Spectra of Infinite Graphs with Tails

Linear and Multilinear Algebra

ISSN: 0308-1087 (Print) 1563-5139 (Online) Journal homepage: http://www.tandfonline.com/loi/glma20

Spectra of infinite graphs with tails

L. Golinskii

To cite this article: L. Golinskii (2016): Spectra of infinite graphs with tails, Linear and Multilinear Algebra, DOI: 10.1080/03081087.2016.1155529

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Published online: 09 Mar 2016.

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Download by: [Leonid Golinskii] Date: 10 March 2016, At: 02:03 Linear and Multilinear Algebra, 2016 http://dx.doi.org/10.1080/03081087.2016.1155529

Spectra of inﬁnite graphs with tails L. Golinskii∗

Mathematics Division, Institute for Low Temperature Physics and Engineering, Kharkov, Ukraine Communicated by S. Pati (Received 17 April 2015; accepted 20 January 2016)

We compute explicitly (modulo solutions of certain algebraic equations) the spectra of infinite graphs obtained by attaching one or several infinite paths to some vertices of given finite graphs. The main result concerns a canonical form for the adjacency matrix of such infinite graphs, and the algorithm of its calculation. The argument relies upon the spectral theory of eventually free Jacobi matrices. We also study some other couplings of infinite graphs (stars and Bethe–Caley trees).

Keywords: inﬁnite graphs; adjacency operator; spectrum; Jacobi matrices of ﬁnite rank; Jost function

AMS Subject Classiﬁcations: Primary: 05C63; Secondary: 05C76; 47B36; 47B15; 47A10

1. Introduction and main results We begin with rudiments of the graph theory. For the sake of simplicity, we restrict ourselves with simple, connected, undirected, finite or infinite (countable) weighted graphs, although the main result holds for weighted multigraphs and graphs with loops as well. We will label V() N ={, ,...} {v} ={ }ω ω ≤∞ the vertex set by positive integers 1 2 , v∈V j j=1, .The symbol i ∼ j means that the vertices i and j are incident, i.e. {i, j} belongs to the edge set E(). A graph is weighted if a positive number dij (a weight) is assigned to each edge Downloaded by [Leonid Golinskii] at 02:03 10 March 2016 {i, j}∈E(). In case dij = 1 for all i, j, the graph is unweighted. The degree (valency) of a vertex v ∈ V() is a number γ(v)of edges emanating from v. A graph is said to be locally finite, if γ(v)<∞ for all v ∈ V(), and uniformly locally finite, if supV γ(v)<∞. The spectral graph theory deals with the study of spectra and spectral properties of certain matrices related to graphs (more precisely, operators generated by such matrices in n 2 2 the standard basis {ek}k∈N and acting on the corresponding Hilbert spaces C or = (N)). One of the most notable among them is the adjacency matrix A() ω d , {i, j}∈E(); A() =[a ] , a = ij (1.1) ij i, j=1 ij 0, otherwise.

∗ Email: [email protected]

The corresponding adjacency operator will be denoted by the same symbol. It acts as A() ek = a jk e j , k ∈ N. (1.2) j∼k Clearly, A() is a symmetric, densely defined linear operator, whose domain contains the set of all finite linear combinations of the basis vectors. The operator A() is bounded and self-adjoint in 2, as long as the graph is uniformly locally finite. Whereas the spectral theory of finite graphs is very well established (see, e.g. [1–4]), the corresponding theory for infinite graphs is in its infancy. We refer to [5–7] for the basics of this theory. In contrast to the general consideration in [6], our goal is to carry out a complete spectral analysis (canonical models for the adjacency operators and computation of the spectrum) for a class of infinite graphs which loosely speaking can be called ‘finite graphs with tails attached to them’. To make the notion precise, we define first an operation of coupling well known for finite graphs (see, e.g. [4, Theorem 2.12]).

Deﬁnition 1.1 Let k, k = 1, 2, be two weighted graphs with no common vertices, with the vertex sets and edge sets V(k) and E(k), respectively, and let vk ∈ V(k). A weighted graph = 1 + 2 will be called a coupling by means of the bridge {v1,v2} of weight d > 0if

V() = V(1) ∪ V(2), E() = E(1) ∪ E(2) ∪{v1,v2}. (1.3)

So we join 2 to 1 by the new edge of weight d between v2 and v1.

( ) ={ , ,..., } ( ) ={ }ω If the graph 1 is ﬁnite, V 1 1 2 n , and V 2 j j=1, we can with no loss of generality put v1 = n, v2 = 1, so the adjacency matrix A() is written as a block matrix ⎡ ⎤ 000... ⎢ . . . ⎥ ( ) ⎢ . . . ⎥ () = A 1 Ed , = ⎢ . . . ⎥ . A ∗ ( ) Ed ⎣ ⎦ (1.4) Ed A 2 000... d 00...

If 2 = P∞({a j }), the one-sided weighted infinite path, we can view the coupling = 1 + P∞({a j }) as a finite graph with the tail. This is exactly the class of graphs we will be dealing with in the paper. Each such graph has a finite number of essential ramification Downloaded by [Leonid Golinskii] at 02:03 10 March 2016 nodes (a vertex v is an essential ramification node if γ(v)≥ 3, see [8]).

A special class of inﬁnite matrices will play a crucial role in what follows. Under Jacobi or tridiagonal matrices, we mean here one-sided inﬁnite matrices of the form Linear and Multilinear Algebra 3 ⎡ ⎤ b1 a1 ⎢ ⎥ ⎢a1 b2 a2 ⎥ = ({ }, { }) = ⎢ . ⎥ , ∈ R, > . J J b j a j ⎢ .. ⎥ b j a j 0 (1.5) ⎣ a2 b3 ⎦ ......

They generate linear operators called the Jacobi operators on the Hilbert space 2(N).The matrix ⎡ ⎤ 01 0 0 ⎢ ⎥ ⎢10 1 0 ⎥ := ⎢ ⎥ J0 ⎣01 0 1 ⎦ (1.6) ......

called a discrete Laplacian or a free Jacobi matrix is of particular interest in the sequel. = ({β(k)}, {α(k)}) = , Given two Jacobi matrices Jk J j j , k 1 2, the matrix J2 is called a truncation of J1 (and J1 is an extension of J2)if β(2) = β(1) ,α(2) = α(1) , ∈ N, j j+q j j+q j

for some q ∈ N. In other words, J2 is obtained from J1 by deleting the ﬁrst q rows and columns. If J2 = J0, J1 is said to be a Jacobi matrix of ﬁnite rank or an eventually free Jacobi matrix. The Jacobi matrices arise in the spectral graph theory, thanks to the relation for the adjacency matrix A(P∞({a j })) of the weighted path

A(P∞({a j })) := J({0}, {a j }). (1.7)

In case of the unweighted path, that is, a j ≡ 1, we have

A(P∞) = J0. (1.8)

Note that the adjacency matrix for the ﬁnite (unweighted) path Pm with m vertices is the ﬁnite Jacobi matrix J({0}, {1}) of order m. The spectrum of this matrix is well known [2, p.9] π j m σ(Pm) = 2 cos . (1.9) m + 1 j=1

Downloaded by [Leonid Golinskii] at 02:03 10 March 2016 It follows from (1.4) that for an arbitrary ﬁnite weighted graph G ( ) ( + P ({ })) = A G Ed . A G ∞ a j ∗ ({ }, { }) (1.10) Ed J 0 a j We suggest here a ‘canonical’ form for such block matrices and the algorithm of their reducing to this form.

