We start with fixing some notations. Throughout the paper, let
G
=
(
V
,
E
)
be a simple graph with vertex set
V
=
{
v
1
,
v
2
,
…
,
v
n
}
and edge set
E
=
{
e
1
,
e
2
,
…
,
e
m
}
. Let
A
(
G
)
=
(
h
i
j
)
n
×
n
be the
(
0,1
)
-adjacency matrix of
G
, and let
D
(
G
)
=
diag
(
d
1
,
d
2
,
…
,
d
n
)
be the diagonal matrix with
d
i
being the degree of the
i
th vertex of
G
. The Laplacian matrix of
G
is defined to be
L
(
G
)
=
D
(
G
)
-
A
(
G
)
, and the corresponding characteristic polynomial of
L
(
G
)
is denoted by
P
(
G
,
λ
)
. Since the matrix
L
(
G
)
is symmetric, all its eigenvalues are real. We assume without loss of generality that they are arranged in the nondecreasing order; that is,
0
=
λ
1
≤
λ
2
≤
⋯
≤
λ
n
. The number of spanning trees of
G
is denoted by
t
(
G
)
. For other terminologies and notations which are not defined here, the reader is referred to [17].
Proof.
Let
L
(
G
)
be the Laplacian matrix of the graph
G
. Let
u
1
,
u
2
,
…
,
u
s
1
,
…
,
u
s
1
+
s
2
,
…
,
u
w
(
w
=
∑
i
=
1
3
s
i
)
be the labels of the graph
K
s
1
,
s
2
,
s
3
, where
u
j
~
u
i
if and only if
i
∈
A
a
and
j
∉
A
a
, or
i
∉
A
a
and
j
∈
A
a
, where
A
a
=
{
x
∣
∑
i
=
0
a
-
1
s
i
+
1
≤
x
≤
∑
i
=
0
a
s
i
}
(assume that
s
0
=
0
), for
a
=
1,2
,
3
. The node
(
u
i
,
v
j
)
of the graph
K
s
1
,
s
2
,
s
3
⊙
G
is the lexicographic order among vertex sequences; then we can represent the Laplacian matrix
L
(
K
s
1
,
s
2
,
s
3
⊙
G
)
in the following block matrix form:
(3)
L
(
K
s
1
,
s
2
,
s
3
⊙
G
)
=
[
D
1
0
⋯
0
0
D
2
⋯
0
⋮
⋮
⋮
0
0
⋯
D
w
]
-
[
A
11
A
12
⋯
A
1
w
A
21
A
22
⋯
A
2
w
⋮
⋮
⋮
A
w
1
A
w
2
⋯
A
w
w
]
,
where the element
a
j
l
i
k
of the block
A
i
k
=
(
a
j
l
i
k
)
n
×
n
denotes the adjacency relation of the two nodes
(
u
i
,
v
j
)
and
(
u
k
,
v
l
)
. By the definition of the composition graphs,
(
u
i
,
v
j
)
~
(
u
k
,
v
l
)
if and only if
u
i
~
u
k
; or
u
i
=
u
k
and
v
j
~
v
l
. Thus, we have
(4)
a
j
l
i
k
=
{
0
i
≠
k
i
,
k
∈
A
a
1
i
≠
k
otherwise
0
i
=
k
j
=
l
h
j
l
i
=
k
j
≠
l
(
a
=
1,2
,
3
)
.
That is,
(5)
A
i
k
=
{
J
0
i
≠
k
,
i
,
k
∈
A
a
J
n
i
≠
k
otherwise
A
(
G
)
i
=
k
otherwise
(
a
=
1,2
,
3
)
.
Similarly, since the sum of the elements of every row (resp., column) of Laplacian matrix
L
(
K
s
1
,
s
2
,
s
3
⊙
G
)
is zero, we have
(6)
D
i
=
{
D
(
G
)
+
n
(
w
-
s
1
)
I
n
1
≤
i
≤
s
1
D
(
G
)
+
n
(
w
-
s
2
)
I
n
s
1
+
1
≤
i
≤
s
1
+
s
2
D
(
G
)
+
n
(
w
-
s
3
)
I
n
s
1
+
s
2
+
1
≤
i
≤
s
1
+
s
2
+
s
3
.
