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Sao Paulo Journal of Mathematical Sciences 7, 1 (2013), 83–104

Dynkin diagrams and spectra of graphs

Michael A. Dokuchaev

Departamento de Matematica Univ. de Sao Paulo, Caixa Postal 66281, Sao Paulo, SP05315-970 – Brazil

Nadiya M. Gubareni

Institute of Econometry & Computer Science, Department of Management TechnicalUniv. of Czestochowa, Dabrowskiego str., 69, 42-200 Czestochowa, Poland

Vyacheslav M. Futorny

Departamento de Matematica Univ. de Sao Paulo, Caixa Postal 66281, Sao Paulo, SP05315-970 – Brazil

Marina A. Khibina

In-t of Eng. Thermophysics of NAS of Ukraine, Zheljabova Str., 2A, 03057, Kiev,Ukraine

Vladimir V. Kirichenko

Faculty of Mechanics and Mathematics, Kiev National, Taras Shevchenko Univ.,Vladimirskaya Str., 64, 01033 Kiev, Ukraine

1. Introduction

Dynkin diagrams first appeared in [20] in the connection with classifica-tion of simple Lie groups. Among Dynkin diagrams a special role is playedby the simply laced Dynkin diagrams An, Dn, E6, E7 and E8. Dynkin dia-grams are closely related to Coxeter graphs that appeared in geometry (see[8]). After that Dynkin diagrams appeared in many braches of mathematicsand beyond, em particular em representation theory.

In [22] P. Gabriel introduced a notion of a quiver (directed graph) andits representations. He proved the famous Gabriel’s theorem on represen-tations of quivers over algebraically closed field.

Let Q be a finite quiver and Q the undirected graph obtained from Q bydeleting the orientation of all arrows.

83

84 M. A. Dokuchaev, N. M. Gubareni, V. M. Futorny, M.A. Khibina, and V. V. Kirichenko

Theorem 1.1. (Gabriel’s Theorem). A connected quiver Q is of finitetype if and only if the graph Q is one of the following simply laced Dynkindiagrams: An, Dn, E6, E7 or E8.

I.N. Bernstein, I.M. Gelfand and V.A. Ponomarev [5] gave a proof ofGabriel’s Theorem using roots, Weyl groups and Coxeter functors.

The terms “tame type” and “wild type” were introduced by P. Donovanand M.R. Freislich [16]. Extended Dynkin diagrams or Euclidean diagrams

are An, Dn, E6, E7 and E8 (see, for example, [2]). Tame quivers in termsof extended Dynkin diagrams were classified by L.A. Nazarova [39] and byP. Donovan–M.R. Freislich [16]. For finite dimensional algebras and someother algebraic structures the tame-wild dichotomy problem was solved byYu.A. Drozd [17]–[19]. The theory of K-species was first considered byP. Gabriel in [23]. He obtained the characterization of K-species of finitetype in a special case. His result was extended by V. Dlab and C.M. Ringel(see [14, Theorem B]).

Theorem 1.2. (Theorem B). A K-species is of a finite type if and onlyif its diagram is a finite disjoint union of Dynkin diagrams.

The problem of the ubiquity of the symply laced Dynkin diagrams An,Dn, En was formulated by V.I. Arnold [1] as follows.

A-D-E classification. The Coxeter-Dynkin graphs An, Dn and Enappear in many independent classification theorems. For instance

(a) the classification of the platonic solids (or finite orthogonal groups ineuclidean 3-space),

(b) the classification of the categories of linear spaces and maps (repre-sentations of quivers,

(c) the classification of the singularities of algebraic hypersurfaces, witha definite intersection form of the neighboring smooth fibre,

(d) the classification of the critical points of functions having no moduli,

(e) the classification of the Coxeter groups generated by reflections, or,of Weyl groups with roots of equal length.

The problem is to find the common origin of all A-D-E classificationtheorems and to substitute a priori proofs to a posteriori verifications ofthe parallelism of the classifications. An introduction to the A-D-E-problemcan be found in [30].

Dynkin diagrams and extended Dynkin diagrams are widely used in thestudy of generalized Cartan matrices and Kac–Moody algebras [2]–[4], [6],[31], [35], [36], [40] and [42].

