THE SPECTRUM OF THE LAPLACIAN MATRIX OF A BALANCED 2 p -ARY TREE

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1 Proyecciones Vol 3, N o, pp , August 004 Universidad Católica del Norte Antofagasta - Chile THE SPECTRUM OF THE LAPLACIAN MATRIX OF A BALANCED p -ARY TREE OSCAR ROJO Universidad Católica del Norte, Chile Received : February 004 Accepted : June 004 Abstract Let p>1 be an integer We consider an unweighted balanced tree B p of levels with a root vertex of degree p, vertices from the level until the level ( 1) of degree p +1 and vertices in the level of degree 1 The case p =1it was studied in [8, 9, 10] We prove that the spectrum of the Laplacian ³ matrix L (B p ) is σ (L (Bp )) = j=1σ T (p) j where, for 1 j 1, T (p) j is the j j principal submatrix of the tridiagonal singular matrix T (p) 1 p 0 0 p p +1 p T (p) = 0 p 0 p +1 p 0 0 p p We derive that the multiplicity of each eigenvalue of T j, as an eigenvalue of L (B p ), is at least (p 1) ( j 1)p Moreover, we show that the multiplicity of the eigenvalue λ =1of L (B p ) is exactly (p 1) ( )p Finally,weprove that 3, 7 σ L B if and only if is a multiple of 3, that5 σ L B if and only if is an even number, and that no others integer eigenvalues exist for L B AMS classification: 5C50, 15A48 Keywords: Tree; balanced tree; binary tree; n ary tree; Laplacian matrix Wor supported by Fondecyt , Chile This research was conducted while the author was visitor at Centro de Modelamiento Matemático, Universidad de Chile, Chile,

2 13 Oscar Rojo 1 Introduction Let G be a graph with vertices 1,,, n Let d i be the degree of the vertex i Let A (G) be the adjacency matrix of G and let D (G) bethe diagonal matrix of vertex degrees The Laplacian matrix of G is L (G) = D (G) A (G) Clearly, L (G) is a real symmetric matrix From this fact and Geršgorin s theorem, it follows that its eigenvalues are nonnegative real numbers Moreover, since its rows sum to 0, 0 is the smallest eigenvalue of L (G) In [7], some of the many results nown for Laplacian matrices are given Fiedler [3] proved that G is a connected graph if and only if the second smallest eigenvalue of L (G) is positive This eigenvalue is called the algebraic connectivity of G and it is denoted by a (G) This concept has been studied by many authors In section 3 of [7], some results concerning a (G) and some of its many applications are presented Denote by B an unweighted balanced binary tree of levels The tree B 4 and our labeling for its vertices are illustrated below The number of vertices in B is n = = 1 The degree of the root vertex is while the rest of the vertices have degree 3, except the vertices in the level with degree 1, and there are two branches with the same number of vertices, say the left branch and right branch Using the labels 1,, 3,, n = 1, in this order, our labeling for the vertices of B [9] follows the steps:

3 The spectrum of the Laplacian matrix Label the vertices on the left branch from the bottom to the root vertex and, in each level, from the left to the right Label the vertices on the right branch from the root vertex to the bottom and, in each level, from the left to the right With this labeling the Laplacian matrix L (B ) becomes a symmetric persymmetric matrix, that is, a symmetric matrix with respect to the main diagonal as well as to the secondary diagonal In [9] by using properties for this type of matrices, among other results, we characterized completely the spectrum of L (B ):σ (L (B )) = j=1 σ (T j)where,for1 j 1, T j is the j j principal submatrix of the tridiagonal singular matrix T, T = Other results concerning to L (B ) can be found in [8] and [10] In [8], quite tight lower and upper bounds for the algebraic connectivity of B are given; and, in [10], we find the integer eigenvalues of L (B ) Here we consider an unweighted balanced tree of levels with a root vertex of degree p, vertices from the level until the level ( 1) of degree p + 1 and vertices in the level of degree 1 Denote by B p such a tree In particular, B 1 = B The number of vertices in B p is n =1+ p + p + + p( 1) = p 1 p 1 For =, B p is the star graph K 1,p and it is nown that the eigenvalues of L (K 1, p)are0, p + 1 and 1 with multiplicity p 1 Henceforth, we assume 3 In this paper, we characterize completely the spectrum of L B p : σ L B p ³ = j=1 σ T (p) j where, for 1 j 1, T (p) j is the j j principal submatrix of the tridiagonal singular matrix T (p),