Theorem 1.2 Let A be a block matrix in 2, ⎡ ⎤ 000... ⎢ . . . ⎥ A ⎢ . . . ⎥ = Ed , = ⎢ . . . ⎥ , > , A ∗ Ed ⎣ ⎦ d 0 (1.11) Ed J 000... d 00... 4 L. Golinskii

A =[ ]n = ({β }, {α }) where aij i, j=1 is a real symmetric matrix of order n, J J j j is a Jacobi matrix. Then, A can be reduced to the block diagonal form A A (1.12) J

where A is a real symmetric matrix of order n − k, for some 1 ≤ k ≤ n, and the Jacobi matrix J is an extension of J. In other words, there is a unitary operator U on 2 such that − U 1 AU = A⊕ J. (1.13)

Corollary 1.3 Given a ﬁnite weighted graph G, the adjacency operator of the coupling = G + P∞({a j }) is unitarily equivalent to the orthogonal sum − U 1 A()U = F() ⊕ J() (1.14)

of a ﬁnite-dimensional operator F() and a Jacobi operator J() which is an extension of J({0}, {a j }).

If P∞ is unweighted, then, by definition, J() is of finite rank. We call F() a finite-dimensional component of the coupling , and J() its Jacobi component. In particular, the spectrum of = G + P∞ is σ() =[− , ]∪σ (), σ () ={λ }ω ,ω<∞, 2 2 d d j j=1 (1.15) the discrete spectrum of , is a union of the eigenvalues of F() and J(), see Section 2. We explain the algorithm of finding the finite-dimensional and Jacobi components in Section 3, and illustrate it on a number of examples in the further sections. The spectral analysis of eventually free Jacobi matrices is discussed in Section 2. To get some flavour of the results on the structure of spectra for certain graphs with tails, consider the following example. Let Cm be a cycle of order m, := Cm + P∞. This graph can be viewed as a ‘kite’ with infinite tail.

Proposition 1.4 The spectrum of the graph = Cm + P∞ is

σ(Cm + P∞) =[−2, 2]∪σd ()

Downloaded by [Leonid Golinskii] at 02:03 10 March 2016 with the discrete component 2πk l m − 1 σd () = 2 cos {λ1(m), λ2(m)}, l = , (1.16) m k=1 2 where 1 λ j (m) = x j (m) + , j = 1, 2, −1 < x2(m)<0 < x1(m)<1 x j (m) are real roots of the polynomial p(x) = xm + 2x2 − 1.

We observe here a number of eigenvalues lying on the essential (absolutely continuous) spectrum [−2, 2] (the hidden spectrum in the terminology of [9]), and two eigenvalues off [−2, 2]. Linear and Multilinear Algebra 5

Remark 1.5 Given a finite graph G, one can attach p ≥ 1 copies of the infinite path P∞ to some vertex v ∈ V(G). Although the graph thus obtained is not exactly the coupling in the sense of Definition 1.1, its adjacency operator acts similarly to one for the coupling. Indeed, it is not hard to see that A(G) E√ p−1 () = p ⊕ ⊕ . A √∗ J0 (1.17) E p J0 i=1 Hence, Theorem 1.2 applies, and the spectral analysis of such graph can be accomplished.

Surprisingly enough, the case when p ≥ 1 inﬁnite rays are attached to each vertex of a ﬁnite graph G is easy to work out, and the spectrum of such graph can be found explicitly in terms of the spectrum of G. Denote such graph by = G + P∞(p).

Theorem 1.6 Given a ﬁnite graph G with vertices 1, 2,...,n and the spectrum σ(G) = {λ }n = + P ( ) ∈ N () j j=1,let G ∞ p ,p . Then, the adjacency operator A is unitarily equivalent to the orthogonal sum n √ (p−1)n A() ⊕ J(λ j , p) ⊕ ⊕ J0 , = = √ j 1 √i 1 J(λ j , p) := J({λ j , 0, 0,...}, { p, 1, 1,...}). (1.18) The spectrum of is n √ σ() =[−2, 2] σd (J(λ j , p)). (1.19) j=1

For p = 1, this result is proved in [10]. The spectral theory of inﬁnite graphs with one or several rays attached to certain ﬁnite graphs was initiated in [10–13], wherein several particular examples of unweighted (background) graphs are examined. The spectral analysis of similar graphs appeared earlier in the study of thermodynamical states on complex networks.[9] We argue in the spirit of [14–16] and supplement to the list of examples. The general canonical form for the adjacency matrices of such graphs and the algorithm of their reducing to this form suggested in the paper apply to a wide class of couplings (not only graphs with tails, see Section 6), and also Downloaded by [Leonid Golinskii] at 02:03 10 March 2016 to Laplacians on graphs of such type.

2. Jacobi matrices of ﬁnite rank and their spectra For the class of eventually free Jacobi matrices, the complete spectral analysis is available at the moment, see [17,18]. A basic object known as the perturbation determinant [19]isa key ingredient of perturbation theory. Given bounded linear operators T0 and T on the Hilbert space such that T − T0 is a nuclear operator, the perturbation determinant is deﬁned as

−1 L(λ; T, T0) := det(I + (T − T0)R(λ, T0)), R(λ, T0) := (T0 − λ) (2.1)

is the resolvent of operator T0, an analytic operator function on the resolvent set ρ(T0). 6 L. Golinskii

The perturbation determinant is designed for the spectral analysis of the perturbed operator T , once the spectral analysis for T0 is available. In particular, the essential spectra of T and T0 agree, and the discrete spectrum of T is exactly the zero set of the analytic function L on ρ(T0), at least if the latter is a domain, i.e. a connected, open set in the complex plane. In the simplest case, rank(T − T0)<∞ the perturbation determinant is the standard ﬁnite-dimensional determinant. Indeed, now p p ∗ (T − T0)h = h,ϕk ψk,(T − T0)R(λ, T0)h = h, R (λ, T0)ϕk ψk, k=1 k=1 so L can be computed by the formula (see, e.g. [19, Section IV.1.3]) (λ; , ) = [δ + (λ, )ψ ,ϕ ]p . L T T0 det ij R T0 i j i, j=1 (2.2)

Our particular concern is T0 = J0, the free Jacobi matrix. Its resolvent matrix in the standard basis in 2 is given by (see, e.g. [20]) |i− j| i+ j 1 ∞ z − z R z + , J =[r (z)] , = , r (z) = , z ∈ D. (2.3) z 0 ij i j 1 ij z − z−1 If T = J is a Jacobi matrix of ﬁnite rank p, we end up with the computation of the ordinary determinant (2.2)oforderp. It is instructive for the further usage to compute two simplest perturbation determinants for rank (J − J0) = 1 and 2.

