THE UPPER BOUND FOR THE LARGEST
SIGNLESS LAPLACIAN EIGENVALUE OF WEIGHTED GRAPHS
Şerife BÜYÜKKÖSE 1 , Nurşah MUTLU 2
1
Gazi University, Faculty of Sciences, Departments Mathematics, 06500, Teknikokullar\Ankara,
Turkey, [email protected]
2
Gazi University, Graduate School of Natural and Applied Sciences, Departments Mathematics,
06500, Teknikokullar\Ankara, Turkey, [email protected]
ABSTRACT
In this study, we find an upper bound for the largest signless Laplacian eigenvalue of simple
connected weighted graphs, where the edge weights are positive definite square matrices. Also
we obtain some results on weighted and unweighted graphs by using this bound.
Keywords: Weighted graph, signless Laplacian eigenvalue, upper bound
1. INTRODUCTION
In this paper, we consider a simple connected weighted graph in which the edge weights are
positive definite square matrices. Let G  V , E  be a simple connected weighted graph on n
vertices. Denote by wij the positive definite weight matrix of order t of the edge ij and assume
that wij  w ji . Let wi 
w
ij
j: j ~i
, for all i  V .
The signless Laplacian matrix QG  of a weighted graph G is a block matrix and defined as
QG   qij ntnt , where
 wi

qij   wij
0

;
if i  j,
;
if i ~ j,
; otherwise.
In the definitions above, the zero denotes the t  t zero matrix. Thus QG  is a square matrix of
order nt . Let q1 denote the largest signless Laplacian eigenvalue of QG  and q1 wij  denote the
largest eigenvalue of wij .
Upper bounds for the largest signless Laplacian eigenvalue for unweighted graphs have been
investigated to great extent in the literature. In Section 2, we give an upper bound for the largest
signless Laplacian eigenvalue of weighted graphs. We also characterize graphs for which equality
holds in the upper bound.
Lemma 1.
Let A be a Hermitian n  n matrix with eigenvalues q1  q2  ...  qn , then for any



x  Rn x  0 , y  Rn y  0
T
T

T
x A y  q1 x x y y .
Equality holds if and only if x is an eigenvector of A corresponding to the largest eigenvalue q1
and y  α x for some α  R [5].
Theorem 1.
If A is a Hermitian n  n matrix with eigenvalues q1  q2  ...  qn , then for any x  C n
T
T
T
qn x x  x A x  q1 x x
T
qmax  q1  max
x Ax
qmin  qn  min
x Ax
x 0
T
x x
T
 max
x Ax
T
x x 1
T
x 0
T
x x
T
 min
x A x [5].
T
x x 1
Corollary 1.
Let A M n have eigenvalues qi . Even if A is not Hermitian, one has the bounds
T
min
x 0 , y 0
x Ay
T
x y
T
 qi  max
x 0 , y 0

x Ay
T
,
x y



for any x  R n x  0 , y  R n y  0 and for i  1, 2,..., n [8].
Corollary 2.
Let A, B  M n are positive definite matrices and k  N . Then, Ak and A  B are also positive
definite matrices [5].
2. MAIN RESULTS
In this section we find an upper bound on the largest signless Laplacian eigenvalue of simple
connected weighted graphs.
Theorem 2.
Let G be a simple connected weighted graph. Then,




2
2
q1  max  q1 wi    q1 wik    q1 wi wis  wis ws     q1 wik wkr   .
iV
k :k ~i
s:s ~i
1i , r n k :k ~i


i r
k ~r


Moreover equality holds in (1) if and only if
(i) G is a bipartite regular graph or a bipartite semiregular graph,
(1)
(ii) wij have a common eigenvector corresponding to the largest eigenvalue q1 wij  for all i, j .
Proof.
Let us consider the matrix Q 2 G  . The i, j   th element of Q 2 G  is


2
2
wi   wij

j: j ~ i


w
w

w
w
 i ij
ij j   wik wkj
k :k ~i

k~ j

w
w
 ik kj

k :k ~i

k~ j

Let x  x1 , x2 ,..., xn
T
T

T T
;
if i  j,
;
if i ~ j,
; otherwise.
be an eigenvector corresponding to the eigenvalue q1 of Q 2 G  and
2
xi be the vector component of x such that


xi xi  max xk xk .
T
kV
T
(2)
Since x is nonezero, so is xi . We have
Q 2 G x  q1 x .
2
(3)
From the i  th equation of (3), we get
q1 xi  wi xi 
2
2
w
ik
2
xi 
k :k ~i
 w w
i
is
 wis ws x s 
s:s ~i
 w
1i , r  n k :k ~i
i r
k ~r
ik
wkr x r ,
i.e.,
q1 xi xi  xi wi xi 
2
T
T
2
x
k :k ~i
T
i
wik xi   xi wi wis  wis ws xs 
2
T
s:s ~i
From Corollary 1 and Lemma 1, we have
 x
1i , r n k :k ~i
i r
k ~r
T
i
wik wkr xr .
(4)
 
