# Exact Inverse Matrices of Fermat and Mersenne Circulant Matrix

Authors: Yanpeng Zheng; Sugoog Shon
Journal: Abstract and Applied Analysis
 (2015)

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## Abstract

The well known circulant matrices are applied to solve networked systems. In this paper, circulant and left circulant matrices with the Fermat and Mersenne numbers are considered. The nonsingularity of these special matrices is discussed. Meanwhile, the exact determinants and inverse matrices of these special matrices are presented.


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## Body

## 1. Introduction

Circulant matrices are an important tool in solving networked systems. In [1], the authors investigated the storage of binary cycles in Hopfield-type and other neural networks involving circulant matrix. In [2], the authors considered a special class of the feedback delay network using circulant matrices. Distributed differential space-time codes that work for networks with any number of relays using circulant matrices were proposed by Jing and Jafarkhani in [3]. Bašić [4] solved the question for when circulant quantum spin networks with nearest-neighbor couplings can give perfect state transfer. Wang et al. considered two-way transmission model ensured that circular convolution between two frequency selective channels in [5]. Li et al. [6] presented a low-complexity binary framewise network coding encoder design based on circulant matrix.Circulant matrices have been applied to various disciplines including image processing, communications, signal processing, and encoding. Circulant type matrices have established the substantial basis with the work in [7–12] and so on.Lately, some authors gave the explicit determinant and inverse of the circulant and skew-circulant involving famous numbers. For example, Yao and Jiang [13] presented the determinants, inverses, norm, and spread of skew circulant type matrices involving any continuous Lucas numbers. Jiang et al. [14] considered circulant type matrices with the k-Fibonacci and k-Lucas numbers and presented the explicit determinant and inverse matrix by constructing the transformation matrices. Dazheng [15] got the determinant of the Fibonacci-Lucas quasi-cyclic matrices. Determinants and inverses of circulant matrices with Jacobsthal and Jacobsthal-Lucas numbers were obtained by Bozkurt and Tam in [16].For any integerm≥0, let Fm=22m+1 be the mth Fermat number. It is well known that Fm is prime for m≤4, but there is no other m for which Fm is known to prime. The Mersenne and Fermat sequences are defined by the following recurrence relations [17, 18], respectively:(1)Mn+1=3Mn-2Mn-1Fn+1=3Fn-2Fn-1with the initial condition M0=0, M1=1, F0=2, F1=3, for n≥1.Letα and β be the roots of the characteristic equation x2-3x+2=0; then the Binet formulas of the sequences {Mk+n} and {Fk+n} have the form (2)Mk+n=αk+n-βk+nα-β,Fk+n=αk+n+βk+n.Lemma 1.
LetMk+n be the (k+n)th Mersenne number and let Fk+n be the (k+n)th Fermat number; then (1)
Mn+1-Mn=2n,Mn+1-2Mn=1,Mn=2n-1,Mn2-Mn+1Mn-1=2n-1.(2)
Fn+1-Fn=2n,Fn+1-2Fn=-1,Fn=2n+1,Fn2-Fn+1Fn-1=-2n-1.We define a Fermat circulant matrix which is ann×n matrix with the following form: (3)Circ(Fk+1,Fk+2,…,Fk+n)=Fk+1Fk+2⋯Fk+nFk+nFk+1⋯Fk+n-1⋮⋮⋮Fk+2Fk+3⋯Fk+1.A Mersenne circulant matrix which is ann×n matrix is defined with the following form: (4)Circ(Mk+1,Mk+2,…,Mk+n)=Mk+1Mk+2⋯Mk+nMk+nMk+1⋯Mk+n-1⋮⋮⋮Mk+2Mk+3⋯Mk+1.Besides, a Fermat left circulant matrix is given by(5)LCirc(Fk+1,Fk+2,…,Fk+n)=Fk+1Fk+2⋯Fk+nFk+2Fk+3⋯Fk+1⋮⋮⋮Fk+nFk+1⋯Fk+n-1.A Mersenne left circulant matrix is given by(6)LCirc(Mk+1,Mk+2,…,Mk+n)=Mk+1Mk+2⋯Mk+nMk+2Mk+3⋯Mk+1⋮⋮⋮Mk+nMk+1⋯Mk+n-1.The main content of this paper is to obtain the results for the exact determinants and inverses of Fermat and Mersenne circulant matrix. In this paper, letk be a nonnegative integer, Ak,n=Circ(Fk+1,Fk+2,…,Fk+n), and Bk,n=Circ(Mk+1,Mk+2,…,Mk+n).