4 134 Oscar Rojo T (p) = 1 p 0 0 p p +1 p 0 p 0 p +1 p 0 0 p p The Laplacian matrix L (B p ) and its eigenvalues We introduce the following notations: If all the eigenvalues of an n n matrix A arerealnumbers,wewrite λ n (A) λ n 1 (A) λ (A) λ 1 (A) J is the reversal matrix, that is, the matrix with ones in the secondary diagonal and zeros elsewhere Observe that J is the identity matrix The order of J will be clear from the context in which it is used I m is the identity matrix of order m m e m is the all ones column vector of dimension m For j =1,, 3,, 1, n j = ( j)p 1 For j =1,,, 3, C j is the bloc diagonal matrix defined by e p 0 0 e C j = p e p with n j+1 diagonal blocs The order of C j is n j n j+1 Since the root vertex degree of B p is an even number, we can distinguish two parts in B p with the same number of vertices, say the left part and the right part The tree B3 and our labeling for its vertices are shown below

5 The spectrum of the Laplacian matrix 135 According to our labeling the vertices on the left part of B3 are1,,3, 4, 5, 6, 7, 8 on the level 3 and 9, 10 on the level, 11 is the root vertex, and the vertices on the right part of B3 are 1, 13 on the level and 14, 15, 16, 17, 18, 19, 0, 1 on the level 3 The Laplacian matrix L B3 becomes where U = " I8 C 1 C 1 5I ³ L B3 = #,C 1 = U b 0 b T 4 b T J 0 Jb JUJ and b = In general, using the labels 1,, 3,, n = p 1 p, in this order, our 1 labeling for the vertices of B p follows the steps: 1 Label the vertices on the left part of B p from the bottom to the root vertex and, in each level, from the left to the right

6 136 Oscar Rojo Label the vertices on the right part of B p from the root vertex to the bottom and, in each level, from the left to the right As we expect, for this labeling the Laplacian matrix L B p becomes a symmetric persymmetric matrix More precisely (1) where L B p = U b 0 b T p b T J 0 Jb JUJ () U = I n1 C C1 T ( p +1)I n C 0 C T 0 ( p +1)I n C 0 0 C T ( p +1)I n 1 and h i T (3) b = 0 0 e p 1 We recall a basic fact on symmetric persymmetric matrices of order (p +1) (p +1)[] Lemma 1 Let A be a complex symmetric persymmetric (p +1) (p +1) matrix Then, A has the form U b VJ A = b T a b T (4) J, JV Rb JUJ where U and V are complex symmetric matrices of order p p, b is a p dimensional complex column vector and a = a p+1,p+1 is a complex scalar Moreover, if then F = " U + V b b T a σ (A) =σ (F ) σ (U V ) #

7 The spectrum of the Laplacian matrix 137 Theorem (5) σ L B p = σ (F ) σ (U) where (6) " F = # U b b T p with U and b as in () and (3), respectively Moreover, the smallest eigenvalue and the largest eigenvalue of U are, respectively, the algebraic connectivity of B (p) ³ and the second largest eigenvalue of L B (p) Also,the ³ largest eigenvalue of F is the largest eigenvalue of L B (p) Proof We apply Lemma 1 to L B p By comparing (4) and (1), we see that U is defined by (),V =0andb is defined by (3) Thus, by Lemma 1, (5) follows immediately The rest of the proof follows from (5) together with the fact that the eigenvalues of U interlace the eigenvalues of F Now, we search for the eigenvalues of U and F The following lemma is a slight variation of Lemma 3 in [9] and, because of this, its proof is omitted Lemma 3 Let a, b and c be real numbers Let β 1 = a, β j = b p β j 1,j=, 3,, 1, β j 1 6=0, and β = c p β 1 Let

8 138 Oscar Rojo M = ai n1 C C1 T bi n C 0 C T 0 bin C and N = 0 0 C bi n 1 ai n1 C C1 T bi n C n 0 C n bin C n 0 C T bi n 1 e p e T p 1 c Then, (a) If β j 6=0for j =1,,,, (7) det M = β ( 1)p 1 1 β ( )p 1 β 3 3p 1 β p 1 β 1 p 1 (b) detm 6= 0if and only if β j 6=0for j =1,,, 1 (c) If β j 6=0for j =1,,, 1, (8) det N = β ( 1)p 1 1 β ( )p 1 β 3 3p 1 β p 1 β 1 p 1 β (d) If β j 6=0for j =1,,, then det N 6= 0 If β j =0for some j, 1 j, thendet N =0 Also, if β 1 6=0and β =0then det N =0 Theorem 4 Let (9) (10) P 0 (λ) =1,P 1 (λ) =λ 1, P j (λ) =(λ p 1) P j 1 (λ) p P j (λ) for j =, 3,, 1

