Circulant matrices play an important role in solving delay differential equations. In this paper, circulant type matrices including the circulant and left circulant and g-circulant matrices with any continuous Fibonacci and Lucas numbers are considered. Firstly, the invertibility of the circulant matrix is discussed and the explicit determinant and the inverse matrices by constructing the transformation matrices are presented. Furthermore, the invertibility of the left circulant and g-circulant matrices is also studied. We obtain the explicit determinants and the inverse matrices of the left circulant and g-circulant matrices by utilizing the relationship between left circulant, g-circulant matrices and circulant matrix, respectively.
1. Introduction
Circulant matrices have important applications in solving various differential equations [1–3]. The use of circulant preconditioners for solving structured linear systems has been studied extensively since 1986; see [4, 5]. Circulant matrices also play an important role in solving delay differential equations. In [6], Chan et al. proposed a preconditioner called the Strang-type block-circulant preconditioner for solving linear systems from IVPs. The Strang-type preconditioner was also used to solve linear systems from differential-algebraic equations and delay differential equations; see [7–14]. In [15], Jin et al. proposed the GMRES method with the Strang-type block-circulant preconditioner for solving singular perturbation delay differential equations.
The g-circulant matrices play an important role in various applications as well; please refer to [16, 17] for details. There are discussions about the convergence in probability and in distribution of the spectral norm of g-circulant matrices in [18, 19]. Ngondiep et al. showed the singular values of g-circulants in [20].
Recently, some scholars have given various algorithms for the determinants and inverses of nonsingular circulant matrices [21, 22]. Unfortunately, the computational complexity of these algorithms is increasing dramatically with the increasing order of matrices. However, some authors gave the explicit determinants and inverse of circulant involving Fibonacci and Lucas numbers. For example, Jaiswal evaluated some determinants of circulant whose elements are the generalized Fibonacci numbers [23]. Lind presented the determinants of circulant involving Fibonacci numbers [24]. Lin gave the determinant of the Fibonacci-Lucas quasicyclic matrices in [25]. Shen et al. considered circulant matrices with Fibonacci and Lucas numbers and presented their explicit determinants and inverses [26]. Bozkurt and Tam gave determinants and inverses of circulant matrices with Jacobsthal and Jacobsthal-Lucas numbers in [27].
The purpose of this paper is to obtain the explicit determinants, explicit inverses of circulant, left circulant, and g-circulant matrices involving any continuous Fibonacci numbers and Lucas numbers. And we generalize the result in [26].
In the following, let r be a nonnegative integer. We adopt the following two conventions 00=1, and for any sequence {an},∑k=inak=0 in the case i>n.
The Fibonacci and Lucas sequences are defined by the following recurrence relations [23–26], respectively:
(1)Fn+1=Fn+Fn-1whereF0=0,F1=1,Ln+1=Ln+Ln-1whereL0=2,L1=1,
for n≥0. The first few values of the sequences are given by the following table:
(2)n012345678⋯Fn01123581321⋯Ln2134711182947⋯.
Let α and β be the roots of the characteristic equation x2-x-1=0; then the Binet formulas of the sequences {Fr+n} and {Lr+n} have the form
(3)Fr+n=αr+n-βr+nα-β,Lr+n=αr+n+βr+n.
Definition 1 (see [21, 22]).
In a right circulant matrix (or simply, circulant matrix)
(4)
Circ
(a1,a2,…,an)=[a1a2⋯anana1⋯an-1⋮⋮⋮a2a3⋯a1],
each row is a cyclic shift of the row above to the right. Right circulant matrix is a special case of a Toeplitz matrix. It is evidently determined by its first row (or column).
Definition 2 (see [22, 28]).
In a left circulant matrix (or reverse circulant matrix )
(5)
LCirc
(a1,a2,…,an)=[a1a2⋯ana2a3⋯a1⋮⋮⋮ana1⋯an-1],
each row is a cyclic shift of the row above to the left. Left circulant matrix is a special Hankel matrix.
Definition 3 (see [19, 29]).
A g-circulant matrix is an n×n complex matrix with the following form:
(6)Ag,n=(a1a2⋯anan-g+1an-g+2⋯an-gan-2g+1an-2g+2⋯an-2g⋮⋮⋱⋮ag+1ag+2⋯ag),
where g is a nonnegative integer and each of the subscripts is understood to be reduced modulo n.
The first row of Ag,n is (a1,a2,…,an); 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 standard circulant matrix. If g=n-1, then we obtain the so called left circulant matrix.
Lemma 4 (see [26]).
Let A=
Circ
(a1,a2,…,an) be circulant matrix; then one has
A is invertible if and only if the eigenvalues of A(7)λk=f(ωk)≠0,(k=0,1,…,n-1),
where f(x)=∑j=1najxj-1 and ω=exp(2πi/n);
if A is invertible, then the inverse A-1 of A is a circulant matrix.
Lemma 5.
Define
(8)Δ≔(100⋯00000⋯01000⋯10⋮⋮⋱⋮001⋯00010⋯00);
the matrix Δ is an orthogonal cyclic shift matrix (and a left circulant matrix). It holds that
LCirc
(a1,a2,…,an)=Δ
Circ
(a1,a2,…,an).
Lemma 6 (see [29]).
The n×n matrix Qg is unitary if and only if (n,g)=1, where Qg is a g-circulant matrix with first row e*=[1,0,…,0].
Lemma 7 (see [29]).
Ag,n is a g-circulant matrix with first row [a1,a2,…,an] if and only if Ag,n=QgC, where C=
Circ
(a1,a2,…,an).
2. Determinant, Invertibility, and Inverse of Circulant Matrix with Any Continuous Fibonacci Numbers
In this section, let Ar,n=
Circ
(Fr+1,Fr+2,…,Fr+n) be a circulant matrix. Firstly, we give the determinant equation of the matrix Ar,n. Afterwards, we prove that Ar,n is an invertible matrix for n>2, and then we find the inverse of the matrix Ar,n. Obviously, when n=2,r≠0, or n=1, Ar,n is also an invertible matrix.
Theorem 8.
