# Invertibility and Explicit Inverses of Circulant-Type Matrices withk-Fibonacci and k-Lucas Numbers

Authors: Zhaolin Jiang; Yanpeng Gong; Yun Gao
Journal: Abstract and Applied Analysis
 (2014)

---

## Abstract

Circulant matrices have important applications in solving ordinary differential equations. In this paper, we consider circulant-type matrices with thek-Fibonacci and k-Lucas numbers. We discuss the invertibility of these circulant matrices and present the explicit determinant and inverse matrix by constructing the transformation matrices, which generalizes the results in Shen et al. (2011).


---

## Body

## 1. Introduction

Circulant matrices play an important role in solving ordinary differential equations. Wilde [1] developed a theory for the solution of ordinary and partial differential equations whose structure involves the algebra of circulants. He showed how the algebra of 2
×
2 circulants relates to the study of the harmonic oscillator, the Cauchy-Riemann equations, Laplace’s equation, the Lorentz transformation, and the wave equation. And he used n
×
n circulants to suggest natural generalizations of these equations to higher dimensions. By using the well-known results on circulant matrices for computation of eigenvalues of the perturbation coefficient matrix, in computing the stability criteria, Voorhees and Nip [2] got that the rate constants were all set equal to one. By using a Strang-type block-circulant preconditioner, Zhang et al. [3] speeded up the convergent rate of boundary-value methods. Joy and Tavsanoglu [4] showed that feedback matrices of ring cellular neural networks, which can be described by the ODE, are block circulants. Delgado et al. [5] developed some techniques to obtain global hyperbolicity for a certain class of endomorphisms of (
R
p
)
n with p
,
n
≥
2; this kind of endomorphisms is obtained from vectorial difference equations where the mapping defining these equations satisfies a circulant matrix condition. In [6], nonsymmetric, large, and sparse linear systems were solved by using the generalized minimal residual (GMRES) method; a circulant-block preconditioner was proposed to speed up the convergence rate of the GMRES method.Circulant-type matrices include the circulant and left circulant andg-circulant matrices. They have been put on the firm basis with the work of Davis [7], Gray [8], and Jiang and Zhou [9]. In [10], the authors pointed out the processes based on the eigenvalue of the circulant-type matrices with i.i.d. entries. There are discussions about the convergence in probability and in distribution of the spectral norm of circulant-type matrices in [11]. Furthermore, the g-circulant matrices are focused on by many researchers; for details, please refer to [12, 13] and the references therein. Ngondiep et al. showed the singular values of g-circulants in [14].Thek-Fibonacci andk-Lucas number sequences are defined by the following recurrence relations [15, 16], respectively,
(1)
F
k
,
n
+
1
=
k
F
k
,
n
+
F
k
,
n
-
1
,
where
F
k
,
0
=
0
,
F
k
,
1
=
1
,
L
k
,
n
+
1
=
k
L
k
,
n
+
L
k
,
n
-
1
,
where
L
k
,
0
=
2
,
L
k
,
1
=
k
.Letα and β be the roots of the characteristic equation x
2
-
k
x
-
1
=
0; then the Binet formulas of the sequences {
F
k
,
n
} and {
L
k
,
n
} have the form
(2)
F
k
,
n
=
α
n
-
β
n
α
-
β
,
L
k
,
n
=
α
n
+
β
n
,
where α
+
β
=
k
,
α
β
=
-
1.Besides, some scholars have given various algorithms for the determinants and inverses of nonsingular circulant matrices [7, 9]. It is worth pointing out that the computational complexity of these algorithms is very amazing with the order of matrix increasing. However, some authors gave the explicit determinant and inverse of the circulant and skew-circulant involving Fibonacci and Lucas numbers. Shen et al. considered circulant matrices with Fibonacci and Lucas numbers and presented their explicit determinants and inverses in [17]. Cambini presented an explicit form of the inverse of a particular circulant matrix in [18]. Bozkurt and Tam gave determinants and inverses of circulant matrices with Jacobsthal and Jacobsthal-Lucas numbers in [19].The purpose of this paper is to obtain better results for the determinants and inverses of circulant-type matrices by some perfect properties of thek-Fibonacci andk-Lucas numbers.In this paper, we adopt the following two conventions0
0
=
1 and, for any sequence {
a
n
}
,
∑
k
=
i
n
‍
a
k
=
0 in the case i
>
n.(1)
Circulant matrix: then
×
n circulant matrix (denoted by Circ
(
a
1
,
a
2
,
…
,
a
n
)) with input {
a
i
} is the matrix whose (
i
,
j
)th entry is a
(
j
-
i
+
n
)
mod
⁡
n.
(2)
Left circulant matrix: this is also a symmetric matrix (denoted byL
Circ
(
a
1
,
a
2
,
…
,
a
n
)) where the (
i
,
j
)th element of the matrix is a
(
i
+
j
-
2
)
mod
⁡
n.
(3)
g-circulant matrix: a g-circulant matrix is an n
×
n matrix as in the following form:
(3)
g
-
Circ
(
a
1
,
a
2
,
…
,
a
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
)
n
×
n
,where g is a nonnegative integer and each of the subscripts is understood to be reduced modulo n.The first row ofg-Circ
(
a
1
,
a
2
,
…
,
a
n
) is (
a
1
,
a
2
,
…
,
a
n
); its (
j
+
1
)th 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 standardcirculant matrix. If g
=
n
-
1, then we obtain the so-calledleft circulant matrix.Lemma 1 (see [7, 9]).
LetA
=
Circ
(
a
1
,
a
2
,
…
,
a
n
) be a circulant matrix; then we have the following.(i)
A is invertible if and only if f
(
ω
k
)
≠
0, (
k
=
0,1
,
2
,
…
,
n
-
1
), where f
(
x
)
=
∑
j
=
1
n
‍
a
j
x
j
-
1 and ω
=
exp
⁡
(
2
π
i
/
n
);
(ii)
IfA is invertible, then the inverse A
-
1 of A is a circulant matrix.Lemma 2.
LetΔ
=
L
Circ
(
1,0
,
…
,
0
); the matrix Δ is an orthogonal cyclic shift matrix. It holds that
(4)
L
Circ
(
a
1
,
a
2
,
…
,
a
n
)
=
Δ
Circ
(
a
1
,
a
2
,
…
,
a
n
)
.Lemma 3 (see [20]).
Then
×
n matrix Q
g is unitary if and only if (
n
,
g
)
=
1, where Q
g is a g-circulant matrix with the first row e
*
=
[
1,0
,
…
,
0
].Lemma 4 (see [20]).
A
g
,
n is a g-circulant matrix with the first row [
a
1
,
a
2
,
…
,
a
n
] if and only if A
g
,
n
=
Q
g
C, where C
=
Circ
(
a
1
,
a
2
,
…
,
a
n
).Lemma 5 (see [20]).
The inverse of a nonsingularg-circulant matrix A
g
,
n is an s-circulant B, where s satisfies g
s
=
k
n
+
1, for some integer k.


