Mysteries around the graph Laplacian eigenvalue 4
MMysteries around the graph Laplacian eigenvalue 4
Yuji Nakatsukasa a , Naoki Saito b, ∗ , Ernest Woei b a School of Mathematics, The University of Manchester, Manchester, M13 9PL, UK b Department of Mathematics, University of California, Davis, CA 95616, USA
Abstract
We describe our current understanding on the phase transition phenomenon of thegraph Laplacian eigenvectors constructed on a certain type of unweighted trees,which we previously observed through our numerical experiments. The eigen-value distribution for such a tree is a smooth bell-shaped curve starting from theeigenvalue 0 up to 4. Then, at the eigenvalue 4, there is a sudden jump. Interest-ingly, the eigenvectors corresponding to the eigenvalues below 4 are semi-global oscillations (like Fourier modes) over the entire tree or one of the branches; onthe other hand, those corresponding to the eigenvalues above 4 are much more lo-calized and concentrated (like wavelets) around junctions / branching vertices. Fora special class of trees called starlike trees , we obtain a complete understandingof such phase transition phenomenon. For a general graph, we prove the numberof the eigenvalues larger than 4 is bounded from above by the number of verticeswhose degrees is strictly higher than 2. Moreover, we also prove that if a graphcontains a branching path, then the magnitudes of the components of any eigen-vector corresponding to the eigenvalue greater than 4 decay exponentially fromthe branching vertex toward the leaf of that branch. Keywords: graph Laplacian, localization of eigenvectors, phase transitionphenomena, starlike trees, dendritic trees, Gerschgorin’s disks ∗ Corresponding author
Email addresses: [email protected] (Yuji Nakatsukasa), [email protected] (Naoki Saito), [email protected] (Ernest Woei)
Preprint submitted to Elsevier November 13, 2018 a r X i v : . [ m a t h . NA ] A ug . Introduction In our previous report [11], we proposed a method to characterize dendrites ofneurons, more specifically retinal ganglion cells (RGCs) of a mouse, and clusterthem into di ff erent cell types using their morphological features, which are derivedfrom the eigenvalues of the graph Laplacians when such dendrites are representedas graphs (in fact literally as “trees”). For the details on the data acquisition andthe conversion of dendrites to graphs, see [11] and the references therein. Whileanalyzing the eigenvalues and eigenvectors of those graph Laplacians, we ob-served a very peculiar phase transition phenomenon as shown in Figure 1.1. Theeigenvalue distribution for each dendritic tree is a smooth bell-shaped curve start-ing from the eigenvalue 0 up to 4. Then, at the eigenvalue 4, there is a suddenjump as shown in Figure 1.1(c, d). Interestingly, the eigenvectors correspondingto the eigenvalues below 4 are semi-global oscillations (like Fourier cosines / sines)over the entire dendrites or one of the dendrite arbors (or branches); on the otherhand, those corresponding to the eigenvalues above 4 are much more localized and concentrated (like wavelets) around junctions / branching vertices, as shownin Figure 1.2.We want to answer the following questions: Q1 Why does such a phase transition phenomenon occur? Q2 What is the significance of the eigenvalue 4? Q3 Is there any tree that possesses an eigenvalue exactly equal to 4? Q4 What about more general graphs that possess eigenvalues exactly equal to 4?As for Q1 and Q2, which are closely related, we have a complete answer for aspecific and simple class of trees called starlike trees as described in Section 3,and a partial answer for more general trees and graphs such as those representingneuronal dendrites, which we discuss in Section 4. For Q3, we identify two classesof trees that have an eigenvalue exactly equal to 4, which is necessarily a simpleeigenvalue, in Section 5.In Section 6, we will prove that the existence of a long path between twosubgraphs implies that the eigenvalues of either of the subgraphs that are largerthan 4 are actually very close to some eigenvalues of the whole graph. Then,in Section 7, we will give a counterexample to the conjecture that the largestcomponent in the eigenvector corresponding to the largest eigenvalue (which islarger than 4) lies on the vertex of the highest degree. Finally, we describe our2
200 −150 −100 −50 0 50 100−150−100−50050100150 X( µ m) Y ( µ m ) (a) RGC −200 −150 −100 −50 0 50 100 150 200 250−250−200−150−100−50050100150200250 X( µ m) Y ( µ m ) (b) RGC λ k (c) Eigenvalues of (a) λ k (d) Eigenvalues of (b)Figure 1.1: Typical dendrites of Retinal Ganglion Cells (RGCs) of a mouse and the graph Lapla-cian eigenvalue distributions. (a) 2D projection of dendrites of an RGC of a mouse; (b) that ofanother RGC revealing di ff erent morphology; (c) the eigenvalue distribution of the RGC shownin (a); (d) that of the RGC shown in (b). Regardless of their morphological features, a phasetransition occurs at the eigenvalue 4.
200 −100 0 100 200−250−200−150−100−50050100150200250 X ( µ m) Y ( µ m ) (a) RGC λ = . −200 −100 0 100 200−250−200−150−100−50050100150200250 X ( µ m) Y ( µ m ) (b) RGC λ = . λ = . λ = . investigation on Q4 in Section 8. But let us first start by fixing our notation andreviewing the basics of graph Laplacians in Section 2.