Let Ar,n=
Circ
(Fr+1,Fr+2,…,Fr+n) be a circulant matrix. Then one has
(9)detAr,n=Fr+1·{∑k=1n-2(Fr+1-Fr+2Fr+1Fr+n)vvvvvvvvv+∑k=1n-2[(Fr+k+2-Fr+2Fr+1Fr+k+1)vvvvvvvvvvvvvv×(Fr+n-FrFr+1-Fr+n+1)n-(k+1)]∑k=1n-2}×(Fr+1-Fr+n+1)n-2,
where Fr+n is the (r+n)th Fibonacci number. Specially, when r=0, this result is the same as Theorem 2.1 in [26].
Proof.
Obviously, detA1=(1-Fn+1)n-1+Fnn-2∑k=1n-1Fk((1-Fn+1)/Fn)k-1 satisfies the formula. In the case n>1, let
(10)Γ=(1-Fr+2Fr+11-11-1001-1-1⋮⋰⋰⋰01⋰⋰01-1⋰001-1-1),Π1=(100⋯000(Fr+n-FrFr+1-Fr+n+1)n-20⋯010(Fr+n-FrFr+1-Fr+n+1)n-30⋯10⋮⋮⋮⋱⋮⋮0Fr+n-FrFr+1-Fr+n+11⋯00010⋯00)
be two n×n matrices; then we have
(11)ΓAr,nΠ1=[Fr+1fr,n′Fr+n-1⋯Fr+20fr,nτn⋯τ300Fr+1-Fr+n+1000Fr-Fr+n0⋮⋮⋱000Fr+1-Fr+n+1],
where
(12)ar=Fr+2Fr+1,fr,n′=∑k=1n-1Fr+k+1(Fr+n-FrFr+1-Fr+n+1)n-(k+1),τk=Fr+k-arFr+k-1,k=3,…,n,fr,n=(Fr+1-Fr+2Fr+1Fr+n)+∑k=1n-2[(Fr+n-FrFr+1-Fr+n+1)n-(k+1)(Fr+k+2-Fr+2Fr+1Fr+k+1)vvvvvvvv×(Fr+n-FrFr+1-Fr+n+1)n-(k+1)].
We obtain
(13)detΓdetAr,ndetΠ1=Fr+1·{(Fr+1-Fr+2Fr+1Fr+n)(Fr+n-FrFr+1-Fr+n+1)n-(k+1)]vvvvvvvv+∑k=1n-2[(Fr+k+2-Fr+2Fr+1Fr+k+1)(Fr+n-FrFr+1-Fr+n+1)n-(k+1)vvvvvvvvvvvvvv×(Fr+n-FrFr+1-Fr+n+1)n-(k+1)]}×(Fr+1-Fr+n+1)n-2;
while
(14)detΓ=(-1)(n-1)(n-2)/2,detΠ1=(-1)(n-1)(n-2)/2,
we have
(15)detAr,n=Fr+1·{(Fr+1-Fr+2Fr+1Fr+n)(Fr+n-FrFr+1-Fr+n+1)n-(k+1)]vvvvvvvv+∑k=1n-2[(Fr+n-FrFr+1-Fr+n+1)n-(k+1)(Fr+k+2-Fr+2Fr+1Fr+k+1)(Fr+n-FrFr+1-Fr+n+1)n-(k+1)vvvvvvvvvvvvvv×(Fr+n-FrFr+1-Fr+n+1)n-(k+1)]}×(Fr+1-Fr+n+1)n-2.
Theorem 9.
Let Ar,n=
Circ
(Fr+1,Fr+2,…,Fr+n) be a circulant matrix; if n>2, then Ar,n is an invertible matrix. Specially, when r=0, one gets Theorem 2.2 in [26].
Proof.
When n = 3 in Theorem 8, then we have detAr,n=(Fr+1+Fr+2+Fr+3)(Fr+12+FrFr+2)≠0; hence Ar,n is invertible. In the case n>3, since Fr+n=(αr+n-βr+n)/(α-β), where α+β=1, αβ=-1. We have
(16)f(ωk)=∑j=1nFr+j(ωk)j-1=1α-β∑j=1n(αr+j-βr+j)(ωk)j-1=1α-β[αr+1(1-αn)1-αωk-βr+1(1-βn)1-βωk]=(αr+1-βr+1)-(αr+n+1-βr+n+1)(α-β)(1-(α+β)ωk+αβω2k)-αβ(αr-βr-αr+n+βr+n)ωk(α-β)(1-(α+β)ωk+αβω2k)=Fr+1-Fr+n+1+(Fr-Fr+n)ωk1-ωk-ω2kvvvvvvvvvvvvvvvvvv(k=1,2,…,n-1).
If there exists ωl(l=1,2,…,n-1) such that f(ωl)=0, we obtain Fr+1-Fr+n+1+(Fr-Fr+n)ωl=0 for 1-ωl-ω2l≠0; thus, ωl=(Fr+1-Fr+n+1)/(Fr+n-Fr) is a real number. While
(17)ωl=exp(2lπin)=cos(2lπn)+isin(2lπn),
hence, sin(2lπ/n)=0; so we have ωl=-1 for 0<2lπ/n<2π. But x=-1 is not the root of the equation Fr+1-Fr+n+1+(Fr-Fr+n)x=0(n>3). We obtain f(ωk)≠0 for any ωk(k=1,2,…,n-1), while f(1)=∑j=1nFr+j=-Fr+1+Fr+n+1-(Fr-Fr+n)=Fr+n+2-Fr+2≠0. By Lemma 4, the proof is completed.
Lemma 10.
Let the entries of the matrix G=[gi,j]i,j=1n-2 be of the form
(18)gi,j={Fr+1-Fr+n+1,i=j,Fr-Fr+n,i=j+1,0,otherwise;
then the entries of the inverse G-1=[gi,j′]i,j=1n-2 of the matrix G are equal to
(19)gi,j′={(Fr+n-Fr)i-j(Fr+1-Fr+n+1)i-j+1,i≥j,0,i<j.
In particular, when r=0, one gets Lemma 2.1 in [26].
Proof.
Let ci,j=∑k=1n-2gi,kgk,j′. Obviously, ci,j=0 for i<j. In the case i=j, we obtain
(20)ci,i=gi,igi,i′=(Fr+1-Fr+n+1)·1Fr+1-Fr+n+1=1.
