# On Jacobsthal and Jacobsthal-Lucas Circulant Type Matrices

Authors: Yanpeng Gong; Zhaolin Jiang; Yun Gao
Journal: Abstract and Applied Analysis
 (2015)

---

## Abstract

Circulant type matrices have become an important tool in solving fractional order differential equations.
In this paper, we consider the circulant and left circulant andg-circulant matrices with
the Jacobsthal and Jacobsthal-Lucas numbers. First, we discuss the invertibility of the circulant
matrix and present the determinant and the inverse matrix. Furthermore, the invertibility of the
left circulant and g-circulant matrices is also discussed. We obtain the determinants and the inverse
matrices of the left circulant and g-circulant matrices by utilizing the relation between left
circulant, g-circulant matrices, and circulant matrix, respectively.


---

## Body

## 1. Introduction

Circulant type matrices possess interesting properties, which have been exploited to obtain the transient solution in a closed form for fractional order differential equations. In [1], Qu et al. derived the discretized linear system from fractional advection-diffusion equations (FADEs) with constant coefficients and introduced the background of circulant and skew-circulant splitting (CSCS) iteration for solving the discretized linear system. In [2], some Routh-Hurwitz stability conditions are generalized to the fractional order case. Ahmed et al. considered the 1-system CML (10) in [2]. They selected a circulant matrix, which reads a tridiagonal matrix. Ahmed et al. used coupled map lattices (CML) as an alternative approach to include spatial effects in fractional order systems (FOS). Consider the 1-system CML (10) in [3]. They claimed that the system is stable if all the eigenvalues of the circulant matrix satisfy (2) in [3]. Lei and Sun [4] proposed the preconditioned CGNR (PCGNR) method with a circulant preconditioner to solve such Toeplitz-like systems. Kloeden et al. adopted the multilevel Monte Carlo method introduced by M. Giles to SDEs with additive fractional noise. They adopted the simplest approximation schemes for (1) in [5] with the Euler method, which reads (5) in [5]. They exploited that the covariance matrix of the increments can be embedded in a circulant matrix. The total loops can be done by fast Fourier transformation, which leads to a total computational cost of O
(
m
log
⁡
⁡
m
), where m is the order of the circulant matrix.Jacobsthal and Jacobsthal-Lucas circulant type matrices are circulant type matrices with Jacobsthal and Jacobsthal-Lucas numbers. Circulant type matrices include the circulant and left circulant andg-circulant matrices. These matrices have important applications in various disciplines including image processing, communications, signal processing, encoding, solving Toeplitz matrix problems, preconditioner, and solving least squares problems. They have been put on the firm basis with the work of Davis [6] and Jiang and Zhou [7].Theg-circulant matrices play an important role in various applications as well; for details please refer to [8, 9] and the references therein. Ngondiep et al. showed the singular values of g-circulants in [10]. There are discussions about the convergence in probability and in distribution of the spectral norm of g-circulant matrices in [11].Forn
>
0, the Jacobsthal number {
J
n
} is defined by J
n
+
1
=
J
n
+
2
J
n
-
1, where J
0
=
0 and J
1
=
1. If we generalize J
n
+
1 to j
n
+
1
=
j
n
+
2
j
n
-
1 and let j
0
=
2, j
1
=
1, then we obtain Jacobsthal-Lucas number {
j
n
}. Let α and β be the roots of the characteristic equation x
2
-
x
-
2
=
0; then the Binet formulas of the sequences {
J
n
} and {
j
n
} have the form [12, 13](1)
J
n
=
α
n
-
β
n
α
-
β
,
j
n
=
α
n
+
β
n
.Besides, some scholars have given various algorithms for the determinants and inverses of nonsingular circulant matrices in [6]. It is worth pointing out that the computational complexity of these algorithms is amazing with the order of matrix increasing. However, some authors gave the explicit determinant and inverse of the circulant and skew-circulant matrix involving Fibonacci and Lucas numbers. For example, authors [14] studied the nonsingularity of the skew-circulant type matrices and presented explicit determinants and inverse matrices of these special matrices. Furthermore, four kinds of norms and bounds for the spread of these matrices are given separately. In [15], the nonsingularity of circulant type matrices with the sum and product of Fibonacci and Lucas numbers is discussed. And the exact determinants and inverses of these matrices are given. In [16], Jiang and Hong presented exact determinants of some special circulant matrices involving four kinds of famous numbers. Jaiswal evaluated some determinants of circulant matrix whose elements are the generalized Fibonacci numbers in [17]. Lind presented the determinants of circulant and skew-circulant matrix involving Fibonacci numbers in [18]. Dazheng gave the determinant of the Fibonacci-Lucas quasicyclic matrices in [19]. Shen et al. considered circulant matrices with Fibonacci and Lucas numbers and presented their explicit determinants and inverses by constructing the transformation matrices in [20].The purpose of this paper is to obtain the better results for the determinants and inverses of circulant type matrices by some properties of Jacobsthal and Jacobsthal-Lucas numbers.Definition 1 (see [21]).
A right circulant matrix (or simply, circulant matrix) is written as(2)
Circ
a
1
,
a
2
,
…
,
a
n
=
a
1
a
2
⋯
a
n
a
n
a
1
⋯
a
n
-
1
⋮
⋮
⋮
a
2
a
3
⋯
a
1
.Definition 2 (see [21]).
A left circulant matrix is written as(3)
LCirc
a
1
,
a
2
,
…
,
a
n
=
a
1
a
2
⋯
a
n
a
2
a
3
⋯
a
1
⋮
⋮
⋮
a
n
a
1
⋯
a
n
-
1
.Definition 3 (see [22]).
Ag-circulant matrix is an n
×
n complex matrix with the following form:(4)
A
g
,
n
=
a
1
a
2
⋯
a
n
a
n
-
g
+
1
a
n
-
g
+
2
⋯
a
n
-
g
a
n
-
2
g
+
1
a
n
-
2
g
+
2
⋯
a
n
-
2
g
⋮
⋮
⋱
⋮
a
g
+
1
a
g
+
2
⋯
a
g
,where g is a nonnegative integer and each of the subscripts is understood to be reduced modulo n.The first row ofA
g
,
n is a
1
,
a
2
,
…
,
a
n; its j
+
1th row is obtained by giving its jth row a right circular shift by g positions (equivalently, g mod n positions). Note that g
=
1 or g
=
n
+
1 yields the standard circulant matrix. If g
=
n
-
1, then we obtain the left circulant matrix.