2. Definitions and Notation
Let G = ( V , E ) be a graph where V = V ( G ) = { v , v , . . . , v n } is a set of verticesin G and E = E ( G ) = { e , e , . . . , e m } is a set of edges where e k connects twovertices v i , v j for some 1 ≤ i , j ≤ n , and we write e k = ( v i , v j ). Let d k = d ( v k )be the degree of the vertex v k . If a graph G is a tree , i.e., a connected graphwithout cycles, then it has m = n − L ( G ) : = D ( G ) − A ( G ) be the Laplacian matrix where D ( G ) : = diag( d , . . . , d n ) is called the degree matrix of G , i.e., the diagonal matrix of vertex degrees, and A ( G ) = ( a i j ) is the adjacencymatrix of G , i.e., a i j = v i and v j are adjacent; otherwise it is 0. Furthermore,let 0 = λ ( G ) ≤ λ ( G ) ≤ · · · ≤ λ n − ( G ) be the eigenvalues of L ( G ), and m G ( λ ) bethe multiplicity of the eigenvalue λ . More generally, if I ⊂ R is an interval of thereal line, then we define m G ( I ) : = { λ k ( G ) ∈ I } .At this point we would like to give a simple yet important example of a tree andits graph Laplacian: a path graph P n consisting of n vertices shown in Figure 2.3.The graph Laplacian of such a path graph can be easily obtained and is instructive.4 igure 2.3: A path graph P n provides a simple yet important example. − − − . . . . . . . . . − − − (cid:124) (cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32) (cid:123)(cid:122) (cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32) (cid:125) L ( G ) = . . . (cid:124) (cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32) (cid:123)(cid:122) (cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32) (cid:125) D ( G ) − . . . . . . . . . (cid:124) (cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32) (cid:123)(cid:122) (cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32) (cid:125) A ( G ) . The eigenvectors of this matrix are nothing but the
DCT Type II basis vectors usedfor the JPEG image compression standard; see e.g., [14]. In fact, we have λ k = (cid:32) π k n (cid:33) ; (2.1) φ j , k = cos (cid:32) π kn (cid:32) j − (cid:33)(cid:33) , j = , . . . , n (2.2)for k = , , . . . , n −
1, where φ k = (cid:0) φ , k , · · · , φ n , k (cid:1) T is the eigenvector correspond-ing to λ k . From these, it is clear that for any finite n ∈ N , λ max = λ n − <
4, andno localization / concentration occurs in the eigenvector φ n − (or any eigenvector),which is simply a global oscillation with the highest possible (i.e., the Nyquist)frequency, i.e., φ n − = (cid:16) ( − j − sin (cid:16) π n (cid:16) j − (cid:17)(cid:17)(cid:17) T ≤ j ≤ n .
3. Analysis of Starlike Trees
As one can imagine, analyzing this phase transition phenomenon for compli-cated dendritic trees turns out to be rather formidable. Hence, we start our analysison a simpler class of trees called starlike trees . A starlike tree is a tree that hasexactly one vertex of degree higher than 2. Examples are shown in Figure 3.4.We use the following notation. Let S ( n , n , . . . , n k ) be a starlike tree that has k ( ≥
3) paths (i.e., branches) emanating from the central vertex v . Let the i th5 a) S (2 , , , , ,
1) (b) S ( n , , , , , , ,
1) a.k.a. cometFigure 3.4: Typical examples of a starlike tree. branch have n i vertices excluding v . Let n ≥ n ≥ · · · ≥ n k . Hence, the totalnumber of vertices is n = + k (cid:88) i = n i .Das proved the following results for a starlike tree S ( n , . . . , n k ) in [3]: λ max = λ n − < k + + k − + (cid:32) π n k + (cid:33) ≤ λ n − ≤ + (cid:32) π n + (cid:33) . (3.3)On the other hand, Grone and Merris [7] proved the following lower bound for ageneral graph G with at least one edge: λ max ≥ max ≤ j ≤ n d ( v j ) + . (3.4)Hence we have the following Corollary 3.1.
A starlike tree has exactly one graph Laplacian eigenvalue greaterthan or equal to 4. The equality holds if and only if the starlike tree is K , = S (1 , , , which is also known as a claw .Proof. The first statement is easy to show. The lower bound in (3.4) is larger thanor equal to 4 for any starlike tree since max ≤ j ≤ n d ( v j ) = d ( v ) ≥
3. On the otherhand, the second largest eigenvalue λ n − is clearly strictly smaller than 4 due to(3.3).To prove the second statement about the necessary condition on the equality(the su ffi ciency is easily verified), first note that by (3.4), d ( v ) = λ max =
4. Let d denote the highest degree of such a starlike tree,i.e., d = d ( v ). Since we only consider starlike trees, the second highest degree d must be either 2 or 1. Now, we use the following6 heorem 3.1 (Das 2004, [4]) . Let G = ( V , E ) be a connected graph and d (cid:44) d where d and d are the highest and the second highest degree, respectively. Then, λ max ( G ) = d + d if and only if G is a star graph. If λ max ( G ) = d =
3, then using this theorem, we must have d = G must be a star graph with d = d =
1, i.e., G = K , .As for the concentration / localization of the eigenvector φ n − corresponding tothe largest eigenvalue λ n − , we prove the following Theorem 3.2.