 q w x
 q1 wi xi xi 
2
T
2
1
ik
i
T
xi   xi wi wis  wis ws xs 
k :k ~i
T
s:s ~i
 x
T
i
wik wkr xr .
(5)
1i , r n k :k ~i
i r
k ~r
Four cases arise;
1. wi wis  wis ws and wik wkr are Hermitian matrices for all s, s ~ i and for all k , k ~ i, k ~ r,
1  i, r  n ,
2. wi wis  wis ws is a Hermitian matrix for all s, s ~ i and wik wkr is not a Hermitian matrix
for all k , k ~ i, k ~ r, 1  i, r  n ,
3. wik wkr is a Hermitian matrix for all k , k ~ i, k ~ r, 1  i, r  n and wi wis  wis ws is not a
Hermitian matrix for all s, s ~ i ,
4. wi wis  wis ws and wik wkr are not Hermitian matrices for all s, s ~ i and for all
k , k ~ i, k ~ r, 1  i, r  n ,
Case 1: wi wis  wis ws and wik wkr are Hermitian matrices. From (5), (2) and using Lemma 1 we
get




2
2
q1  max  q1 wi    q1 wik    q1 wi wis  wis ws     q1 wik wkr  
iV
k :k ~i
s:s ~i
1i , r n k :k ~i


i r
k ~r


Case 2: wi wis  wis ws is a Hermitian matrix and wik wkr is not a Hermitian matrix. Let us take the
ratio of
T
x  wk wkr x r
T
x xr
for 1  , r  n . If N   N r  0 , this ratio is zero, where N   N r is the number of common
neighbors of  and r . So let us consider N   N r  0 . From (2) and using the Cauchy-Schwarz
inequality, we have
T
T
T
x wk wkr x r
x wk wkr x r
x wk wkr x r
.


T
T
T
T
T
x x r
x x x r x r
xi xi x r x r
(6)
Since (6) implies for each x and xr
 x T wk wkr x r 
min 

T
x  0 , x r  0
 x xr

T
xi wik wkr x r
T
.
T
(7)
xi xi x r x r
From (2), (7) and using Corollary 1, we get
 x
T
i
wik wkr xr 
1i , r n k :k ~i
i r
k ~r
  q w
1
ik
T
wkr xi xi .
(8)
1i , r n k :k ~i
i r
k ~r
Since wi wis  wis ws is a Hermitian matrix, from (2), (5), (8) and using Lemma 1, we get




2
2
q1  max  q1 wi    q1 wik    q1 wi wis  wis ws     q1 wik wkr   .
iV
k :k ~i
s:s ~i
1i , r n k :k ~i


i r
k ~r


Case 3: wik wkr is a Hermitian matrix and wi wis  wis ws is not a Hermitian matrix. By a similar
argument to Case 2 we have
 x w w
T
i
s:s ~i
i
is
 wis ws xs 
 q w w
1
i
is
 wis ws xi xi .
T
(9)
s:s ~i
Since wik wkr is a Hermitian matrix, from (2), (5), (9) and using Lemma 1, we get




2
2
q1  max  q1 wi    q1 wik    q1 wi wis  wis ws     q1 wik wkr  
iV
k :k ~i
s:s ~i
1i , r n k :k ~i


i r
k ~r


Case 4: wi wis  wis ws and wik wkr are not Hermitian matrices. By applying the same methods as
Case 2 and Case 3, we can show that
 
 q w x
q1 xi xi  q1 wi xi xi 
2
T
2
T
2
1
ik
T
i
xi   xi wi wis  wis ws xs 
T
k :k ~i
 
 q1 wi xi xi 
2
T
s:s ~i
 q w x
2
1
ik
T
i
xi 
k :k ~i


 q w w
1
i
is
 x
i
T
wik wkr xr
1i , r n k :k ~i
i r
k ~r
 wis ws  xi xi x s x s
T
T
s:s ~i
 q1 wik wkr  xi xi xr xr .
T
T
(10)
1i , r  n k :k ~i
i r
k ~r
From (2), we get




2
2
q1  max  q1 wi    q1 wik    q1 wi wis  wis ws     q1 wik wkr   .
iV
k :k ~i
s:s ~i
1i , r n k :k ~i


i r
k ~r


Now suppose that equality holds in (1). Then all inequalities in the above argument must be
equalities. From equality in (10), we have
xk xk  xi xi ,
T
T
(11)
for all k , k ~ i and for all k , k ~ p, p ~ i . From equality in (10) and using Lemma 1, we get
x s  ais xi and x r  bir xi ,
(12)
for any s, s ~ i and for any r, k ~ r, k ~ i, 1  i, r  n , where ais , bir  R . From (11) and (12),
we get
a
2
is