## 2. Circulant Matrix with the Jacobsthal Numbers

In this section, letA
n
=
Circ
J
1
,
J
2
,
…
,
J
n be a circulant matrix. Firstly, we give a determinant formula for the matrix A
n. Afterwards, we prove that A
n is an invertible matrix for n
>
2, and then we find the inverse of the matrix A
n.Theorem 4.
LetA
n
=
Circ
J
1
,
J
2
,
…
,
J
n be a circulant matrix; then we have(5)
det
⁡
A
n
=
1
-
J
n
+
1
n
-
1
+
2
J
n
n
-
2
∑
k
=
1
n
-
1
‍
2
J
k
1
-
J
n
+
1
2
J
n
k
-
1
,where J
n is the nth Jacobsthal number.Proof.
Obviously, whenn
=
1, det
⁡
A
1
=
1 satisfies the formula (5). In the case n
>
1, let(6)
Γ
=
1
-
1
1
-
2
1
-
1
0
0
1
-
1
-
2
⋮
⋰
⋰
⋰
0
1
⋰
⋰
0
1
-
1
⋰
0
0
1
-
1
-
2
,
Π
1
=
1
0
0
⋯
0
0
0
2
J
n
J
1
-
J
n
+
1
n
-
2
0
⋯
0
1
0
2
J
n
J
1
-
J
n
+
1
n
-
3
0
⋯
1
0
⋮
⋮
⋮
⋱
⋮
⋮
0
2
J
n
J
1
-
J
n
+
1
1
⋯
0
0
0
1
0
⋯
0
0be two n
×
n matrices; then the product Γ
A
n
Π
1 has the form(7)
J
1
δ
n
′
J
n
-
1
⋯
J
3
J
2
0
δ
n
2
J
n
-
2
⋯
2
J
2
2
J
1
0
0
J
1
-
J
n
+
1
0
0
-
2
J
n
⋱
0
⋮
⋮
⋱
⋱
0
0
0
0
0
-
2
J
n
J
1
-
J
n
+
1
,where(8)
δ
n
=
J
1
-
J
n
+
∑
k
=
1
n
-
2
‍
2
J
k
2
J
n
J
1
-
J
n
+
1
n
-
k
+
1
,
δ
n
′
=
∑
k
=
1
n
-
1
‍
J
k
+
1
2
J
n
J
1
-
J
n
+
1
n
-
k
+
1
.So we obtain(9)
det
⁡
Γ
det
⁡
A
n
det
⁡
Π
1
=
J
1
×
J
1
-
J
n
+
∑
k
=
1
n
-
2
‍
2
J
k
2
J
n
J
1
-
J
n
+
1
n
-
k
+
1
×
J
1
-
J
n
+
1
n
-
2
=
J
1
×
J
1
-
J
n
+
1
+
∑
k
=
1
n
-
1
‍
2
J
k
2
J
n
J
1
-
J
n
+
1
n
-
k
+
1
×
J
1
-
J
n
+
1
n
-
2
=
1
-
J
n
+
1
n
-
1
+
2
J
n
n
-
2
∑
k
=
1
n
-
1
‍
2
J
k
1
-
J
n
+
1
2
J
n
k
-
1
,while(10)
det
⁡
Γ
=
det
⁡
Π
1
=
-
1
n
-
1
n
-
2
/
2
,and we have (11)
det
⁡
A
n
=
1
-
J
n
+
1
n
-
1
+
2
J
n
n
-
2
∑
k
=
1
n
-
1
‍
2
J
k
1
-
J
n
+
1
2
J
n
k
-
1
.The proof is completed.Theorem 5.
LetA
n
=
Circ
J
1
,
J
2
,
…
,
J
n be a circulant matrix. If n
>
2, then A
n is an invertible matrix.Proof.
Whenn
=
3 in Theorem 4, then we have det
⁡
A
3
=
20
≠
0; hence A
3 is invertible. In the case n
>
3, since J
n
=
α
n
-
β
n
/
α
-
β, where α
+
β
=
1,  α
β
=
-
2, we have(12)
f
ω
k
=
∑
i
=
1
n
‍
J
i
ω
k
i
-
1
=
1
α
-
β
∑
i
=
1
n
‍
α
i
-
β
i
ω
k
i
-
1
=
1
α
-
β
α
1
-
α
n
1
-
α
ω
k
-
β
1
-
β
n
1
-
β
ω
k
=
1
α
-
β
α
-
β
-
α
n
+
1
-
β
n
+
1
1
-
α
+
β
ω
k
+
α
β
ω
2
k
k
k
k
k
k
k
k
k
i
+
α
-
β
-
α
n
+
1
-
β
n
+
1
1
-
α
+
β
ω
k
+
α
β
ω
2
k
α
β
α
n
-
β
n
ω
k
1
-
α
+
β
ω
k
+
α
β
ω
2
k
=
1
-
J
n
+
1
-
2
J
n
ω
k
1
-
ω
k
-
2
ω
2
k
k
=
1,2
,
…
,
n
-
1
.If there exists ω
l  l
=
1,2
,
…
,
n
-
1 such that f
ω
l
=
0, then we obtain 1
-
J
n
+
1
-
2
J
n
ω
l
=
0 for 1
-
ω
l
-
2
ω
2
l
≠
0; thus, ω
l
=
1
-
J
n
+
1
/
2
J
n is a real number. While(13)
ω
l
=
exp
⁡
2
l
π
i
n
=
cos
⁡
2
l
π
n
+
i
sin
2
l
π
n
,
sin
⁡
(
2
l
π
/
n
)
=
0, so we have ω
l
=
-
1 for 0
<
2
l
π
/
n
<
2
π. But x
=
-
1 is not the root of the equation 1
-
J
n
+
1
-
2
J
n
x
=
0  n
>
3. We obtain f
ω
k
≠
0 for any ω
k  k
=
1,2
,
…
,
n
-
1, while f
1
=
∑
i
=
1
n
‍
J
i
=
-
1
/
2
1
-
J
n
+
1
-
2
J
n
≠
0. By Lemma 4 in [15], the proof is completed.Lemma 6.
Let the matrixG
=
[
g
i
j
]
i
,
j
=
1
n
-
2 be of the form(14)
g
i
j
=
J
1
-
J
n
+
1
,
i
=
j
,
-
2
J
n
,
i
=
j
+
1
,
0
,
o
t
h
e
r
w
i
s
eand then the inverse G
-
1
=
[
g
i
j
′
]
i
,
j
=
1
n
-
2 is equal to (15)
g
i
j
′
=
2
J
n
i
-
j
J
1
-
J
n
+
1
i
-
j
+
1
,
i
≥
j
,
0
,
i
<
j
.Proof.
Letc
i
j
=
∑
k
=
1
n
-
2
‍
g
i
k
g
k
j
′. Obviously, c
i
j
=
0 for i
<
j. In the case i
=
j, we obtain(16)
c
i
i
=
g
i
i
g
i
i
′
=
J
1
-
J
n
+
1
·
1
J
1
-
J
n
+
1
=
1
.For i
≥
j
+
1, we obtain(17)
c
i
j
=
∑
k
=
1
n
-
2
‍
g
i
k
g
k
j
′
=
g
i
,
i
-
1
g
i
-
1
,
j
′
+
g
i
i
g
i
j
′
=
-
2
J
n
·
2
J
n
i
-
j
-
1
J
1
-
J
n
+
1
i
-
j
+
J
1
-
J
n
+
1
·
2
J
n
i
-
j
J
1
-
J
n
+
1
i
-
j
+
1
=
0
.Hence, we verify G