## 2. Determinant and Inverse of Fermat Circulant Matrix

In this section, letAk,n=Circ(Fk+1,Fk+2,…,Fk+n) be a Fermat circulant matrix. Firstly, we obtain the exact form determinant of the matrix Ak,n. Afterwards, we find the exact form inverse of the matrix Ak,n.Theorem 2.
LetAk,n=Circ(Fk+1,Fk+2,…,Fk+n) be a Fermat circulant matrix. Then one has (7)det⁡Ak,n=Fk+1·∑j=1n-2‍Fj+k+2-τkFj+k+1·yn-j-1ddddddd+Fk+1-τkFk+n∑j=1n-2‍·-fn-2,where y=-e/f, e=2(Fk-Fk+n), f=Fk+n+1-Fk+1, τk=Fk+2/Fk+1, and Fk+n is the (k+n)th Fermat number. Moreover, Ak,n is singular if and only if (1-ακl)(1-βκl)≠0 and Fk+1-2κlFk-Fk+n+1+2κlFk+n=0, for k∈N, n∈N+, where κl=cos⁡(2lπ/n)+isin(2lπ/n), l=1,2,…,n.Proof.
It is clear that detA0,n=F1·[∑j=1n-2‍(Fj+2-τ0Fj+1)[2(Fn-F0)/(Fn+1-F1)]n-j-1 + F1-τ0Fn]·[F1-Fn+1]n-2 satisfies (7). In the following, let(8)Σ=1-τk121-3001-32⋮⋰⋰⋰01⋰⋰01-3⋰001-32,Ω1=100⋯000yn-20⋯000yn-30⋯0-1⋮⋮⋮⋱⋮⋮0y0⋯0001-1⋯00be two n×n matrices; we have (9)ΣAk,nΩ1=Fk+1hk,n′-Fk+n-Fk+n-1⋯-Fk+30hk,na3a4⋯an00ef0000e0⋮⋮⋮⋮⋱⋮0000⋱f0000e,where(10)τk=Fk+2Fk+1,y=-ef,a3=τkFk+n-Fk+1,aj=τkFk+n+3-j-Fk+n+4-jj=4,5,…,n,hk,n′=∑t=1n-1‍Ft+k+12Fk+n-FkFk+n+1-Fk+1n-t-1,hk,n=∑t=1n-2‍Ft+k+2-τkFt+k+1yn-t-1+Fk+1-τkFk+n.We obtain(11)det⁡Σdet⁡Ak,ndet⁡Ω1=Fk+1·∑t=1n-2‍Ft+k+2-τkFt+k+1yn-t-1ddddddddd+Fk+1-τkFk+n∑t=1n-2‍·en-2,while (12)det⁡Σ=-1(n-1)(n-2)/2,det⁡Ω1=-1(n-1)(n-2)/22Fk-Fk+nFk+1-Fk+n+1n-2.We have(13)det⁡Ak,n=Fk+1·∑t=1n-2‍Ft+k+2-τkFt+k+1·yn-t-1ddddi+Fk+1-τkFk+n∑t=1n-2‍·-fn-2.Next, we discuss the singularity of the matrix Ak,n.
The roots of polynomialg(x)=xn-1 are κl(l=1,2,…,n), where κl=cos⁡⁡2lπ/n)+isin(2lπ/n. We have (14)fκl=Fk+1+Fk+2κl+⋯+Fk+nκln-1=Fk+1-2κlFk-Fk+n+1+2κlFk+n(1-ακl)(1-βκl).By Lemma  1 in [14], the matrix Ak,n is nonsingular if and only if f(κl)≠0; that is, when (1-ακl)(1-βκl)≠0, Ak,n is nonsingular if and only if Fk+1-2κlFk-Fk+n+1+2κlFk+n≠0; when (1-ακl)(1-βκl)=0, we obtain κl=1/α or κl=1/β.
Letκl=1/α; then the eigenvalue of Ak,n is (15)f(κl)=nαk+n-βk+1Fnαn-1(α-β)≠0,for α=2, β=1, k∈N, n∈N+, l=1,2,…,n, so Ak,n is nonsingular. The arguments for κl=1/β are similar. Thus, the proof is completed.Lemma 3.
Let the matrixM=[mi,l′]i,l=1n-2 be of the form(16)mi,l=2(Fk-Fk+n)=e,i=l,Fk+n+1-Fk+1=f,l=i+1,0,otherwise.Then the inverse M-1=[mi,l′]i,l=1n-2 of the matrix M is equal to (17)mi,l′=Fk+1-Fk+n+1l-i2Fk-Fk+nl-i+1=-fl-iel-i+1,l≥i,0,l<i.Proof.