T
2
T
 1 xi xi  0 and bir  1 xi xi  0 ,
i.e.,
ais , bir  1, as xi xi  0 .
T
(13)
On the other hand from Corollary 2 wi 2 and wik 2 are positive definite matrices for a 1  i, k  n .
Thus, we get
xi wi xi  0 and xi wik xi  0 .
T
2
T
2
(14)
From (4), (13) and (14), we have
 a
s:s ~i

x
is i
T
wi wis  wis ws xi 
  b
1i , r  n k :k ~i
i r
k ~r
xi wi wis  wis ws xi
T


(15)
x wik wkr xi  xi wik wkr xi  0.
T
T
ir i
Four cases arise;
i.
xi wi wis  wis ws xi  0 and xi wik wkr xi  0 ,
ii.
xi wi wis  wis ws xi  0 and xi wik wkr xi  0 ,
iii.
xi wi wis  wis ws xi  0 and xi wik wkr xi  0 ,
iv.
xi wi wis  wis ws xi  0 and xi wik wkr xi  0 .
T
T
T
T
T
T
T
T
Case i: xi T wi wis  wis ws xi  0 and xi T wik wkr xi  0 .
From
(15),
we
ais  1 and bir  1 ,
get
for
all
s, s ~ i and for all 1  i, r  n .
Let
U  k : xk  xi , k ~ i and W={k : xk  xi , k≁i}. Since U W  Ø and U W  V, G is
bipartite. Now we have
QG xi  q1 xi ,
i.e.,
q1 xi  wi xi 
w
ik
xk .
k :k ~i
For i, j  U ,
q1 xi  q1 wi xi 
 q w x
,
(16)
 q w x .
(17)
1
ik
i
k :k ~i
q1 xi  q1 w j xi 
1
k :k ~ j
jk
i
From (16), (17) and xi  0 , we get q1 wi   q1 w j . Therefore q1 wi  is constant for all i  U .
Similarly we can also show that q1 wi  is constant for all i  W . Hence G is a bipartite regular
graph.
Case ii: xi T wi wis  wis ws xi  0 and xi T wik wkr xi  0 .
From
(15),
we
get
ais  1 and bir  1 ,
for
all
s, s ~ i and for all 1  i, r  n .
Let
U  k : xk  xi  and W  k : xk   xi . Since U W  Ø and U W  V, G is bipartite. By a
similar argument Case i, we can show that q1 wi  is constant for all i  U and q1 w j  is constant
for all j  W . Hence G is a bipartite semiregular graph.
Case iii: xi T wi wis  wis ws xi  0 and xi T wik wkr xi  0 ,
Case iv: xi T wi wis  wis ws xi  0 and xi T wik wkr xi  0 .
By applying the same methods as Case i and Case ii, we can show that G is a bipartite regular
graph or a bipartite semiregular graph.
Coversely, suppose that conditions (i)-(ii) of theorem hold for the graph G .
G is a bipartite regular graph. Let U and W be the vertex classes of G . Also, let
q1 wi   q1 w j   α for i  U and for j  W . The following equation can be easily verified:
2
2

w1   w1k
 w1 w1n  w1n wn   w1k wkn 

k:k~ 1
k:k~ 1
 x 
 x 
k~n
 
2
   .
4α     




 
2
2
 x
wn   wnk
 x 
   wn wn1  wn1 w1   wnk wk 1 
k:k~n
k:k~n


k~ 1


Thus 4α 2 is an eigenvalue of Q 2 G  . So,
2α  q1 .
On the other hand, we have




2
2
q1  max  q1 wi    q1 wik    q1 wi wis  wis ws     q1 wik wkr    2α .
iV
k:k~i
s:s~i
1i,r n k:k~i


i r
k~r


Thus




2
2
q1  max  q1 wi    q1 wik    q1 wi wis  wis ws     q1 wik wkr   .
iV
k :k ~i
s:s ~i
1i , r  n k :k ~i


i r
k ~r


Hence the theorem is proved.
Corollary 3.
Let G be a simple connected weighted graph where each edge weight wij is a positive number.
Then




2
2
q1  max  wi   wik   wi wis  wis ws     wik wkr  .
iV
k :k ~i
s:s ~i
1i , r n k :k ~i


i r
k ~r


(18)
Moreover equality holds in (18) if and only if G is a bipartite regular graph or a bipartite
semiregular graph.
Proof.
For weighted graph where the edge weights wij are positive number, we have
q1 wij   wij and q1 wi   wi ,
for all i, j . Using Theorem 2 we get the required result.
Corollary 4.
Let G be a simple connected unweighted graph. Then


2
q1  max  d i  d i 
iV



d i  d j    N i  N j  ,

j: j ~i
1i , j  n

i j

(19)
where d i is the degree of vertex i and N i  N j is the number of common neighbors of i and
j . Moreover, equality holds in (19) if and only if G is a bipartite regular graph or a bipartite
semiregular graph.
Proof.
For an unweighted graph, wij  1 and wi  d i for all i, j and i ~ j . Using Corollary 3 we get the
required result.
CONFLICT OF INTEREST
No conflict of interest was declared by the authors.
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