G
-
1
=
I
n
-
2, where I
n
-
2 is n
-
2
×
n
-
2 identity matrix. Similarly, we can verify G
-
1
G
=
I
n
-
2. Thus, the proof is completed.Theorem 7.
LetA
n
=
Circ
(
J
1
,
J
2
,
…
,
J
n
)  (
n
>
2
) be a circulant matrix; then one has(18)
A
n
-
1
=
1
δ
n
Circ
x
1
′
,
x
2
′
,
…
,
x
n
′
,where(19)
x
1
′
=
1
+
2
∑
i
=
1
n
-
2
‍
J
n
-
i
2
J
n
i
-
1
J
1
-
J
n
+
1
i
,
x
2
′
=
-
1
+
4
∑
i
=
1
n
-
2
‍
J
n
-
1
-
i
2
J
n
i
-
1
J
1
-
J
n
+
1
i
,
x
k
′
=
-
2
2
J
n
k
-
3
J
1
-
J
n
+
1
k
-
2
,
k
=
3,4
,
…
,
n
,
δ
n
=
J
1
-
J
n
+
∑
k
=
1
n
-
2
‍
2
J
k
2
J
n
J
1
-
J
n
+
1
n
-
(
k
+
1
)
.Proof.
Let(20)
Π
2
=
1
-
δ
n
′
2
δ
n
′
δ
n
J
n
-
2
-
J
n
-
1
⋯
2
δ
n
′
δ
n
J
1
-
J
2
0
1
-
2
J
n
-
2
δ
n
⋯
-
2
J
1
δ
n
0
0
1
⋯
0
0
0
0
⋯
0
⋮
⋮
⋮
⋱
⋮
0
0
0
⋯
1
,where(21)
δ
n
=
J
1
-
J
n
+
∑
k
=
1
n
-
2
‍
2
J
k
2
J
n
J
1
-
J
n
+
1
n
-
(
k
+
1
)
,
δ
n
′
=
∑
k
=
1
n
-
1
‍
J
k
+
1
2
J
n
J
1
-
J
n
+
1
n
-
(
k
+
1
)
.Then we have(22)
Γ
A
n
Π
1
Π
2
=
D
1
⊕
G
,where D
1
=
diag
J
1
,
δ
n is a diagonal matrix, and D
1
⊕
G is the direct sum of D
1 and G. If we denote Π
=
Π
1
Π
2, then we obtain(23)
A
n
-
1
=
Π
D
1
-
1
⊕
G
-
1
Γ
,since the last row entries of the matrix Π are 0,1
,
-
2
(
J
n
-
2
/
δ
n
)
,
-
2
(
J
n
-
3
/
δ
n
)
,
…
,
-
2
(
J
2
/
δ
n
)
,
-
2
(
J
1
/
δ
n
).  By Lemma 6, if we let A
n
-
1
=
Circ
(
x
1
,
x
2
,
…
,
x
n
), then its last row entries are given by the following equations:(24)
x
2
=
-
1
δ
n
+
4
δ
n
∑
i
=
1
n
-
2
‍
J
n
-
1
-
i
2
J
n
i
-
1
J
1
-
J
n
+
1
i
,
x
3
=
-
2
J
1
δ
n
J
1
-
J
n
+
1
,
x
4
=
-
2
δ
n
∑
i
=
1
2
‍
J
3
-
i
2
J
n
i
-
1
J
1
-
J
n
+
1
i
+
2
J
1
δ
n
J
1
-
J
n
+
1
,
x
5
=
-
2
δ
n
∑
i
=
1
3
‍
J
4
-
i
2
J
n
i
-
1
J
1
-
J
n
+
1
i
+
2
δ
n
∑
i
=
1
2
‍
J
3
-
i
2
J
n
i
-
1
J
1
-
J
n
+
1
i
+
4
J
1
δ
n
J
1
-
J
n
+
1
,
⋮
x
n
=
-
2
δ
n
∑
i
=
1
n
-
2
‍
J
n
-
1
-
i
2
J
n
i
-
1
J
1
-
J
n
+
1
i
+
2
δ
n
∑
i
=
1
n
-
3
‍
J
n
-
2
-
i
2
J
n
i
-
1
J
1
-
J
n
+
1
i
+
4
δ
n
∑
i
=
1
n
-
4
‍
J
n
-
3
-
i
2
J
n
i
-
1
J
1
-
J
n
+
1
i
,
x
1
=
1
δ
n
+
2
δ
n
∑
i
=
1
n
-
2
‍
J
n
-
1
-
i
2
J
n
i
-
1
J
1
-
J
n
+
1
i
+
4
δ
n
∑
i
=
1
n
-
3
‍
J
n
-
2
-
i
2
J
n
i
-
1
J
1
-
J
n
+
1
i
.Let C
n
(
k
)
=
∑
i
=
1
k
‍
(
J
k
+
1
-
i
2
J
n
i
-
1
/
J
1
-
J
n
+
1
i
), (
j
=
1,2
,
…
,
n
-
2
); then we have(25)
C
n
2
-
C
n
1
=
∑
i
=
1
2
‍
J
3
-
i
2
J
n
i
-
1
J
1
-
J
n
+
1
i
-
J
1
J
1
-
J
n
+
1
=
2
J
n
J
1
-
J
n
+
1
2
,
C
n
n
-
2
+
2
C
n
n
-
3
=
∑
i
=
1
n
-
2
‍
J
n
-
1
-
i
2
J
n
i
-
1
J
1
-
J
n
+
1
i
+
2
∑
i
=
1
n
-
3
‍
J
n
-
2
-
i
2
J
n
i
-
1
J
1
-
J
n
+
1
i
=
J
1
2
J
n
n
-
3
J
1
-
J
n
+
1
n
-
2
+
∑
i
=
1
n
-
3
‍
J
n
-
i
2
J
n
i
-
1
J
1
-
J
n
+
1
i
=
∑
i
=
1
n
-
2
‍
J
n
-
i
2
J
n
i
-
1
J
1
-
J
n
+
1
i
,
C
n
k
+
2
-
C
n
k
+
1
-
2
C
n
k
=
∑
i
=
1
k
+
2
‍
J
k
+
3
-
i
2
J
n
i
-
1
J
1
-
J
n
+
1
i
-
∑
i
=
1
k
+
1
‍
J
k
+
2
-
i
2
J
n
i
-
1
J
1
-
J
n
+
1
i
-
2
∑
i
=
1
k
‍
J
k
+
1
-
i
2
J
n
i
-
1
J
1
-
J
n
+
1
i
=
J
2
2
J
n
k
J
1
-
J
n
+
1
k
+
1
+
J
1
2
J
n
k
+
1
J
1
-
J
n
+
1
k
+
2
-
J
1
2
J
n
k
J
1
-
J
n
+
1
k
+
1
+
∑
i
=
1
k
‍
J
k
+
3
-
i
-
J
k
+
2
-
i
-
2
J
k
+
1
-
i
2
J
n
i
-
1
J
1
-
J
n
+
1
i
=
2
J
n
k
+
1
J
1
-
J
n
+
1
k
+
2
k
=
1,2
,
…
,
n
-
4
.We obtain(26)
x
1
=
1
+
2
C
n
n
-
2
+
2
C
n
n
-
3
δ
n
,
x
2
=
4
C
n
n
-
2
-
1
δ
n
,
x
3
=
-
2
C
n
1
δ
n
,
x
4
=
-
2
C
n
2
-
C
n
1
δ
n
,
x
k
=
-
2
C
n
(
k
-
2
)
-
C
n
(
k
-
3
)
-
2
C
n
(
k
-
4
)
δ
n
,
k
k
k
k
k
k
k
k
k
i
k
k
k
k
k
k
k
k
=
5,6
,
…
,
n
,
A
n
-
1
=
1
δ
n
Circ
x
1
′
,
x
2
′
,
…
,
x
n
′
,where(27)
x
1
′
=
1
+
2
∑
i
=
1
n
-
2
‍
J
n
-
i
2
J
n
i
-
1
J
1
-
J
n
+
1
i
,
x
2
′
=
-
1
+
4
∑
i
=
1
n
-
2
‍
J
n
-
1
-
i
2
J
n
i
-
1
J
1
-
J
n
+
1
i
,
x
k
′
=
-
2
2
J
n
k
-
3
J
1
-
J
n
+
1
k
-
2
,
k
=
3,4
,
…
,
n
.The proof is completed.