Let
D
i
=
D
(
s
i
)
, where
D
(
s
i
)
is an
n
×
n
matrix,
i
=
1,2
,
…
,
w
. Then we get
(7)
L
(
K
s
1
,
s
2
,
s
3
⊙
G
)
=
|
C
1
-
J
s
1
×
s
2
-
J
s
1
×
s
3
-
J
s
2
×
s
1
C
2
-
J
s
2
×
s
3
-
J
s
3
×
s
1
-
J
s
3
×
s
2
C
3
|
w
×
w
,
w
=
∑
i
=
1
3
s
i
,
where
J
s
i
×
s
j
is an
s
i
by
s
j
block matrix with each block being equal to
J
n
,
i
,
j
=
1,2
,
3
, and
(8)
C
i
=
[
D
(
s
i
)
-
A
(
G
)
0
0
0
D
(
s
i
)
-
A
(
G
)
0
0
0
D
(
s
i
)
-
A
(
G
)
]
s
i
×
s
i
,
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
i
=
1,2
,
3
.
So the characteristic polynomial of the Laplacian matrix is
(9)
P
(
K
s
1
,
s
2
,
s
3
⊙
G
,
λ
)
=
det
(
λ
I
p
×
p
-
L
(
K
s
1
,
s
2
,
s
3
⊙
G
)
)
(10)
=
|
B
1
J
s
1
×
s
2
J
s
1
×
s
3
J
s
2
×
s
1
B
2
J
s
2
×
s
3
J
s
3
×
s
1
J
s
3
×
s
2
B
3
|
w
×
w
,
w
=
∑
i
=
1
n
s
i
,
where
(11)
B
i
=
[
λ
(
s
i
)
-
L
(
G
)
0
0
0
λ
(
s
i
)
-
L
(
G
)
0
0
0
λ
(
s
i
)
-
L
(
G
)
]
s
i
×
s
i
,
and
λ
(
s
i
)
=
(
λ
-
n
(
w
-
s
i
)
)
I
n
, for
i
=
1,2
,
3
.
In (10), according to the property of the Laplacian matrix, add all the columns to the first column; then every element of the first column is equal to
λ
. Extract
λ
; then every element of the first column is equal to 1. Using the notation
B
(
G
,
λ
)
from Lemma 1, we have
(12)
P
(
K
s
1
,
s
2
,
s
3
⊙
G
,
λ
)
=
det
(
λ
I
w
×
w
-
L
(
K
s
1
,
s
2
,
s
3
⊙
G
)
)
(13)
=
λ
|
B
*
J
s
1
×
s
2
J
s
1
×
s
3
J
s
2
×
s
1
B
2
J
s
2
×
s
3
J
s
3
×
s
1
J
s
3
×
s
2
B
3
|
w
×
w
,
w
=
∑
i
=
1
3
s
i
,
where
(14)
B
*
=
[
B
(
G
,
λ
-
n
(
w
-
s
1
)
)
0
⋯
0
C
λ
(
s
1
)
-
L
(
G
)
⋯
0
⋮
⋮
⋮
C
0
⋯
λ
(
s
1
)
-
L
(
G
)
]
s
1
×
s
2
C
=
[
1
0
⋯
0
1
0
⋯
0
⋮
⋮
⋮
1
0
⋯
0
]
.
In order to compute the determinant in (13), we start by subtracting the first column from the
(
n
s
1
+
1
)
th column to the last column, getting
(15)
P
(
K
s
1
,
s
2
,
s
3
⊙
G
,
λ
)
=
λ
|
B
*
0
0
J
s
2
×
s
1
A
2
0
J
s
3
×
s
1
J
s
3
×
s
2
A
3
|
w
×
w
,
w
=
∑
i
=
1
3
s
i
,
where
(16)
A
j
=
[
λ
(
s
j
)
-
L
(
G
)
-
J
n
-
J
n
⋯
-
J
n
-
J
n
λ
(
s
j
)
-
L
(
G
)
-
J
n
⋯
-
J
n
⋮
⋮
⋮
-
J
n
-
J
n
⋯
λ
(
s
j
)
-
L
(
G
)
-
J
n
]
,
s
j
×
s
j
for
j
=
2,3
.