Example 2.1 Let J = J({b j }, {1}) : b j = 0, j = q,

so J − J0 = ·, eq bq eq .By(2.3) and (2.2), 2q 1 z − 1 L(z) := L z + ; J, J = 1 + bqrqq(z) = 1 − bq z . (2.4) z 0 z2 − 1 Similarly, let J = J({0}, {a j }) : a j = 1, j = q, Downloaded by [Leonid Golinskii] at 02:03 10 March 2016 so J − J0 = ·, eq (aq − 1) eq+1 + ·, eq+1 (aq − 1) eq , and again 1 + (aq − 1) rq,q+1(z)(aq − 1) rqq(z) L(z) = (aq − 1) rq+1,q+1(z) 1 + (aq − 1) rq+1,q (z) z2q − 1 = 1 + (1 − a2)z2 . (2.5) q z2 − 1 In the Jacobi matrices setting, there is yet another way of computing perturbation determinants based on the so-called Jost solution and Jost function (see, e.g. [7, Section 3.7]). Consider the basic recurrence relation for the Jacobi matrix J 1 a − y − + b y + a y + = z + y , z ∈ D, n ∈ N, (2.6) n 1 n 1 n n n n 1 z n Linear and Multilinear Algebra 7

where we put a0 = 1. Its solution yn = un(z) is called the Jost solution if −n lim z un(z) = 1, z ∈ D. (2.7) n→∞

In this case, the function u = u0 is called the Jost function. The Jost solution certainly exists for ﬁnite rank Jacobi matrices. The Jost function is now an algebraic polynomial, called the Jost polynomial. Indeed, let

aq+1 = aq+2 = ...= 1, bq+1 = bq+2 = ...= 0.

k One can put uk(z) = z , k = q + 1, q + 2,... and then determine uq , uq−1,...,u0 consecutively from (2.6). So,

q aq uq (z) = z , ( ) = α q+1 − q + q−1,α:= − 2, aq−1aq uq−1 z q z bq z z q 1 aq (2.8) etc., and in general, k q+ j uq−k(z) = βq, j z , k = 0, 1,...,q,βq, j ∈ R,βq,−k = 1. j=−k In particular, for q = 1 2 a1 u(z) = α1 z − b1z + 1, (2.9) and for q = 2

4 3 2 a1a2 u(z) = α2 z − (b2 + b1α2) z + (α1 + α2 + b1b2) z − (b1 + b2) z + 1. (2.10) The relation between the perturbation determinant and the Jost function is given by ∞ ( ) = −1 · ( ), u z a j L z (2.11) j=1 see, e.g. [20], and such recursive way of computing perturbation determinants is sometimes far easier than computing ordinary determinants (2.2), especially for large enough ranks of perturbation. Downloaded by [Leonid Golinskii] at 02:03 10 March 2016

Example 2.2 Let J = J({b j }, {a j }) be a Jacobi matrix such that

b j = 0, j = 1, a j = 1, j = q.

q+ j We have uq+ j (z) = z , j = 1, 2 ...,

q q+1 q−1 aq uq (z) = z , aq uq−1(z) = αq z + z , q+2 q q−2 aq uq−2 = αq (z + z ) + z , and, by induction, 2k − q−k+2 z 1 aq uq−k(z) = αq z , k = 1, 2,...,q − 1. (2.12) z2 − 1 8 L. Golinskii

Next, for q = 1, we have exactly (2.9), so let q ≥ 2. The recurrence relation (2.6) with n = 1 gives 1 a u(z) + b a u (z) + a u (z) = z + a u (z), q 1 q 1 q 2 z q 1 and so we come to the following expression for the Jost polynomial ( )( 2 − ) = α ( − ) 2q+1 − 2 3 + 2 2 + − . aq u z z 1 q z b1 z b1aq z aq z b1z 1 (2.13)

Similarly, for the Jacobi matrix J = J({b j }, {a j }) with

b j = 0, j = q; a j = 1, j = 1

one has 2q+1 + α 2q−1 − α 3 − z 1z 1z z 2 a u(z) =−bq + α z + 1. (2.14) 1 z2 − 1 1

For the Jacobi matrix J = J({0}, {a j }) with a j = 1, j = 1, q, the Jost polynomial is given by 2q+2 + α 2q − α 4 − 2 z 1z 1z z 2 a aq u(z) = αq + α z + 1. (2.15) 1 z2 − 1 1 The spectral theorem for ﬁnite rank Jacobi matrices due to Damanik and Simon [17] provides a complete description of the spectral measure of such matrices.

Theorem DS Let J = J({b j }, {a j }) be a Jacobi matrix of ﬁnite rank

aq+1 = aq+2 = ...= 1, bq+1 = bq+2 = ...= 0,

and u = u0(J) be its Jost polynomial. Then,

• u is a real polynomial of degree deg u ≤ 2q, deg u = 2q if and only if aq = 1. • All roots of u in the unit disk D are real and simple, u(0) = 0. A number λ j is an eigenvalue of J if and only if 1 λ j = z j + , z j ∈ (−1, 1), u(z j ) = 0. (2.16) Downloaded by [Leonid Golinskii] at 02:03 10 March 2016 z j • The spectral measure σ(J) is of the form

N σ(J, dx) = σac(J, dx) + σd (J, dx) = w(x) dx + σ j δ(λj ), (2.17) j=1 where √ −2 2 4 − x2 z j (1 − z ) w(x) := , x = t,σ= j . it 2 2 cos j 2π|u(e )| u (z j )u(1/z j )

Note that |u(eit)|2 = Q(x),x= 2 cos t, Q is a real polynomial of the same degree as the Jost polynomial u. Linear and Multilinear Algebra 9

Sometimes, two-sided Jacobi matrices ⎡ ⎤ ...... ⎢ . . . ⎥ ⎢ ⎥ ⎢ a−1 b0 a0 ⎥ ⎢ ⎥ J = J({b j } j∈Z, {a j } j∈Z) = ⎢ a0 b1 a1 ⎥ , (2.18) ⎢ ⎥ ⎣ a1 b2 a2 ⎦ ......

2 b j ∈ R, a j > 0, which generate linear operators on the two-sided (Z), arise in the canonical models for certain infinite graphs. Such matrices are said to have a finite rank if there are integers N ±, N − < N + so that − + − + bn = 0, n ∈[/ N , N ], an = 1, n ∈[/ N , N − 1]. { ±} In this case, one has a pair of Jost solutions un of the basic recurrence relation (2.6) (with n ∈ Z) which satisfies +( ) = n, ≥ +, −( ) = −n, ≤ −. un z z n N un z z n N (2.19)

The Wronskian of two solutions { fn(z)}n∈Z and {gn(z)}n∈Z of (2.6) are deﬁned as [ f, g]:=an fn(z)gn+1(z) − fn+1(z)gn(z) (2.20) (the right-hand side does not actually depend on n). We denote by + − w(z) := [u (z), u (z)], z ∈ D (2.21) the Wronskian of two Jost solutions of (2.6). It is well known (see, e.g. [21, Theorem 10.4]) that the spectrum of a two-sided Jacobi matrix J of the ﬁnite rank is σ( ) =[− , ]∪σ ( ), σ ( ) ={λ }N , J 2 2 d J d J j j=1

and for the eigenvalues λ j , the relation 1 ∃ h ∈ 2(Z) : Jh = ζ + h ⇐⇒ w(ζ) = ,ζ∈ D, ζ 0 (2.22) holds. So the eigenvalues of J are exactly the Zhukovsky images of zeros of Wronskian w

Downloaded by [Leonid Golinskii] at 02:03 10 March 2016 (2.21) in the unit disk.