9 The spectrum of the Laplacian matrix 139 and (11) P (λ) =(λ p ) P 1 (λ) p P (λ) Hence, (a) If λ R is such that P j (λ) 6= 0for j =1,,,, then (1) det (λi U) = P (p 1) ( )p 1 1 (λ) P (p 1) ( 3)p 1 (λ) P (p 1) p 1 3 (λ) P (p 1) p 1 (λ) P p 1 1 (b) det(λi U) 6= 0 if and only if λ R is such that P j (λ) 6= 0 for j =1,,, 1 (c) If λ R is such that P j (λ) 6= 0for j =1,,, 1, then (λ) (13) det (λi F ) = P (p 1) ( )p 1 1 (λ) P (p 1) ( 3)p 1 (λ) P (p 1) p 1 3 (λ) P (p 1) p 1 (λ) P p (λ) P (λ) (d) If λ R is such that P j (λ) 6= 0,forj =1,,, then det (λi F ) 6= 0 If λ R is such that P j (λ) =0, for some j, 1 j, orifλ R is such that P (λ) =0then det (λi F )=0 Proof Let λ R WeapplyLemma3tothematricesM = λi U and N = λi F Then, a = λ 1, b= λ p 1andc = λ p (a) Suppose that P j (λ) 6= 0forj =1,,, For brevity, we write P j (λ) =P j Then, β 1 = λ 1= P 1 P 0 6=0andβ j =(λ p 1) p P j P j 1 = P j P j 1 6=0

10 140 Oscar Rojo From (7), det (λi U) = P ( 1)p 1 1 P ( )p 1 P ( )p 1 1 P 3 3p 1 P 4 3p 1 = P (p 1) ( )p 1 1 P (p 1) ( 3)p 1 P p 1 P p 1 3 P p 1 1 P p 1 P (p 1) p 1 3 P (p 1) p 1 P 1 p 1 Thus, (1) is proved (b) Suppose that P j (λ) 6= 0forj =1,,, 1 Then, β j = P j(λ) P j 1 (λ) 6= 0, for j =1,,, 1 From Lemma 3, part (b), it follows that det (λi U) 6= 0 Conversely, suppose that det (λi U) 6= 0 and P j (λ) = 0 for some j, 1 j 1 Since P 0 (λ) =16= 0, we may assume P j 1 (λ) 6= 0and P j(λ) P j 1 (λ) P j (λ) =0 Then β j = =0andthusdet(λI U) =0, which is a contradiction (c) Suppose that P j (λ) 6= 0forj =1,,, 1 Then, as in part (a), β j = P j(λ) P j 1 (λ) 6=0forj =1,,, 1 Moreover, β =(λ p ) From (7),(8) and (1), det (λi F ) = P (p 1) ( )p 1 1 P (p 1) ( 3)p 1 = P (p 1) ( )p 1 1 P (p 1) ( 3)p 1 p =(λ p ) p P (λ) = P (λ) β 1 P 1 (λ) P 1 (λ) P (p 1) p 1 3 P (p 1) p 1 3 P (p 1) p 1 P p 1 1 P P 1 P (p 1) p 1 P p P Thus, (15) is proved (d) Suppose P j (λ) 6= 0, for j =1,,, Then, β j = P j(λ) P j 1 (λ) 6=0 for j =1,,, 1andβ = P (λ) P 1 (λ) 6=0 From Lemma 3, part (d), det (λi F ) 6= 0 Suppose P j (λ) =0forsomej, 1 j Since P 0 (λ) = 1 6= 0, we may suppose P j 1 (λ) 6= 0 and P j (λ) = 0 Thus, β j = P j(λ) P j 1 (λ) = 0 and therefore det (λi F )=0 Finally, if P 1 (λ) 6= 0 and P (λ) =0thenβ = = 0 and hence det (λi F )=0 P (λ) P 1 (λ) An immediate consequence of theorem and theorem 4 is