It follows from (15) that
(17)
P
(
K
s
1
,
s
2
,
s
3
⊙
G
,
λ
)
=
λ
B
(
G
,
λ
-
n
(
w
-
s
1
)
)
×
[
det
(
(
λ
-
n
(
w
-
s
1
)
)
I
n
-
L
(
G
)
)
]
s
1
-
1
×
F
,
where
(18)
F
=
|
A
2
0
0
A
3
|
(
w
-
s
1
)
×
(
w
-
s
1
)
.
Let
H
i
=
det
(
A
i
)
,
i
=
2,3
. According to the property of the determinant, we have
(19)
F
=
H
2
×
H
3
=
∏
i
=
2
3
H
i
,
where
(20)
H
i
=
|
λ
(
s
i
)
-
L
(
G
)
-
J
n
-
J
n
⋯
-
J
n
-
J
n
λ
(
s
i
)
-
L
(
G
)
-
J
n
⋯
-
J
n
⋮
⋮
⋮
-
J
n
-
J
n
⋯
λ
(
s
i
)
-
L
(
G
)
-
J
n
|
s
i
×
s
i
,
for
i
=
2,3
.
Adding all the columns to the first column and extracting the term
λ
-
n
∑
i
=
1
3
s
i
, let
(21)
X
=
[
1
h
12
-
1
⋯
h
1
n
-
1
1
λ
-
n
(
w
-
s
i
)
-
d
2
-
1
⋯
h
2
n
-
1
⋮
⋮
⋮
1
h
n
2
-
1
⋯
λ
-
n
(
w
-
s
i
)
-
d
n
-
1
]
n
×
n
,
Y
=
[
1
-
1
⋯
-
1
1
-
1
⋯
-
1
⋮
⋮
⋮
1
-
1
⋯
-
1
]
n
×
n
then
(22)
H
i
=
(
λ
-
n
∑
i
=
1
3
s
i
)
|
X
-
J
n
⋯
-
J
n
Y
Z
⋯
-
J
n
⋮
⋮
⋮
Y
-
J
n
⋯
Z
|
s
i
×
s
i
,
where
Z
=
λ
-
n
(
w
-
s
i
)
I
n
-
L
(
G
)
-
J
n
, for
i
=
2,3
.
Adding the first column to all the other columns, we get
(23)
H
i
=
(
λ
-
n
w
)
|
B
(
G
,
λ
-
n
(
w
-
s
i
)
)
0
⋯
0
C
Z
+
J
n
⋯
0
⋮
⋮
⋮
C
0
⋯
Z
+
J
n
|
s
i
×
s
i
(24)
=
(
λ
-
n
w
)
·
B
(
G
,
λ
-
n
(
w
-
s
i
)
)
·
[
det
(
λ
-
n
(
w
-
s
i
)
I
n
-
L
(
G
)
)
]
s
i
-
1
,
for
i
=
2,3
.
From Lemma 1 and the property of the determinant, notice that
B
(
G
,
λ
)
=
∏
i
=
2
n
(
λ
-
λ
i
(
G
)
)
,
det
(
λ
I
n
-
L
(
G
)
)
=
λ
∏
i
=
2
n
(
λ
-
λ
i
(
G
)
)
; substituting these into equality (24), we get the value of the determinant
H
i
:
(25)
H
i
=
(
λ
-
n
w
)
·
∏
i
=
2
n
(
λ
-
n
(
w
-
s
i
)
-
λ
i
(
G
)
)
×
[
(
λ
-
n
(
w
-
s
i
)
)
·
∏
i
=
2
n
[
λ
-
n
(
w
-
s
i
)
-
λ
i
(
G
)
]
]
s
i
-
1
for
i
=
2,3
.
Combining (17), (19), and (25), we obtain
(26)
P
(
K
s
1
,
s
2
,
s
3
⊙
G
,
λ
)
=
λ
(
λ
-
n
w
)
2
∏
i
=
1
n
(
λ
-
n
(
w
-
s
i
)
)
s
i
-
1
×
∏
i
=
1
3
∏
k
=
2
n
[
λ
-
n
(
w
-
s
i
)
-
λ
k
(
G
)
]
s
i
.
Then we get the eigenvalues of
K
s
1
,
s
2
,
s
3
⊙
G
from above equality immediately. This completes the proof.