## 2. Circulant Matrix with thek-Fibonacci Numbers

In this section, letA
k
,
n
=
Circ
(
F
k
,
1
,
F
k
,
2
,
…
,
F
k
,
n
) be a circulant matrix. Firstly, we give the determinant equation of the matrix A
k
,
n. Afterwards, we prove that A
k
,
n is an invertible matrix for n
>
2, and then we find the inverse of the matrix A
k
,
n.In the following, letμ
=
F
k
,
1
-
F
k
,
n
+
1, ν
=
F
k
,
n
/
(
F
k
,
1
-
F
k
,
n
+
1
), s
=
L
k
,
1
-
L
k
,
n
+
1, and t
=
L
k
,
n
-
2.Theorem 6.
LetA
k
,
n
=
Circ
(
F
k
,
1
,
F
k
,
2
,
…
,
F
k
,
n
) be a circulant matrix; then we have
(5)
det
⁡
A
k
,
n
=
μ
n
-
1
+
F
k
,
n
n
-
2
∑
i
=
1
n
-
1
‍
F
k
,
i
ν
i
-
1
,
where F
k
,
n is the nth k-Fibonacci number. Specially, when k
=
1, this result is the same as Theorem  2.1 in [17].Proof.
Obviously,det
⁡
A
k
,
1
=
1 satisfies formula (5). In the case n
>
1, let
(6)
Γ
=
(
1
-
k
1
-
1
1
-
k
0
0
1
-
k
-
1
⋮
⋰
⋰
⋰
0
1
⋰
⋰
0
1
-
k
⋰
0
0
1
-
k
-
1
)
,
Π
1
=
(
1
0
0
…
0
0
0
ν
n
-
2
0
…
0
1
0
ν
n
-
3
0
…
1
0
⋮
⋮
⋮
⋱
⋱
⋮
0
ν
1
…
0
0
0
1
0
⋯
0
0
)
be two n
×
n matrices; then we have
(7)
Γ
A
k
,
n
Π
1
=
(
F
k
,
1
f
k
,
n
′
F
k
,
n
-
1
F
k
,
n
-
2
…
F
k
,
2
0
f
k
,
n
F
k
,
n
-
2
F
k
,
n
-
3
…
F
k
,
1
0
0
μ
0
0
-
F
k
,
n
μ
⋮
⋮
⋱
⋱
0
0
0
-
F
k
,
n
μ
)
,
where
(8)
f
k
,
n
=
F
k
,
1
-
k
F
k
,
n
+
∑
i
=
1
n
-
2
‍
F
k
,
i
ν
n
-
(
i
+
1
)
,
f
k
,
n
′
=
∑
i
=
1
n
-
1
‍
F
k
,
i
+
1
ν
n
-
(
i
+
1
)
.
So we obtain
(9)
det
⁡
Γ
det
⁡
A
k
,
n
det
⁡
Π
1
=
F
k
,
1
[
∑
i
=
1
n
-
2
‍
F
k
,
1
-
k
F
k
,
n
+
∑
i
=
1
n
-
2
‍
F
k
,
i
ν
n
-
(
i
+
1
)
]
μ
n
-
2
=
F
k
,
1
[
∑
i
=
1
n
-
1
‍
F
k
,
1
-
F
k
,
n
+
1
+
∑
i
=
1
n
-
1
‍
F
k
,
i
ν
n
-
(
i
+
1
)
]
μ
n
-
2
=
μ
n
-
1
+
F
k
,
n
n
-
2
∑
i
=
1
n
-
1
‍
F
k
,
i
ν
i
-
1
,
while
(10)
det
⁡
Γ
=
det
⁡
Π
1
=
(
-
1
)
(
n
-
1
)
(
n
-
2
)
/
2
;
hence, we have
(11)
det
⁡
A
k
,
n
=
μ
n
-
1
+
F
k
,
n
n
-
2
∑
i
=
1
n
-
1
‍
F
k
,
i
ν
i
-
1
.
The proof is completed.Theorem 7.
LetA
k
,
n
=
Circ
(
F
k
,
1
,
F
k
,
2
,
…
,
F
k
,
n
) be a circulant matrix; if n
>
2, then A
k
,
n is an invertible matrix. Specially, when k
=
1, we get Theorem  2.2 in [17].Proof.
Whenn
=
3 in Theorem 6, then we have det
⁡
A
k
,
3
=
[
k
2
(
k
-
1
)
+
(
-
k
2
+
k
-
1
)
2
]
(
k
2
+
k
+
2
)
≠
0; hence, A
k
,
3 is invertible. In the case n
>
3, since F
k
,
n
=
(
α
n
-
β
n
)
/
(
α
-
β
), where α
+
β
=
k, α
·
β
=
-
1, we have
(12)
f
(
ω
t
)
=
∑
j
=
1
n
‍
F
k
,
j
(
ω
t
)
j
-
1
=
1
α
-
β
∑
j
=
1
n
‍
(
α
j
-
β
j
)
(
ω
t
)
j
-
1
=
1
α
-
β
[
α
(
1
-
α
n
)
1
-
α
ω
t
-
β
(
1
-
β
n
)
1
-
β
ω
t
]
(
because
1
-
α
ω
t
≠
0
,
1
-
β
ω
t
≠
0
)
=
1
α
-
β
[
(
α
-
β
)
-
(
α
n
+
1
-
β
n
+
1
)
+
α
β
(
α
n
-
β
n
)
ω
t
1
-
(
α
+
β
)
ω
t
+
α
β
ω
2
t
]
=
1
-
F
k
,
n
+
1
-
F
k
,
n
ω
t
1
-
k
ω
t
-
ω
2
t
(
t
=
1,2
,
…
,
n
-
1
)
.
If there exists ω
l
(
l
=
1,2
,
…
,
n
-
1
) such that f
(
ω
l
)
=
0, then we obtain 1
-
F
k
,
n
+
1
-
F
k
,
n
ω
l
=
0 for 1
-
k
ω
l
-
ω
2
l
≠
0; thus, ω
l
=
(
1
-
F
k
,
n
+
1
)
/
F
k
,
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
,
hence, 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
-
F
k
,
n
+
1
-
F
k
,
n
x
=
0
(
n
>
3
). Hence, we obtain f
(
ω
t
)
≠
0 for any ω
t
(
t
=
1,2
,
…
,
n
-
1
), while f
(
1
)
=
∑
j
=
1
n
‍
F
k
,
j
=
-
(
1
/
k
)
(
1
-
F
k
,
n
+
1
-
F
k
,
n
)
≠
0. Hence, by Lemma 1, the proof is completed.Lemma 8.
Let the matrixG
=
[
g
i
,
j
]
i
,
j
=
1
n
-
2 be of the form
(14)
g
i
,
j
=
{
μ
,
i
=
j
,
-
F
k
,
n
,
i
=
j
+
1
,
0
,
otherwise
.
Then the inverse G
-
1
=
[
g
i
,
j
′
]
i
,
j
=
1
n
-
2 of the matrix G is equal to
(15)
g
i
,
j
′
=
{
F
k
,
n
i
-
j
μ
i
-
j
+
1
,
i
≥
j
,
0
,
i
<
j
.
In particular, whenk
=
1, we get Lemma  2.1 in [17].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
′
=
μ
·
1
μ
=