11 The spectrum of the Laplacian matrix 141 Corollary 5 σ (U) = 1 j=1 {λ R : P j (λ) =0}, σ (F )= j=1 {λ R : P j (λ) =0} and (14) det λi L B p = P (p 1) ( )p 1 (λ) P (p 1) ( 3)p P (p 1) p 3 (λ) P (p 1) p (λ) (λ) P p 1 1 (λ) P (λ) Lemma 6 For j =1,, 3,, 1, let T j be the j j principal submatrix of the tridiagonal matrix T 1 p 0 0 p p +1 p 0 p (15) T = 0 p +1 p 0 0 p p Then, det (λi T j )=P j (λ),j=1,,, Proof It is well nown (see for instance [1, page 9]) that the characteristic polynomials p j of the j j principal submatrix of the symmetric tridiagonal matrix T, a 1 b b 1 a b 0 b T =, 0 a 1 b b 1 a

12 14 Oscar Rojo satisfy the three-term recursion formula p j (λ) =(λ a j ) p j 1 (λ) b j 1p j (λ) with p 0 (λ) =1andp 1 (λ) =λ a 1 In our case, a 1 =1,a j = p +1 for j =, 3,, 1, a = p and b j = p for j =1,,, 1 For these values, the above recursion formula gives the polynomials P j, j =0, 1,,, Theorem 7 Let p LetT j, j =1,,, definedinlemma6then, (a) σ (U) = 1 j=1 σ (T j) (b) σ (F )= j=1 σ (T j) (c) σ L B p = j=1 σ (T j ) (d) For j =1,,, 1, the multiplicity of each eigenvalue of the matrix T j, as an eigenvalue of L B p,isatleast ( p 1) ( j 1)p Proof (a), (b) and(c) are immediate consequences of Corollary 5 and Lemma 6 Moreover, since ³ ³ det λi L B (p) = P (p 1) ( )p 1 (λ) P (p 1) ( 3)p (λ) P (p 1) p 3 (λ) P (p 1) p (λ) P (p 1) 1 (λ) P (λ) and det (λi T j )=P j (λ),j=1,,, 1, we have that the multiplicity of each eigenvalue of the matrix T j, as an eigenvalue of L (B ), is at least (p 1) ( j 1)p We recall the following interlacing property [4]: Let T be a symmetric tridiagonal matrix with nonzero codiagonal entries and λ (j) i be the i th smallest eigenvalue of its j j principal submatrix Then, λ (j+1) j+1 <λ (j) j <λ (j+1) j <<λ (j+1) i+1 <λ (j) i <λ (j+1) i < <λ (j+1) <λ (j) 1 <λ (j+1) 1

13 The spectrum of the Laplacian matrix 143 From this interlacing property and theorem 7, we have Theorem 8 (a) σ (T j 1 ) σ (T j )=φ for j =, 3,, (b) The largest eigenvalue of T is the largest eigenvalue of L B p (c) The smallest eigenvalue of T 1 is the algebraic connectivity of B p (d) The largest eigenvalue of T 1 is the second largest eigenvalue of L B p Example 1 Let p =and =6 Then, n = 1365 The eigenvalues of L B 6, rounded to four decimal places, are given in the following table: T 1 : 1 T : T 3 : T 4 : T 5 : S 6 : The integer eigenvalues of L (B p ) The purpose of this section is to search for the integer eigenvalues of L B p The case p = 1 was studied in [10] Then, we assume p Since B p is connected, 0 is a simple eigenvalue of B p From theorem 7, Geršgorin s theorem and the fact that L B p is a positive semidefinite matrix, it follows that if λ is an eigenvalue of L B p then 0 λ p +1+ p Since P 1 (1) = 0, it follows that 1 is an eigenvalue of L B p Next, we prove that P j (1) 6= 0forj =1,,, Keep in mind that we are assuming p From (10), we obtain the second order homogeneous linear difference equation (31) P j (1) + p P j 1 (1) + p P j (1) = 0 for j =1,,, 1, with the initial conditions P 0 (1) = 1 and P 1 (1) = 0 The characteristic equation of (31) is (3) u + p u + p =0