Sao Paulo J.Math.Sci. 7, 1 (2013), 83–104

Dynkin diagrams and spectra of graphs 85

Let G be a finite graph without loops and multiple edges (G is a finitesimple graph). J.H. Smith [41] formulated the following result:

Theorem 1.3. Let G be a finite simple graph with the spectral radius (in-dex) rG. Then rG = 2 if and only if each connected component of G is one

of the extended Dynkin diagram An, Dn, E6, E7, E8. Moreover, rG < 2 ifand only if each connected component of G is one of Dynkin diagrams An,Dn, E6, E7, E8.

For the full proofs of this Smith’s theorem see, for example, [27, chap-ter I and Appendix I], [37] and [21, Theorem 2.12]. Note that Theorem1.3 was obtained also in [33, Theorem 5.1] and [7]. In 1975 (see [11])D.M. Cvetkovich and I. Gutman introduced for extended Dynkin diagramsof type A and D the symbols Cn and Wn. Moreover, they used the follow-ing notations: Pn for An; Zn for Dn+2, T1 for E6, T2 for E7, T3 for E8, T4for E6, T5 for E7 and T6 for E8.

The following terminology is used in [12, pp. 77-79]: “Smith’s graphs”means extended Dynkin diagrams and “reduced Smith’s graphs” meanssimply laced Dynkin diagrams An, Dn, E6, E7, E8 (see also [9] and [10]).

In this paper we consider spectral properties of graphs based on Perron-Frobenius theory of non-negative matrices. We will use terminology andresults from [29, Section 6.5] and [25].

2. Symmetric non-negative matrices

Let G be an undirected finite graph without loops and multiple edges,i.e., G is a finite simple graph.

Let V G = {1, . . . , n} be the vertex set of G and EG be the edge set of G.Two vertices i and j are called adjacent if they are connected by an edge.

The adjacency matrix [G] of a simple graph with n vertices is a squarematrix [G] = (αj) of order n, whose (i, j)-entry αij is 1, if the verticesi and j are adjacent, otherwise αij = 0. Therefore, [G] is a symmetric(0, 1)-matrix with zero main diagonal.

Denote by Mn(R) the ring of all n×n matrices with real entries. Let A =(aij) ∈ Mn(R) be a non-negative symmetric permutationally irreduciblematrix.

From the Perron-Frobenius Theorem it follows that A has the largestpositive eigenvalue rA such that any eigenvalue λ of A one has that |λ| ≤ rA,and there exists a positive eigenvector ~z = (z1, . . . , zn)T with A~z = rA~z.We give the next.



Theorem 2.1. Let A = (aij) ∈ Mn(R) be a nonnegative symmetric per-mutationally irreducible matrix and B be its proper main submatrix. ThenrB < rA.

Before the proof of the theorem we give necessary information about theproperties of A.

Lemma 2.1. [26] Eigenvectors of a matrix belonging to different eigenval-ues are orthogonal.

Corollary 2.1. Let A ∈Mn(R) be a permutationally irreducible symmetricmatrix and ~z = (z1, . . . , zn)T be its positive eigenvector, then A~z = rA~z.

Proof. Suppose that A~z = λ~z and λ 6= rA. Let ~w = (w1, . . . , wn)T be apositive eigenvector of A with eigenvalue rA. Then by Lemma 2.1 the innerproduct (~z, ~w) is zero. We obtain a contradiction:

n∑i=1

ziwi > 0.

�

Now we give a proof of Theorem 2.1.

Proof. Let B be a proper principal m×m-submatrix of A. We enumeratethe rows and columns of A such that:

A =

B1 . . . 0 X1...

. . ....

...0 . . . Bt Xt

XT1 . . . XT

t C

,

where B =

B1 . . . 0...

. . ....

0 . . . Bt

and the matrices B1, . . . , Bt are permutation-

ally irreducible.

We may assume that rB = rB1 , B1 ∈ Mm1(R), . . . , Bt ∈ Mmt(R),

m1 + . . . + mt = m. Then, C ∈ Mn−m(R) and X =

X1...Xt

, where

Xi ∈Mmi×(n−m)(R).

The matrix A is permutationally irreducible, so X1 6= 0.



Let ~z = (z1, . . . , zn)T be the Perron-Frobenius positive eigenvector of A,i.e., A~z = rA~z. Denote by ~zs = (z1, . . . , zm1) th evector formed by the firstm1 coordinates of ~z and by ~ze = (zn−m+1, . . . , zn).