Letei,l=∑k=1n-2‍mi,kmk,l′. Distinctly, ci,l=0 for l<i. In the case i=l, we obtain (18)ei,i=mi,imi,i′=(Fk+1-Fk+n+1)·1(Fk+1-Fk+n+1)=1.For l≥i+1, we get (19)ei,l=∑k=1n-2‍mi,kmk,l′=mi,imi,l′+mi,i+1mi+1,l′=e·-fl-iel-i+1+f·-fl-i-1el-i=0.We check on MM-1=In-2, where In-2 is (n-2)×(n-2) identity matrix. Similarly, we can verify M-1M=In-2. Thus, the proof is completed.Theorem 4.
LetAk,n=Circ(Fk+1,Fk+2,…,Fk+n) be a Fermat circulant matrix. Then one acquires Ak,n-1=Circ(v1,v2,…,vn), where (20)v1=1hk,n+(Fk+1-τkFk+n-hk,n)·-Fk+n+1+3Fk+n+Fk+1-3Fk2hk,nFk-Fk+n2+(Fk+n-τkFk+n-1)hk,n(Fk-Fk+n),v2=-2k-hk,nFk+1Fk+1hk,n(Fk-Fk+n),v3=Fk+1-τkFk+n-hk,nhk,n·Fk+1-Fk+n+1n-32Fk-Fk+nn-2+1hk,n·∑i=4n‍Fk+n+4-i-τkFk+n+3-iddd·Fk+1-Fk+n+1n-i2Fk-Fk+nn-i+1,v4=Fk+n+2-Fk+2hk,n×(Fk+1-τkFk+n-hk,n)Fk+1-Fk+n+1n-i-12Fk-Fk+nn-i+1ddd×Fk+1-Mk+n+1n-42Mk-Mk+nn-2ddd+∑i=4n‍(Fk+n+4-i-τkFk+n+3-i)ddddddd·Fk+1-Fk+n+1n-i-12Fk-Fk+nn-i+1,vs=0(s=5,6,…,n).Proof.
Let(21)Ω2=1-hk,n′Fk+1x3′x4′⋯xn′01y3′y4′⋯yn′0010⋯00001⋯0⋮⋮⋮⋮⋱⋮0000⋯1,where(22)τk=Fk+2Fk+1,x3′=Fk+nFk+1+hk,n′hk,n·(τkFk+n-Fk+1)Fk+1,y3′=Fk+1-τkFk+nhk,n,xi′=Fk+n+3-iFk+1+hk,n′hk,n·τkFk+n+3-i-Fk+n+4-iFk+1(i=4,…,n),yi′=Fk+n+4-i-τkFk+n+3-ihk,n(i=4,…,n),hk,n′=∑i=1n-1‍Fi+k+12Fk+n-FkFk+n+1-Fk+1n-i-1,hk,n=∑i=1n-2‍(Fi+k+2-τkFi+k+1)yn-i-1ddd+Fk+1-τkFk+n.We have (23)ΣAk,nΩ1Ω2=D2⊕M,where D2=diag(Fk+1,hk,n) is a diagonal matrix, and D2⊕M is the direct sum of D2 and M. If we denote Ω=Ω1Ω2, then we obtain (24)Ak,n-1=Ω(D2-1⊕M-1)Σ.
LetAk,n-1=Circ(v1,v2,…,vn). Since the last row elements of the matrix Ω are 0,1,y3′-1,y4′,…,yn-1′,yn′, according to Lemma 3, then the last row elements of Ak,n-1 are given by the following equations:(25)v2=-τkhk,n+y3′-1Fk-Fk+n,v3=(y3′-1)-fn-3en-2+∑i=4n‍yi′·-fn-ien-i+1,v4=y3′-1-fn-4-fn-3-3-fn-3en-2+∑i=4n‍yi′·-fn-i-1en-i-3-fn-ien-i+1dddddddddddiit<0,-ft=0,vs=y3′-1-fn-sen-s+1-3-fn-s+1en-s+2+2-fn-s+2en-s+3+∑i=4n-s+5‍yi′·-fn-i-s+3en-i-s+4-3-fn-i-s+4en-i-s+5ddddddddddd+2-fn-i-s+5en-i-s+6dddddddddddd(s=5,6,…,n;t<0,-ft=0),v1=1hk,n+-2f-3ee2(y3-1)+2ey4,where f=Fk+n+1-Fk+1, e=2(Fk-Fk+n), according to Lemma 1; then we have (i)
e+f=0,(ii)
e+2f=2k+n+1-2k+1.Hence, we obtain(26)v1=1hk,n+Fk+1-τkFk+n-hk,n·-Fk+n+1+3Fk+n+Fk+1-3Fk2hk,nFk-Fk+n2+(Fk+n-τkFk+n-1)hk,n(Fk-Fk+n),v2=-2k-hk,nFk+1Fk+1hk,n(Fk-Fk+n),v3=Fk+1-τkFk+n-hk,nhk,n·Fk+1-Fk+n+1n-32Fk-Fk+nn-2+1hk,n·∑i=4n‍(Fk+n+4-i-τkFk+n+3-i)sssss·Fk+1-Fk+n+1n-i2Fk-Fk+nn-i+1,v4=Fk+n+2-Fk+2hk,n×(Fk+1-τkFk+n-hk,n)Fk+1-Fk+n+1n-i-12Fk-Fk+nn-i+1ddd·Fk+1-Fk+n+1n-42Fk-Fk+nn-2ddd+∑i=4n‍Fk+n+4-i-τkFk+n+3-iddddddd×Fk+1-Fk+n+1n-i-12Fk-Fk+nn-i+1,vs=0(s=5,6,…,n).Thus, the proof is completed.