For i≥j+1, we obtain
(21)ci,j=∑k=1n-2gi,kgk,j′=gi,i-1gi-1,j′+gi,igi,j′=(Fr-Fr+n)·(Fr+n-Fr)i-j-1(Fr+1-Fr+n+1)i-j+(Fr+1-Fr+n+1)·(Fr+n-Fr)i-j(Fr+1-Fr+n+1)i-j+1=0.
Hence, we verify GG-1=In-2, where In-2 is (n-2)×(n-2) identity matrix. Similarly, we can verify G-1G=In-2. Thus, the proof is completed.
Theorem 11.
Let Ar,n=Circ(Fr+1,Fr+2,…,Fr+n)(n>2) be a circulant matrix.
Then one has
(22)Ar,n-1=1fr,n×
Circ
(1+∑i=1n-2(∑i=1n-2(Fr+n+2-i-Fr+2Fr+1Fr+n+1-i)vvvvvvvvvvvvvvvvv×(Fr+n-Fr)i-1∑i=1n-2)vvvvvvvvvvvvvvvv×((Fr+1-Fr+n+1)i)-1,-Fr+2Fr+1vvvvvvvvv+∑i=1n-2(22(Fr+n+1-i-(Fr+2/Fr+1)Fr+n-i)vvvvvvvvvvvvvvv×(Fr+n-Fr)i-122)vvvvvvvvvvvvvv×((Fr+1-Fr+n+1)i),vvvvvvvvv-Fr+3-(Fr+2/Fr+1)Fr+2Fr+1-Fr+n+1,vvvvvvvvv-(Fr+3-(Fr+2/Fr+1)Fr+2)(Fr+n-Fr)(Fr+1-Fr+n+1)2,vvvvvvvvv-(Fr+3-(Fr+2/Fr+1)Fr+2)(Fr+n-Fr)2(Fr+1-Fr+n+1)3,…,vvvvvvvvv-(Fr+3-(Fr+2/Fr+1)Fr+2)(Fr+n-Fr)n-3(Fr+1-Fr+n+1)n-2∑i=1n-2),
where
(23)fr,n=(Fr+1-Fr+2Fr+1Fr+n)+∑k=1n-2(Fr+k+2-Fr+2Fr+1Fr+k+1)(Fr+n-FrFr+1-Fr+n+1)n-(k+1).
Specially, when r=0, this result is the same as Theorem 2.3 in [26].
Proof.
Let
(24)Π2=(1-fr,n′Fr+1x3x4⋯xn01y3y4⋯yn0010⋯00001⋯0⋮⋮⋮⋮⋱⋮0000⋯1),
where ar=Fr+2/Fr+1,
(25)xi=fr,n′fr,nFr+n+3-i-(Fr+2/Fr+1)Fr+n+2-iFr+1-Fr+n+2-iFr+1bbbbbbbbbbbbbbbbbbbbbbbbb(i=3,4,…,n),yi=-Fr+n+3-i-(Fr+2/Fr+1)Fr+n+2-ifr,n(i=3,4,…,n),fr,n′=∑k=1n-1Fr+k+1(Fr+n-FrFr+1-Fr+n+1)n-(k+1),fr,n=(Fr+1-Fr+2Fr+1Fr+n)+∑k=1n-2(Fr+k+2-Fr+2Fr+1Fr+k+1)(Fr+n-FrFr+1-Fr+n+1)n-(k+1).
We have
(26)ΓAr,nΠ1Π2=D1⊕G,
where D1=diag(Fr+1,fr,n) is a diagonal matrix and D1⊕G is the direct sum of D1 and G. If we denote Π=Π1Π2, then we obtain
(27)Ar,n-1=Π(D1-1⊕G-1)Γ,
and the last row elements of the matrix Π are 0,1,y3,y4,…,yn-1,yn. By Lemma 10, if let Ar,n-1=
Circ
(u1,u2,…,un), then its last row elements are given by the following equations:
(28)u2=-1fr,nFr+2Fr+1+1fr,nCn(n-2),u3=-1fr,nCn(1),u4=-1fr,nCn(2)+1fr,nCn(1),u5=-1fr,nCn(3)+1fr,nCn(2)+1fr,nCn(1),⋮un=-1fr,nCn(n-2)+1fr,nCn(n-3)+1fnCn(n-4),u1=1fr,n+1fr,nCn(n-2)+1fr,nCn(n-3).
Let
(29)Cn(j)=∑i=1j(Fr+3+j-i-(Fr+2/Fr+1)Fr+2+j-i)(Fr+n-Fr)i-1(Fr+1-Fr+n+1)i=∑i=1jaj,r′(m1,r)i(m2,r)i-1(j=1,2,…,n-2);
we have
(30)Cn(2)-Cn(1)=∑i=12a2,r′(m2,r)i-1(m1,r)i-a1,r′m1,r=a1,r′(m1,r)2m2,r,Cn(n-2)+Cn(n-3)=∑i=1n-2an-2,r′(m2,r)i-1(m1,r)i+∑i=1n-3an-3,r′(m2,r)i-1(m1,r)i=∑i=1n-3(Fr+n+2-i-Ar,nFr+n+1-i)(m2,r)i-1(m1,r)i+a1,r′(m1,r)n-2(m2,r)n-3=∑i=1n-2(Fr+n+2-i-(Fr+2/Fr+1)Fr+n+1-i)(m2,r)i-1(m1,r)i,Cn(j+2)-Cn(j+1)-Cn(j)=∑i=1j+2aj+2,r′(m2,r)i-1(m1,r)i-∑i=1j+1aj+1,r′(m2,r)i-1(m1,r)i-∑i=1jaj,r′(m2,r)i-1(m1,r)i=(Fr+4-Ar,nFr+3)(m2,r)j(m1,r)j+1+(Fr+3-Ar,nFr+2)(m2,r)j+1(m1,r)j+2-(Fr+3-Ar,nFr+2)(m2,r)j(m1,r)j+1=(Fr+3-Ar,nFr+2)(m2,r)j+1(m1,r)j+2vvvvvvvvvv(j=1,2,…,n-4).