Then we obtain: B1~zs+X1~ze = rA~zs. Obviously the non-negative vectorX1~ze is nonzero (vector ~ze is positive and X1 6= 0 and non-negative). Wehave yi ≥ 0 for i = 1, . . . ,m1. Therefore yi ≤ rAzi for i = 1, . . . ,m1 and

there exists 1 ≤ k ≤ m1 such that yk < rAzk. Let ~f = (f1, . . . , fm1)T

be a Perron-Frobenius vector of B1, so B1~f = rB ~f . Then (~zs, B1

~f) =

(~zs, rB1~f) = rB1(~zs, ~f) = (B1~zs, ~f) < (rA~zs, ~f) = rA(~zs, ~f), i.e., rB1(~zs, ~f) <

rA(~zs, ~f). Then ~zs, ~f) > 0 as inner product of positive vectors. ThereforerB = rB1 < rA. Theorem is proved. �

3. Spectra of Dynkin diagrams and extended Dynkin dia-grams

In this section we give a list of characteristic polynomials and spectra ofDynkin diagrams and of extended Dynkin diagrams.

Theorem 3.1. (L. Kronecker, [32]) Suppose that all the real roots of amonic polynomial with integer coefficients belong to the interval [−2, 2] andare given in the form

2 cosα, 2 cosβ, 2 cos γ, . . . .

Then the angles α, β, γ, . . . are rational multiples of π/2.

The following simple graphs are simply laced Dynkin diagrams:

An, n ≥ 1 : • • • . . . • • •

Dn, n ≥ 4 :

