Extending a theorem by Fiedler and applications to graph energy

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Extending a theorem by Fiedler and

applicatications to graph energy

María Robbiano Departamento de Matemáticas Universidad Católica del Norte Avenida Angamos 0610 Antofagasta, Chile

Departamento de Matemática, Universidade de Aveiro, 3810-193 Aveiro, Portugal. [email protected] Ivan Gutman University of Kragujevac, Kragujevac, Serbia. [email protected] April 10, 2011 Abstract

We use a lemma due to Fiedler to obtain eigenspaces of some graphs and we apply these results to graph energy..

1 Notations and preliminaries

Let A be an n n matrix.The scalars _{and vectors v 6= 0 satisfying Av = v} we call eigenvalues and eigenvectors of A; respectively and, any such pair, ( ; v); is called an eigenpair for A:

The set of distinct eigenvalues (including multiplicies), denoted by (A); is called the spectrum of A: The eigenvectors of the adjacency matrix of a graph G, together with the eigenvalues, provide a useful tool in the investiga-tion of the structure of the graph. In this work, having a result presented in

Research supported by Centre for Research on Optimization and Control (CEOC) from the "Fundação para a Ciência e a Tecnologia - FCT", co…nanced by the European Community Fund FEDER/POCI 2010.

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[1] as motivation, we obtain, for graphs with a special type of adjacency ma-trix, a lower bound for the energy of these graphs. In 1974 Fiedler obtained the following result [1] .

Let A, B be m m and n n symmetric matrices with corresponding eigenpairs ( i; ui) ; i = 1; : : : ; m, ( l; vl) ; l = 1; : : : ; n; respectively.

Lemma 1 [1] Let A, B be m m and n n symmetric matrices with corresponding eigenpairs ( i; ui) ; i = 1; : : : ; m, ( i; vi) ; i = 1; : : : ; n;

re-spectively. Suppose that ku1k = 1 = kv1k : Then, for any the matrix

C = A u1v

T 1

v1uT1 B

has eigenvalues 2; : : : ; n; 2; : : : ; m; 1; 2, where 1; 2 are eigenvalues

of

b

C = 1

1

:

We now o¤er a generalization of Fiedler’s lemma.

Suppose now that u1; : : : ; um (resp. v1; : : : ; vn) constitute an

ortho-normal system of eigenvectors of A (resp. B). Let ( u1j ; : : : ; jum) and

( v1j ; : : : ; jvn) be the matrices whose columns consist of the cordinates of

the eigenvectors ui and vi associated to the eigenvalues i and i;

respec-tively. In fact, ( u1j ; : : : ; jum) = 0 B B B @ u11 u12 u1m u21 u22 u2m .. . ... ... um1 um2 umm 1 C C C A; where, up = 0 B B B @ u1p u2p .. . ump 1 C C C A for p = 1; : : : ; m and ( v1j ; : : : ; jvn) = 0 B B B @ v11 v12 v1n v21 v22 v2n .. . ... ... vn1 vn2 vnn 1 C C C A, where vp= 0 B B B @ v1p v2p .. . vnp 1 C C C A for p = 1; : : : ; n:

Lemma 2 Let k _{min fm; ng and U = (u}1j ; : : : ; juk) and V = ( v1j ; : : : ; jvk) : Then,

for any ; the matrix

C = A U V

T

V UT _B

has eigenvalues k+1; : : : ; m; k+1; : : : ; n; 1j; 2j; j = 1; 2; : : : ; k; where

for s = 1; 2; _sj is an eigenvalue of b

Cj = j j

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Proof. For j = 1; 2; : : : ; k let _sj;wcsj ; s = 1; 2 be an eigenpair of the matrix b Cj = j j ; wherew_bsj = w1sj w2sj T

: Then, for j = 1; 2; : : : ; k; s = 1; 2; we have,

j j w1sj w2sj = sj w1sj w2sj : let w1sjuj w2sjvj be an (m + n) vector :Then, A U VT V UT B w1sjuj w2sjvj = w1sjAuj+ w2sjU V T_v j w_1sjV UTuj+w2sjBvj as w1sjAuj+ w2sjU VTvj = w1sj j 0 B B B @ u1j u2j . . . umj 1 C C C A+ w2sjU 0 B B B B B B @ 0 . . . 1 . . . 0 1 C C C C C C A and w_1sjV UTuj+w2sjBvj = w1sjV 0 B B B B B B @ 0 . . . 1 . . . 0 1 C C C C C C A + w2sj j 0 B B B @ v1j v2j . . . vnj 1 C C C A: Therefore, A U VT V UT B w1sjuj w2sjvj = w1sj juj+ w2sjuj w_1sjvj+w2sj jvj = (w1sj j + w2sj) uj w1sj + w2sj j vj = sjw1sjuj sjw2sjvj = _sj w1sjuj w2sjvj :

Recall that _sj;w_c_sj ; s = 1; 2 is an eigenpair of

b

Cj = j j

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wherew_bsj = w1sj w2sj T

: Then, for s = 1; 2; we have,

j j w1sj w2sj = sj w1sj w2sj : Therefore, for j = 1; 2; : : : ; k; s = 1; 2; _sj; w1sjuj w2sjvj are 2k eigenpairs

for C: For i = k + 1; : : : ; m; we have

A U VT V UT B ui 0 = Aui 0 = i ui 0 and for t = k + 1; : : : ; n we set

A U VT V UT _B 0 vt = 0 Bvt = _t 0 vt : Therefore, for i = k+1; : : : ; m i; ui 0 and for t = k+1; : : : ; n t; 0 vt are

eigenpairs for C: Thus the result is proved.