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
′
=
-
F
k
,
n
·
F
k
,
n
i
-
j
-
1
μ
i
-
j
+
μ
·
F
k
,
n
i
-
j
μ
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.Theorem 9.
LetA
k
,
n
=
Circ
(
F
k
,
1
,
F
k
,
2
,
…
,
F
k
,
n
)
(
n
>
2
) be a circulant matrix; then we have
(18)
A
k
,
n
-
1
=
1
f
k
,
n
Circ
(
1
+
∑
i
=
1
n
-
2
‍
F
k
,
n
-
i
F
k
,
n
i
-
1
μ
i
,
kkkkkkkkkk
hh
-
k
+
∑
i
=
1
n
-
2
‍
F
k
,
n
-
1
-
i
F
k
,
n
i
-
1
μ
i
,
-
1
μ
,
-
F
k
,
n
μ
2
,
kkkkkkkkkk
h
h
-
F
k
,
n
2
μ
3
,
…
,
-
F
k
,
n
n
-
3
μ
n
-
2
)
,
where f
k
,
n
=
F
k
,
1
-
k
F
k
,
n
+
∑
i
=
1
n
-
2
‍
F
k
,
i
ν
n
-
(
i
+
1
). Specially, when k
=
1, this result is the same as Theorem2.3 in [17].Proof.
Let(19)
Π
2
=
(
1
-
f
k
,
n
′
ρ
n
-
2
…
ρ
1
0
1
-
F
k
,
n
-
2
f
k
,
n
…
-
F
k
,
1
f
k
,
n
0
0
1
…
0
⋮
⋮
⋮
⋱
⋮
0
0
0
…
1
)
,
where
(20)
f
k
,
n
=
F
k
,
1
-
k
F
k
,
n
+
∑
i
=
1
n
-
2
‍
F
k
,
i
ν
n
-
(
i
+
1
)
,
f
k
,
n
′
=
∑
i
=
1
n
-
1
‍
F
k
,
i
+
1
ν
n
-
(
i
+
1
)
,
ρ
1
=
f
k
,
n
′
f
k
,
n
F
k
,
1
-
F
k
,
2
,
…
,
ρ
n
-
2
=
f
k
,
n
′
f
k
,
n
F
k
,
n
-
2
-
F
k
,
n
-
1
.
Then we have
(21)
Γ
A
k
,
n
Π
1
Π
2
=
D
1
⊕
G
,
where D
1
=
diag
⁡
(
F
k
,
1
,
f
k
,
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
(22)
A
k
,
n
-
1
=
Π
(
D
1
-
1
⊕
G
-
1
)
Γ
.
Since the last row elements of matrix Π are 0
,
1
,
-
(
F
k
,
n
-
2
/
f
k
,
n
)
,
-
(
F
k
,
n
-
3
/
f
k
,
n
)
,
…
,
-
(
F
k
,
2
/
f
k
,
n
), -
(
F
k
,
1
/
f
k
,
n
), by Lemma 8, if A
k
,
n
-
1
=
Circ
(
x
1
,
x
2
,
…
,
x
n
), then its last row elements are given by the following equations:
(23)
x
2
=
-
k
f
k
,
n
+
1
f
k
,
n
∑
i
=
1
n
-
2
‍
F
k
,
n
-
1
-
i
F
k
,
n
i
-
1
μ
i
,
x
3
=
-
1
f
k
,
n
F
k
,
1
μ
,
x
4
=
-
1
f
k
,
n
∑
i
=
1
2
‍
F
k
,
3
-
i
F
k
,
n
i
-
1
μ
i
+
k
F
k
,
1
f
k
,
n
μ
,
x
5
=
-
1
f
k
,
n
∑
i
=
1
3
‍
F
k
,
4
-
i
F
k
,
n
i
-
1
μ
i
+
k
f
k
,
n
∑
i
=
1
2
‍
F
k
,
3
-
i
F
k
,
n
i
-
1
μ
i
+
F
k
,
1
f
k
,
n
μ
,
x
5
i
⋮
x
n
=
-
1
f
k
,
n
∑
i
=
1
n
-
2
‍
F
k
,
n
-
1
-
i
F
k
,
n
i
-
1
μ
i
+
k
f
k
,
n
∑
i
=
1
n
-
3
‍
F
k
,
n
-
2
-
i
F
k
,
n
i
-
1
μ
i
+
1
f
k
,
n
∑
i
=
1
n
-
4
‍
F
k
,
n
-
3
-
i
F
k
,
n
i
-
1
μ
i
,
x
1
=
1
f
k
,
n
+
k
f
k
,
n
∑
i
=
1
n
-
2
‍
F
k
,
n
-
1
-
i
F
k
,
n
i
-
1
μ
i
+
1
f
k
,
n
∑
i
=
1
n
-
3
‍
F
k
,
n
-
2
-
i
F
k
,
n
i
-
1
μ
i
.
LetC
n
(
j
)
=
∑
i
=
1
j
‍
(
F
k
,
j
+
1
-
i
F
k
,
n
i
-
1
/
μ
i
)
(
j
=
1,2
,
…
,
n
-
2
); then we have
(24)
C
n
(
2
)
-
k
C
n
(
1
)
=
∑
i
=
1
2
‍
F
k
,
3
-
i
F
k
,
n
i
-
1
μ
i
-
k
F
k
,
1
μ
=
F
k
,
n
μ
2
,
k
C
n
(
n
-
2
)
+
C
n
(
n
-
3
)
=
k
∑
i
=
1
n
-
2
‍
F
k
,
n
-
1
-
i
F
k
,
n
i
-
1
μ
i
+
∑
i
=
1
n
-
3
‍
F
k
,
n
-
2
-
i
F
k
,
n
i
-
1
μ
i
=
k
F
k
,
1
F
k
,
n
n
-
3
μ
n
-
2
+
∑
i
=
1
n
-
3
‍
(
k
F
k
,
n
-
1
-
i
+
F
k
,
n
-
2
-
i
)
F
k
,
n
i
-
1
μ
i
=
∑
i
=
1
n
-
2
‍
F
k
,
n
-
i
F
k
,
n
i
-
1
μ
i
,
C
n
(
j
+
2
)
-
k
C
n
(
j
+
1
)
-
C
n
(
j
)
=
∑
i
=
1
j
+
2
‍
F
k
,
j
+
3
-
i
F
k
,
n
i
-
1
μ
i
-
k
∑
i
=
1
j
+
1
‍
F
k
,
j
+
2
-
i
F
k
,
n
i
-
1
μ
i
-
∑
i
=
1
j
‍
F
k
,
j
+
1
-
i
F
k
,
n
i
-
1
μ
i
=
F
k
,
2
F
k
,
n
j
μ
j
+
1
+
F
k
,
1
F
k
,
n
j
+
1
μ
j
+
2
-
k
F
k
,
1
F
k
,
n
j
μ
j
+
1
+
∑
i
=
1
j
‍
(
F
k
,
j
+
3
-
i
-
k
F
k
,
j
+
2
-
i
-
F
k
,
j
+
1
-
i
)
F
k
,
n
i
-
1
μ
i
=
F
k
,
n
j
+
1
μ
j
+
2
(
j
=
1,2
,
…
,
n
-
4
)
.
Hence, we obtain(25)
A
k
,
n
-
1
=
Circ
(
1
+
k
C
n
(
n
-
2
)
+
C
n
(
n
-
3
)
f
k
,
n
,
C
n
(
n
-
2
)
-
k
f
k
,
n
,
ggggggg
g
-
C
n
(
1
)
f
k
,
n
,
-
C
n
(
2
)
-
k
C
n
(
1
)
f
k
,
n
,
-
C
n
(
3
)
-
k
C
n
(
2
)
-
C
n
(
1
)
f
k
,
n
,
…
,
ggggggg
g
-
C
n
(
n
-
2
)
-
k
C
n
(
n
-
3
)
-
C
n
(
n
-
4
)
f
k
,
n
)
=
1
f
k
,
n
Circ
(
1
+
∑
i
=
1
n
-
2
‍
F
k
,
n
-
i
F
k
,
n
i
-
1
μ
i
,
ggggggg
gg
kk
-
k
+
∑
i
=
1
n
-
2
‍
F
k
,
n
-
1
-
i
F
k
,
n
i
-
1
μ
i
,
-
1
μ
,
-
F
k
,
n
μ
2
,
ggggggg
gg
kk
-
F
k
,
n
2
μ
3
,
…
,
-
F
k
,
n
n
-
3
μ
n
-
2
)
.