•

• • • . . . • • •

~~~~~

@@@@

@

•

E6 :•

• • • • •



E7 :•

• • • • • •

E8 :•

• • • • • • •

The following simple graphs are extended versions of simply laced Dynkindiagrams:

An (n ≥ 2) :

•

•

qqqqqqqqqqq •

• •

qqqqqq

qqqqq

•

MMMMMMMMMMM

Dn (n ≥ 4):

• •

•

KKKKKKKK

ssssss

ss• • •

ssssssss

KKKKKK

KK

• •

E6 :

•

•

• • • • •

E7 :•

• • • • • • •



E8 :•

• • • • • • • •

Often extended Dynkin diagrams are called Euclidean diagrams.

Proposition 3.1. For the Dynkin diagram An (n ≥ 1) we have

χAn(x) =∏

1≤k≤n

(x− 2 cos

kπ

n+ 1

)Consequently,

S(An) =

{2 cos

kπ

n+ 1| k = 1, . . . , n

}and rAn = 2 cos π

n+1 , where S(An) denotes the spectrum of An.

Proposition 3.2. For the Dynkin diagram Dn (n ≥ 4) we have

χDn(x) = x

∏0≤k≤n−2

(x− 2 cos(1 + 2k)π

2(k − 1)

.

Consequently, S(Dn) consists of zero and of the following set:{2 cos

(1 + 2k)π

2(n− 1)| k = 0, . . . , n− 2

}and rDn = 2 cos π

2(n−1) .

Proposition 3.3. For the Dynkin diagram E6 we have

χE6(x) = x6 − 5x4 + 5x2 − 1 =∏

1≤k≤6

(x− 2 cos

mkπ

12

),

where mk = 1, 4, 5, 7, 8, 11. Then

S(E6) ={

2 cosmkπ

12|mk = 1, 4, 5, 7, 8, 11

}and rE6 = 2 cos π

12 .


χE7(x) = x(x6 − 6x4 + 9x2 − 3) =∏

1≤k≤7

(x− 2 cos

mkπ

18

),



where mk = 1, 5, 7, 9, 11, 13, 17. Then

S(E7) ={

2 cosmkπ

18|mk = 1, 5, 7, 9, 11, 13, 17

}and rE7 = 2 cos π

18 .


χE8(x) = x8 − 7x6 + 14x4 − 8x2 + 1 =∏

1≤k≤8

(x− 2 cos

mkπ

30

),

where mk = 1, 7, 11, 13, 17, 19, 23, 29. Then

S(E8) ={

2 cosmkπ

30|mk = 1, 7, 11, 13, 17, 19, 23, 29

}and rE8 = 2 cos π

30 .

Proposition 3.6. For the extended Dynkin diagram An (n ≥ 2) we have

χAn(x) = µn+1 + µ−n−1 − 2 =

∏1≤k≤n

(x− 2 cos

2kπ

n+ 1

),

where x = µ+ 1µ . Then consequently,

S(An) =

{2 cos

2kπ

n+ 1| k = 0, . . . , n

}and rAn

= 2.

Proposition 3.7. For the extended Dynkin diagram Dn (n ≥ 4) we have

χDn(x) = χA3

(x)χn−3(x) = x2(x2 − 4)∏

0≤k≤n−3

(x− 2 cos

kπ

n− 2

).

Then

S(Dn) =

{2 cos

kπ

n− 2| k = 1, . . . , n− 3

}∪ [−2, 0, 0, 2]

and rDn= 2.



Proposition 3.8. For the extended Dynkin diagrams E6, E7, E8 we have

χE6(x) = x(x2 − 1)2(x2 − 4);

χE7(x) = x(x2 − 1)(x2 − 4)

∏1≤k≤3

(x− 2 cos kπ4 );

χE8(x) = x(x2 − 1)(x2 − 4)

∏1≤k≤4

(x− 2 cos kπ5 ).

Then

S(E6) = [0,±1,±1,±2] and rE6= 2.

S(E7) ={

2 cos kπ4 | k = 1, 2, 3}∪ {0,±1,±2} and rE7

= 2.

S(E8) ={

2 cos kπ5 | k = 1, 2, 3, 4}∪ {0,±1,±2} and rE8

= 2.

4. Perron-Frobenius vectors of extended Dynkin diagrams

We consider simply laced extended Dynkin diagrams and its Perron-Frobenius vectors.

We give the list of these graphs with the numbering of vertices suitablefor us:

E6 :

• 6

• 3

• • • • •5 2 1 4 7

E7 :

• 3

• • • • • • •7 5 2 1 4 6 8