2 Applications

A simple graph G is a pair of sets (V; E) such that V is a nonempty …nite set of n vertices and E is the set of m edges. We say that G is a simple (n; m) graph. Let A(G) be the adjacency matrix of the graph G. Its eigenvalues

1; : : : ; n form the spectrum of G. (cf. [2]).

The notion of energy of an (n; m) graph G (written E(G)) is a nowadays much studied spectral invariant. This concept is of great interest in a vast range of …elds, aspecially in chemistry since it can be used to approximate the total -electron energy of a molecule (cf. [3], [4], [5] and [6]). It is de…ned by E(G) = n X j=1 j jj :

Given a complex m n matrix C; we index its singular values by s1(C); s2(C); : : : :

The value

E (C) =X

j

sj(C);

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en-ergy. Consequently, if the matrix C 2 Rn n is symmetric with eigenvalues

1(C); : : : ; n(C); its energy is given by

E (C) =

n

X

i=1

j i(C)j :

In this section, using Lemma 2, we present an application of the concept of energy to some special kinds of graphs.

We …rst formulate an auxiliary result.

Let Q = (qij) be an n1 n2 matrix with real-valued elements. Without

loss of generality we may assume that n1 n2: Let the singular values of Q

be s1; s2; :::; sn1: Then E(Q) =

Pn1

i=1si:

Lemma 3

E(Q) pn1kQkF

where kQkF is the Frobenius norm of Q; de…ned as,

kQkF = v u u t n1 X i=1 n2 X j=1 q2 ij: [17].

Proof. Let ₁; ₂; :::; _pbe real numbers. Their variance1_p

p X i=1 2 i 1p p X i=1 i !2

is known to be non-negative. Therefore,

p X i=1 i pp v u u t p X i=1 2 i.

Setting in the above inequality p = n1 and i = si; we obtain

E(Q) = n1 X i=1 si pn1 v u u t n1 X i=1 s2 i:

Now, the singularities of the matrix Q are just the square roots of the eigenvalues of QQT. Therefore,

n1

X

i=1

s2_i is equal to the sum of the eigenvalues of QQT, which in turn is the trace of QQT. Lemma 3 follows from

T r(QQT) = n1 X i=1 n2 X j=1 qijqTji= n1 X i=1 n2 X j=1 q2_ij _{= kQk}2_F .

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Let

A(G) = B X XT C

be a partition of the adjacency matrix of a graph G; where B and C have orders n1 n1and n2 n2; respectively. Thus X has order n1 n2: Morever,

let k _{min fn}1; n2g and consider the eigenpairs f( i; ui) : 1 i kg of B

f( i; vi) : 1 i kg of C such that the sets fu1; : : : ; ukg and fv1; : : : ; vkg

are orthonormal vectors. Let U = ( u1j : : : juk) and V = ( v1j : : : jvk).

Theorem 4 Let G be a graph of order n = n1+ n2 , such that its adjacency

matrix is given by A(G) = B X

XT C , where B and C represent adjacency matrices of graphs. Morever, let k _{min fn}1; n2g and consider the

eigen-pairs f( i; ui) : 1 i kg of B f( i; vi) : 1 i kg of C such that the sets

fu1; : : : ; ukg and fv1; : : : ; vkg are orthonormal vectors. Let U = (u1j : : : juk)

and V = ( v1j : : : jvk). Then E(G) 1₂ k P i=1 i+ i+ q ( i i)2+ 4 +12 k P i=1 i+ i q ( i i)2+ 4 + p n1 k s kBk2F k P i=1j ij2+ p n2 k s kCk2F k P i=1j ij 2₊ 2p_{min fn}1; n2g X U VT _F: Proof. A(G) = B X XT C = B U VT V UT C + 0 X U VT XT V UT 0 = M + N: Thus, according by Ky Fan theorem [16], E(G) E(M ) + E(N ):

Note that the spectrum of the matrix cMi = i

1 1 _i is Mci = 1 2 i+ i q ( i i) 2

+ 4 : Then appling Lemma 2,

E(M ) = 1 2 Pk i=1 i+ i+ q ( i i)2+ 4 + 1 2 Pk i=1 i+ i q ( i i)2+ 4 +Pn1 i=k+1j ij +Pni=k+12 j ij : By Lemma 3; Pn1 i=k+1j ij+Pni=k+12 j ij p n1 kqPni=k+11 j ij2=pn1 k q kBk2F Pk i=1j ij2 +pn2 k q kCk2F Pk i=1j ij2:

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Moreover, also by Lemma 3, E(N ) = 2E(X U VT) 2p_{min fn}1; n2g X U VT _F:

Consider the following special case of Theorem 4. Let both submatrices B abd C be equal, say equal to A, the adjacency matrix of an (n; m)-graph G .Further, let k = n1 = n2 = n: By setting in Lemma 2; A = B = A(G);

and considering ( i; ui) ; i = 1; : : : ; n, such that U = ( u1j ; : : : ; jun) is an

orthonormal matrix and = 1; the matrix

C = A U U T U UT A = A In In A ;

has eigenvalues _s1; _s2; : : : ; _sn, s = 1; 2; where for j = 1; 2; : : : ; n; s = 1; 2; _sj is an eigenvalue of b Cj = j 1 1 j ;

respectively. Thus _1j = j 1; 2j = j+ 1 and for the symmetric matrix

C; we have

E(C) =Pn_i=1_j i 1j+Pni=1j i+ 1j jPni=1( i 1)j+jPni=1( i+ 1)j = n+n = 2n:

In conclusion, the energy of the graph whose adjacency matrix is C is greater than its number of vertices.

At this point it is worth noting that the graph whose adjacency matrix is C is just the sum of the graphs G and K2 , denoted by G + K2 , where

K2 is the complete graph on two vertices. The graph operation marked

by + is described in detail in textbooks [quote Cvetkovic]. For the present consideration it is important that the spectrum of G1+ G2 consists of the

sums of eigenvalues of G1and G2 [2]. In view of this, the …nding that 1j = j 1; 2j = j+1 is an immediate consequence of the fact that the spectrum

of K2 consists of the numbers +1 and 1:

The result E(C) 2n can now be generalized as follows.

Let 1; :::; n1 and 1; :::; n2 be, respectively, the eigenvalues of G1 and

G2. Then the eigenvaues of G1+G2are of the form i+ j , i = 1; :::; n1; j =

1; :::; n2; and E(G1+ G2) = n1 P i=1 n2 P j=1 i+ j : We have, E(G1+ G2) n1 P i=1 n2 P j=1 i+ j = n1 P i=1jn 2 ij = n2 n1 P i=1j ij = n2E(G1)

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because of n2 P j=1 i = n2 i and n2 P j=1 j = 0:

In an analogous manner it can be shown that E(G1+ G2) n1E(G2):

The sum G1+ G2 has n1n2 vertices. Bearing this in mind, we see that

if either the energy of G1 exceeds or is equal to the number of vertices of

G1 or the energy of G2 exceeds or is equal to the number of vertices of G2 ,

then the energy of G1+ G2 exceeds the number of vertices of G1+ G2:

Let A; B; C; X; U and V be the same matrices as in the formulation and proof of Theorem 4. Let, as before M = B U V

T

V UT C and Q = 0 U VT

V UT 0 :

Theorem 5 E(A) E(M ) E(Q):

Proof. Let bA = B X

XT _C ; then as well known E(A) = E( bA): We

have A = M + P and bA = M + bP where P = 0 X U V T XT V UT 0 and bP = 0 X U VT XT V UT 0 : Therefore A + bA = 2M 2Q: By Ky Fan theorem [16],

2E(M ) E(A) + E( bA) + 2E(Q) implying the theorem.

As a special case of Theorem 5, for k = 1,

M = B u1v T 1 v1uT1 C and Q = 0 u1v T 1 v1uT1 0 :

Hence, E(Q) = 2 v₁Tu1 and E(A) E(M ) 2 v1Tu1 :

By Fiedler’s Lemma 1, E(M ) = E(B)+E(C)+1₂ 1+ 1+

q ( 1 1)2+ 4 + 1 2 1+ 1 q ( 1 1) 2 + 4 _(j 1j + j 1j):

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Therefore E(A) E(B) + E(C) + ";where " = 1₂ 1+ 1+ q ( 1 1) 2 + 4 + 1₂ 1+ 1 q ( 1 1) 2 + 4 (j 1j + j 1j) 2 vT1u1 :

The inequality E(A) E(B) + E(C) has been reported earlier [18]. Therefore our result will be an improvement of that inequality only if " > 0: If 1 > 0, 1 > 0 and 1 1 1; then 12 1+ 1+ q ( 1 1)2+ 4 + 1 2 1+ 1 q ( 1 1) 2

+ 4 _(j 1j + j 1j) = 0 and therefore " < 0: Thus,

in order to get an improvement, we must choose 1and 1such that 1 1<

1: For instance, for 1 = 1 = 0; " = 2 2 vT1u1 which is positive because

of vT₁u1 kv1k ku1k = 1 1 = 1:

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