We obtain
(31)Ar,n-1=
Circ
(1+Cn(n-2)+Cn(n-3)fr,n,Cn(n-2)-Fr+2/Fr+1fr,n,cccvcccccc-Cn(1)fr,n,-Cn(2)-Cn(1)fr,n,cccvcccccc-Cn(3)-Cn(2)-Cn(1)fr,n,…,cccvcccccc-Cn(n-2)-Cn(n-3)-Cn(n-4)fr,n)=1fr,n
Circ
(1a1,r′(m1,r)n-2+∑i=1n-2((Fr+n+2-i-Fr+2Fr+1Fr+n+1-i)vvcccccccccccccccvvvv×(Fr+n+2-i-Fr+2Fr+1Fr+n+1-i)∑i=1n-2(m2,r)i-1)×((m1,r)i)-1,cccvccccvvccc-Fr+2Fr+1+∑i=1n-2an-2,r′(m2,r)i-1(m1,r)i,cccvccccvvccc-a1,r′m1,r,-a1,r′(m1,r)2m2,r,cccvccccvvccc-a1,r′(m1,r)3(m2,r)2,…,cccvccccvvccc-a1,r′(m1,r)n-2(m2,r)n-3)=1fr,n
Circ
(1a1,r′(m1,r)n-2+∑i=1n-2((Fr+n+2-i-Fr+2Fr+1Fr+n+1-i)vvvvvvvvvvvvvvvvv×Fr+2Fr+1(Fr+n-Fr)i-1∑i=1n-2)cccvccccvvcvvvvcc×((Fr+1-Fr+n+1)i)-1,-Fr+2Fr+1cccvccccvvccc+∑i=1n-2((Fr+n+1-i-Fr+2Fr+1Fr+n-i)vvvvvvvvvvvvvvvvv×(Fr+n+1-i-Fr+2Fr+1Fr+n-i)(Fr+n-Fr)i-1)ccvccccvvvvvvvvcc×((Fr+1-Fr+n+1)i)-1cccvccccvvcccc-Fr+3-(Fr+2/Fr+1)Fr+2Fr+1-Fr+n+1,cccvccccvvcccc-(Fr+3-(Fr+2/Fr+1)Fr+2)(Fr+n-Fr)(Fr+1-Fr+n+1)2,cccvccccvvcccc-(Fr+3-(Fr+2/Fr+1)Fr+2)(Fr+n-Fr)2(Fr+1-Fr+n+1)3,…,cccvccccvvcccc-(Fr+3-(Fr+2/Fr+1)Fr+2)(Fr+n-Fr)n-3(Fr+1-Fr+n+1)n-2),
where
(32)fr,n=(Fr+1-Fr+2Fr+1Fr+n)+∑k=1n-2(Fr+k+2-Fr+2Fr+1Fr+k+1)(Fr+n-FrFr+1-Fr+n+1)n-(k+1).
3. Determinant, Invertibility, and Inverse of Circulant Matrix with Any Continuous Lucas Numbers
In this section, let Br,n=
Circ
(Lr+1,Lr+2,…,Lr+n) be a circulant matrix. Firstly, we give a determinant formula for the matrix Br,n. Afterwards, we prove that Br,n is an invertible matrix for any positive integer n, and then we find the inverse of the matrix Br,n.
Theorem 12.
Let Br,n=
Circ
(Lr+1,Lr+2,…,Lr+n) be a circulant matrix; then one has
(33)detBr,n=Lr+1·{(Lr+1-Lr+2Lr+1Lr+n)(Lr+n-LrLr+1-Lr+n+1)n-(k+1)]vvvvvvvvvvv+∑k=1n-2[(Lr+k+2-Lr+2Lr+1Lr+k+1)(Lr+n-LrLr+1-Lr+n+1)n-(k+1)vvvvvvvvvvvvvvvvvv×(Lr+n-LrLr+1-Lr+n+1)n-(k+1)]}×(Lr+1-Lr+n+1)n-2,
where Lr+n is the (r+n)th Lucas number. In particular, when r=0, one gets Theorem 3.1 in [26].
Proof.
Obviously, detB1=(1-Ln+1)n-1+(Ln-2)n-2∑k=1n-1(Lk+2-3Lk+1)((1-Ln+1)/(Ln-2))k-1 satisfies the formula, when n>1; let
(34)Σ=(1-Lr+2Lr+11-11-1001-1-1⋮⋰⋰⋰01⋰⋰01-1⋰001-1-1),Ω1=(100⋯000(Lr+n-LrLr+1-Lr+n+1)n-20⋯010(Lr+n-LrLr+1-Lr+n+1)n-30⋯10⋮⋮⋮⋱⋮⋮0Lr+n-LrLr+1-Lr+n+11⋯00010⋯00),
be two n×n matrices, we have
(35)ΣBr,nΩ1=[Lr+1lr,n′Lr+n-1⋯Lr+20lr,nLr+n-brLr+n-1⋯Lr+3-brLr+200Lr+1-Lr+n-1000Lr-Lr+n0⋮⋮⋱000Lr+1-Lr+n-1],
where
(36)br=Lr+2Lr+1,lr,n=(Lr+1-Lr+2Lr+1Lr+n)+∑k=1n-2(Lr+k+2-Lr+2Lr+1Lr+k+1)vvvvvvv×(Lr+n-LrLr+1-Lr+n+1)n-(k+1),lr,n′=∑k=1n-1Lr+k+1(Lr+n-LrLr+1-Lr+n+1)n-(k+1).
We obtain
(37)detΣdetBr,ndetΩ1=Lr+1·{(Lr+1-Lr+2Lr+1Lr+n)(Lr+n-LrLr+1-Lr+n+1)n-(k+1)]vvvvvvvvv+∑k=1n-2[(Lr+k+2-brLr+k+1)(Lr+n-LrLr+1-Lr+n+1)n-(k+1)vvvvvvvvvvvvvvv×(Lr+n-LrLr+1-Lr+n+1)n-(k+1)]}×(Lr+1-Lr+n+1)n-2,
while
(38)detΣ=detΩ1=(-1)(n-1)(n-2)/2.
We have
(39)detBr,n=Lr+1·{(Lr+1-Lr+2Lr+1Lr+n)(Lr+n-LrLr+1-Lr+n+1)n-(k+1)]vvvvvvvvv+∑k=1n-2[(Lr+k+2-Lr+2Lr+1Lr+k+1)(Lr+n-LrLr+1-Lr+n+1)n-(k+1)vvvvvvvvvvvvvvv×(Lr+n-LrLr+1-Lr+n+1)n-(k+1)]}×(Lr+1-Lr+n+1)n-2.