E8 :

• 3

• • • • • • • •5 2 1 4 6 7 8 9



An (n ≥ 2) :

3

•

2 •

ooooooooooooo • n− 1

1 • •

oooooo

oooooo

o n

•

OOOOOOOOOOOOO

n+ 1

Dn (n ≥ 4):

n− 1 n+ 1

• •

•

OOOOOOOOOO

tttttt

ttttt

• • •

mmmmmmmmmmmm

MMMMMM

MMMMMM

M

1 2 n− 4 n− 3

• •n− 2 n

Case E6.

The adjacency matrix is

[E6] =

0 1 1 1 0 0 01 0 0 0 1 0 01 0 0 0 0 1 01 0 0 0 0 0 10 1 0 0 0 0 00 0 1 0 0 0 00 0 0 1 0 0 0

Let ~z = (z1, z2, z3, z4, z5, z6, z7)

T be a positive eigenvector of E6.



0 1 1 1 0 0 01 0 0 0 1 0 01 0 0 0 0 1 01 0 0 0 0 0 10 1 0 0 0 0 00 0 1 0 0 0 00 0 0 1 0 0 0

z1z2z3z4z5z6z7

=

λz1λz2λz3λz4λz5λz6λz7

z2 + z3 + z4 = λz1z1 + z5 = λz2z1 + z6 = λz3z1 + z7 = λz4

z2 = λz5z3 = λz6z4 = λz7

z2 + z3 + z4 = λ(z5 + z6 + z7) = λz1;

z1 = z5 + z6 + z7, 4z1 = λ2z1, i.e., λ = 2.

z2 = 2z5, z3 = 2z6 and z2 = z3 = z4 = 2, 2z1 = 6, z1 = 3.

We obtain ~z = (3, 2, 2, 2, 1, 1, 1)T .

Case E7.


[E7] =

0 1 1 1 0 0 0 01 0 0 0 1 0 0 01 0 0 0 0 0 0 01 0 0 0 0 1 0 00 1 0 0 0 0 1 00 0 0 1 0 0 0 10 0 0 0 1 0 0 00 0 0 0 0 1 0 0

.

As above

0 1 1 1 0 0 0 01 0 0 0 1 0 0 01 0 0 0 0 0 0 01 0 0 0 0 1 0 00 1 0 0 0 0 1 00 0 0 1 0 0 0 10 0 0 0 1 0 0 00 0 0 0 0 1 0 0

z1z2z3z4z5z6z7z8

= λ

z1z2z3z4z5z6z7z8



z1 + z3 + z4 = λz1z1 + z5 = λz2

z1 = λz3z1 + z6 = λz4z2 + z7 = λz5z4 + z8 = λz6

z5 = λz7z6 = λz8

We have z6 = λz8, z4 = (λ2 − 1)z8, z1 = (λ3 − 2λ)z8,

z3 = (λ2 − 2)z8, z2 = (λ4 − 4λ2 + 3)z8,

z5 = (λ5 − 5λ3 + 5λ)z8, z7 = (λ4 − 5λ2 + 5)z8.

Let z8 = 1. Then from z2 + z7 = λz5 it follows that

λ4 − 4λ2 + 3 + λ4 − 5λ2 + 5 = λ6 − 5λ4 + 5λ2, i.e.,

λ6 − 7λ4 + 14λ2 − 8 = 0.

Obviously, 26− 7 · 24 + 14 · 22− 23 = 22(16− 28 + 14− 2) = 0, i.e., λ = 2is a root. Therefore ~z = (4, 3, 2, 3, 2, 2, 1, 1)T .

Case E8.


[E8] =

0 1 1 1 0 0 0 0 01 0 0 0 1 0 0 0 01 0 0 0 0 0 0 0 01 0 0 0 0 1 0 0 00 1 0 0 0 0 0 0 00 0 0 1 0 0 1 0 00 0 0 0 0 1 0 1 00 0 0 0 0 0 1 0 10 0 0 0 0 0 0 1 0

.

As above

0 1 1 1 0 0 0 0 01 0 0 0 1 0 0 0 01 0 0 0 0 0 0 0 01 0 0 0 0 1 0 0 00 1 0 0 0 0 0 0 00 0 0 1 0 0 1 0 00 0 0 0 0 1 0 1 00 0 0 0 0 0 1 0 10 0 0 0 0 0 0 1 0

z1z2z3z4z5z6z7z8z9

= λ

z1z2z3z4z5z6z7z8z9



z2 + z3 + z4 = λz1z1 + z5 = λz2

z1 = λz3z1 + z6 = λz4

z2 = λz5z4 + z7 = λz6z6 + z8 = λz7z7 + z9 = λz8

z8 = λz9

We have z8 = λz9, z7 = (λ2 − 1)z9, z6 = (λ3 − 2λ)z9,

z4 = (λ4 −−3λ2 + 1)z9, z1 = (λ5 − 4λ3 + 3λ)z9,

z3 = (λ4 − 4λ2 + 3)z9, z2 = (λ6 − 6λ4 + 10λ2 − 4)z9,

z5 = 9λ7 − 7λ5 = 14λ3 − 7λ)z9.

Let z9 = 1. From z2 = λz5 we obtain

λ6−6λ4+10λ2−4 = λ8−7λ6+14λ4−7λ2, i.e., λ8−8λ6+20λ4−17λ2+4 =0. Obviously, 28 + 8 · 26 + 20 · 24 − 17 · 22 + 4 = 4(26 − 27 + 20 · 4− 16) =4(64−128 + 80−16) = 0, i.e., 2 is a root. Then ~z = (6, 4, 3, 5, 2, 4, 3, 2, 1)T .

Case An, (n ≥ 2).

A2 :

2•

•