14 144 Oscar Rojo We have p p = p p 1 0 because p Thus, the roots of (3) are α = p 1 + p p β = p 1 p p Let p = Then α = β = and the general solution of (31) is [6]: P j (1) = c 1 ( ) j + c j ( ) j From the initial conditions, we get c 1 =1andc = 1 Hence Moreover P j (1) = ( ) j (1 j),j=1,,, 1 P (1) = 3P 1 (1) 4P (1) = 3( ) 1 ( ) 4( ) (3 ) = 3( ) 1 ( ) ( ) 1 (3 ) = ( ) 1 (5 6) 6= 0 Therefore, if p =thenp j (1) = 0 if and only if j =1 Suppose now p> In this case, β<α<0 and the general solution of (31) is [6]: P j (1) = c 1 α j + c β j From the initial conditions, c 1 + c =1andc 1 α + c β =0,weget c 1 = β β α and c = α β α Hence Moreover β β α αj β α α βj = β α 1 α j β αβ j P j (1) = P (1) = (1 p ) P 1 (1) p P (1) = P 1 (1) p (P 1 (1) + P (1)) 6= 0 Hence, if p>1thenp j (1) = 0 if and only if j =1 Theorem 9 If p then the multiplicity of λ =1as an eigenvalue of L B p is ( p 1) ( )p

15 The spectrum of the Laplacian matrix 145 Proof It is an immediate consequence of (14) and the fact that P j (1) 6= 0forallj ifp We recall an important fact concerning an integer eigenvalue of a tree [5]: If λ>1 is an integer eigenvalue of the Laplacian matrix of a tree with n vértices then λ is a simple eigenvalue and exactly divides n Since, in addition, n = p 1 p is odd for any and p, the only possible 1 integer eigenvalues of L B p greater than 1 are restricted to the odd positive integers not exceeding p +1+ p Lemma 10 Let p If j =1,,, 1thenP j (λ) 6= 0 for all integer λ>1 Proof We already now that the multiplicity of each eigenvalue of the matrix T j, as an eigenvalue of L (B ), is at least (p 1) ( j 1)p and that det (λi T j )=P j (λ) Moreover, for j =1,,, 1, (p 1) ( j 1)p > 1 because p Hence, if λ>1 is an integer number such that P j (λ) =0 then λ is an eigenvalue of T j with multiplicity greater than 1 This is a contradiction because the integer eigenvalues greater than 1 of a tree are simple Therefore in searching for the integer eigenvalues of L B p greater than 1, we have to loo for the integer roots of the equation P (λ) =0 From now on, we consider p = In this case, we may restrict the values of λ to the interval (1, 9] From the recursion formula P j (λ) =(λ 5) P 1 (λ) 4P (λ),j=,, 1 we obtain the second order homogeneous linear difference equation (33) P j (λ) (λ 5) P 1 (λ)+4p (λ) =0 with P 0 (λ) = 1 and P 1 (λ) = λ 1 The corresponding characteristic equation is (34) u (λ 5) u +4=0 Let λ =9 For this value the equation (34) becomes (u ) =0

16 146 Oscar Rojo and the solution of (33) is Hence P j (9) = j +3j j = j (1 + 3j) P (9) = 4P 1 (9) 4P (9) = +1 (3 ) (3 5) = (3 +1)6= 0 Thus, we may consider λ (0, 9) Then, λ 10λ +9 < 0 and the roots of (34) is a pair of conjugate complex root : u 1 = λ 5 and u = λ λ λ 9 i 10λ λ 9 i where i = 1 Therefore, the general solution of (33) is [6, p 9] : where P j (λ) =ρ j (c 1 cos φ (λ) j + c sin φ (λ) j) µ ρ = λ 5 Ã! 10λ λ = and cos φ (λ) = λ 5 10λ λ and sin φ (λ) = 9 (35) 4 4 From the initial conditions, we obtain c 1 =1and Hence, for j =, 3,, 1, (cosφ (λ)+c sin φ (λ)) = λ 1 P j (λ) = j µcos φ (λ) j + λ +3 10λ λ 9 sin φ (λ) j We already now that the possible integer eigenvalues greater than 1 of L B are 3, 5and7 Theorem 11 (a) 3 L B if and only if is a multiple of 3 (b) 7 L B if and only if is a multiple of 3 (b) 5 L B if and only if is even