Example 2.3 Let

J = J({0} j∈Z, {a j } j∈Z), a j = 1, j = 0. (2.23) We now have N − = 0, N + = 1. A simple computation gives 1 w(z) = − a0z, a0z and so 1 σd (J) =± a0 + , a0 > 1, a0 and σd (J) is empty for 0 < a0 ≤ 1. 10 L. Golinskii

Example 2.4 Let

J = J({0} j∈Z, {a j } j∈Z), a j = 1, j =±1. (2.24)

We now have N − =−1, N + = 2. A simple computation gives α α 3 1 −1 1 1 w(z) = α1α−1z + + − z + , a−1 a1 a1a−1 a1a−1z 1 αk := − ak, k =±1, ak

and so to determine the discrete spectrum σd (J), one has to analyse the roots of the quadratic equation ( ) = 2 − + = , := ( 2 − )( 2 − ), := 2 + 2 − . Q y Ay By 1 0 A a−1 1 a1 1 B a1 a−1 1 (2.25)

Precisely, each root y0 of this equation in (0, 1) generates two symmetric eigenvalues √ 1 λ± := ± y0 + √ . y0

The algebraic equations which we encounter later on cannot in general be solved explicitly. By means of the following well-known result,[22, p.41] we can determine how many roots (if any) they have in (−1, 1).

n Theorem (Descarte’s rule) Let a(x) = a0x + ...+ an be a real polynomial. Denote by μ(a) the number of its positive roots, and ν(a) the number of the sign changes in the sequence {a0,...,an} of its coefﬁcients (the zero coefﬁcients are not taken into account). Then, ν(a) − μ(a) is a nonnegative even number.

3. Canonical form for adjacency matrix Before starting the proof, we note that the algorithm of constructing a new basis suggested below is even more important than the result itself. To ﬁnd the spectrum of graphs in the next Downloaded by [Leonid Golinskii] at 02:03 10 March 2016 sections, we will have to carry it out by hand. The result of Theorem 1.2 only guarantees that the procedure can be accomplished with enough paper and patience. The algorithm applies in a number of situations beyond graphs with one tail (for instance, for the chain of cycles, the ladder with missing rungs, etc.). We will elaborate on this topic in our forthcoming papers.

Proof of Theorem 1.2 One of the versions of the Spectral Theorem for self-adjoint linear operator on a separable Hilbert space claims that each such operator A is unitarily equivalent to an inﬁnite orthogonal sum of Jacobi operators (operators with the simple spectrum), see, e.g. [23, Section VII.2, Lemma 2]. As a matter of fact, the proof is just a consecutive n application of the Gram–Schmidt orthogonalization process to the sequence {A hk}n≥1 with an appropriate sequence of vectors {hk}k≥1. Our algorithm is actually based on this method. Linear and Multilinear Algebra 11

2 We construct an orthogonal basis {ek}k≥1 in so that the matrix of A in this basis has the canonical form (1.12). We take n−1 ek = ek, k = n, n + 1,..., ek = bikei , k = 1, 2,...,n − 1, (3.1) i=1 =[ ]n−1 where an orthogonal matrix B bij i, j=1 (more precisely, the matrix with orthogonal columns) is to be determined. It is clear that

Aen+k = αk−1en+k−1 + βken+k + αken+k+1, k = 2, 3,... Aen+1 = den + β1en+1 + α1en+2. (3.2) The following notation will be convenient throughout the proof n−1 w jk := bikaij, j = 1, 2,...,n, k = 1, 2,...,n − 1. (3.3) i=1

(1) Put bi,n−1 := ain, i = 1, 2,...,n − 1, so n−1 Aen = Aen = ainei + annen + den+1 = en−1 + annen + den+1. (3.4) i=1 Next, n−1 n−1 n n Aen−1 = ain Aei = ain aije j = w j,n−1e j i=1 i=1 j=1 j=1 n−1 n−1 = w + 2 · . j,n−1e j ain en j=1 i=1 2 = 2 > Note that i ain i bi,n−1 0, for, otherwise, the original matrix already has the canonical form. Deﬁne n−1 w j=1 j,n−1b j,n−1 y := , b , − := w , − − y b , − . (3.5) 1 n−1 2 j n 2 j n 1 1 j n 1 j=1 b j,n−1 Downloaded by [Leonid Golinskii] at 02:03 10 March 2016 Then, n−1 = + + , := 2 , Aen−1 en−2 y1en−1 z1en z1 bi,n−1 i=1 and orthogonality of the columns n−1 bi,n−2bi,n−1 = 0 (3.6) i=1

follows directly from the choice of y1 (3.5). If en−2 = 0, that is, bi,n−2 = 0 for all i = 1,...,n − 1, the algorithm terminates since

Aen−1 = y1en−1 + z1en, 12 L. Golinskii

the subspace Ln−1 := span{e j } j≥n−1 is invariant for A and so is its orthogonal complement L⊥ = 2 L { } n−1 n−1, which is ﬁnite dimensional. Thus, we can extend the basis e j j≥n−1 2 in Ln−1 in an arbitrary way to the basis in the whole . The latter is exactly the basis we are looking for. If en−2 = 0, we proceed to the next step. (2) We have n−1 n−1 n n Aen−2 = bi,n−2 Aei = bi,n−2 aije j = w j,n−2e j i=1 i=1 j=1 j=1 n−1 n−1 = w j,n−1e j + wn,n−2en = w j,n−1e j j=1 j=1

(wn,n−2 = 0by(3.6)). Put n−1 w n−1 w j= j,n−2b j,n−2 j= j,n−2b j,n−1 y := 1 , z := 1 (3.7) 2 n−1 2 2 n−1 2 j=1 b j,n−2 j=1 b j,n−1 and deﬁne

b j,n−3 := w j,n−2 − y2b j,n−2 − z2b j,n−1, j = 1, 2,...,n − 1, (3.8) which leads to Aen−2 = en−3 + y2en−2 + z2en−1. (3.9) The orthogonality relations n−1 n−1 bi,n−3bi,n−2 = bi,n−3bi,n−1 = 0 (3.10) i=1 i=1

stem directly from the deﬁnition of y2,z2 and (3.6). Again, if en−3 = 0, we are done. If not, we proceed to the next step. { }n−1,... In such a way, we determine consecutively pairwise orthogonal columns bi,n−1 i=1 { }n−1 ( , ) bi,n−k i=1 , a pair of numbers yk zk and put

Downloaded by [Leonid Golinskii] at 02:03 10 March 2016 b j,n−k−1 := w j,n−k − ykb j,n−k − zkb j,n−k+1, j = 1, 2,...,n − 1 (3.11) with the new column orthogonal to all previous ones, and

Aen−k = en−k−1 + yken−k + zken−k+1. (3.12)