## 3. Determinant and Inverse of Mersenne Circulant Matrix

In this section, letBk,n=Circ(Mk+1,Mk+2,…,Mk+n) be a Mersenne circulant matrix. Firstly, we obtain the determinant of the matrix Bk,n. Afterwards, we seek out the inverse of the matrix Bk,n.Theorem 5.
LetBk,n=Circ(Mk+1,Mk+2,…,Mk+n) be a Mersenne circulant matrix. Then one obtains(27)det⁡Bk,n=Mk+1·∑k=1n-2‍(Mk+k+2-μkMk+k+1)xn-k-1dddd+Mk+1-μkMk+n∑k=1n-2‍·-dn-2,where x=-c/d, c=2(Mk-Mk+n), d=Mk+n+1-Mk+1, μk=Mk+2/Mk+1, and Mk+n is the (k+n)th Mersenne number. Furthermore, Bk,n is singular if and only if (1-ακl)(1-βκl)≠0 and Mk+1-2κlMk-Mk+n+1+2κlMk+n=0, for k∈N, n∈N+, where κl=cos⁡(2lπ/n)+isin(2lπ/n), l=1,2,…,n.Proof.
Obviously,(28)det⁡A0,n=M1·∑i=1n-2‍(Mi+2-μ0Mi+1)·2Mn-M0Mn+1-M1n-k-1dd+M1-μ0Mn2Mn-M0Mn+1-M1n-k-1·2M0-Mnn-2satisfies (27). In the following, let(29)Γ=1-μk121-3001-32⋮⋰⋰⋰01⋰⋰01-3⋰001-32,Π1=100⋯000xn-20⋯000xn-30⋯0-1⋮⋮⋮⋱⋮⋮0x0⋯0001-1⋯00be two n×n matrices; then we have (30)ΓBk,nΠ1=Mk+1fk,n′-Mk+n⋯-Mk+30fk,nh3⋯hn00c⋯0000⋯0⋮⋮⋮⋱⋮000⋯d000⋯c,where (31)c=2(Mk-Mk+n),d=Mk+n+1-Mk+1,x=-cd,μk=Mk+2Mk+1,h3=μkMk+n-Mk+1,hj=μkMk+n+3-j-Mk+n+4-jj=4,5,…,n,fk,n′=∑i=1n-1‍Mi+k+12Mk+n-MkMk+n+1-Mk+1n-i-1,fk,n=∑i=1n-2‍Mi+k+2-μkMi+k+1,xn-i-1+Mk+1-μkMk+n.We get(32)det⁡Γdet⁡Bk,ndet⁡Π1=Mk+1·∑i=1n-2‍(Mi+k+2-μkMi+k+1)xn-i-1ddddddddd+Mk+1-μkMk+n∑i=1n-2‍·cn-2;besides (33)det⁡Γ=-1n-1n-2/2,det⁡Π1=-1n-1n-2/22Mk+n-MkMk+n+1-Mk+1n-2.We have (34)det⁡Bk,n=Mk+1·∑i=1n-2‍(Mi+k+2-μkMi+k+1)xn-i-1dddddddd+Mk+1-μkMk+n∑i=1n-2‍·-dn-2.Now, we discuss the singularity of the matrix Bk,n.