## 3. Circulant Matrix with thek-Lucas Numbers

In this section, letB
k
,
n
=
Circ
(
L
k
,
1
,
L
k
,
2
,
…
,
L
k
,
n
) be a circulant matrix. Firstly, we give a determinant formula for the matrix B
k
,
n. Afterwards, we prove that B
k
,
n is an invertible matrix for any positive integer n, and then we find the inverse of the matrix B
k
,
n.Theorem 10.
LetB
k
,
n
=
Circ
(
L
k
,
1
,
L
k
,
2
,
…
,
L
k
,
n
) be a circulant matrix; then we have
(26)
det
⁡
B
k
,
n
=
k
s
n
-
1
+
k
t
n
-
2
∑
i
=
1
n
-
1
‍
[
(
L
k
,
i
+
2
-
k
2
+
2
k
L
k
,
i
+
1
)
(
s
t
)
i
-
1
]
,
where L
k
,
n is the nth k-Lucas number. In particular, when k
=
1, we get Theorem  3.1 in [17].Proof.
Obviously,det
⁡
B
k
,
1
=
k satisfies formula (26); when n
>
1, let
(27)
Σ
=
(
1
-
k
2
+
2
k
1
-
1
1
-
k
0
0
1
-
k
-
1
⋮
⋰
⋰
⋰
0
1
⋰
⋰
0
1
-
k
⋰
0
0
1
-
k
-
1
)
,
Ω
1
=
(
1
0
0
⋯
0
0
0
(
t
s
)
n
-
2
0
⋯
0
1
0
(
t
s
)
n
-
3
0
⋯
1
0
⋮
⋮
⋮
⋱
⋮
⋮
0
t
s
1
⋯
0
0
0
1
0
⋯
0
0
)
be two n
×
n matrices; then the form of Σ
B
k
,
n
Ω
1 is given as follows:
(28)
(
L
k
,
1
l
k
,
n
′
L
k
,
n
-
1
L
k
,
n
-
2
…
L
k
,
2
0
l
k
,
n
δ
n
δ
n
-
1
…
δ
3
0
0
s
0
0
-
t
s
⋮
⋮
⋱
⋱
0
0
…
…
-
t
s
)
,
where
(29)
δ
i
=
L
k
,
i
-
k
2
+
2
k
L
k
,
i
-
1
,
i
=
3
,
…
,
n
,
l
k
,
n
=
L
k
,
1
-
k
2
+
2
k
L
k
,
n
+
∑
i
=
1
n
-
2
‍
δ
i
+
2
(
t
s
)
n
-
(
i
+
1
)
,
l
k
,
n
′
=
∑
i
=
1
n
-
1
‍
L
k
,
i
+
1
(
t
s
)
n
-
(
i
+
1
)
.
Hence, we obtain
(30)
det
⁡
Σ
det
⁡
B
k
,
n
det
⁡
Ω
1
=
L
k
,
1
[
L
k
,
1
-
k
2
+
2
k
L
k
,
n
+
∑
i
=
1
n
-
2
‍
δ
i
+
2
(
t
s
)
n
-
(
i
+
1
)
]
s
n
-
2
=
L
k
,
1
[
s
+
∑
i
=
1
n
-
1
‍
δ
i
+
2
(
t
s
)
n
-
(
i
+
1
)
]
s
n
-
2
=
k
s
n
-
1
+
k
t
n
-
2
∑
i
=
1
n
-
1
‍
(
L
k
,
i
+
2
-
k
2
+
2
k
L
k
,
i
+
1
)
(
s
t
)
i
-
1
,
while
(31)
det
⁡
Σ
=
det
⁡
Ω
1
=
(
-
1
)
(
n
-
1
)
(
n
-
2
)
/
2
;
thus, we have
(32)
det
⁡
B
k
,
n
=
k
s
n
-
1
+
k
t
n
-
2
∑
i
=
1
n
-
1
‍
[
(
L
k
,
i
+
2
-
k
2
+
2
k
L
k
,
i
+
1
)
(
s
t
)
i
-
1
]
.Theorem 11.
LetB
k
,
n
=
Circ
(
L
k
,
1
,
L
k
,
2
,
…
,
L
k
,
n
) be a circulant matrix; then B
k
,
n is invertible for any positive integer n. Specially, when k
=
1, we get Theorem  3.2 in [17].Proof.
SinceL
k
,
n
=
α
n
+
β
n, where α
+
β
=
k, α
β
=
-
1, we have
(33)
f
(
ω
t
)
=
∑
j
=
1
n
‍
L
k
,
j
(
ω
t
)
j
-
1
=
∑
j
=
1
n
‍
(
α
j
+
β
j
)
(
ω
t
)
j
-
1
=
α
(
1
-
α
n
)
1
-
α
ω
t
+
β
(
1
-
β
n
)
1
-
β
ω
t
(
because
1
-
α
ω
t
≠
0
,
1
-
β
ω
t
≠
0
)
=
(
α
+
β
)
-
(
α
n
+
1
+
β
n
+
1
)
+
α
β
(
α
n
+
β
n
)
ω
t
-
2
α
β
ω
t
1
-
(
α
+
β
)
ω
t
+
α
β
ω
2
t
=
k
-
L
k
,
n
+
1
+
(
2
-
L
k
,
n
)
ω
t
1
-
k
ω
t
-
ω
2
t
(
t
=
1,2
,
…
,
n
-
1
)
.