If en−k−1 = 0, then Aen−k = yken−k + zken−k+1, Ln−k := span{e j } j≥n−k is A- invariant, and so is its orthogonal complement which is ﬁnite dimensional. Otherwise, the algorithm can be extended to yet another step. { }n−1 When the algorithm continues till e1, we come to the orthogonal set e j j=1 in span {e j }1≤ j≤n−1, so necessarily e0 = 0. In this case, the ﬁnite-dimensional component is −1 missing. The operator A in the normalized basis {ek}k≥1, ek =ek ek is given by the Jacobi matrix which agrees with the original matrix J from some point on. The proof is complete. Linear and Multilinear Algebra 13

4. Spectra of trees with tails As we have already mentioned in the Introduction, the adjacency matrix A() for the coupling = G + P∞({a j }) is (1.11). The algorithm suggested in Theorem 1.2 provides a way to implement the spectral analysis of the adjacency operator A() which consists of two stages. First, one has to apply the above algorithm by hand to obtain the canonical form (1.12). Second, the solutions of two algebraic equations (the characteristic equation for the ﬁnite-dimensional component F() and the Jost equation for the Jacobi component J()) provide the spectrum of .

Example 4.1 ‘A weighted star’. We begin with the coupling = Sn(w) + P∞, where Sn(w) is a simple weighted star graph of order n + 1, n ≥ 2, with vertices 1,...,n of degree 1 and the weight of the edge (k, n + 1) is wk,1≤ k ≤ n. The canonical basis {ek}k∈N looks as follows. We put

e j := e j , j ≥ n + 1 =⇒ A()e j = e j−1 +e j+1, j ≥ n + 2. w := (w ,w ,...,w ) w= w2 + ...+ w2 Next, let 1 2 n , 1 n, and let n 1 e := w e . n w k j j=1 Then, A()en+1 =wen +en+2, A()en =wen+1. So the Jacobi subspace and Jacobi component of are LJ = span{e j } j≥n, J() = J {0}, {w, 1, 1,...} . (4.1) To find the finite-dimensional component, we construct an orthonormal basis in Cn by ξ =[ξ ]n means of a unitary matrix kj k, j=1 with the specified last column n w f := ξ e (n), ξ = k , k = ,...,n, j kj k kn w 1 (4.2) k=1 { ( )}n Cn where ek n k=1 is the standard basis in .Put

Downloaded by [Leonid Golinskii] at 02:03 10 March 2016 e j := { f j , 0, 0,...}, j = 1,...,n. (4.3)

The orthogonality relations ek,en=0, 1 ≤ k ≤ n − 1, give n A()ek = ξkjξkn ·en+1 = 0. (4.4) j=1 () = O { }n−1 Hence, the ﬁnite-dimensional component F n−1 on the subspace span e j j=1.So the canonical form is A() On−1 ⊕ J {0}, {w, 1, 1,...} . (4.5) The Jost polynomial is now given by (2.9) w u(z) = (1 −w2)z2 + 1. 14 L. Golinskii √ Clearly, u > 0forw≤1, and it has zeros inside (−1, 1) if and only if w > 2. In this case, the spectrum is σ() =[−2, 2]∪σd () with the discrete component (n−1) 2 1 σd (Sn(w) + P∞) = 0 , ± w − 1 + . (4.6) w2 − 1

For the unweighted star Sn, we have √ (n−1) 1 σd (Sn + P∞) = 0 , ± n − 1 + √ , n ≥ 3, n − 1 (1) σd (S2 + P∞) ={0 }. (4.7)

The spectrum σ(Sn + P∞) was found in [13]. Note that Sn is a complete bipartite graph, Sn = K1,n. For the general complete bipartite graph K p,n+1−p, see Example 5.6 below.

ξ =[ξ ]n Remark 4.2 Although an explicit form of the matrix kj k, j=1 in (4.2) is immaterial, −1/2 it is worth noting that in the unweighted case, ξkn = n ,1≤ k ≤ n, and one can take

π 1 kj n 2 i ξ = Fn := √ [ε ] , = ,εn := e n , (4.8) n n k j 1

which is known as the Fourier matrix. Clearly, there are lots of options for ξ to be a real orthogonal matrix (rotation in Rn with appropriate Euler’s angles, orthogonal polynomials, etc.).

Example 4.3 ‘A multiple star’. Consider an unweighted star-like graph Sn,p with n rays, n ≥ 2, each of which contains p + 1 vertices, p ≥ 2. The vertices are numbered as

{1, n + 1,...,(p − 1)n + 1}, {2, n + 2,...,(p − 1)n + 2}, ... {n, 2n,...,pn},

and the root is pn + 1, so Sn,1 = Sn.Let = Sn,p + P∞ with the path attached to the root. As above, we put e j := e j , j = pn + 1,..., and Downloaded by [Leonid Golinskii] at 02:03 10 March 2016

n 1 ep(n− )+i := √ e(i− )n+q , i = 1, 2,...,p. (4.9) 1 n 1 q=1

We have A()e j = e j−1 +e j+1, j = pn + 2 ..., √ √ A()e + = ne +e + , A()e = e − + ne + , pn 1 pn √ pn 2 pn pn 1 pn 1 A()epn = epn−1 + nepn+1, A()ep(n−1)+i = ep(n−1)+i−1 +ep(n−1)+i+1, i = 2,...,p − 1, A()ep(n−1)+1 = ep(n−1)+2, Linear and Multilinear Algebra 15

so the Jacobi subspace and Jacobi component of are

L = span{e } ≥ ( − )+ , J j j p n 1 1 √ n, j = p; J() = J({0}, {a }), a = (4.10) j j 1, j = p.

To ﬁnd the ﬁnite-dimensional component note that, by the construction, ep(n−1)+k ∈ span{e(k−1)n+1,...,ekn}. As in the above example, we supplement each ep(n−1)+i to the basis in this subspace by means of the Fourier matrix (4.8) n n (k) := ξ , (k) := ξ = f j qje(k−1)n+q fn qne(k−1)n+q ep(n−1)+k q=1 q=1 for 1 ≤ j ≤ n − 1, 1 ≤ k ≤ p.Asin(4.4) we have () (1) = (2), () (2) = (1) + (3),..., A f j f j A f j f j f j () (p−1) = (p−2) + (p), () (p) = (p−1). A f j f j f j A f j f j (4.11) H := { (1),..., (p)} () Relations (4.11) mean that the subspace j span f j f j is A -invariant, and A()|H j = A(Pp). There are exactly n − 1 such subspaces for j = 1,...,n − 1. Finally, we come to the following canonical form for the adjacency matrix

n−1 A() F() ⊕ J(), F() = ⊕ A(Pp). (4.12) j=1 The Jost polynomial is computed in (2.5) √ z2p − 1 (n − 1)z2p+2 − nz2 + 1 − nu(z) = (n − 1)z2 − 1 = . z2 − 1 z2 − 1 It is easy to see that the polynomial q(x) = (n − 1)x p+1 − nx + 1 has exactly two positive < ( , )< ( , ) = roots 0 x1 p n x1 p n 1, so the ﬁrst one has the spectral meaning. Hence, σ() =[−2, 2]∪σd () with the discrete component π j p 1 σd () = 2 cos ± x (p, n) + √ , (4.13) = 1 p + 1 j 1 x1(p, n)

Downloaded by [Leonid Golinskii] at 02:03 10 March 2016 so we have p eigenvalues, each of multiplicity n −1, on the absolutely continuous spectrum [−2, 2] and two simple eigenvalues off [−2, 2]. The spectrum of the multiple star was studied in [12], but no explicit formulae were provided.