The roots of polynomialg(x)=xn-1 are κl(l=1,2,…,n), where κl=cos⁡(2lπ/n)+isin(2lπ/n). So we have (35)fκl=Mk+1+Mk+2κl+⋯+Mk+nκln-1=Mk+1-2κlMk-Mk+n+1+2κlMk+n1-ακl1-βκl.By Lemma  1 in [14], the matrix Bk,n is nonsingular if and only if f(κl)≠0. That is when (1-ακl)(1-βκl)≠0, Bk,n is nonsingular if and only if Mk+1-2κlMk-Mk+n+1+2κlMk+n≠0, for k∈N, n∈N+, l=1,2,…,n. When (1-ακl)(1-βκl)=0, we obtain κl=1/α or κl=1/β. Let κl=1/α; then the eigenvalue of Bk,n is (36)f(κl)=nαk+n-βk+1Mnαn-1(α-β)≠0,for α=2, β=1, k∈N, n∈N+, l=1,2,…,n, so Bk,n is nonsingular. The arguments for κl=1/β are similar. Thus, the proof is completed.Lemma 6.
Let the matrixG=[gi,j]i,j=1n-2 be of the form (37)gi,j=2(Mk-Mk+n)=c,i=j,Mk+n+1-Mk+1=d,j=i+1,0,otherwise.Then the inverse G-1=[gi,j′]i,j=1n-2 of the matrix G is equal to (38)gi,j′=Mk+1-Mk+n+1j-i2Mk-Mk+nj-i+1=-dj-icj-i+1,j≥i,0,j<i.Proof.
Letci,j=∑k=1n-2‍gi,kgk,j′. Distinctly, ci,j=0 for j<i. When i=j, we obtain (39)ci,i=gi,igi,i′=-d·1-d=1.For j≥i+1, we obtain (40)ci,j=∑k=1n-2‍gi,kgk,j′=gi,igi,j′+gi,i+1gi+1,j′=c·-dj-icj-i+1+d·-dj-i-1cj-i=0.We verify GG-1=In-2, where In-2 is (n-2)×(n-2) identity matrix. Similarly, we check on G-1G=In-2. Thus, the proof is completed.Theorem 7.
LetBk,n=Circ(Mk+1,Mk+2,…,Mk+n) be a Mersenne circulant matrix. Then one acquires (41)Bk,n-1=Circ(u1,u2,…,un),where (42)u1=1fk,n+(Mk+1-μkMk+n-fk,n)·-Mk+n+1+3Mk+n+Mk+1-3Mk2fk,nMk-Mk+n2+(Mk+n-μkMk+n-1)fk,n(Mk-Mk+n),u2=2k-fk,nMk+1Mk+1fk,n(Mk-Mk+n),u3=Mk+1-μkMk+n-fk,nfk,n·Mk+1-Mk+n+1n-32Mk-Mk+nn-2+1fk,n·∑i=4n‍(Mk+n+4-i-μkMk+n+3-i)ddd·Mk+1-Mk+n+1n-i2Mk-Mk+nn-i+1,u4=Mk+n+2-Mk+2fk,n×(Mk+1-μkMk+n-fk,n)·Mk+1-Mk+n+1n-42Mk-Mk+nn-2ddd+∑i=4n‍(Mk+n+4-i-μkMk+n+3-i)dddddid·Mk+1-Mk+n+1n-i-12Mk-Mk+nn-i+1,us=0s=5,6,…,n,where (43)μk=Mk+2Mk+1,fk,n=∑i=1n-2‍(Mi+k+2-μkMi+k+1)xn-i-1+Mk+1-μkMk+n.Proof.