Theorem 13.
Let Br,n=
Circ
(Lr+1,Lr+2,…,Lr+n) be a circulant matrix; then Br,n is invertible for any positive integer n. Specially, when r=0, one gets Theorem 3.2 in [26].
Proof.
Since Ln+r=αn+r+βn+r, where α+β=1, α·β=-1. Hence we have
(40)f(ωk)=∑j=1nLn+r(ωk)j-1=∑j=1n(αr+j+βr+j)(ωk)j-1=αr+1(1-αn)1-αωk+βr+1(1-βn)1-βωk(because 1-αωk≠0 and 1-βωk≠0)=(αr+1+βr+1)-(αr+n+1+βr+n+1)1-(α+β)ωk+αβω2k-αβ(αr+βr-αr+n-βr+n)ωk1-(α+β)ωk+αβω2k=Lr+1-Lr+n+1+(Lr-Lr+n)ωk1-ωk-ω2kvvvvvvvvvvvvv(k=1,2,…,n-1).
If there exists ωl(l=1,2,…,n-1) such that f(ωl)=0, we obtain Lr+1-Lr+n+1+(Lr-Lr+n)ωl=0 for 1-ωl-ω2l≠0; thus, ωl=(Lr+1-Lr+n+1)/(Lr+n-Lr) is a real number, while
(41)ωl=exp(2lπin)=cos2lπn+isin2lπn.
Hence, sin(2lπ/n)=0; we have ωl=-1 for 0<2lπ/n<2π. But x=-1 is not the root of the equation Lr+1-Lr+n+1+(Lr-Lr+n)x=0 for any positive integer n. We obtain f(ωk)≠0 for any ωk(k=1,2,…,n-1), while f(1)=∑j=1nLr+j=Lr+n+2-Lr+2≠0. By Lemma 4, the proof is completed.
Lemma 14.
Let the entries of the matrix H=[hi,j]i,j=1n-2 be of the form
(42)hi,j={Lr+1-Lr+n+1,i=j,Lr-Lr+n,i=j+1,0,otherwise;
then the entries of the inverse H-1=[hi,j′]i,j=1n-2 of the matrix H are equal to
(43)hi,j′={(Lr+n-Lr)i-j(Lr+1-Lr+n+1)i-j+1,i≥j,0,i<j.
Specially, when r=0, one gets Lemma 3.1 in [26].
Proof.
Let ri,j=∑k=1n-2hi,khk,j′. Obviously, ri,j=0 for i<j. In the case i=j, we obtain
(44)ri,i=hi,ihi,i′=(Lr+1-Lr+n+1)·1Lr+1-Lr+n+1=1.
For i≥j+1, we obtain
(45)ri,j=∑k=1n-2hi,khk,j′=hi,i-1hi-1,j′+hi,ihi,j′=(Lr-Lr+n)·(Lr+n-Lr)i-j-1(Lr+1-Lr+n+1)i-j+(Lr+1-Lr+n+1)·(Lr+n-Lr)i-j(Lr+1-Lr+n+1)i-j+1=0.
Hence, we verify HH-1=In-2, where In-2 is (n-2)×(n-2) identity matrix. Similarly, we can verify H-1H=In-2. Thus, the proof is completed.
Theorem 15.
Let Br,n=
Circ
(Lr+1,Lr+2,…,Lr+n) be a circulant matrix; then we have
(46)Br,n-1=1lr,n
Circ
(1(Lr+3-(Lr+2/Lr+1)Lr+2)(Lr+n-Lr)n-3(Lr+1-Lr+n+1)n-2vvvvvvvvvvv+∑i=1n-2(∑k=1n-2(Lr+n+2-i-Lr+2Lr+1Lr+n+1-i)vvvvvvvvvvvvvvvvv×(Lr+n-Lr)i-1∑k=1n-2)vvvvvvvvvvvvvvvv×((Lr+1-Lr+n+1)i)-1,-Lr+2Lr+1vvvvvvvvvvv+∑i=1n-2((Lr+n+1-i-Lr+2Lr+1Lr+n-i)(Lr+n-Lr)i-1vvvvvvvvvvvvvvvvv×(Lr+n+1-i-Lr+2Lr+1Lr+n-i)(Lr+n-Lr)i-1)vvvvvvvvvvvvvvvv×((Lr+1-Lr+n+1)i)-1,vvvvvvvvvvv-Lr+3-(Lr+2/Lr+1)Lr+2Lr+1-Lr+n+1,vvvvvvvvvvv-(Lr+3-(Lr+2/Lr+1)Lr+2)(Lr+n-Lr)(Lr+1-Lr+n+1)2,vvvvvvvvvvv-(Lr+3-(Lr+2/Lr+1)Lr+2)(Lr+n-Lr)2(Lr+1-Lr+n+1)3,…,vvvvvvvvvvv-(Lr+3-(Lr+2/Lr+1)Lr+2)(Lr+n-Lr)n-3(Lr+1-Lr+n+1)n-2),
where
(47)lr,n=(Lr+1-Lr+2Lr+1Lr+n)+∑k=1n-2(Lr+k+2-Lr+2Lr+1Lr+k+1)(Lr+n-LrLr+1-Lr+n+1)n-(k+1).
In particular, when r=0, the result is the same as Theorem 3.3 in [26].
Proof.
Let Ω2 be the form of
(48)(1-ln′Lr+1x3′x4′⋯xn′01y3′y4′⋯yn′0010⋯00001⋯0⋮⋮⋮⋮⋱⋮0000⋯1),
where
(49)br=Lr+2Lr+1,xi′=lr,n′lr,nLr+n+3-i-(Lr+2/Lr+1)Lr+n+2-iLr+1-Lr+n+2-iLr+1(i=3,4,…,n),yi′=-Lr+n+3-i-(Lr+2/Lr+1)Lr+n+2-ilr,nvvvvvvvvvvvvvvv(i=3,4,…,n),lr,n′=∑k=1n-1Lr+k+1(Lr+n-LrLr+1-Lr+n+1)n-(k+1),lr,n=(Lr+1-Lr+2Lr+1Lr+n)+∑k=1n-2(Lr+k+2-Lr+2Lr+1Lr+k+1)(Lr+n-LrLr+1-Lr+n+1)n-(k+1).