The problem becomes harder (in the sense of computation) if the original ﬁnite star-like graph is nonsymmetric (the rays are different). 16 L. Golinskii

Example 4.4 ‘A sword’.

This graph (also known as T (1, 2, 2, ∞)) can be viewed as a coupling = T (1, 2, 2)+ P∞. We put ek = ek, k ≥ 6, so A()ek = ek−1 +ek+1 for k ≥ 7. Next, take

e1 + e3 + e5 e2 + e4 −2e1 + e3 + e5 e5 := √ , e4 := √ , e3 := √ . 3 2 6 A direct computation leads to the Jacobi subspace and Jacobi component of 1 2 √ LJ = span{e j } j≥3, J() = J {0}, √ , , 3, 1, 1 ... . (4.14) 3 3 To compute the ﬁnite-dimensional component, we put

e3 − e5 e2 − e4 e2 := √ , e1 := √ , 2 2 2 so A()e2 = e1, A()e1 = e2, and {e j } j≥1 turns out to be the canonical basis in for A(). The adjacency operator is unitarily equivalent to 01 A() F() ⊕ J(), F() = A(P ) = (4.15) 2 10 The Jost polynomial can be computed directly from relations (2.6) Downloaded by [Leonid Golinskii] at 02:03 10 March 2016 3 u(z) = (1 − z2 − 3z4 − 2z6). 2 It is not hard to see that the cubic polynomial p(x) = 2x3 + 3x2 + x − 1 has the only real root x1,0< x1 < 1. Hence, √ 1 σ() =[−2, 2]∪σd (), σd () = ±1, ± x1 + √ . (4.16) x1

5. Spectra of graphs with cycles and tails Proof of Proposition 1.4 Consider the coupling of m-cycle and the inﬁnite path = Cm + P∞. We call it the kite graph with inﬁnite tail. Linear and Multilinear Algebra 17

The construction depends heavily on the parity of m. We provide the detailed proof for the case of odd m (the case of even m goes through in the same fashion). So, let m = 2n + 1, n ≥ 1. With ek := ek, k ≥ 2n + 1, we put

en− j+1 + en+ j en− j+1 − en+ j en+ j := √ , e j := √ , j = 1, 2,...,n, (5.1) 2 2

so A()en+1 = en+1 +en+2,

A()e = e − +e + , j = n + 2,...,2n − 1, 2n + 2, 2n + 3,... j √j 1 j 1 √ A()e2n+1 = 2e2n +e2n+2, A()e2n = e2n−1 + 2e2n+1, A()e j = e j−1 + e j+1, j = n + 2,...,2n − 1, (with the obvious modiﬁcation for n = 1, 2). The Jacobi subspace and Jacobi components are now LJ = span{e j } j≥n+1, 1, j = n; J() := J({1, 0, 0,...}, {a }), a = √ (5.2) j j 2, j = n.

The ﬁnite-dimensional component is LJ = span{e j }1≤ j≤n, and relations

A()e1 =−e1 +e2, A()e j = e j−1 + e j+1, j = 2,...,n − 1, A()en = en−1, provide the canonical form

A() F() ⊕ J(), F() := J({−1, 0,...,0}, {1}) (5.3)

is a Jacobi matrix of order n.

Downloaded by [Leonid Golinskii] at 02:03 10 March 2016 The Jost polynomial is given in (2.13) √ + − 2(z + 1)u(z) = z2n 1 + 2z2 − 1.

2n+1 2 The Descarte’s rule applied to the polynomial p2n+1(x) = x + 2x − 1 shows that there are two real roots such that −1 < x2(n)<0 < x1(n)<1. Both of them contribute to the discrete spectrum of the Jacobi component

1 1 σd (J()) = x1(n) + , x2(n) + . (5.4) x1(n) x2(n) To ﬁnd the spectrum of F() in (5.3), we expand the characteristic determinant over the ﬁrst row x x det |F() + x|=det |A(Pn) + x|−det |A(Pn− + x|=Un − Un− , 1 2 1 2 18 L. Golinskii

where Un is the Chebyshev polynomial of the second kind, + ( + ) cos 2n 1 t ( ) := sin n 1 t , ( ) − ( ) = 2 . Un cos t Un cos t Un−1 cos t t sin t cos 2 So the spectrum of F() is

2π j n σ(F()) = 2 cos (5.5) 2n + 1 j=1

and σ() =[−2, 2]∪σd (J()) ∪ σ(F()). The proof is complete.

Example 5.1 ‘A cycle with a double tail’. Given the cycle C2n+1, we attach two inﬁnite paths to the last vertex (see Remark 1.5). Denote this graph by = C2n+1 + 2P∞.

We proceed as in the case of the kite graph above. As in (5.1), put

en− j+1 + en+ j en− j+1 − en+ j en+ j := √ , e j := √ , j = 1, 2,...,n, 2 2 and e2n+2k + e2n+2k+1 e2n+1 = e2n+1, e2n+1+k = √ , k = 1, 2,.... 2 Clearly, the subspace L = span{e j } j≥1 is A()-invariant and A() |L= F() ⊕ J(), F() Downloaded by [Leonid Golinskii] at 02:03 10 March 2016 where is given in (5.3), and 1, j = n, n + 1; J() := J({1, 0, 0,...}, {a }), a = √ j j 2, j = n, n + 1.

The orthogonal complement 2 Lis A()-invariant, spanned by the system of vectors − e2n+2k e2n+2k+1 fk = √ , k = 1, 2,..., 2 2 and A() | L = J0. Finally,

A() F() ⊕ J() ⊕ J0. (5.6) We omit the lengthy computation of the Jost polynomial for J() and the discrete spectrum of J() based on the general Theorem DS. Linear and Multilinear Algebra 19

In exactly the same way, we can compute the spectrum of a cycle with multiple tail, i.e. p ≥ 2 tails attached to a vertex of the cycle.

Example 5.2 ‘A propeller with equal blades’.

The graph is composed of two equal cycles C2n+1 having one common vertex, with the inﬁnite path attached to it. We put ek = ek, k ≥ 4n + 1,

e j+1 + e2n− j + e2n+ j+1 + e4n− j e n− j := , j = 0, 1,...,n − 1. 4 2

Then, A()e3n+1 = e3n+1 +e3n+2,

A()e4n− j = e4n− j−1 +e4n− j+1, j = 1, 2,...,n − 2, A()e4n = e4n−1 + 2e4n+1, A()e4n+1 = 2e4n +e4n+2.