## 3. Circulant Matrix with the Jacobsthal-Lucas Numbers

In this section, letB
n
=
Circ
j
1
,
j
2
,
…
,
j
n be a circulant matrix. Firstly, we give a determinant formula for the matrix B
n. Then, we prove that B
n is an invertible matrix for any positive integer n, and finally we find the inverse of the matrix B
n.Theorem 8.
LetB
n
=
Circ
j
1
,
j
2
,
…
,
j
n be a circulant matrix; then one has(28)
det
⁡
B
n
=
1
-
j
n
+
1
n
-
1
+
2
j
n
-
4
n
-
2
×
∑
k
=
1
n
-
1
‍
j
k
+
2
-
5
j
k
+
1
1
-
j
n
+
1
2
j
n
-
4
k
-
1
,where j
n is the nth Jacobsthal-Lucas number.Proof.
Obviously, whenn
=
1, det
⁡
B
1
=
1 satisfies the formula (28). When n
>
1, let(29)
Σ
=
1
-
5
1
-
2
1
-
1
0
0
1
-
1
-
2
⋮
⋰
⋰
⋰
0
1
⋰
⋰
0
1
-
1
⋰
0
0
1
-
1
-
2
,
Ω
1
=
1
0
0
⋯
0
0
0
2
j
n
-
4
j
1
-
j
n
+
1
n
-
2
0
⋯
0
1
0
2
j
n
-
4
j
1
-
j
n
+
1
n
-
3
0
⋯
1
0
⋮
⋮
⋮
⋱
⋮
⋮
0
2
j
n
-
4
j
1
-
j
n
+
1
1
⋯
0
0
0
1
0
⋯
0
0be two n
×
n matrices; then the product Σ
B
n
Ω
1 has the form(30)
j
1
γ
n
′
j
n
-
1
⋯
j
3
j
2
0
γ
n
j
n
-
5
j
n
-
1
⋯
j
4
-
5
j
3
j
3
-
5
j
2
0
0
j
1
-
j
n
+
1
0
0
4
-
2
j
n
⋱
0
⋮
⋮
⋱
⋱
0
0
4
-
2
j
n
j
1
-
j
n
+
1
,where(31)
γ
n
=
j
1
-
5
j
n
+
∑
k
=
1
n
-
2
‍
j
k
+
2
-
5
j
k
+
1
2
j
n
-
4
j
1
-
j
n
+
1
n
-
k
+
1
,
γ
n
′
=
∑
k
=
1
n
-
1
‍
j
k
+
1
2
j
n
-
4
j
1
-
j
n
+
1
n
-
k
+
1
.We obtain(32)
det
⁡
Σ
det
⁡
B
n
det
⁡
Ω
1
=
j
1
×
j
1
-
5
j
n
+
∑
k
=
1
n
-
2
‍
j
k
+
2
-
5
j
k
+
1
×
j
1
-
5
j
n
+
∑
k
=
1
n
-
2
‍
j
k
+
2
-
5
j
k
+
1
2
j
n
-
4
j
1
-
j
n
+
1
n
-
k
+
1
×
j
1
-
j
n
+
1
n
-
2
=
j
1
×
j
1
-
j
n
+
1
+
∑
k
=
1
n
-
1
‍
j
k
+
2
-
5
j
k
+
1
×
∑
k
=
1
n
-
1
‍
j
k
+
2
-
5
j
k
+
1
2
j
n
-
4
j
1
-
j
n
+
1
n
-
k
+
1
×
j
1
-
j
n
+
1
n
-
2
=
1
-
j
n
+
1
n
-
1
+
2
j
n
-
4
n
-
2
∑
k
=
1
n
-
1
‍
j
k
+
2
-
5
j
k
+
1
×
1
-
j
n
+
1
2
j
n
-
4
k
-
1
,while(33)
det
⁡
Σ
=
det
⁡
Ω
1
=
-
1
n
-
1
n
-
2
/
2
,and we have(34)
det
⁡
B
n
=
1
-
j
n
+
1
n
-
1
+
2
j
n
-
4
n
-
2
×
∑
k
=
1
n
-
1
‍
j
k
+
2
-
5
j
k
+
1
1
-
j
n
+
1
2
j
n
-
4
k
-
1
.Theorem 9.
LetB
n
=
Circ
j
1
,
j
2
,
…
,
j
n be a circulant matrix; then B
n is invertible for any positive integer n.Proof.
Sincej
n
=
α
n
+
β
n, where α
+
β
=
1,  α
β
=
-
2. We have(35)
f
ω
k
=
∑
i
=
1
n
‍
j
i
ω
k
i
-
1
=
∑
i
=
1
n
‍
α
i
+
β
i
ω
k
i
-
1
=
α
1
-
α
n
1
-
α
ω
k
+
β
1
-
β
n
1
-
β
ω
k
=
α
+
β
-
2
α
β
ω
k
-
α
n
+
1
+
β
n
+
1
1
-
α
+
β
ω
k
+
α
β
ω
2
k
+
α
β
α
n
+
β
n
ω
k
1
-
α
+
β
ω
k
+
α
β
ω
2
k
=
1
-
j
n
+
1
+
4
-
2
j
n
ω
k
1
-
ω
k
-
2
ω
2
k
,
k
k
i
k
k
k
k
k
=
1,2
,
…
,
n
-
1
.If there exists ω
l  l
=
1,2
,
…
,
n
-
1 such that f
ω
l
=
0, we obtain 1
-
j
n
+
1
+
4
-
2
j
n
ω
l
=
0 for 1
-
ω
l
-
2
ω
2
l
≠
0; thus, ω
l
=
1
-
j
n
+
1
/
2
j
n
-
4 is a real number. While(36)
ω
l
=
exp
⁡
2
l
π
i
n
=
cos
⁡
2
l
π
n
+
i
sin
2
l
π
n
,
sin
⁡
(
2
l
π
/
n
)
=
0, so we have ω
l
=
-
1 for 0
<
2
l
π
/
n
<
2