We have
(50)ΣBr,nΩ1Ω2=D2⊕H,
where D2=diag(Lr+1,lr,n) is a diagonal matrix and D2⊕H is the direct sum of D2 and H. If we denote Ω=Ω1Ω2, we obtain
(51)Br,n-1=Ω(D2-1⊕H-1)Σ,
and the last row elements of the matrix Ω are 0,1,y3′,y4′,⋯,yn′. By Lemma 14, if let Br,n-1=
Circ
(v1,v2,…,vn), then its last row elements are given by the following equations:
(52)v2=-1lr,nLr+2Lr+1+1lr,nDn(n-2),v3=-1lr,nDn(1),v4=-1lr,nDn(2)+1lr,nCn(1),v5=-1lr,nDn(3)+1lr,nDn(2)+1lr,nDn(1),⋮vn=-1lr,nDn(n-2)+1lr,nDn(n-3)+1lnDn(n-4),v1=1lr,n+1lr,nDn(n-2)+1lr,nDn(n-3).
Let
(53)Dn(j)=∑i=1j((Lr+3+j-i-Lr+2Lr+1Lr+2+j-i)∑i=1jvvvvvv×∑i=1j(Lr+n-Lr)i-1)×((Lr+1-Lr+n+1)i)-1=∑i=1jbj,r′(h1,r)i(h2,r)i-1(j=1,2,…,n-2);
we have
(54)Dn(2)-Dn(1)=∑i=12b2,r′(h2,r)i-1(h1,r)i-b1,r′h1,r=b1,r′(h1,r)2h2,r,Dn(n-2)+Dn(n-3)=∑i=1n-2bn-2,r′(h2,r)i-1(h1,r)i+∑i=1n-3(bn-3,r′)(h2,r)i-1(h1,r)i=∑i=1n-3(Lr+n+2-i-Br,nLr+n+1-i)(h2,r)i-1(h1,r)i+b1,r′(h1,r)n-2(h2,r)n-3=∑i=1n-2(Lr+n+2-i-Br,nLr+n+1-i)(h2,r)i-1(h1,r)i,Dn(j+2)-Dn(j+1)-Dn(j)=∑i=1j+2(bj+2,r′)(h2,r)i-1(h1,r)i-∑i=1j+1(bj+1,r′)(h2,r)i-1(h1,r)i-∑i=1j(bj,r′)(h2,r)i-1(h1,r)i=(Lr+4-Br,nLr+3)(h2,r)j(h1,r)j+1+(Lr+3-Br,nLr+2)(h2,r)j+1(h1,r)j+2-(Lr+3-Br,nLr+2)(h2,r)j(h1,r)j+1=(Lr+3-Br,nLr+2)(h2,r)j+1(h1,r)j+2=b1,r′(h2,r)j+1(h1,r)j+2(j=1,2,…,n-4).
We obtain
(55)Br,n-1=
Circ
(1+Dn(n-3)+Dn(n-2)lr,n,Dn(n-2)-Lr+2/Lr+1lr,n,vvvvvvvv-Dn(1)lr,n,Dn(1)-Dn(2)lr,n,Dn(1)+Dn(2)-Dn(3)lr,n,…,vvvvvvvvDn(n-4)+Dn(n-3)-Dn(n-2)lr,n)=1lr,n
Circ
(1b1,r′(h1,r)n-2(h2,r)n-3vvvvvvvvvvv+∑i=1n-2(Lr+n+2-i-Br,nLr+n+1-i)(h2,r)i-1(h1,r)i,vvvvvvvvvvv-Lr+2Lr+1+∑i=1n-2bn-2,r′(h2,r)i-1(h1,r)i,vvvvvvvvvvv-b1,r′h1,r,-b1,r′(h1,r)2h2,r,vvvvvvvvvvv-b1,r′(h1,r)3(h2,r)2,…,-b1,r′(h1,r)n-2(h2,r)n-3)=1lr,n
Circ
(1(Lr+3-(Lr+2/Lr+1)Lr+2)(Lr+n-Lr)n-3(Lr+1-Lr+n+1)n-2vvvvvvvvvvv+∑i=1n-2(∑i=1n-2(Lr+n+2-i-Lr+2Lr+1Lr+n+1-i)vvvvvvvvvvvvvvvvvv×(Lr+n-Lr)i-1∑i=1n-2)vvvvvvvvvvvvvvvv×((Lr+1-Lr+n+1)i)-1,vvvvvvvvvvv-Lr+2Lr+1+∑i=1n-2(∑i=1n-2(Lr+n+1-i-Lr+2Lr+1Lr+n-i)vvvvvvvvvvvvvvvvvvvvvvvv×(Lr+n-Lr)i-1∑i=1n-2)vvvvvvvvvvvvvvvvvvvvvvv×((Lr+1-Lr+n+1)i)-1,vvvvvvvvvvv-Lr+3-(Lr+2/Lr+1)Lr+2Lr+1-Lr+n+1,vvvvvvvvvvv-(Lr+3-(Lr+2/Lr+1)Lr+2)(Lr+n-Lr)(Lr+1-Lr+n+1)2,vvvvvvvvvv-(Lr+3-(Lr+2/Lr+1)Lr+2)(Lr+n-Lr)2(Lr+1-Lr+n+1)3,…,vvvvvvvvvv-(Lr+3-(Lr+2/Lr+1)Lr+2)(Lr+n-Lr)n-3(Lr+1-Lr+n+1)n-2).
4. Determinant, Invertibility, and Inverse of Left Circulant Matrix with Any Continuous Fibonacci and Lucas Numbers
In this section, let Ar,n′=
LCirc
(Fr+1,Fr+2,…,Fr+n) and Br,n′=
LCirc
(Lr+1,Lr+2,…,Lr+n) be left circulant matrices. By using the obtained conclusions, we give a determinant formula for the matrices Ar,n′ and Br,n′. Afterwards, we prove that Ar,n′ is an invertible matrix for n>2 and Br,n′ is an invertible matrix for any positive integer n. The inverse of the matrices Ar,n′ and Br,n′ is also presented.
According to Lemma 5, Theorem 8, Theorem 9, and Theorem 11, we can obtain the following theorems.
Theorem 16.