So the Jacobi subspace and Jacobi components are now LJ = span{e j } j≥3n+1, 1, j = n; J() := J({1, 0, 0,...}, {a }), a = (5.7) j j 2, j = n. The ﬁnite-dimensional component arises by putting

en− j+1 − en+ j e3n− j+1 − e3n+ j f j := √ , g j := √ , Downloaded by [Leonid Golinskii] at 02:03 10 March 2016 2 2 en− j+1 + en+ j − e3n− j+1 − e3n+ j h j := , j = 1,...,n. 2

Each of the sequences { f j }, {g j }, {h j } satisﬁes

A() y1 =−y1 + y2, A() yk = yk−1 + yk+1, k = 2,...,n − 1, A() yn = yn−1, so each of the linear spans is n-dimensional and A()-invariant subspace, and A() acts as (cf. (5.3)) 3 F() = ⊕ J({−1, 0,...,0}, {1}). k=1 The Jost polynomial + −2(z + 1)u(z) = 3z2n 1 + 4z2 − 1, 20 L. Golinskii

has two real roots such that −1 < x2(n)<0 < x1(n)<1. Both of them contribute to the discrete spectrum of the Jacobi component

1 1 σd (J()) = x1(n) + , x2(n) + . x1(n) x2(n) The spectrum of F() is given in (5.5), but now each eigenvalue has multiplicity 3. For spectral properties of ﬁnite propeller graphs, see [24].

Remark 5.3 If the blades are C2n, the argument goes through in exactly the same way. Moreover, the algorithm applies equally well to propellers with p equal blades for p ≥ 2 with the inﬁnite path attached to their common vertex (ﬂowers).

For the rest three examples, the argument is close to one above, so we provide only the answer. For detailed computation, see [25].

Example 5.4 ‘An umbrella’.

The canonical from is A() F() ⊕ J(), (5.8)

where Downloaded by [Leonid Golinskii] at 02:03 10 March 2016 01 1 3 1 F() := , J() := J − , , 0, 0,... , , 2, 1, 1,... . 1 −1 2 2 2

4 3 2 The Jost polynomial is u(z) =−3z − 3z − 3z − z + 1, having two real roots −1 < x2 < 0 < x1 < 1 with the spectral meaning. The spectrum is √ −1 ± 5 1 1 σ() =[−2, 2] σd (), σd () = , x1 + , x2 + . (5.9) 2 x1 x2 Example 5.5 ‘A wheel’. The wheel Wn is a graph consisting of a cycle on n vertices, n ≥ 3, and a vertex n + 1, adjacent to each of 1, 2,...,n (cf. [1, p. 49]). Consider the coupling = Wn + P∞ with the path attached to the root n + 1. Linear and Multilinear Algebra 21

The canonical form is A() F() ⊕ J(), (5.10) where

2π 2π(n − 1) F() = diag 2 cos ,...,2 cos , n √ n J() = J({2, 0, 0,...}, { n, 1, 1,...}). √ The Jost polynomial is computed in (2.9): − nu(z) = (n − 1)z2 + 2z − 1. It has two roots √ √ n + 1 n − 1 x (n) =− < 0 < x (n) = , 2 n − 1 1 n − 1

such that x1(n) ∈ (0, 1) for all n ≥ 3. For x2(n), we have

x2(n) ≤−1, n = 3, 4; x2(n)>−1, n ≥ 5.

Hence, σ() =[−2, 2]∪σd (), where the discrete spectrum is either

2kπ n−1 1 σd () = 2 cos x1(n) + , n = 3, 4, n k=1 x1(n) or 2kπ n−1 1 1 σd () = 2 cos x1(n) + , x2(n) + , n ≥ 5. n k=1 x1(n) x2(n) Example 5.6 ‘The complete bipartite graph’. Let G = K p,n+1−p be the complete bipartite graph of order n + 1[4, p.16], = K p,n+1−p + P∞. The canonical form is

A() F() ⊕ J(), F() := On−2 (5.11)

and √ J() := J({0}, { p(n − p), p, 1, 1,...}). The Jost polynomial is given in (2.10) √ −p n − pu(z) = (p − 1)z4 + (p(n − p) + p − 2)z2 − 1 = q(z2). Downloaded by [Leonid Golinskii] at 02:03 10 March 2016 It is easy to see that the quadratic polynomial q has the only root x1(p, n) in (−1, 1), 0 < x1(p, n)<1, so σ() =[−2, 2]∪σd (), (n−2) 1 σd (K p,n+1−p + P∞) ={0 } ± x1(p, n) + √ . (5.12) x1(p, n) For p = 1, we have Example 4.1.

The situation when an equal number of tails are attached to each vertex of a given ﬁnite Graph, G turns out to be relatively simple.

Proof of Theorem 1.6 We label the vertices along the rays (off G)as ={ + }∞ , := + ( − ) , = , ,..., . Ri m j i j=1 m j n j 1 pn i 1 2 pn 22 L. Golinskii

Put p 1 g (k) := √ e +( − ) + , k = 1, 2,...,n, j ∈ N, j p m j k 1 p l l=1 2 so {g j (k)} j,k is an orthonormal sequence in . It is easy to see from the way of labelling of the vertices that the adjacency operator A() acts on these vectors as

A() g (k) = g − (k) + g + (k), j = 2, 3,... j √j 1 j 1 A() g1(k) = pek + g2(k). (5.13) Next, by the deﬁnition, √ () ={ ( ) (n), , ,...}+ ( ), = , ,..., , A ek A G ek 0 0 pg1 k k 1 2 n (5.14) { (n)}n Cn where ek k=1 is the standard basis in . To construct the canonical basis, we invoke the eigenvectors of A(G) n (n) := ζ (n) ,ζ=[ζ ]n fk qk eq qk q,k=1 q=1 ( ) (n) = λ (n) = ,..., which is a unitary matrix, so A G fk k fk , k 1 n. With such a notation at hand, we put n ( ) := ζ = ( (n), , ...), e0 k qk eq fk 0 0 q=1 n e j (k) := ζqk g j (q), j ∈ N, k = 1,...,n. q=1

It is clear from (5.13) that A() e j (k) = e j−1(k) +e j+1(k), j ≥ 2, n √ √ A()e1(k) = ζqk ( peq + g2(q)) = pe0(k) +e2(k), q=1

and ⎛ ⎞ n n Downloaded by [Leonid Golinskii] at 02:03 10 March 2016 √ () ( ) = ζ () = ⎝ ζ ( ) (n), , ,...⎠ + ( ) A e0 k qk A eq qk A G eq 0 0 p e1 k = = q 1 √ q 1 = λke0(k) + pe1(k). Hence, the subspace L := span{e j (k)}1≤k≤n, j=0,1... is A()-invariant and n √ A() |L ⊕ J(λk, p). k=1 It is not hard to see that on the orthogonal complement (p−1)n 2 A() | L ⊕ J0. k=1 The proof is complete. Linear and Multilinear Algebra 23 √ To ﬁnd the discrete spectrum of J(λk, p), one has to solve the Jost Equation (2.9)

2 (p − 1)x + λk x − 1 = 0, (5.15) pick its spectral roots, that is the roots in (−1, 1), and then take their Zhukovsky images. = λ |λ | > λ + λ−1 The simplest case is p 1, when each k with k 1 generates one eigenvalue k k of A() .Forp ≥ 3, an elementary analysis of Equation (5.15) shows that there are two |λ | < p(p−2) spectral roots for k p−1 , and there is one such root otherwise. The case p = 2 is particularly simple. Equation (5.15) has the roots λ ± λ2 + k k 4 x± = , x− < 0 < x+. 2