π. But x
=
-
1 is not the root of the equation 1
-
j
n
+
1
+
4
-
2
j
n
x
=
0 for any positive integer n. We obtain f
ω
k
≠
0 for any ω
k  k
=
1,2
,
…
,
n
-
1, while f
1
=
∑
i
=
1
n
‍
j
i
=
-
1
/
2
5
-
j
n
+
1
-
2
j
n
≠
0. By Lemma 4 in [15], the proof is completed.Lemma 10.
Let the matrixH
=
[
h
i
j
]
i
,
j
=
1
n
-
2 be of the form(37)
h
i
j
=
j
1
-
j
n
+
1
,
i
=
j
,
4
-
2
j
n
,
i
=
j
+
1
,
0
,
o
t
h
e
r
w
i
s
eand then the inverse H
-
1
=
[
h
i
j
′
]
i
,
j
=
1
n
-
2 of the matrix H is equal to(38)
h
i
j
′
=
2
j
n
-
4
i
-
j
j
1
-
j
n
+
1
i
-
j
+
1
,
i
≥
j
,
0
,
i
<
j
.Proof.
Letr
i
j
=
∑
k
=
1
n
-
2
‍
h
i
k
h
k
j
′. Obviously, r
i
j
=
0 for i
<
j. In the case i
=
j, we obtain(39)
r
i
i
=
h
i
i
h
i
i
′
=
j
1
-
j
n
+
1
·
1
j
1
-
j
n
+
1
=
1
.For i
≥
j
+
1, we obtain(40)
r
i
j
=
∑
k
=
1
n
-
2
‍
h
i
k
h
k
j
′
=
h
i
,
i
-
1
h
i
-
1
,
j
′
+
h
i
i
h
i
j
′
=
4
-
2
j
n
·
2
j
n
-
4
i
-
j
-
1
j
1
-
j
n
+
1
i
-
j
+
j
1
-
j
n
+
1
×
2
j
n
-
4
i
-
j
j
1
-
j
n
+
1
i
-
j
+
1
=
0
.Hence, we verify H
H
-
1
=
I
n
-
2, where I
n
-
2 is n
-
2
×
n
-
2 identity matrix. Similarly, we can verify H
-
1
H
=
I
n
-
2. Thus, the proof is completed.Theorem 11.
LetB
n
=
Circ
j
1
,
j
2
,
…
,
j
n be a circulant matrix; then we have(41)
B
n
-
1
=
1
γ
n
Circ
y
1
′
,
y
2
′
,
…
,
y
n
′
,
where(42)
y
1
′
=
1
+
∑
i
=
1
n
-
2
‍
j
n
+
2
-
i
-
5
j
n
+
1
-
i
2
j
n
-
4
i
-
1
j
1
-
j
n
+
1
i
,
y
2
′
=
-
5
+
2
∑
i
=
1
n
-
2
‍
j
n
+
1
-
i
-
5
j
n
-
i
2
j
n
-
4
i
-
1
j
1
-
j
n
+
1
i
,
y
k
′
=
18
2
j
n
-
4
k
-
3
j
1
-
j
n
+
1
k
-
2
,
k
=
3,4
,
…
,
n
,
γ
n
=
j
1
-
5
j
n
+
∑
k
=
1
n
-
2
‍
j
k
+
2
-
5
j
k
+
1
2
j
n
-
4
j
1
-
j
n
+
1
n
-
k
+
1
.Proof.
Let(43)
Ω
2
=
1
-
γ
n
′
ω
13
⋯
ω
1
n
0
1
ω
23
⋯
ω
2
n
0
0
1
⋯
0
0
0
0
⋯
0
⋮
⋮
⋮
⋱
⋮
0
0
0
⋯
1
,where(44)
ω
1
j
=
γ
n
′
j
n
+
3
-
j
-
5
j
n
+
2
-
j
γ
n
-
j
n
+
2
-
j
,
ω
2
j
=
-
j
n
+
3
-
j
-
5
j
n
+
2
-
j
γ
n
,
k
k
k
k
i
k
k
k
j
=
3,4
,
…
,
n
,
γ
n
=
j
1
-
5
j
n
+
∑
k
=
1
n
-
2
‍
j
k
+
2
-
5
j
k
+
1
2
j
n
-
4
j
1
-
j
n
+
1
n
-
k
+
1
,
γ
n
′
=
∑
k
=
1
n
-
1
‍
j
k
+
1
2
j
n
-
4
j
1
-
j
n
+
1
n
-
k
+
1
.We have(45)
Σ
B
n
Ω
1
Ω
2
=
D
2
⊕
H
,where D
2
=
diag
j
1
,
γ
n is a diagonal matrix and D
2
⊕
H is the direct sum of D
2 and H. If we denote Ω
=
Ω
1
Ω
2, then we obtain(46)
B
n
-
1
=
Ω
D
2
-
1
⊕
H
-
1
Σ
,
since the last row entries of the matrixΩ are 0,1
,
5
j
n
-
1
-
j
n
/
γ
n
,
5
j
n
-
2
-
j
n
-
1
/
γ
n
,
…
,
5
j
2
-
j
3
/
γ
n. By Lemma 10, if we let B
n
-
1
=
Circ
y
1
,
y
2
,
…
,
y
n, then its last row entries are given by the following equations:(47)
y
2
=
-
5
γ
n
+
2
γ
n
∑
i
=
1
n
-
2
‍
j
n
+
1
-
i
-
5
j
n
-
i
2
j
n
-
4
i
-
1
j
1
-
j
n
+
1
i
,
y
3
=
-
j
3
-
5
j
2
γ
n
j
1
-
j
n
+
1
,
y
4
=
j
3
-
5
j
2
γ
n
j
1
-
j
n
+
1
-
1
γ
n
∑
i
=
1
2
‍
j
5
-
i
-
5
j
4
-
i
2
j
n
-
4
i
-
1
j
1
-
j
n
+
1
i
,
y
5
=