If there existsω
l
(
l
=
1,2
,
…
,
n
-
1
) such that f
(
ω
l
)
=
0, then we obtain k
-
L
k
,
n
+
1
+
(
2
-
L
k
,
n
)
ω
l
=
0 for 1
-
k
ω
l
-
ω
2
l
≠
0; thus, ω
l
=
(
k
-
L
k
,
n
+
1
)
/
(
L
k
,
n
-
2
) is a real number. While
(34)
ω
l
=
exp
⁡
(
2
l
π
i
n
)
=
cos
⁡
2
l
π
n
+
i
sin
2
l
π
n
,
hence, 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 k
-
L
k
,
n
+
1
+
(
2
-
L
k
,
n
)
x
=
0 for any positive integer n. Hence, we obtain f
(
ω
k
)
≠
0 for any ω
k
(
k
=
1,2
,
…
,
n
-
1
), while f
(
1
)
=
∑
j
=
1
n
‍
L
k
,
j
=
(
-
1
/
k
)
(
k
+
2
-
L
k
,
n
+
1
-
L
k
,
n
)
≠
0. Thus, by Lemma 1, the proof is completed.Lemma 12.
Let the matrixH
=
[
h
i
,
j
]
i
,
j
=
1
n
-
2 be of the form
(35)
h
i
,
j
=
{
s
,
i
=
j
,
t
,
i
=
j
+
1
,
0
,
otherwise
;
then the inverse H
-
1
=
[
h
i
,
j
′
]
i
,
j
=
1
n
-
2 of the matrix H is equal to
(36)
h
i
,
j
′
=
{
t
i
-
j
s
i
-
j
+
1
i
≥
j
,
0
,
i
<
j
.
Specially, whenk
=
1, we get Lemma  3.1 in [17].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
(37)
r
i
,
i
=
h
i
,
i
h
i
,
i
′
=
s
·
1
s
=
1
.
For i
≥
j
+
1, we obtain
(38)
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
′
=
(
-
t
)
·
t
i
-
j
-
1
s
i
-
j
+
s
·
t
i
-
j
s
i
-
j
+
1
=
0
.
Hence, we verify H
H
-
1
=
I
n
-
2, where I
n
-
2 is an (
n
-
2
)
×
(
n
-
2
) identity matrix. Similarly, we can verify H
-
1
H
=
I
n
-
2. Thus, the proof is completed.Theorem 13.
LetB
k
,
n
=
Circ
(
L
k
,
1
,
L
k
,
2
,
…
,
L
k
,
n
) be a circulant matrix; then we have
(39)
B
k
,
n
-
1
=
1
l
k
,
n
Circ
(
1
+
∑
i
=
1
n
-
2
‍
(
L
k
,
n
+
2
-
i
-
(
(
k
2
+
2
)
/
k
)
L
k
,
n
+
1
-
i
)
t
i
-
1
s
i
,
gggggg
jjjjjjjjjjjjjj
-
k
2
+
2
k
gggggg
jjjjjjjjjjjjjj
+
∑
i
=
1
n
-
2
‍
(
L
k
,
n
+
1
-
i
-
(
(
k
2
+
2
)
/
k
)
L
k
,
n
-
i
)
t
i
-
1
s
i
,
gggggg
jjjjjjjjjjjjjj
k
2
+
4
k
s
,
(
k
2
+
4
)
t
k
s
2
,
(
k
2
+
4
)
t
2
k
s
3
,
…
,
gggggg
jjjjjjjjjjjjjj
(
k
2
+
4
)
t
n
-
3
k
s
n
-
2
)
.
In particular, whenk
=
1, the result is the same as Theorem  3.3 in [17].Proof.
Let(40)
Ω
2
=
(
1
-
l
k
,
n
′
k
λ
n
…
λ
3
0
1
-
δ
n
l
k
,
n
…
-
δ
3
l
k
,
n
0
0
1
…
0
⋮
⋮
⋮
⋱
⋮
0
0
0
…
1
)
,
where
(41)
λ
i
=
1
k
[
l
k
,
i
′
(
L
k
,
i
-
(
(
k
2
+
2
)
/
k
)
L
k
,
i
-
1
)
l
k
,
n
-
L
k
,
i
-
1
]
,
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
j
i
=
3
,
4
,
…
,
n
,
δ
i
=
L
k
,
i
-
k
2
+
2
k
L
k
,
i
-
1
,
i
=
3
,
…
,
n
l
k
,
n
=
L
k
,
1
-
k
2
+
2
k
L
k
,
n
+
∑
i
=
1
n
-
2
‍
(
L
k
,
i
+
2
-
k
2
+
2
k
L
k
,
i
+
1
)
(
t
s
)
n
-
(
i
+
1
)
,
l
k
,
n
′
=
∑
i
=
1
n
-
1
‍
L
k
,
i
+
1
(
t
s
)
n
-
(
i
+
1
)
.
Then we have
(42)
Σ
B
k
,
n
Ω
1
Ω
2
=
D
2
⊕
H
,