Let φ n − = (cid:0) φ , n − , · · · , φ n , n − (cid:1) T , where φ j , n − is the value of theeigenvector corresponding to the largest eigenvalue λ n − at the vertex v j , j = , . . . , n. Then, the absolute value of this eigenvector at the central vertex v can-not be exceeded by those at the other vertices, i.e., | φ , n − | > | φ j , n − | , j = , . . . , n . To prove this theorem, we use the following lemma, which is simply a corol-lary of Gerschgorin’s theorem [15, Theorem 1.1]:
Lemma 3.1.
Let A be a square matrix of size n × n, λ k ( A ) be any eigenvalueof A, and φ k = ( φ , k , . . . , φ n , k ) T be the corresponding eigenvector. Let k ∗ denotethe index of the largest eigenvector component in φ k , i.e., (cid:12)(cid:12)(cid:12) φ k ∗ , k (cid:12)(cid:12)(cid:12) = max j ∈ N (cid:12)(cid:12)(cid:12) φ j , k (cid:12)(cid:12)(cid:12) where N : = { , . . . , n } . Then, we must have λ k ( A ) ∈ Γ k ∗ ( A ) , where Γ i ( A ) : = (cid:110) z ∈ C : | z − a ii | ≤ (cid:80) j ∈ N \{ i } | a i j | (cid:111) is the ith Gerschgorin disk of A. In other words,for the index of the largest eigenvector component, the corresponding Gerschgorindisk must contain the eigenvalue.Proof. Recall the proof of Gerschgorin’s theorem. The k ∗ th row of A φ k = λ k φ k yields | λ k − a k ∗ k ∗ | ≤ (cid:88) j ∈ N \{ k ∗ } (cid:12)(cid:12)(cid:12) a k ∗ j (cid:12)(cid:12)(cid:12) (cid:12)(cid:12)(cid:12) φ j , k (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) φ k ∗ , k (cid:12)(cid:12)(cid:12) ≤ (cid:88) j ∈ N \{ k ∗ } (cid:12)(cid:12)(cid:12) a k ∗ j (cid:12)(cid:12)(cid:12) . This implies λ k ∈ Γ k ∗ ( A ), which proves the lemma. Proof of Theorem 3.2.
First of all, by Corollary 3.1 we have λ n − ≥
4. However, λ n − = K , . In that case, it is easy to see that this theorem holdsby directly examining the eigenvector φ n − = φ ∝ (3 , − , − , − T . Hence, let usexamine the case λ n − >
4. In this case, Lemma 3.1 indicates 4 < λ n − ∈ Γ ( n − ∗ ( L )where ( n − ∗ ∈ N is the index of the largest component in φ n − . Now, note thatthe disk Γ i ( L ) for any vertex v i that has degree 2 is { z ∈ C : | z − | ≤ } (and7 z ∈ C : | z − | ≤ } for a degree 1 vertex). This means that the Gerschgorindisk Γ ( n − ∗ containing the eigenvalue λ n − > φ n − must correspond toan index for which the vertex has degree 3 or higher. In our starlike-tree case,there is only one such vertex, v , i.e., ( n − ∗ = ff erent proofs without using Gerschgorin’s theorem, see Das [3, Lemma 4.2]and E. Woei’s dissertation [16]. We note that our proof using Gerschgorin’s disksis more powerful than those other proofs and can be used for more general situa-tions than the starlike trees as we will see in Section 4.R emark Let φ = ( φ , φ , . . . , φ n ) T be an eigenvector of a starlike tree S ( n , . . . , n k ) corresponding to the Laplacian eigenvalue λ . Without loss of generality, let v , . . . , v n + be the n vertices along a branch emanating from the central vertex v with v n + being the leaf (or pendant) vertex. Then, along this branch, the eigenvector com-ponents satisfy the following equations: λφ n + = φ n + − φ n , (3.5) λφ j = φ j − φ j − − φ j + ≤ j ≤ n . (3.6) From Eq. (3.6) , we have the following recursion relation: φ j + + ( λ − φ j + φ j − = , j = , . . . , n . This recursion can be explicitly solved using the roots of the characteristic equa-tion r + ( λ − r + = , (3.7) and when (3.7) has distinct roots r , r , the general solution can be written as φ j = Ar j − + Br j − , j = , . . . , n + , (3.8) where A , B are appropriate constants derived from the boundary condition (3.5) .Now, let us consider these roots of (3.7) in detail. The discriminant of (3.7) is D ( λ ) : = ( λ − − = λ ( λ − . Since we know that λ ≥ , this discriminant changes its sign depending on λ < or λ > . (Note that λ = occurs only for the claw K , on which we explicitlyknow everything; hence we will not discuss this case further in this remark.) If < , then D ( λ ) < and it is easy to show that the roots are complex valued withmagnitude 1. This implies that (3.8) becomes φ j = A (cid:48) cos( ω ( j − + B (cid:48) sin( ω ( j − , j = , . . . , n + , where ω satisfies tan ω = √ λ (4 − λ ) / (2 − λ ) , and A (cid:48) , B (cid:48) are appropriate constants.In other words, if λ < , the eigenvector along this branch