~~~~~•

@@@@@

1 3

The adjacency matrix

[A2] =

[0 1 11 0 11 1 0

]and

[A2]

(111

)= 2

(111

). Therefore, rA2

= 2.

For A3 :2 • • 3

1 • • 4



the adjacency matrix is [A3] =

0 1 0 11 0 1 00 1 0 11 0 1 0

and [A3]

1111

= 2

1111

.

Therefore, rA3= 2.

In general case, obviously, [An]~z = 2~z, ~z = (1, . . . , 1)T and rAn= 2.

Case D4 :

3•

2 • • • 4

1

•5

Clearly, the adjacency matrix of D4 is:

[D4] =

0 1 1 1 11 0 0 0 01 0 0 0 01 0 0 0 01 0 0 0 0

and

0 1 1 1 11 0 0 0 01 0 0 0 01 0 0 0 01 0 0 0 0

z1z2z3z4z5

= λ

z1z2z3z4z5

.

Therefore,

z2 + z3 + z4 + z5 = λz1;

z1 = λz2;

z1 = λz3;

z1 = λz4;

z1 = λz5.

If λ ≤ 0, then ~z is a non-positive eigenvector. So, λ > 0 and z2 = z3 =z4 = z5. Let z5 = 1. We obtain z1 = λ and λ2 = 4. Thus, λ = 2 and~z = (2, 1, 1, 1, 1). We have rD4

= 2.

For D5 :



4 6• •

•

KKKKKKKK

xxxxxxxxx •

ssssssss

FFFF

FFFF

F1 2

• •3 5

and [D5]~z = λ~z, where ~z = (z1, z2, z3, z4, z5, z6)T . We have z5 = z6 and

z3 = z4. z1 = λz3, z2 = (λ2 − 2)z3, z5 = λ3−3λ2 z3.

Let z3 = 1. Then ~z = (λ, λ2 − 2, 1, 1, λ3−3λ2 , λ

3−3λ2 )T . From z2 = λz5 we

obtain: λ2 − 2 = λ4−3λ2

2 and λ4 − 5λ2 + 4 = 0. 24 − 5 · 4 + 4 = 0. So, 2 is aroot and

~z = (2, 2, 1, 1, 1, 1)T . Therefore, rD5= 2.

Consider D8 :

7 9• •

•

KKKKKKKK

xxxxxxxxx • • • •

ssssssss

FFFF

FFFF

F1 2 3 4 5

• •6 8

We have

0 1 0 0 0 1 1 0 01 0 1 0 0 0 0 0 00 1 0 1 0 0 0 0 00 0 1 0 1 0 0 0 00 0 0 1 0 0 0 1 11 0 0 0 0 0 0 0 01 0 0 0 0 0 0 0 00 0 0 0 1 0 0 0 00 0 0 0 1 0 0 0 0

222221111

= 2

222221111

.

Therefore, rD8= 2.

Consider the general case Dn:



n− 1 n+ 1

• •

•

OOOOOOOOOO

tttttt

ttttt

• • •

mmmmmmmmmmmm

MMMMMM

MMMMMM

M

1 2 n− 4 n− 3

• •n− 2 n

1 0 1 0 0 . . . 0 1 1 0 02 1 0 1 0 . . . 0 0 0 0 03 0 1 0 1 . . . 0 0 0 0 04 0 0 1 0 . . . 0 0 0 0 0...

......

......

......

......

......

n-3 0 0 0 0 . . . 0 0 0 1 1n-2 1 0 0 0 . . . 0 0 0 0 0n-1 1 0 0 0 . . . 0 0 0 0 0n 0 0 0 0 . . . 1 0 0 0 0

n+1 0 0 0 0 . . . 1 0 0 0 01 2 3 4 . . . n-3 n-2 n-1 n n+1

2222...21111

= 2

2222...21111

Thus, rDn

= 2.

Corollary 4.1.

(a) For each extended Dynkin diagram G ∈ {An, Dn, E6, E7, E8} rG = 2.

(b) For each Dynkin diagram G ∈ {An, Dn, E6, E7, E8} we have rG < 2.

Proof. (a) For any extended Dynkin diagram G we already gave a positiveeigenvector with eigenvalue 2. Therefore, rG = 2.

(b) We have the following inclusions: An ⊂ An, Dn ⊂ Dn, E6 ⊂ E6, E7 ⊂E7, E8 ⊂ E8. By Theorem 2.1 rG < 2 for any G ∈ {An, Dn, E6, E7, E8}.

�

Proof of Smith’s theorem. Corollary 4.1 gives the “if” part of Smith’s the-orem.

Conversely, let G be a connected finite simple graph with rG ≤ 2. If Gis not a tree, then G must be the extended Dynkin diagram An. So, G is atree. It is easy to see G must be a tree of the form Tp,q,r (see [31, Exercise4.3]). Using Theorem 2.1 we obtain that Tp,q,r is either one of simply lacedDynkin diagrams or one of simply laced extended Dynkin diagrams.



5. Some examples

Let E6 be given in the form:

E6 :

• 3

• • • • •5 2 1 4 6

Then0 1 1 1 0 01 0 0 0 1 01 0 0 0 0 01 0 0 0 0 10 1 0 0 0 00 0 0 1 0 0

z1z2z3z4z5z6

= λ

z1z2z3z4z5z6

z2 + z3 + z4 = λz1

z1 + z5 = λz2z1 = λz3

z1 + z6 = λz4z2 = λz5z4 = λz6

Let z6 = 1. Then z4 = λ, z)1 = λ2 − 1 and z3 = λ2−1λ . Obviously,

z2 = λ4−3λ2+1λ . Therefore, z5 = λ4−3λ2+1

λ2. On the other hand, z5 =

λz2−z1 = λ4−4λ2 +2. Consequently, λ4−3λ2+1λ2

= λ4−4λ2 +2. We obtain

that λ6 − 5λ4 + 5λ2 − 1 = 0.