Let(44)Π2=1-fk,n′Mk+1x3x4⋯xn01y3y4⋯yn0010⋯00001⋯0⋮⋮⋮⋮⋱⋮0000⋯1,where (45)μk=Mk+2Mk+1,x3=Mk+nMk+1+fk,n′fk,n·(μkMk+n-Mk+1)Mk+1,y3=Mk+1-μkMk+nfk,n,xi=Mk+n+3-iMk+1+fk,n′fk,n·μkMk+n+3-i-Mk+n+4-iMk+1ddddddddd(i=4,…,n),yi=Mk+n+4-i-μkMk+n+3-ifk,ni=4,…,n,fk,n′=∑i=1n-1‍Mi+k+12Mk+n-MkMk+n+1-Mk+1n-i-1,fk,n=∑i=1n-2‍(Mi+k+2-μkMi+k+1)xn-i-1+Mk+1-μkMk+n.We have (46)ΓBk,nΠ1Π2=D1⊕G,where D1=diag(Mk+1,fk,n) is a diagonal matrix, and D1⊕G is the direct sum of D1 and G. If we denote Π=Π1Π2, then we obtain (47)Bk,n-1=Π(D1-1⊕G-1)Γ.
LetBk,n-1=Circ(u1,u2,…,un). Since the last row elements of the matrix Π are 0,1,y3-1,y4,…,yn-1,yn, according to Lemma 6, then the last row elements of Bk,n-1 are given by the following equations:(48)u2=-μkfk,n+2(y3-1)c,u3=(y3-1)·-dn-3cn-2+∑i=4n‍yi·-dn-icn-i+1,u4=(y3-1)·-dn-4cn-3-3-dn-3cn-2+∑i=4n‍yi·-dn-i-1cn-i-3-dn-icn-i+1=(y3-1)·-dn-4cn-2(c+3d)+∑i=4n‍yi·-dn-i-1cn-i+1(c+3d)(t<0,-dt=0),us=y3-1·-dn-scn-s+1-3-dn-s+1cn-s+2+2-dn-s+2cn-s+3+∑i=4n-s+5‍yi·-dn-i-s+3cn-i-s+4ddddddddd-3-dn-i-s+4cn-i-s+5+2-dn-i-s+5cn-i-s+6=y3-1·-dn-scn-s+3+∑i=4n-s+5‍yi·-dn-i-s+3cn-i-s+6×(c+2d)(c+d)(s=5,6,…,n;t<0,-dt=0),u1=1fk,n+-2d-3cc2(y3-1)+2cy4,where d=Mk+n+1-Mk+1,c=2(Mk-Mk+n), according to Lemma 1; then we have (i)
c+d=0,(ii)
c+2d=2k+n+1-2k+1.We get (49)u1=1fk,n+Mk+1-μkMk+n-fk,n·-Mk+n+1+3Mk+n+Mk+1-3Mk2fk,nMk-Mk+n2+(Mk+n-μkMk+n-1)fk,n(Mk-Mk+n),u2=2k-fk,nMk+1Mk+1fk,n(Mk-Mk+n),u3=Mk+1-μkMk+n-fk,nfk,n·Mk+1-Mk+n+1n-32Mk-Mk+nn-2+1fk,n·∑i=4n‍(Mk+n+4-i-μkMk+n+3-i)ddd·Mk+1-Mk+n+1n-i2Mk-Mk+nn-i+1,u4=Mk+n+2-Mk+2fk,n×(Mk+1-μkMk+n-fk,n)Mk+1-Mk+n+1n-42Mk-Mk+nn-2ddd+∑i=4n‍(Mk+n+4-i-μkMk+n+3-i)ddd·Mk+1-Mk+n+1n-i-12Mk-Mk+nn-i+1,us=0(s=5,6,…,n).Thus, the proof is completed.