is of oscillatory na-ture. On the other hand, if λ > , then D ( λ ) > and it is easy to show thatboth r and r are real valued with − < r = (cid:16) − λ + √ λ ( λ − (cid:17) / < whiler = (cid:16) − λ − √ λ ( λ − (cid:17) / < − . On the surface, the term Br j − looks like adominating part in (3.8) ; however, we see from (3.5) that | φ n | > | φ n + | , whichmeans the real dominating part in (3.8) for j = , . . . , n + is the term Ar j − .Hence we conclude that | φ j | decays exponentially with j, that is, the eigenvectorcomponent decays rapidly towards the leaves. The siuation is the same for theother branches. In summary, we have shown that a starlike tree has only one eigenvalue ≥
4. The Localization Phenomena on General Graphs
Unfortunately, actual dendritic trees are not exactly starlike. However, ournumerical computations and data analysis on totally 179 RGCs indicate that:0 ≤ { j ∈ N | d ( v j ) > } − m G ([4 , ∞ )) n ≤ . starlikeliness S (cid:96) ( T ) of a given tree T as S (cid:96) ( T ) : = − { j ∈ N | d ( v j ) > } − m T ([4 , ∞ )) n We note that S (cid:96) ( T ) ≡ S (cid:96) ( T ) ≡ S (cid:96) ( T ) <
40 −20 0 20 40 60 80−220−210−200−190−180−170−160−150−140−130−120 X ( µ m) Y ( µ m ) (a) RGC S (cid:96) ( T ) ≡ −50 −40 −30 −20 −10 0 10 20 30 40 50−100−95−90−85−80−75−70−65−60−55−50 X ( µ m) Y ( µ m ) (b) RGC S (cid:96) ( T ) = . < Theorem 4.1.
For any graph G of finite volume, i.e., (cid:80) nj = d ( v j ) < ∞ , we have ≤ m G ([4 , ∞ )) ≤ { j ∈ N | d ( v j ) > } and each eigenvector corresponding to λ ≥ has its largest component (in abso-lute value) on the vertex whose degree is higher than 2. We refer the interested readers to [10, Sec. 2] that reviews various relationshipsbetween the multiplicity of certain eigenvalues and the graph structural propertiesdi ff erent from our Theorem 4.1. Proof.
The second statement follows from Lemma 3.1, because the Gerschgorindisks corresponding to vertices of degree 1 or 2 do not include λ > L be a Laplacian matrix of G . We canapply a permutation P such that P T LP = (cid:34) L E T E L (cid:35) , (4.9)where the diagonals of L are 3 or larger (correspond to vertices of degree > L are 2 or 1. Suppose L is (cid:96) -by- (cid:96) . By Gerschgorin’s theoremall the eigenvalues of L must be 4 or below.In fact, we can prove the eigenvalues of L are strictly below 4. By [15, The-orem 1.12], for an irreducible matrix (a Laplacian of a connected graph is irre-ducible) an eigenvalue can exist on the boundary of the union of the Gerschgorin10isks only if it is the boundary of all the disks. Furthermore, if there is such aneigenvalue, then the corresponding eigenvector has the property that all its com-ponents have the same absolute value.Suppose on the contrary that L x = x . Suppose without loss of generalitythat L is irreducible; if not, we can apply a permutation so that PL P T is blockdiagonal and treat each block separately.Now if L has a diagonal 1, then the corresponding Gerschgorin disk lies on[0 , L are 2, and the sum of the absolute values ofthe rows of L are all 4 (this happens only if L is disjoint from L ). So we need (cid:96) ≥ (cid:96) = L = − − − − − − ). Now the i th (1 ≤ i ≤ (cid:96) ) rowof L x = x and the fact | x | = | x | = · · · = | x (cid:96) | force x i = − x j for all j (cid:44) i .This needs to hold for all i , which clearly cannot happen for (cid:96) ≥
3. Therefore theeigenvalues of L must be strictly below 4.By the min-max characterization of the eigenvalues of P T LP , denoting by λ (cid:96) ( P T LP ) the (cid:96) th smallest eigenvalue, we have λ (cid:96) ( P T LP ) = min dim S = (cid:96) max y ∈ span( S ) , (cid:107) y (cid:107) = y T ( P T LP ) y . Hence letting S be the last (cid:96) column vectors of the identity I n and noting S T P T LPS = L , we have λ (cid:96) ( P T LP ) ≤ max y ∈ S , (cid:107) y (cid:107) = y T ( P T LP ) y = λ max ( S T P T LPS ) = λ max ( L ) . Since λ max ( L ) <
4, we conclude that P T LP (and hence L ) has at least (cid:96) eigen-values smaller than 4, i.e., m G ([0 , ≥ (cid:96) . Hence, m G ([4 , ∞ )) = n − m G ([0 , ≤ n − (cid:96) = { j ∈ N | d ( v j ) > } , which proves the first statement.To give a further explanation for the eigenvector localization behavior ob-served in Introduction, we next show that eigenvector components of λ > Theorem 4.2.