Let Ar,n′=
LCirc
(Fr+1,Fr+2,…,Fr+n) be a left circulant matrix; then one has
(56)detAr,n′=(-1)(n-1)(n-2)/2·Fr+1·[(Fr+1-Fr+2Fr+1Fr+n)(Fr+n-FrFr+1-Fr+n+1)n-(k+1)vvvvvvv+∑k=1n-2(Fr+k+2-Fr+2Fr+1Fr+k+1)vvvvvvv×(Fr+n-FrFr+1-Fr+n+1)n-(k+1)]×(Fr+1-Fr+n+1)n-2,
where Fr+n is the (r+n)th Fibonacci number.
Theorem 17.
Let Ar,n′=
LCirc
(Fr+1,Fr+2,…,Fr+n) be a left circulant matrix; if n>2, then Ar,n′ is an invertible matrix.
Theorem 18.
Let Ar,n′=
LCirc
(Fr+1,Fr+2,…,Fr+n)(n>2) be a left circulant matrix; then one has
(57)Ar,n′-1=1fr,n×
LCirc
(1∑21+∑i=1n-2((Fr+n+2-i-Fr+2Fr+1Fr+n+1-i)vvvvvvvvvvvvvvvvvv×(Fr+n+2-i-Fr+2Fr+1Fr+n+1-i)(Fr+n-Fr)i-1)vvvvvvvvvvvvvvvvv×((Fr+1-Fr+n+1)i)-1,vvvvvvvvv-(Fr+3-(Fr+2/Fr+1)Fr+2)(Fr+n-Fr)n-3(Fr+1-Fr+n+1)n-2,…,vvvvvvvvv-(Fr+3-(Fr+2/Fr+1)Fr+2)(Fr+n-Fr)2(Fr+1-Fr+n+1)3,vvvvvvvvv-(Fr+3-(Fr+2/Fr+1)Fr+2)(Fr+n-Fr)(Fr+1-Fr+n+1)2,vvvvvvvvv-Fr+3-(Fr+2/Fr+1)Fr+2Fr+1-Fr+n+1,vvvvvvvvv-Fr+2Fr+1vvvvvvvvv+∑i=1n-2((Fr+n+1-i-Fr+2Fr+1Fr+n-i)vvvvvvvvvvvvvvv×(Fr+n+1-i-Fr+2Fr+1Fr+n-i)(Fr+n-Fr)i-1)vvvvvvvvvvvvvv×((Fr+1-Fr+n+1)i)-1∑21),
where
(58)fr,n=(Fr+1-Fr+2Fr+1Fr+n)+∑k=1n-2(Fr+k+2-Fr+2Fr+1Fr+k+1)(Fr+n-FrFr+1-Fr+n+1)n-(k+1).
By Lemma 5, Theorem 12, Theorem 13, and Theorem 15, the following conclusions can be attained.
Theorem 19.
Let Br,n′=
LCirc
(Lr+1,Lr+2,…,Lr+n) be a left circulant matrix; then one has
(59)detBr,n′=(-1)(n-1)(n-2)/2·Lr+1·[(Lr+1-Lr+2Lr+1Lr+n)(Lr+n-LrLr+1-Lr+n+1)n-(k+1)vvvvvvv+∑k=1n-2(Lr+k+2-Lr+2Lr+1Lr+k+1)vvvvvvvvvvv×(Lr+n-LrLr+1-Lr+n+1)n-(k+1)]×(Lr+1-Lr+n+1)n-2,
where Lr+n is the (r+n)th Lucas number.
Theorem 20.
Let Br,n′=
LCirc
(Lr+1,Lr+2,…,Lr+n) be a left circulant matrix; then Br,n′ is invertible for any positive integer n.
Theorem 21.
Let Br,n′=
LCirc
(Lr+1,Lr+2,…,Lr+n) be a left circulant matrix; then one can obtain
(60)Br,n′-1=1lr,n×
LCirc
(1+∑i=1n-2((Lr+n+2-i-Lr+2Lr+1Lr+n+1-i)vvvvvvvvvvvvvvvvvv×(Lr+n+2-i-Lr+2Lr+1Lr+n+1-i)(Lr+n-Lr)i-1)vvvvvvvvvvvvvvvvv×((Lr+1-Lr+n+1)i)-1,vvvvvvvvv-(Lr+3-(Lr+2/Lr+1)Lr+2)(Lr+n-Lr)n-3(Lr+1-Lr+n+1)n-2,…,vvvvvvvvv-(Lr+3-(Lr+2/Lr+1)Lr+2)(Lr+n-Lr)2(Lr+1-Lr+n+1)3,vvvvvvvvv-(Lr+3-(Lr+2/Lr+1)Lr+2)(Lr+n-Lr)(Lr+1-Lr+n+1)2,vvvvvvvvv-Lr+3-(Lr+2/Lr+1)Lr+2Lr+1-Lr+n+1,vvvvvvvvv-Lr+2Lr+1vvvvvvvvv+∑i=1n-2((Lr+n+1-i-Lr+2Lr+1Lr+n-i)∑k=1n-2vvvvvvvvvvvvvvvv×∑k=1n-2(Lr+n-Lr)i-1)vvvvvvvvvvvvvvv∑i=1n-2((Lr+n+2-i-Lr+2Lr+1Lr+n+1-i)×((Lr+1-Lr+n+1)i)-1),
where
(61)lr,n=(Lr+1-Lr+2Lr+1Lr+n)+∑k=1n-2(Lr+k+2-Lr+2Lr+1Lr+k+1)(Lr+n-LrLr+1-Lr+n+1)n-(k+1).
5. Determinant, Invertibility, and Inverse of g-Circulant Matrix with Any Continuous Fibonacci and Lucas Numbers
In this section, let Ag,r,n=g-Circ
(Fr+1,Fr+2,…,Fr+n) and Bg,r,n=g-Circ
(Lr+1,Lr+2,…,Lr+n) be g-circulant matrices. By using the obtained conclusions, we give a determinant formula for the matrices Ag,r,n and Bg,r,n. Afterwards, we prove that Ag,r,n is an invertible matrix for n>2 and Bg,r,n is an invertible matrix if (n,g)=1. The inverse of the matrices Ag,r,n and Bg,r,n is also presented.
From Lemma 6, Lemma 7, Theorem 8, Theorem 9, and Theorem 11, we deduce the following results.
Theorem 22.