2
j
3
-
5
j
2
γ
n
j
1
-
j
n
+
1
+
1
γ
n
∑
i
=
1
2
‍
j
5
-
i
-
5
j
4
-
i
2
j
n
-
4
i
-
1
j
1
-
j
n
+
1
i
-
1
γ
n
∑
i
=
1
3
‍
j
6
-
i
-
5
j
5
-
i
2
j
n
-
4
i
-
1
j
1
-
j
n
+
1
i
,
⋮
y
n
=
2
γ
n
∑
i
=
1
n
-
4
‍
j
n
-
1
-
i
-
5
j
n
-
2
-
i
2
j
n
-
4
i
-
1
j
1
-
j
n
+
1
i
+
1
γ
n
∑
i
=
1
n
-
3
‍
j
n
-
i
-
5
j
n
-
1
-
i
2
j
n
-
4
i
-
1
j
1
-
j
n
+
1
i
-
1
γ
n
∑
i
=
1
n
-
2
‍
j
n
+
1
-
i
-
5
j
n
-
i
2
j
n
-
4
i
-
1
j
1
-
j
n
+
1
i
,
y
1
=
1
γ
n
+
2
γ
n
∑
i
=
1
n
-
3
‍
j
n
-
i
-
5
j
n
-
1
-
i
2
j
n
-
4
i
-
1
j
1
-
j
n
+
1
i
+
1
γ
n
∑
i
=
1
n
-
2
‍
j
n
+
1
-
i
-
5
j
n
-
i
2
j
n
-
4
i
-
1
j
1
-
j
n
+
1
i
.Let(48)
D
n
k
=
∑
i
=
1
k
‍
j
k
+
3
-
i
-
5
j
k
+
2
-
i
2
j
n
-
4
i
-
1
j
1
-
j
n
+
1
i
,
k
k
k
k
k
k
k
k
i
k
k
k
k
k
k
k
j
=
1,2
,
…
,
n
-
2and we have(49)
D
n
1
-
D
n
2
=
j
3
-
5
j
2
j
1
-
j
n
+
1
-
∑
i
=
1
2
‍
j
5
-
i
-
5
j
4
-
i
2
j
n
-
4
i
-
1
j
1
-
j
n
+
1
i
=
18
2
j
n
-
4
j
1
-
j
n
+
1
2
,
2
D
n
n
-
3
+
D
n
n
-
2
=
2
∑
i
=
1
n
-
3
‍
j
n
-
i
-
5
j
n
-
i
-
1
2
j
n
-
4
i
-
1
j
1
-
j
n
+
1
i
+
∑
i
=
1
n
-
2
‍
j
n
+
1
-
i
-
5
j
n
-
i
2
j
n
-
4
i
-
1
j
1
-
j
n
+
1
i
=
2
∑
i
=
1
n
-
3
‍
j
n
-
i
-
5
j
n
-
i
-
1
2
j
n
-
4
i
-
1
j
1
-
j
n
+
1
i
+
∑
i
=
1
n
-
3
‍
j
n
+
1
-
i
-
5
j
n
-
i
2
j
n
-
4
i
-
1
j
1
-
j
n
+
1
i
=
∑
i
=
1
n
-
3
‍
j
n
+
2
-
i
-
5
j
n
+
1
-
i
2
j
n
-
4
i
-
1
j
1
-
j
n
+
1
i
+
j
3
-
5
j
2
2
j
n
-
4
n
-
3
j
1
-
j
n
+
1
n
-
2
=
∑
i
=
1
n
-
2
‍
j
n
+
2
-
i
-
5
j
n
+
1
-
i
2
j
n
-
4
i
-
1
j
1
-
j
n
+
1
n
-
2
,
2
D
n
k
+
D
n
k
+
1
-
D
n
k
+
2
=
2
∑
i
=
1
k
‍
j
k
+
3
-
i
-
5
j
k
+
2
-
i
2
j
n
-
4
i
-
1
j
1
-
j
n
+
1
i
+
∑
i
=
1
k
+
1
‍
j
k
+
4
-
i
-
5
j
k
+
3
-
i
2
j
n
-
4
i
-
1
j
1
-
j
n
+
1
i
-
∑
i
=
1
k
+
2
‍
j
k
+
5
-
i
-
5
j
k
+
4
-
i
2
j
n
-
4
i
-
1
j
1
-
j
n
+
1
i
=
∑
i
=
1
k
‍
2
j
k
+
3
-
i
-
10
j
k
+
2
-
i
2
j
n
-
4
i
-
1
j
1
-
j
n
+
1
i
+
j
k
+
4
-
i
-
5
j
k
+
3
-
i
2
j
n
-
4
i
-
1
j
1
-
j
n
+
1
i
-
j
k
+
5
-
i
-
5
j
k
+
4
-
i
2
j
n
-
4
i
-
1
j
1
-
j
n
+
1
i
+
j
3
-
5
j
2
2
j
n
-
4
k
j
1
-
j
n
+
1
k
+
1
-
j
4
-
5
j
3
2
j
n
-
4
k
j
1
-
j
n
+
1
k
+
1
-
j
3
-
5
j
2
2
j
n
-
4
k
+
1
j
1
-
j
n
+
1
k
+
2
=
-
18
2
j
n
-
4
k
+
1
j
1
-
j
n
+
1
k
+
2
j
=
1,2
,
…
,
n
-
4
.We obtain(50)
y
1
=
1
+
2
D
n
(
n
-
3
)
+
D
n
(
n
-
2
)
γ
n
,
y
2
=
2
D
n
(
n
-
2
)
-
5
γ
n
,
y
3
=
-
D
n
(
1
)
γ
n
,
y
4
=
-
D
n
(
2
)
-
D
n
(
1
)
γ
n
,
y
k
=
-
D
n
(
i
-
2
)
-
D
n
(
i
-
3
)
-
2
D
n
(
i
-
4
)
γ
n
,
k
=
5,6
,
…
,
n
,
B
n
-
1
=
1
γ
n
Circ
y
1
′
,
y
2
′
,
…
,
y
n
′
,where(51)
y
1
′
=
1
+
∑
i
=
1
n
-
2
‍
j
n
+
2
-
i
-
5
j
n
+
1
-
i
2
j
n
-
4
i
-
1
j
1
-
j
n
+
1
i
,
y
2
′
=
-
5
+
2
∑
i
=
1
n
-
2
‍
j
n
+
1
-
i
-
5
j
n
-
i
2
j
n
-
4
i
-
1
j
1
-
j
n
+
1
i
,
y
k
′
=
18
2
j
n
-
4
k
-
3
j
1
-
j
n
+
1
k
-
2
,
k
=
3,4
,
…
,
n
,
γ
n
=
j
1
-
5
j
n
+
∑
k
=
1
n
-
2
‍
j
k
+
2
-
5
j
k
+
1
2
j
n
-
4
j
1
-
j
n
+
1
n
-
k
+
1
.