Let E7 be given as follows:

E7 :

• 3

• • • • • •5 2 1 4 6 7

Then

0 1 1 1 0 0 01 0 0 0 1 0 01 0 0 0 0 0 01 0 0 0 0 1 00 1 0 0 0 0 00 0 0 1 0 0 10 0 0 0 0 1 0

z1z2z3z4z5z6z7

= λ

z1z2z3z4z5z6z7


100M. A. Dokuchaev, N. M. Gubareni, V. M. Futorny, M.A. Khibina, and V. V. Kirichenko

z2 + z3 + z4 = λz1z1 + z5 = λz2

z1 = λz3z1 + z6 = λz4

z2 = λz5z4 + z7 = λz6

z6 = λz7

Let z7 = 1 and z6 = λ. Then z4 = λz6 − z7. We obtain z4 = λ2 − 1.Therefore, z1 = λ3 − 2λ. Obviously, z3 = λ2 − 2. We have z2 = λz1 −z3 − z4 and z2 = λ4 − 4λ2 + 3. From the equality z2 = λz5 it follows that

z5 = λ4−4λ2+3λ . On the other hand, z5 = λz2 − z1 = λ5 − 5λ3 + 5λ. So,

λ4−4λ2+3λ = λ5 − 5λ3 + 5λ and λ6 − 6λ4 + 9λ2 − 3 = 0.

Let E8 be given in the following form:

E8 :

• 3

• • • • • • •5 2 1 4 6 7 8

Then

0 1 1 1 0 0 0 01 0 0 0 1 0 0 01 0 0 0 0 0 0 01 0 0 0 0 1 0 00 1 0 0 0 0 0 00 0 0 1 0 0 1 00 0 0 0 0 1 0 10 0 0 0 0 0 1 0

z1z2z3z4z5z6z7z8

= λ

z1z2z3z4z5z6z7z8

We obtain the following system of linear equations

z2 + z3 + z4 = λz1z1 + z5 = λz2

z1 = λz3z1 + z6 = λz4

z2 = λz5z4 + z7 = λz6z6 + z8 = λz7

z7 = λz8

Let z8 = 1. Then z7 = λ and z6 = λ2 − 1. Obviously, we have: z4 =

λ3 − 2λ, z1 = λ4 − 3λ2 + 1, z3 = λ4−3λ2+1λ , z2 = λ6−5λ4+6λ2−1

λ and z5 =λ6−5λ4+6λ2−1

λ .



Hence, z5 = λz2 − z1 = λ6 − 6λ4 + 9λ2 − 2. Consequently, λ6 − 6λ4 +

9λ2 − 2 = λ6−5λ4+6λ2−1λ2

and λ8 − 7λ6 + 14λ4 − 8λ2 + 1 = 0

In conclusion we consider the following simple graph G5:

4

3 • 5• •

•

JJJJJJJJ

tttttttt

qqqqqqqqq

MMMMMM

MMM1

• •2 6

with the adjacency matrix [G5]:0 1 1 1 1 11 0 0 0 0 01 0 0 0 0 01 0 0 0 0 01 0 0 0 0 01 0 0 0 0 0

We have

[G5]~z = λ~z, (1)

where ~z = (z1, z2, z3, z4, z5, z6)T . From 1 we obtain

z2 + z3 + z4 + z5 + z6 = λz1z1 = λz2z1 = λz3z1 = λz4z1 = λz5z1 = λz6

.

Consequently, 5z1 = λ(z2 + z3 + z4 + z5 + z6) = λ2z1. Since, z1 6= 0, weobtain λ =

√5 and ~z = (

√5, 1, 1, 1, 1, 1) and rG5 =

√5 > 2.

Acknowledgements

The first and the third authors were partially supported by Fapespand CNPq (Brazil). The second and the last authors were supported by



FAPESP (Brazil), and they thank the Department of Mathematics of theUniversity of Sao Paulo for its warm hospitality during their visit in 2010.

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