Let Ag,r,n=g-Circ
(Fr+1,Fr+2,…,Fr+n) be a g-circulant matrix; then one has
(62)detAg,r,n=detQg·Fr+1·[(Fr+1-Fr+2Fr+1Fr+n)(Fr+n-FrFr+1-Fr+n+1)n-(k+1)vvvvv+∑k=1n-2(Fr+k+2-Fr+2Fr+1Fr+k+1)vvvvv×(Fr+n-FrFr+1-Fr+n+1)n-(k+1)]×(Fr+1-Fr+n+1)n-2,
where Fr+n is the (r+n)th Fibonacci number.
Theorem 23.
Let Ag,r,n=g-Circ
(Fr+1,Fr+2,…,Fr+n) be a g-circulant matrix and (g,n)=1; if n>2, then Ag,r,n is an invertible matrix.
Theorem 24.
Let Ag,r,n=g-Circ
(Fr+1,Fr+2,…,Fr+n)(n>2) be a g-circulant matrix and (g,n)=1; then
(63)Ag,r,n-1=[1fr,n(Fr+3-(Fr+2/Fr+1)Fr+2)(Fr+n-Fr)n-3(Fr+1-Fr+n+1)n-2)vvvv×
Circ
(1+∑i=1n-2((Fr+n+2-i-Fr+2Fr+1Fr+n+1-i)vvvvvvvvvvvvvvvvvvv×(Fr+n+2-i-Fr+2Fr+1Fr+n+1-i)(Fr+n-Fr)i-1)vvvvvvvvvvvvvvvvvv×((Fr+1-Fr+n+1)i)-1,vvvvvvvvvvv-Fr+2Fr+1vvvvvvvvvvv+∑i=1n-2((Fr+n+1-i-Fr+2Fr+1Fr+n-i)vvvvvvvvvvvvvvvvvv×(Fr+n+1-i-Fr+2Fr+1Fr+n-i)(Fr+n-Fr)i-1)vvvvvvvvvvvvvvvvv×((Fr+1-Fr+n+1)i)-1,vvvvvvvvvvv-Fr+3-(Fr+2/Fr+1)Fr+2Fr+1-Fr+n+1,vvvvvvvvvvv-(Fr+3-(Fr+2/Fr+1)Fr+2)(Fr+n-Fr)(Fr+1-Fr+n+1)2,vvvvvvvvvvv-(Fr+3-(Fr+2/Fr+1)Fr+2)(Fr+n-Fr)2(Fr+1-Fr+n+1)3,…,vvvvvvvvvvv-(Fr+3-(Fr+2/Fr+1)Fr+2)(Fr+n-Fr)n-3(Fr+1-Fr+n+1)n-2)]vvvvvvvvvvv×QgT,
where
(64)fr,n=(Fr+1-Fr+2Fr+1Fr+n)+∑k=1n-2(Fr+k+2-Fr+2Fr+1Fr+k+1)(Fr+n-FrFr+1-Fr+n+1)n-(k+1).
Taking Lemma 6, Lemma 7, Theorem 12, Theorem 13, and Theorem 15 into account, one has the following theorems.
Theorem 25.
Let Bg,r,n=g-Circ
(Lr+1,Lr+2,…,Lr+n) be a g-circulant matrix; then one has
(65)detBg,r,n=detQg·Lr+1·[(Lr+1-Lr+2Lr+1Lr+n)(Lr+n-LrLr+1-Lr+n+1)n-(k+1)vvvv+∑k=1n-2(Lr+k+2-Lr+2Lr+1Lr+k+1)vvvv×(Lr+n-LrLr+1-Lr+n+1)n-(k+1)]×(Lr+1-Lr+n+1)n-2,
where Lr+n is the (r+n)th Lucas number.
Theorem 26.
Let Bg,r,n=g-Circ
(Lr+1,Lr+2,…,Lr+n) be a g-circulant matrix and (g,n)=1; if n>2, then Bg,r,n is an invertible matrix.
Theorem 27.
Let Bg,r,n=g-Circ
(Lr+1,Lr+2,…,Lr+n)(n>2) be a g-circulant matrix and (g,n)=1; then
(66)Bg,r,n-1=[1lr,n(Lr+3-(Lr+2/Lr+1)Lr+2)(Lr+n-Lr)n-3(Lr+1-Lr+n+1)n-2)vvv×
Circ
(1+∑i=1n-2(∑k=1n-2(Lr+n+2-i-Lr+2Lr+1Lr+n+1-i)vvvvvvvvvvvvvvvvvv×(Lr+n-Lr)i-1∑k=1n-2)vvvvvvvvvvvvvvvvv×((Lr+1-Lr+n+1)i)-1,vvvvvvvvvv-Lr+2Lr+1vvvvvvvvvv+∑i=1n-2((Lr+n+1-i-Lr+2Lr+1Lr+n-i)vvvvvvvvvvvvvvvvv×(Lr+n+1-i-Lr+2Lr+1Lr+n-i)(Lr+n-Lr)i-1)vvvvvvvvvvvvvvvv×((Lr+1-Lr+n+1)i)-1,vvvvvvvvvv-Lr+3-(Lr+2/Lr+1)Lr+2Lr+1-Lr+n+1,vvvvvvvvvv-(Lr+3-(Lr+2/Lr+1)Lr+2)(Lr+n-Lr)(Lr+1-Lr+n+1)2,vvvvvvvvvv-(Lr+3-(Lr+2/Lr+1)Lr+2)(Lr+n-Lr)2(Lr+1-Lr+n+1)3,…,vvvvvvvvvv-(Lr+3-(Lr+2/Lr+1)Lr+2)(Lr+n-Lr)n-3(Lr+1-Lr+n+1)n-2)]vvvvvvvvvv×QgT,
where
(67)lr,n=(Lr+1-Lr+2Lr+1Lr+n)+∑k=1n-2(Lr+k+2-Lr+2Lr+1Lr+k+1)(Lr+n-LrLr+1-Lr+n+1)n-(k+1).
Conflict of Interests
The authors declare that there is no conflict of interests regarding the publication of this paper.
Acknowledgments
The research was supported by the Development Project of Science & Technology of Shandong Province (Grant no. 2012GGX10115) and NSFC (Grant no. 11301251) and the AMEP of Linyi University, China.
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