π2 π β 2 π 2 π + π 2 π = π π π β π . ο Completing the Proof ο By the lemma, we have π¦ π π΄π¦ β₯ ππ . π¦π π¦ ο We have ο Thus π¦ π π΄π¦ = β π 2 ππ. π¦π π¦ = π π π β π . β π 2 ππ ππ β€ π π πβ π ππ π β π β€ βπ π β = βπ π . πβ π π β€ π 1 β π/ππ 16 3/1/2015 The End 17