17 The spectrum of the Laplacian matrix 147 Proof (a) Letλ =3 Then, cos φ (3) = 1 and sin φ (3) = 3 Hence, φ (3) = π 3 Moreover P (3) = P 1 ³ (3) 4P (3) = 1 cos ( 1) φ (3) + 3sin( 1) φ (3) ³ cos ( ) φ (3) + 3sin( ) φ (3) Ã! = 1 cos ( 1) φ (3) + 3sin( 1) φ (3) + cos ( ) φ (3) + 3sin( ) φ (3) cos φ (3) cos φ (3) + sin φ (3) sin φ (3) = 1 + 3sinφ (3) cos φ (3) 3cosφ (3) sin φ (3) + cos φ (3) cos φ (3) + sin φ (3) sin φ (3) + 3sinφ (3) cos φ (3) 3cosφ (3) sin φ (3) 1 = 1 cos φ (3) + 1 3sinφ (3) 3 sin φ (3) 3 cos φ (3) cos φ (3) 3sinφ (3) 3sinφ (3) + 3 cos φ (3) = 3sinφ (3) Consequently, P (3) = 0 if and only if = πl integer l φ(3) =3l some positive (b) Letλ =7 Then, cos φ (7) = 1 and sin φ (7) = 3 Hence, φ (7) = π 3 Moreover P (7) = 3P 1 (7) 4P (7) = µ3cos( 1 1) φ (7) + 5 sin ( 1) φ (7) 3 µcos ( ) φ (7) + 5 sin ( ) φ (7) 3 Ã! 3cos( 1) φ (7) + 5 3sin( 1) φ (7) = 1 cos( ) φ (7) 10 3 sin ( ) φ (7) 3cosφ (7) cos φ (7) + 3 sin φ (7) sin φ (7) = sinφ (7) cos φ (7) 5 3cosφ (7) sin φ (7) cosφ (7) cos φ (7) sinφ (7) sin φ (7) 10 3 sin φ (7) cos φ (7) cos φ (7) sin φ (7)

18 148 Oscar Rojo = 1 3 cos φ (7) + 3 3sinφ (7) sin φ (7) 15 cos φ (7) + cos φ (7) 3sinφ (7) sin φ (7) + 5 cos φ (7) = sinφ (7) πl φ(7) It follows, P (7) = 0 if and only if = =3l some for positive integer l (c) Letλ =5 Then, cos φ (5) = 0 and sin φ (5) = 1 Hence, φ (5) = π Moreover P (5) = P 1 (5) 4P (5) = 1 (cos ( 1) φ (5) + sin ( 1) φ (5)) (cos ( ) φ (5) + sin ( ) φ (5)) Ã! = 1 cos ( 1) φ (5) + sin ( 1) φ (5) cos( ) φ (5) 4sin( ) φ (5) cos φ (5) cos φ (5) + sin φ (5) sin φ (5) = 1 + sin φ (5) cos φ (5) cosφ (5) sin φ (5) cosφ (5) cos φ (5) sinφ (5) sin φ (5) 4sinφ (5) cos φ (5) + 4 cos φ (5) sin φ (5) = 1 (sin φ (5) cosφ (5) + cos φ (5) + 4 sin φ (5)) = 1 5sinφ (5) It follows, P (5) = 0 if and only if = l πl φ(5) =l for some positive integer References [1] L N Trefethen and D Bau, III, Numerical Linear Algebra, Society for Industrial and Applied Mathematics, (1997) [] A Cantoni and P Butler, Eigenvalues and eigenvectors of symmetric centrosymmetric matrices, Linear Algebra Appl 13, pp 75-88, (1976)

19 The spectrum of the Laplacian matrix 149 [3] M Fiedler, Algebraic connectivity of graphs, Czechoslova Math J, 3: pp , (1973) [4] G H Golub and C F Van Loan, Matrix Computations, d ed, Baltimore: Johns Hopins University Press, (1989) [5] R Grone, R Merris and V S Sunder, The Laplacian spectrum of a graph, SIAM J Matrix Ana Appl 11 (), pp 18-38, (1990) [6] F B Hildebrand, Finite-Difference Equations and Simulations, Prentice-Hall,Inc, Englewood Cliffs, NJ, (1968) [7] R Merris, Laplacian Matrices of Graphs: A Survey, Linear Algebra Appl a Appl 197, 198: pp , (1994) [8] J J Molitierno, M Neumann and B L Shader, Tight bounds on the algebraic connectivity of a balanced binary tree, Electronic Journal of Linear Algebra, Vol 6, pp 6-71, March (000) [9] O Rojo, The spectrum of the Laplacian matrix of a balanced binary tree, Linear Algebra Appl 349, pp 03-19, (00) [10] O Rojo and M Peña, A note on the integer eigenvalues of the Laplacian matrix of a balanced binary tree, Linear Algebra Appl 36, pp (003) Oscar Rojo Departamento de Matemáticas Universidad Católica del Norte Casilla 180 Antofagasta Chile orojo@ucncl