It is easy to see that there is exactly one spectral root for λk = 0, and no such roots for λk = 0. The discrete spectrum now looks as follows σ () = λ λ2 + . d sign k k 4 λk =0

6. Stars and Bethe lattices Consider two examples wherein two-sided Jacobi matrices arise naturally.

Example 6.1 ‘A double infinite star graph’. Denote by Sk,∞ the infinite star-like graph with k ≥ 2 infinite rays emanating from the common root (cf. Example 4.3). The main object under consideration is the coupling of two such graphs with the bridge connecting their roots, so = Sp,∞ + Sq,∞. We label the vertices of the ‘right’ star Sp,∞ along the rays by positive integers as {1, 2, p + 2,...}, {1, 3, p + 3,...},..., {1, p + 1, 2p + 1,...},

so the root is 1. Similarly, we number the vertices of the ‘left’ graph Sq,∞ by negative integers, so the root is −1 and the vertices along the rays are

{−1, −2, −q − 2,...}, {−1, −3, −q − 3,...},..., {−1, −q − 1, −2q − 1,...}.

Downloaded by [Leonid Golinskii] at 02:03 10 March 2016 2 2(Z ) Z := Z\{ } { } The underlying space is 0 , 0 0 , and the standard basis is e j j∈Z0 . { } We construct an orthonormal system e j j∈Z0 by

p+1 1 e := e , e := √ e( − ) + , k ≥ 2; 1 1 k p k 2 p i i=2 q+1 1 e− := e− , e := √ e( + ) − , j ≤−2. 1 1 j q j 2 q i i=2 It is easy to see that A() acts as √ √ A()e = e− + pe , A()e− = q e− +e , 1 √ 1 2 1 2√ 1 A()e2 = pe1 +e3, A()e−2 = e−3 + q e−1, 24 L. Golinskii

A()ek = ek−1 +ek+1, k ≥ 3, A()e− j = e− j−1 +e− j+1, j ≥ 3. L= { } () Hence, the subspace span e j j∈Z0 is A -invariant and ⎧ √ ⎨ , = ; √p j 1 () |L= ( , ) = ({ }, { } ), = , =− ; A J p q J 0 a j j∈Z a j ⎩ q j 1 (6.1) 1, j =±1. { } To supplement the system e j j∈Z0 to the orthonormal basis, we proceed in a standard way p+1 (l) := ξ , = ,..., , = , ,..., [ξ ]=F ; ek i−1,l e(k−1)p+i l 1 p k 1 2 rs p i=2 q+1 (l) := η , = ,..., , =− , − ,..., [η ]=F . e j i−1,l e( j+1)q−i l 1 q j 1 2 rs q i=2

Finally, p+q−2 A() J(p, q) ⊕ ⊕ J0 . (6.2) i=1 √ √ To ﬁnd the discrete spectrum, we apply Example 2.4 with a−1 = q, a1 = p,so characteristic Equation (2.25) looks

Q(y) = (p − 1)(q − 1)y2 − (p + q − 1)y + 1 = 0.

It is easy to see that this equation always has two positive roots 0 < y− < y+. In the cases min(p, q) = 2 and p = q = 3, we have y+ ≥ 1, so only the ﬁrst root has the spectral meaning. Otherwise, 0 < y− < y+ < 1, so both of them are the spectral roots. Hence, √ 1 σd (J(p, q)) =± y− + √ , min(p, q) = 2or p = q = 3, y−

or √ 1 √ 1 σd (J(p, q)) = ± y− + √ , ± y+ + √ y− y+ for the rest of the values p, q ≥ 2. Here, Downloaded by [Leonid Golinskii] at 02:03 10 March 2016 p + q − 1 ± (p − q)2 + 2(p + q) − 3 y± = . 2(p − 1)(q − 1) Note that if the bridge has weight d, the above argument ends up with the Jacobi matrix ⎧ √ ⎨ , = ; √p j 1 ( , , ) = ({ }, { } ), = , =− ; = , = , ± , J p q d J 0 a j j∈Z a j ⎩ q j 1 a j 1 j 0 1 d, j = 0, for which the spectrum can be found explicitly (see Introduction). The example was studied in [10] using a different model based on one-sided Jacobi matrices (1.5), with no explicit expressions provided. Linear and Multilinear Algebra 25

Example 6.2 ‘Infinite regular trees’. The Bethe–Cayley trees Bd of degree d ∈ N (or Bethe lattices in physical literature) are among the most notable infinite trees. For the introduction to the spectral theory on such trees, see [7, Chapter 10]. We consider here the coupling of two copies of Bd with the bridge connecting their roots, = Bd + Bd . Note that is an infinite regular tree of degree d + 1. We number the vertices of as in the previous example (the vertices of the ‘right’ copy by positive integers, and of the ‘left’ copy by negative integers). So, the vertices of k’s generation in the right copy (the root forms the 1st generation) are labelled with

k−1 k − j d 1 {lk− + 1, lk− + 2,...,lk}, lk := d = , k ∈ N. 1 1 d − 1 j=1

2(Z ) { } Again, the underlying Hilbert space is 0 with the basis e j j∈Z0 . Put e±1 := e±1,

l l 1 k 1 k ek := √ ei , e−k := √ e− j , k = 2, 3,.... dk−1 dk−1 i=lk−1+1 j=lk−1+1

The operator A() acts as √ √ A() e = e− + d e , A() e− = d e− +e , 1 √ 1 √2 1 2 1 A() ek = d ek−1 + d ek+1, |k|≥2. L= { } () So the subspace span e j j∈Z0 is A -invariant and √ √ 1/ d, j = 0; A() |]L = J() = dJ({0}, {a } ∈Z), a = (6.3) j j j 1, j = 0

The structure of operator A() on the orthogonal complement is the same as in the case of Bd (see [14], [7, Theorem 10.2.2]), so ﬁnally ω √ d ∞, d ≥ 2; A() J() ⊕ ⊕ dJ0 ,ωd = (6.4) i=1 0, d = 1. Downloaded by [Leonid Golinskii] at 02:03 10 March 2016 To ﬁnd the discrete spectrum of A() , note that J() (6.3) is the subject of Example 2.3. It follows that the discrete spectrum of J() is empty, and so √ √ σ() =[−2 d, 2 d] (6.5)

and is pure absolutely continuous of inﬁnite multiplicity.

Remark 6.3 Since B1 = P∞, the question arises naturally whether it is possible to ﬁnd spectra of the couplings = G + Bd . The above algorithm works equally well in this situation and leads to the canonical form √ ωd A(G + Bd ) A(G + P∞) ⊕ ⊕ dJ0 . i=1 26 L. Golinskii

Acknowledgements I thank Professor Yu. Samoilenko for drawing my attention to the spectral graph theory and putting forward some particular problems related to inﬁnite graphs with tails and their spectra.

Disclosure statement No potential conﬂict of interest was reported by the authors.

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