where D
2
=
diag
⁡
(
L
k
,
1
,
f
k
,
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
(43)
B
k
,
n
-
1
=
Ω
(
D
2
-
1
⊕
H
-
1
)
Σ
.
Since the last row elements of the matrix Ω are 0
,
1
,
(
(
(
k
2
+
2
)
/
k
)
L
k
,
n
-
1
-
L
k
,
n
)
/
l
k
,
n
,
(
(
(
k
2
+
2
)
/
k
)
L
k
,
n
-
2
-
L
k
,
n
-
1
)
/
l
k
,
n
,
…
,
(
(
(
k
2
+
2
)
/
k
)
L
k
,
2
-
L
k
,
3
)
/
l
k
,
n
, by Lemma 12, if B
k
,
n
-
1
=
Circ
(
y
1
,
y
2
,
…
,
y
n
), then its last row elements are given by the following equations:
(44)
y
2
=
-
k
2
+
2
k
l
k
,
n
+
1
l
k
,
n
∑
i
=
1
n
-
2
‍
δ
n
+
1
-
i
t
i
-
1
s
i
,
y
3
=
-
1
l
k
,
n
δ
3
s
,
y
4
=
k
δ
3
s
l
k
,
n
-
1
l
k
,
n
∑
i
=
1
2
‍
δ
5
-
i
t
i
-
1
s
i
,
y
5
=
δ
3
s
l
k
,
n
+
k
l
k
,
n
∑
i
=
1
2
‍
δ
5
-
i
t
i
-
1
s
i
-
1
l
k
,
n
∑
i
=
1
3
‍
δ
6
-
i
t
i
-
1
s
i
,
y
5
i
⋮
y
n
=
1
l
k
,
n
∑
i
=
1
n
-
4
‍
δ
n
-
1
-
i
t
i
-
1
s
i
+
k
l
k
,
n
∑
i
=
1
n
-
3
‍
δ
n
-
i
t
i
-
1
s
i
-
1
l
k
,
n
∑
i
=
1
n
-
2
‍
δ
n
+
1
-
i
t
i
-
1
s
i
,
y
1
=
1
l
k
,
n
+
1
l
k
,
n
∑
i
=
1
n
-
3
‍
δ
n
-
i
t
i
-
1
s
i
+
k
l
k
,
n
∑
i
=
1
n
-
2
‍
δ
n
+
1
-
i
t
i
-
1
s
i
.
Let D
n
(
j
)
=
∑
i
=
1
j
‍
(
(
L
k
,
j
+
3
-
i
-
(
(
k
2
+
2
)
/
k
)
L
k
,
j
+
2
-
i
)
t
i
-
1
/
s
i
)
(
j
=
1,2
,
…
,
n
-
2
); then we have
(45)
k
D
n
(
1
)
-
D
n
(
2
)
=
k
δ
3
s
-
∑
i
=
1
2
‍
δ
5
-
i
t
i
-
1
s
i
=
(
k
2
+
4
)
t
k
s
2
,
D
n
(
n
-
3
)
+
k
D
n
(
n
-
2
)
=
∑
i
=
1
n
-
3
‍
δ
n
-
i
t
i
-
1
s
i
+
∑
i
=
1
n
-
2
‍
δ
n
+
1
-
i
t
i
-
1
s
i
=
k
δ
3
t
n
-
3
s
n
-
2
+
∑
i
=
1
n
-
3
‍
[
δ
n
-
i
t
i
-
1
s
i
+
k
δ
n
+
1
-
i
t
i
-
1
s
i
]
=
δ
4
t
n
-
3
s
n
-
2
+
∑
i
=
1
n
-
3
‍
δ
n
+
2
-
i
t
i
-
1
s
i
=
∑
i
=
1
n
-
2
‍
δ
n
+
2
-
i
t
i
-
1
s
i
,
D
n
(
j
)
+
k
D
n
(
j
+
1
)
-
D
n
(
j
+
2
)
=
∑
i
=
1
j
‍
δ
j
+
3
-
i
t
i
-
1
s
i
+
k
∑
i
=
1
j
+
1
‍
δ
j
+
4
-
i
t
i
-
1
s
i
-
∑
i
=
1
j
+
2
‍
δ
j
+
5
-
i
t
i
-
1
s
i
=
k
2
+
4
k
t
j
+
1
s
j
+
2
(
j
=
1,2
,
…
,
n
-
4
)
.
Hence, we obtain
(46)
B
k
,
n
-
1
=
Circ
(
1
+
D
n
(
n
-
3
)
+
k
D
n
(
n
-
2
)
l
k
,
n
,
D
n
(
n
-
2
)
-
(
(
k
2
+
2
)
/
k
)
l
k
,
n
,
j
j
j
j
j
j
ggg
-
D
n
(
1
)
l
k
,
n
,
k
D
n
(
1
)
-
D
n
(
2
)
l
k
,
n
,
D
n
(
1
)
+
k
D
n
(
2
)
-
D
n
(
3
)
l
k
,
n
,
…
,
j
j
j
j
j
j
j
j
j
j
j
j
D
n
(
n
-
4
)
+
k
D
n
(
n
-
3
)
-
D
n
(
n
-
2
)
l
k
,
n
)
=
1
l
k
,
n
×
Circ
(
1
+
∑
i
=
1
n
-
2
‍
(
L
k
,
n
+
2
-
i
-
(
(
k
2
+
2
)
/
k
)
L
k
,
n
+
1
-
i
)
t
i
-
1
s
i
,
j
j
j
j
j
hh
h
-
k
2
+
2
k
+
∑
i
=
1
n
-
2
‍
(
L
k
,
n
+
1
-
i
-
(
(
k
2
+
2
)
/
k
)
L
k
,
n
-
i
)
t
i
-
1
s
i
,
j
j
j
j
j
hh
h
k
2
+
4
k
s
,
(
k
2
+
4
)
t
k
s
2
,
(
k
2
+
4
)
t
2
k
s
3
,
…
,
(
k
2
+
4
)
t
n
-
3
k
s
n
-
2
)
.