Suppose that a graph G has a branch that consists of a path oflength k, whose indices are { i , i , . . . , i k } where i is connected to the rest of thegraph and i k is the leaf of that branch. Then for any eigenvalue λ greater than 4,the corresponding eigenvector φ = ( φ , · · · , φ n ) T satisfies | φ i j + | ≤ γ | φ i j | for j = , , . . . , k − , (4.10)11 here γ : = λ − < . (4.11) Hence | φ i j | ≤ γ j − | φ i | for j = , . . . , k, that is, the magnitude of the components ofan eigenvector corresponding to any λ > along such a branch decays exponen-tially toward its leaf with rate at least γ .Proof. There exists a permutation P such that (cid:98) L : = P T LP = (cid:34) L E T E L (cid:35) , where L = − − − − − . . . . . . . . . − − − ∈ R k × k and E has a -1 in the top-right corner and 0 elsewhere. The diagonals of L correspond to the vertices v i , . . . , v i k of the branch under consideration.Let L φ = λ φ with λ >
4. We have (cid:98) L y = λ y where y = ( y , y , · · · , y n ) T = P T φ .Note that ( y n − k + , y n − k + , · · · , y n ) = ( φ i , φ i , . . . , φ i k ). The last row of (cid:98) L y = λ y gives − y n − + y n = λ y n , hence | y n | = λ − | y n − | ≤ γ | y n − | . (4.12)The ( n − (cid:98) L y = λ y gives − y n − + y n − − y n = λ y n − . Using | y n | ≤ | y n − | we get | y n − | = | y n − + y n | λ − ≤ | y n − | + | y n − | λ − , (4.13)from which we get | y n − | ≤ | y n − | . Therefore | y n | ≤ | y n − | ≤ | y n − | , and so | y n − | = | y n − + y n | λ − ≤ | y n − | λ − = γ | y n − | . Repeating this argument k − γ defined in (4.11). We also note thatthe larger the eigenvalue λ >
4, the smaller the decay rate γ is, i.e., the faster theamplitude decays along the branching path.Also note that the above result holds for any branching path of a tree. In partic-ular, if a tree has k branches consisting of paths, they must all have the exponentialdecay in eigenvector components if λ >
4. This gives a partial explanation for theeigenvector localization behavior observed in Introduction. However, the theoremcannot compare the eigenvector components corresponding to branches emanat-ing from di ff erent vertices of degrees higher than 2, so a complete explanationremains an open problem.R emark Let us briefly consider the case λ = . In this case we have γ = λ − = , suggesting the corresponding eigenvector components along a branching pathmay not decay. However, we can still prove that unless φ i = φ i = · · · = φ i k = ,we must have | φ i k | < | φ i k − | < · · · < | φ i | . (4.14) In other words, the eigenvector components must decay along the branch, al-though not necessarily exponentially. To see this, we first note that if y n = , thenthe last row of (cid:98) L y = λ y forces y n − = . Then, y n = y n − = together with the ( n − st row gives y n − = . Repeating this argument we conclude that y j must bezero for all j = n − k + , . . . , n. Now suppose that | y n | > . Following the abovearguments we see that the inequality in (4.12) with γ = must be strict, that is, | y n | < | y n − | . Using this we see that the inequality in (4.13) must also be strict,hence | y n − | < | y n − | . Repeating this argument proves (4.14) .
5. A Class of Trees Having the Eigenvalue 4
As raised in Introduction, we are interested in answering Q3: Is there any treethat possesses an eigenvalue exactly equal to 4? To answer this question, we usethe following result of Guo [9] (written in our own notation).
Theorem 5.1 (Guo 2006, [9]) . Let T be a tree with n vertices. Then, λ j ( T ) ≤ (cid:38) nn − j (cid:39) , j = , . . . , n − , and the equality holds if and only if all of the following hold: a) j (cid:44) ; b) n − jdivides n; and c) T is spanned by n − j vertex disjoint copies of K , jn − j . a) λ j (b)Figure 5.6: (a) A tree spanned by multiple copies of K , connected via their central vertices. Thistree has an eigenvalue equal to 4 with multiplicity 1. (b) The eigenvalue distribution of such a treespanned by 5 copies of K , . We note that S (cid:96) ( T ) = Here, a tree T = T ( V , E ) is said to be spanned by (cid:96) vertex disjoint copies ofidentical graphs K i ( V i , E i ) for i = , . . . , (cid:96) if V = (cid:83) (cid:96) i = V i and V i ∩ V j = ∅ for all i (cid:44) j . Figure 5.6(a) shows an example of such vertex disjoint copies for K i = K , by connecting their central vertices. We note that there are many other ways toform disjoint vertex copies of K , .This theorem implies the following Corollary 5.1.