## 4. Determinants and Inverses of Fermat and Mersenne Left Circulant Matrix

In this section, letAk,n′=LCirc(Fk+1,Fk+2,…,Fk+n) and Bk,n′=LCirc(Mk+1,Mk+2,…,Mk+n) be Mersenne and Fermat left circulant matrices, respectively. By using the obtained conclusions, we give a determinant formula for the matrix Ak,n′ and Bk,n′. In addition, the inverse matrices of Ak,n′ and Bk,n′ are derived.According to Lemma  2 in [14] and Theorems 2, 4, 5, and 7, we can obtain the following theorems.Theorem 8.
LetAk,n′=LCirc(Fk+1,Fk+2,…,Fk+n) be a Fermat left circulant matrix; then one has (50)det⁡Ak,n′=-1n-1n-2/2·Fk+1·∑j=1n-2‍(Fj+k+2-τkFj+k+1)pn-j-1+Fk+1-τkFk+n·Fk+1-Fk+n+1n-2,where τk=Fk+2/Fk+1, p=2(Fk+n-Fk)/(Fk+n+1-Fk+1), and Fk+n is the (k+n)th Fermat number. Moreover, Ak,n′ is singular if and only if (1-ακl)(1-βκl)≠0 and Fk+1-2κlFk-Fk+n+1+2κlFk+n=0, for k∈N, n∈N+, where κl=cos⁡(2lπ/n)+isin(2lπ/n), l=1,2,…,n.Theorem 9.
LetAk,n′=LCirc(Fk+1,Fk+2,…,Fk+n) be a Fermat left circulant matrix; then(51)Ak,n′-1=Circ-1⁡(Fk+1,Fk+2,…,Fk+n)·Δ=Circ⁡(v1,v2,…,vn)·Δ=LCircv1,vn,…,v2,where v1,v2,…,vn were given by Theorem 4 and Δ=LCirc(1,0,…,0) was given by Lemma  2 in [14].Theorem 10.
LetBk,n′=LCirc(Mk+1,Mk+2,…,Mk+n) be a Mersenne left circulant matrix; then one has (52)det⁡Bk,n′=-1n-1n-2/2·Mk+1·Mk+1-μkMk+n∑j=1n-2‍di+∑j=1n-2‍Mj+k+2-μkMj+k+1zn-j-1·Mk+1-Mk+n+1n-2,where μk=Mk+2/Mk+1, z=2(Mk+n-Mk)/(Mk+n+1-Mk+1), and Mk+n is the (k+n)th Mersenne number. Furthermore, Bk,n′ is singular if and only if (1-ακl)(1-βκl)≠0 and Mk+1-2κlMk-Mk+n+1+2κlMk+n=0, for k∈N, n∈N+, where κl=cos⁡(2lπ/n)+isin(2lπ/n), l=1,2,…,n.Theorem 11.
LetBk,n′=LCirc(Mk+1,Mk+2,…,Mk+n) be a Mersenne left circulant matrix; then one has (53)Bk,n′-1=Circ-1Mk+1,Mk+2,…,Mk+n·Δ=Circu1,u2,…,un·Δ=LCircu1,un,…,u2,where u1,u2,…,un were given by Theorem 7 and Δ=LCirc(1,0,…,0) was given by Lemma  2 in [14].


## 5. Conclusion

In this paper, we present the exact determinants and the inverse matrices of Fermat and Mersenne circulant matrix, respectively. Furthermore, we give the exact determinants and the inverse matrices of Fermat and Mersenne left circulant matrix. Meanwhile, the nonsingularity of these special matrices is discussed. On the basis of circulant matrices technology, we will develop solving the problems in [19–22].



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