## 4. Left Circulant Matrix with thek-Fibonacci and k-Lucas Numbers

In this section, letA
k
,
n
′
=
L
Circ
(
F
k
,
1
,
F
k
,
2
,
…
,
F
k
,
n
) and B
k
,
n
′
=
L
Circ
(
L
k
,
1
,
L
k
,
2
,
…
,
L
k
,
n
) be left circulant matrices. By using the obtained conclusions, we give a determinant formula for the matrices A
k
,
n
′ and B
k
,
n
′. Afterwards, we prove that A
k
,
n
′ is an invertible matrix for n
>
2 and B
k
,
n
′ is an invertible matrix for any positive integer n. The inverse of the matrices A
k
,
n
′ and B
k
,
n
′ is also presented.According to Lemma2, Theorem 6, Theorem 7, and Theorem 9, we can obtain the following theorems.Theorem 14.
LetA
k
,
n
′
=
L
Circ
(
F
k
,
1
,
F
k
,
2
,
…
,
F
k
,
n
) be a left circulant matrix; then we have
(47)
det
⁡
A
k
,
n
′
=
(
-
1
)
(
n
-
1
)
(
n
-
2
)
/
2
[
μ
n
-
1
+
F
k
,
n
n
-
2
∑
i
=
1
n
-
1
‍
F
k
,
i
ν
i
-
1
]
,
where F
k
,
n is the nth k-Fibonacci number.Theorem 15.
LetA
k
,
n
′
=
L
Circ
(
F
k
,
1
,
F
k
,
2
,
…
,
F
k
,
n
) be a left circulant matrix; if n
>
2, then A
k
,
n
′ is an invertible matrix.Theorem 16.
LetA
k
,
n
′
=
L
Circ
(
F
k
,
1
,
F
k
,
2
,
…
,
F
k
,
n
)
(
n
>
2
) be a left circulant matrix; then we have(48)
A
k
,
n
′
-
1
=
1
f
k
,
n
L
Circ
(
1
+
∑
i
=
1
n
-
2
‍
F
k
,
n
-
i
F
k
,
n
i
-
1
μ
i
,
-
F
k
,
n
n
-
3
μ
n
-
2
,
…
,
f
f
f
f
f
f
f
f
f
f
f
f
hhhh
-
F
k
,
n
μ
2
,
-
1
μ
,
-
k
+
∑
i
=
1
n
-
2
‍
F
k
,
n
-
1
-
i
F
k
,
n
i
-
1
μ
i
)
,
where
(49)
f
k
,
n
=
F
k
,
1
-
k
F
k
,
n
+
∑
i
=
1
n
-
2
‍
F
k
,
i
ν
n
-
(
i
+
1
)
.By Lemma2, Theorem 10, Theorem 11, and Theorem 13, the following conclusions can be attained.Theorem 17.
LetB
k
,
n
′
=
L
Circ
(
L
k
,
1
,
L
k
,
2
,
…
,
L
k
,
n
) be a left circulant matrix; then we have
(50)
det
⁡
B
k
,
n
′
=
(
-
1
)
(
n
-
1
)
(
n
-
2
)
/
2
×
[
∑
i
=
1
n
-
1
‍
(
L
k
,
i
+
2
-
(
(
k
2
+
2
)
/
k
)
L
k
,
i
+
1
)
(
s
t
)
i
-
1
k
s
n
-
1
+
k
t
n
-
2
g
g
g
g
g
×
∑
i
=
1
n
-
1
‍
(
L
k
,
i
+
2
-
(
(
k
2
+
2
)
/
k
)
L
k
,
i
+
1
)
(
s
t
)
i
-
1
]
,
where L
k
,
n is the nth k-Lucas number.Theorem 18.
LetB
k
,
n
′
=
L
Circ
(
L
k
,
1
,
L
k
,
2
,
…
,
L
k
,
n
) be a left circulant matrix; then B
k
,
n
′ is invertible for any positive integer n.Theorem 19.
LetB
k
,
n
′
=
L
Circ
(
L
k
,
1
,
L
k
,
2
,
…
,
L
k
,
n
) be a left circulant matrix; then we have
(51)
B
k
,
n
′
-
1
=
L
l
k
,
n
×
Circ
(
∑
i
=
1
n
-
2
‍
1
+
∑
i
=
1
n
-
2
‍
(
L
k
,
n
+
2
-
i
-
(
(
k
2
+
2
)
/
k
)
L
k
,
n
+
1
-
i
)
t
i
-
1
s
i
,
jjjjjjjjjj
g
g
g
g
j
j
(
k
2
+
4
)
t
n
-
3
k
s
n
-
2
,
…
,
(
k
2
+
4
)
t
2
k
s
3
,
jjjjjjjjjj
g
g
g
g
j
j
(
k
2
+
4
)
t
k
s
2
,
k
2
+
4
k
s
,
-
k
2
+
2
k
jjjjjjjjjj
g
g
g
g
j
j
+
∑
i
=
1
n
-
2
‍
(
L
k
,
n
+
1
-
i
-
(
(
k
2
+
2
)
/
k
)
L
k
,
n
-
i
)
t
i
-
1
s
i
)
.