A tree has an eigenvalue exactly equal to 4 if it is spanned bym ( = n / ∈ N ) vertex disjoint copies of K , ≡ S (1 , , . Figure 5.6(b) shows the eigenvalue distribution of a tree spanned by m = K , as shown in Figure 5.6(a). Regardless of m , the eigenvector corre-sponding to the eigenvalue 4 has only two values: one constant value at the centralvertices, and the other constant value of the opposite sign at the leaves, as shownin Figure 5.7(a). By contrast, the eigenvector corresponding to the largest eigen-value is again concentrated around the central vertex as shown in Figure 5.7(b).Theorem 5.1 asserts that a general tree T with n vertices can have at most (cid:98) n / (cid:99) Laplacian eigenvalues ≥
4. We also know by Theorem 2.1 of [8] that anytree T possessing the eigenvalue 4 must have multiplicity m T (4) = n = m for some m ∈ N . Hence, Theorem 5.1 also asserts that trees spanned by m vertexdisjoint copies of K , form the only class of trees for which 4 is the 3 n / = m )th14 φ (a) φ φ (b) φ Figure 5.7: (a) The eigenvector φ corresponding to λ = φ corresponding to the maximum eigenvalue λ = . eigenvalue of T ; in other words, trees in this class are the only ones that have exactly n / = m ) eigenvalues ≥ K , , however, are not the only onesthat have an eigenvalue exactly equal to 4. For instance, Example 2.9 of [8],which has n = = · Z as shown in Figure 5.8, is non-isomorphic to any tree spanned by 9 vertex disjoint copies of K , ; yet it has λ = λ = (cid:44) Proposition 5.1.
If n ≤ , any tree possessing an eigenvalue exactly equal to 4must be spanned by vertex disjoint copies of K , .Proof. First of all, 4 must divide n , hence n = m with m = m =
2. If m = λ max = λ ≤ λ ≤
2. Hence, λ = T = K , using the same theorem. If m = λ max = λ ≤ λ ≤
4; and λ ≤
3; . . . If λ =
4, thenthe necessary and su ffi cient conditions for the equality in Theorem 5.1 state that T must be spanned by two vertex disjoint copies of K , , so we are done. Now westill need to show that λ cannot be 4. Let d be the degree of the highest degreevertex of a tree T under consideration. Then, we have d + ≤ λ < d + (cid:112) d − . (5.15)Here, the lower bound is due to Grone and Merris [7] and the upper bound is dueto Stevanovi´c [13]. Now, we need to check a few cases of the values of d .15 igure 5.8: Yet another tree Z (Example 2.9 of [8] with k =
4) that has an eigenvalue exactlyequal to 4. This is non-isomorphic to any tree spanned by vertex disjoint copies of K , such as theone shown in Figure 5.6(a). • If d =
2, then the upper bound in (5.15) is 4. Hence, λ = d ≥ • If d >
3, of course, the lower bound in (5.15) is greater than 4. Hence, λ cannot be 4 either. • Finally, if d =
3, then the above bounds are: 4 ≤ λ < . · · · . Can λ = T . That is, the degree of that vertex is n − =
7. Since d =
3, this cannot happen.Hence, for n =
8, the only possibility for a tree T to have an eigenvalue exactlyequal to 4 is the case when λ =
4, which happens if and only if T is spanned bytwo vertex disjoint copies of K , .It turns out, however, that proving the necessity for n >
11 using similar argu-ments quickly becomes cumbersome, even for the next step n =
12. At this point,we do not know whether there are other classes of trees than Z discussed above orthose spanned by vertex disjoint copies of K , that can have an eigenvalue exactlyequal to 4. Hence, identifying every possible tree that has an eigenvalue exactlyequal to 4 is an open problem. 16 . Implication of a Long Path on Eigenvalues In Section 4 we saw that for a graph that has a branch consisting of a long path,its Laplacian eigenvalue greater than 4 has the property that the correspondingeigenvector components along the branch must decay exponentially.Here we discuss a consequence of such a structure in terms of the eigenvalues.We consider a graph G formed by connecting two graphs G and G with a path G . Note that this is a more general graph than in Section 4 (which can be regardedas the case without G ). We show that if G is a long path then any eigenvaluegreater than 4 of the Laplacian of either of the two subgraphs G ∪ G and G ∪ G must be nearly the same as an eigenvalue of the Laplacian of the whole graph G . Theorem 6.1.