## 4. Left Circulant Matrix with the Jacobsthal and Jacobsthal-Lucas Numbers

In this section, letA
n
′
=
LCirc
J
1
,
J
2
,
…
,
J
n and B
n
′
=
LCirc
j
1
,
j
2
,
…
,
j
n be two left circulant matrices. By using the obtained conclusions, we give the determinant for the matrices A
n
′ and B
n
′. Then, we prove that A
n
′ is an invertible matrix for n
>
2 and B
n
′ is an invertible matrix for any positive integer n. The inverses of the matrices A
n
′ and B
n
′ are also presented.According to Lemma 5 in [15] and Theorems 4, 5, and 7, we can obtain the following theorems.Theorem 12.
LetA
n
′
=
LCirc
J
1
,
J
2
,
…
,
J
n be a left circulant matrix; then one can prove that(52)
det
⁡
A
n
′
=
-
1
n
-
1
n
-
2
/
2
×
1
-
J
n
+
1
n
-
1
+
2
J
n
n
-
2
∑
k
=
1
n
-
1
‍
2
J
k
1
-
J
n
+
1
2
J
n
k
-
1
.Theorem 13.
LetA
n
′
=
LCirc
J
1
,
J
2
,
…
,
J
n be a left circulant matrix; if n
>
2, then A
n
′ is an invertible matrix.Theorem 14.
LetA
n
′
=
LCirc
(
J
1
,
J
2
,
…
,
J
n
), (
n
>
2
) be a left circulant matrix; then one has(53)
A
n
′
-
1
=
1
δ
n
LCirc
z
1
,
z
2
,
…
,
z
n
,where(54)
z
1
=
1
+
2
∑
i
=
1
n
-
2
‍
J
n
-
i
2
J
n
i
-
1
J
1
-
J
n
+
1
i
,
z
k
=
-
2
2
J
n
n
-
1
-
k
J
1
-
J
n
+
1
n
-
k
,
k
=
2,3
,
…
,
n
-
1
,
z
n
=
-
1
+
4
∑
i
=
1
n
-
2
‍
J
n
-
1
-
i
2
J
n
i
-
1
J
1
-
J
n
+
1
i
,
δ
n
=
J
1
-
J
n
+
∑
k
=
1
n
-
2
‍
2
J
k
2
J
n
J
1
-
J
n
+
1
n
-
k
+
1
.By Lemma 5 in [15] and Theorems 8, 9, and 11, the following conclusions can be obtained.Theorem 15.
LetB
n
′
=
LCirc
j
1
,
j
2
,
…
,
j
n be a left circulant matrix; then one has(55)
det
⁡
B
n
′
=
-
1
n
-
1
n
-
2
/
2
×
1
-
j
n
+
1
n
-
1
+
2
j
n
-
4
n
-
2
∑
k
=
1
n
-
1
‍
j
k
+
2
-
5
j
k
+
1
1
-
j
n
+
1
2
j
n
-
4
k
-
1
×
∑
k
=
1
n
-
1
‍
j
k
+
2
-
5
j
k
+
1
1
-
j
n
+
1
2
j
n
-
4
k
-
1
,where j
n is the nth Jacobsthal-Lucas number.Theorem 16.
LetB
n
′
=
LCirc
j
1
,
j
2
,
…
,
j
n be a left circulant matrix; then B
n
′ is invertible for any positive integer n.Theorem 17.
LetB
n
′
=
LCirc
j
1
,
j
2
,
…
,
j
n be a left circulant matrix; then one has(56)
B
n
′
-
1
=
1
γ
n
LCirc
z
1
′
,
z
2
′
,
…
,
z
n
′
,where(57)
z
1
′
=
1
+
∑
i
=
1
n
-
2
‍
j
n
+
2
-
i
-
5
j
n
+
1
-
i
2
j
n
-
4
i
-
1
j
1
-
j
n
+
1
i
,
z
k
′
=
18
2
j
n
-
4
n
-
1
-
k
j
1
-
j
n
+
1
n
-
k
,
k
=
2,3
,
…
,
n
-
1
,
z
n
′
=
-
5
+
2
∑
i
=
1
n
-
2
‍
j
n
+
1
-
i
-
5
j
n
-
i
2
j
n
-
4
i
-
1
j
1
-
j
n
+
1
i
,
γ
n
=
j
1
-
5
j
n
+
∑
k
=
1
n
-
2
‍
j
k
+
2
-
5
j
k
+
1
2
j
n
-
4
j
1
-
j
n
+
1
n
-
k
+
1
.