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

In this section, letA
g
,
k
,
n
=
g-Circ
(
F
k
,
1
,
F
k
,
2
,
…
,
F
k
,
n
) and B
g
,
k
,
n
=
g-Circ
(
L
k
,
1
,
L
k
,
2
,
…
,
L
k
,
n
) be g-circulant matrices. By using the obtained conclusions, we give a determinant formula for the matrices A
g
,
k
,
n and B
g
,
k
,
n. Afterwards, we prove that A
g
,
k
,
n is an invertible matrix for n
>
2 and B
g
,
k
,
n is an invertible matrix if (
n
,
g
)
=
1. The inverse of the matrices A
g
,
k
,
n and B
g
,
k
,
n is also presented.From Lemmas3, 4, and 5 and Theorem 6, Theorem 7, and Theorem 9, we deduce the following results.Theorem 20.
LetA
g
,
k
,
n
=
g-Circ
(
F
k
,
1
,
F
k
,
2
,
…
,
F
k
,
n
) be a g-circulant matrix and (
n
,
g
)
=
1; then we have
(52)
det
⁡
A
g
,
k
,
n
=
det
⁡
Q
g
[
μ
n
-
1
+
F
k
,
n
n
-
2
∑
i
=
1
n
-
1
‍
F
k
,
i
ν
i
-
1
]
,
where F
k
,
n is the nth k-Fibonacci number.Theorem 21.
LetA
g
,
k
,
n
=
g-Circ
(
F
k
,
1
,
F
k
,
2
,
…
,
F
k
,
n
) be a g-circulant matrix and (
n
,
g
)
=
1; if n
>
2, then A
g
,
k
,
n is an invertible matrix.Theorem 22.
LetA
g
,
k
,
n
=
g-Circ
(
F
k
,
1
,
F
k
,
2
,
…
,
F
k
,
n
)
(
n
>
2
) be a g-circulant matrix and (
n
,
g
)
=
1; then we have
(53)
A
g
,
k
,
n
-
1
=
[
F
k
,
n
n
-
3
μ
n
-
2
1
f
k
,
n
Circ
(
1
+
∑
i
=
1
n
-
2
‍
F
k
,
n
-
i
F
k
,
n
i
-
1
μ
i
,
-
k
HHHHHHHHH
+
∑
i
=
1
n
-
2
‍
F
k
,
n
-
1
-
i
F
k
,
n
i
-
1
μ
i
,
-
1
μ
,
HHHHHHHHH
-
F
k
,
n
μ
2
,
-
F
k
,
n
2
μ
3
,
…
,
-
F
k
,
n
n
-
3
μ
n
-
2
)
]
Q
g
T
,
where f
k
,
n
=
F
k
,
1
-
k
F
k
,
n
+
∑
i
=
1
n
-
2
‍
F
k
,
i
ν
n
-
(
i
+
1
).Taking Lemmas3, 4, and 5 and Theorem 10, Theorem 11 and ,Theorem 13 into account, we have the following theorems.Theorem 23.
LetB
g
,
k
,
n
=
g-Circ
(
L
k
,
1
,
L
k
,
2
,
…
,
L
k
,
n
) be a g-circulant matrix and (
n
,
g
)
=
1; then we have
(54)
det
⁡
B
g
,
k
,
n
=
det
⁡
Q
g
[
(
×
(
s
t
)
i
-
1
)
k
s
n
-
1
+
k
t
n
-
2
∑
i
=
1
n
-
1
(
(
L
k
,
i
+
2
-
k
2
+
2
k
L
k
,
i
+
1
)
‍
v
v
v
v
v
v
v
v
v
v
v
j
j
j
k
k
k
k
k
k
k
k
k
k
k
k
×
(
s
t
)
i
-
1
)
]
,
where L
k
,
n is the nth k-Lucas number.Theorem 24.
LetB
g
,
k
,
n
=
g-Circ
(
L
k
,
1
,
L
k
,
2
,
…
,
L
k
,
n
) be a g-circulant matrix and (
n
,
g
)
=
1; then B
g
,
k
,
n is invertible matrix.Theorem 25.
LetB
g
,
k
,
n
=
g-Circ
(
L
k
,
1
,
L
k
,
2
,
…
,
L
k
,
n
) be a g-circulant matrix and (
n
,
g
)
=
1; then we have
(55)
B
g
,
k
,
n
-
1
=
[
(
∑
i
=
1
n
-
2
‍
(
L
k
,
n
+
1
-
i
-
(
(
k
2
+
2
)
/
k
)
L
k
,
n
-
i
)
t
i
-
1
s
i
)
1
l
k
,
n
d
×
Circ
(
1
+
∑
i
=
1
n
-
2
‍
(
L
k
,
n
+
2
-
i
-
(
(
k
2
+
2
)
/
k
)
L
k
,
n
+
1
-
i
)
t
i
-
1
s
i
,
g
g
g
g
g
g
g
g
h
h
(
k
2
+
4
)
t
n
-
3
k
s
n
-
2
,
…
,
(
k
2
+
4
)
t
2
k
s
3
,
g
g
g
g
g
g
g
g
h
h
(
k
2
+
4
)
t
k
s
2
,
k
2
+
4
k
s
,
-
k
2
+
2
k
g
g
g
g
g
g
g
g
h
h
+
∑
i
=
1
n
-
2
‍
(
L
k
,
n
+
1
-
i
-
(
(
k
2
+
2
)
/
k
)
L
k
,
n
-
i
)
t
i
-
1
s
i
)
]
×
Q
g
T
.


## 6. Conclusion

Circulant-type matrices have a very nice structure, and thek-Fibonacci and k-Lucas numbers also have amazing properties. The related problem of circulant-type matrices and some famous numbers are studied in this paper. We not only study invertibility of circulant-type matrices with the k-Fibonacci and k-Lucas numbers but also give the explicit determinants and explicit inverses. We would get a lot of good results if we combine famous numbers with circulant-type matrices, and the results would be used in solving ordinary differential equations. We will still focus our attentions on specific matrices and famous numbers in the future.



---

*Source: 238953-2014-05-20.xml*