Let G be a graph obtained by connecting two graphs with a path,whose Laplacian L can be expressed asL = L E T E L E T E L , where E and E have -1 in the top-right corner and 0 elsewhere. L i is (cid:96) i × (cid:96) i fori = , , and L represents the path G , that is, a tridiagonal matrix with 2 onthe diagonals and -1 on the o ff -diagonals.Let (cid:101) λ > be any eigenvalue of the top-left ( (cid:96) + (cid:96) ) × ( (cid:96) + (cid:96) ) (or bottom-right ( (cid:96) + (cid:96) ) × ( (cid:96) + (cid:96) ) ) submatrix of L. Then there exists an eigenvalue λ of L suchthat | λ − (cid:101) λ | ≤ (cid:101) γ (cid:96) , (6.16) where (cid:101) γ : = (cid:101) λ − < .Proof. We treat the case where (cid:101) λ is an eigenvalue of the top-left ( (cid:96) + (cid:96) ) × ( (cid:96) + (cid:96) )part of L , which we denote by L . The other case is analogous.As in Theorem 4.2, we can show that any eigenvalue (cid:101) λ > L has itscorresponding eigenvector components decay exponentially along the path G .This means that the bottom eigenvector component is smaller than (cid:101) γ (cid:96) in absolutevalue (we normalize the eigenvector so that it has unit norm) where (cid:101) γ : = (cid:101) λ − < L = Q Λ Q T be an eigendecomposition where Q T Q = I and the eigen-values are arranged so that (cid:101) λ appears in the top diagonal of Λ . For notationalconvenience let (cid:96) : = (cid:96) + (cid:96) . Then, consider the matrix (cid:98) L = (cid:34) Q T I (cid:35) L (cid:34) Q I (cid:35) = (cid:34) Λ ve T e v T L (cid:35) , (6.17)17 igure 7.9: Counterexample graph for the conjecture. where e = (1 , , . . . , T ∈ R (cid:96) and v = ( v , . . . , v (cid:96) ) T ∈ R (cid:96) . Direct calculationsshow that v i = − q (cid:96) , i where q (cid:96) , i is the bottom component of the eigenvector q i of L corresponding to the i th eigenvalue. In particular, by the above argument wehave | q (cid:96) , | = | v | ≤ (cid:101) γ (cid:96) ( (cid:28) (cid:98) L , the only nonzeros are the diagonal(which is (cid:101) λ ), and the (1 , (cid:96) +
1) and ( (cid:96) + ,
1) entries, both of which are equal to v . Now, viewing the (1 , (cid:96) +
1) and ( (cid:96) + ,
1) entries of (cid:98) L as perturbations (write (cid:98) L = (cid:98) L + (cid:98) L where (cid:98) L is obtained by setting the (1 , (cid:96) +
1) and ( (cid:96) + ,
1) entriesof (cid:98) L to 0) and using Weyl’s theorem [6, Theorem 8.1.5] we see that there exists aneigenvalue λ of (cid:98) L (and hence of L ) that lies in the interval [ (cid:101) λ − (cid:107) (cid:98) L (cid:107) , (cid:101) λ + (cid:107) (cid:98) L (cid:107) ] = [ (cid:101) λ − | v | , (cid:101) λ + | v | ]. Together with | v | ≤ (cid:101) γ (cid:96) we obtain (6.16).Recall that (cid:101) γ (cid:96) decays exponentially with (cid:96) , and it can be negligibly small formoderate (cid:96) ; for example, for ( λ, (cid:96) ) = (5 ,
30) we have (cid:101) γ (cid:96) = . × − . Weconclude that the existence of a subgraph consisting of a long path implies thatthe eigenvalues λ >
7. On the Eigenvector of the Largest Eigenvalue
In view of the results in Section 4 it is natural to ask whether it is alwaystrue that the largest component of the eigenvector corresponding to the largesteigenvalue of a Laplacian matrix of a graph lies on the vertex of the highest degree.Here we show by a counterexample that this is not necessarily true.Consider for example a tree as in Figure 7.9, which is generated as follows:first we connect m copies of K , (equal to P ) as shown in Figure 5.6(a); then addto the right a comet S ( (cid:96), , , ,
1) as in Figure 3.4(b).Now for su ffi ciently large m and (cid:96) ( m , (cid:96) ≥ ffi cient), the largest compo-nent in the eigenvector φ corresponding to the largest eigenvalue of the resulting18aplacian L occurs at one of the central vertices of K , , not at the vertex of degree5 belonging to the comet.Let us explain how we came up with this counterexample. The idea is basedon two facts. The first is the discussion in Section 6, where we noted that a longpath G implies any eigenvalue larger than 4 must be close to an eigenvalue ofa subgraph G ∪ G or G ∪ G . Therefore, in the notation of Section 6, byconnecting two graphs ( G = mK , and G = K , , a star) with a path G such thatthe largest eigenvalue (cid:101) λ of L is larger than that of L , we ensure that the largesteigenvalue λ of L is very close to (cid:101) λ . The second is the Davis-Kahan sin θ theorem[5], which states that a small perturbation of size (cid:101) γ (cid:96) in the matrix (cid:98) L (recall theproof of Theorem 6.1) can only induce small perturbation also in the eigenvector:its angular perturbation is bounded by (cid:101) γ (cid:96) /δ , where δ is the distance between λ and the eigenvalues of (cid:98) L after removing its first row and column. Furthermore,the eigenpair ( (cid:101) λ, (cid:101) φ ) of (cid:98) L satisfies (cid:101) φ = (1 , , . . . , T , and the eigenvectors (cid:98) φ ( (cid:39) (cid:101) φ by Davis-Kahan) of (cid:98) L and φ of L corresponding to λ are related by φ = (cid:104) Q I (cid:105) (cid:98) φ ,which follows from (6.17). Therefore, φ has its large components at the verticesbelonging to G . In view of these our approach was to find two graphs G and G such that the highest degrees of the vertices of G and G are 4 and 5, respectively,and the largest eigenvalue of the Laplacian of G is larger than that of G .