## 5.g-Circulant Matrix with the Jacobsthal and Jacobsthal-Lucas Numbers

In this section, letA
g
,
n
=
g
-Circ
J
1
,
J
2
,
…
,
J
n and B
g
,
n
=
g
-Circ
j
1
,
j
2
,
…
,
j
n be two g-circulant matrices. By using the obtained conclusions, we give a determinant formula for the matrices A
g
,
n and B
g
,
n. Then, we prove that A
g
,
n is an invertible matrix for n
>
2 and B
g
,
n is an invertible matrix for any positive integer n if (
n
,
g
)
=
1. The inverses of the matrices A
g
,
n and B
g
,
n are also presented.From Lemmas 6 and 7 in [15] and Theorems 4, 5, and 7, we deduce the following results.Theorem 18.
LetA
g
,
n
=
g
-Circ
J
1
,
J
2
,
…
,
J
n be a g-circulant matrix. Then one has(58)
det
⁡
A
g
,
n
=
det
⁡
Q
g
·

1
-
J
n
+
1
n
-
1
+
2
J
n
n
-
2
∑
k
=
1
n
-
1
‍
2
J
k
1
-
J
n
+
1
2
J
n

k
-
1
,where J
n is the nth Jacobsthal number.Theorem 19.
LetA
g
,
n
=
g
-Circ
J
1
,
J
2
,
…
,
J
n be a g
-circulant matrix and (
n
,
g
)
=
1, when n
>
2, A
g
,
n is an invertible matrix.Theorem 20.
LetA
g
,
n
=
g
-Circ
(
J
1
,
J
2
,
…
,
J
n
)  (
n
>
2
) be a g
-circulant matrix and (
n
,
g
)
=
1; one has(59)
A
g
,
n
-
1
=
1
δ
n
Circ
λ
1
,
λ
2
,
…
,
λ
n
Q
g
T
,where(60)
λ
1
=
1
+
2
∑
i
=
1
n
-
2
‍
J
n
-
i
2
J
n
i
-
1
J
1
-
J
n
+
1
i
,
λ
2
=
-
1
+
4
∑
i
=
1
n
-
2
‍
J
n
-
1
-
i
2
J
n
i
-
1
J
1
-
J
n
+
1
i
,
λ
k
=
-
2
2
J
n
k
-
3
J
1
-
J
n
+
1
k
-
2
,
k
=
3,4
,
…
,
n
,
δ
n
=
J
1
-
J
n
+
∑
k
=
1
n
-
2
‍
2
J
k
2
J
n
J
1
-
J
n
+
1
n
-
k
+
1
.Taking Lemmas 6 and 7 in [15] and Theorems 8, 9, and 11 into account, we have the following theorems.Theorem 21.
LetB
g
,
n
=
g
-Circ
j
1
,
j
2
,
…
,
j
n be a g
-circulant matrix; then one has(61)
det
⁡
B
g
,
n
=
det
⁡
Q
g
·
1
-
j
n
+
1
n
-
1
+
2
j
n
-
4
n
-
2
∑
k
=
1
n
-
1
‍
j
k
+
2
-
5
j
k
+
1
1
-
j
n
+
1
2
j
n
-
4
k
-
1
×
∑
k
=
1
n
-
1
‍
j
k
+
2
-
5
j
k
+
1
1
-
j
n
+
1
2
j
n
-
4
k
-
1
,where j
n is the nth Jacobsthal-Lucas number.Theorem 22.
LetB
g
,
n
=
g
-Circ
j
1
,
j
2
,
…
,
j
n be a g-circulant matrix and (
n
,
g
)
=
1; then B
g
,
n is invertible for any positive integer n.Theorem 23.
LetB
g
,
n
=
g
-Circ
j
1
,
j
2
,
…
,
j
n be a g-circulant matrix and (
n
,
g
)
=
1; then one has(62)
B
g
,
n
-
1
=
1
γ
n
Circ
λ
1
′
,
λ
2
′
,
…
,
λ
n
′
Q
g
T
,where(63)
λ
1
′
=
1
+
∑
i
=
1
n
-
2
‍
j
n
+
2
-
i
-
5
j
n
+
1
-
i
2
j
n
-
4
i
-
1
j
1
-
j
n
+
1
i
,
λ
k
′
=
18
2
j
n
-
4
n
-
1
-
k
j
1
-
j
n
+
1
n
-
k
,
k
=
2,3
,
…
,
n
-
1
,
λ
n
′
=
-
5
+
2
∑
i
=
1
n
-
2
‍
j
n
+
1
-
i
-
5
j
n
-
i
2
j
n
-
4
i
-
1
j
1
-
j
n
+
1
i
,
γ
n
=
j
1
-
5
j
n
+
∑
k
=
1
n
-
2
‍
j
k
+
2
-
5
j
k
+
1
2
j
n
-
4
j
1
-
j
n
+
1
n
-
k
+
1
.


## 6. Conclusion

This paper obtains better results for the determinants and inverses of circulant type matrices by some properties of Jacobsthal and Jacobsthal-Lucas numbers. The reason we focus our attentions on circulant type matrices is to explore the application of circulant type matrices in solving fractional order differential equations.



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*Source: 418293-2015-03-22.xml*