8. Discussion
In this paper, we obtained precise understanding of the phase transition phe-nomenon of the combinatorial graph Laplacian eigenvalues and eigenvectors forstarlike trees. For a more complicated class of graphs including those representingdendritic trees of RGCs, we proved in Theorem 4.1 that the number of the eigen-values greater than or equal to 4 is bounded from above by the number of verticeswhose degrees are strictly higher than 2. In Theorem 4.2, we proved that if agraph has a branching path, the magnitude of the components of an eigenvectorcorresponding to any eigenvalue greater than 4 along such a branching path de-cays exponentially toward its leaf. In Remark 4.1, we also extended Theorem 4.2for the case of λ = n vertices ( n = m forsome m ∈ N ) spanned by m vertex disjoint copies of K , possesses an eigenvalueexactly equal to 4 in Corollary 5.1, and that such class of trees are the only onesthat can have an eigenvalue exactly equal to 4 if n ≤
11 in Proposition 5.1. Onthe other hand, for larger n , we pointed out that not only those spanned by vertex19isjoint copies of K , , but also a tree called Z discovered in [8] and shown inFigure 5.8 have an eigenvalue exactly equal to 4. A challenging yet interestingquestion is whether or not one can identify every possible tree that has an eigen-value 4.Another quite interesting question is Q4 raised in Introduction: “Can a simpleand connected graph, not necessarily a tree, have eigenvalues equal to 4?” Theanswer is a clear “Yes.” For example, the d -cube ( d > d -fold Carte-sian product of K with itself is known to have the Laplacian eigenvalue 4 withmultiplicity d ( d − /
2; see e.g., [1, Sec. 4.3.1].Another interesting example is a regular finite lattice graph in R d , d >
1, whichis simply the d -fold Cartesian product of a path P n shown in Figure 2.3 with itself.Such a lattice graph has repeated eigenvalue 4. In fact, each eigenvalue and thecorresponding eigenvector of such a lattice graph can be written as λ j ,..., j d = d (cid:88) i = sin (cid:18) j i π n (cid:19) (8.18) φ j ,..., j d ( x , . . . , x d ) = d (cid:89) i = cos j i π ( x i + ) n , (8.19)where j i , x i ∈ Z / n Z for each i , as shown by Burden and Hedstrom [2]. Note that(8.18) and (8.19) are also valid for d =
1. In that case these reduce to (2.2) thatwe already examined in Section 2.Now, determining m G (4), i.e., the multiplicity of the eigenvalue 4 of this latticegraph, is equivalent to finding the number of the integer solutions ( j , . . . , j d ) ∈ ( Z / n Z ) d to the following equation: d (cid:88) i = sin (cid:18) j i π n (cid:19) = . (8.20)For d =
1, there is no solution as we mentioned in Section 2. For d =
2, it is easy toshow that m G (4) = n − d = m G (4) behaves in a much more complicated manner, whichis deeply related to number theory. We expect that more complicated situationsoccur for d >
3. We are currently investigating this on regular finite lattices. Onthe other hand, it is clear from (8.19) that the eigenvectors corresponding to theeigenvalues greater than or equal to 4 on such lattice graphs cannot be localized orconcentrated on those vertices whose degree is higher than 2 unlike the tree case.20heorem 4.2 and Remark 4.1 do not apply either since such a finite lattice graphdo not have branching paths.Finally, we would like to note that even a simple path, such as the one shownin Figure 2.3, exhibits the eigenfunction localization phenomena if it has nonuni-form edge weights , which we recently observed numerically. We will report ourprogress on investigation of localization phenomena on such weighted graphs at alater date.
Acknowledgments
We thank the referees for their remarks and suggestions. This research waspartially supported by the following grants from the O ffi ce of Naval Research:N00014-09-1-0041; N00014-09-1-0318. A preliminary version of a part of thematerial in this paper [12] was presented at the workshop on “Recent developmentand scientific applications in wavelet analysis” held at the Research Institute forMathematical Sciences (RIMS), Kyoto University, Japan, in October 2010, and atthe 7th International Congress on Industrial and Applied Mathematics (ICIAM),held in Vancouver, Canada, in July 2011. References [1] Bıyıko˘glu T., Leydold J., Stadler P. F. (2007)
Laplacian Eigenvectors ofGraphs . Lecture Notes in Mathematics 1915. Springer, New York.[2] Burden R. L., Hedstrom G. W. (1972) The distribution of the eigenvalues ofthe discrete Laplacian.
BIT
Italian J. Pure Appl. Math.
Linear Algebra Appl.
SIAM J. Numer. Anal.
Matrix Computations . The JohnsHopkins Univ. Press, Baltimore, MD, 3rd edition.217] Grone R., Merris R. (1994) The Laplacian spectrum of a graph II.
SIAM J.Discrete Math.
SIAM J. Matrix Anal. Appl. k th Laplacian eigenvalue of a tree. J. Graph Theor.
Linear AlgebraAppl. / JSIAM Letters
RIMS Kokyuroku
Linear Algebra Appl.
SIAM Review
Gerˇsgorin and His Circles , Springer, New York.[16] Woei E. (2012) Ph.D. dissertation, Dept. Math., Univ. California, Davis. Inpreparation.[17] Zhang, X.-D., Luo, R. (2002) The spectral radius of triangle-free graphs.