Hindawi Publishing Corporation Journal of Applied Mathematics Volume 2014, Article ID 239693, 10 pages http://dx.doi.org/10.1155/2014/239693
Research Article The Determinants, Inverses, Norm, and Spread of Skew Circulant Type Matrices Involving Any Continuous Lucas Numbers
Jin-jiang Yao and Zhao-lin Jiang
Department of Mathematics, Linyi University, Linyi, Shandong 276000, China
Correspondence should be addressed to Zhao-lin Jiang; [email protected]
Received 6 January 2014; Accepted 17 February 2014; Published 17 April 2014
Academic Editor: Feng Gao
Copyright © 2014 J.-j. Yao and Z.-l. Jiang. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
We consider the skew circulant and skew left circulant matrices with any continuous Lucas numbers. Firstly, we discuss the invertibility of the skew circulant matrices and present the determinant and the inverse matrices by constructing the transformation matrices. Furthermore, the invertibility of the skew left circulant matrices is also discussed. We obtain the determinants and the inverse matrices of the skew left circulant matrices by utilizing the relationship between skew left circulant matrices andskew circulant matrix, respectively. Finally, the four kinds of norms and bounds for the spread of these matrices are given, respectively.
1. Introduction of skew circulant matrix firstly and then dealt with the optimal backward perturbation analysis for the linear system Circulant and skew-circulant matrices are appearing increas- with skew circulant coefficient matrix. In [3], a new fast ingly often in scientific and engineering applications. Briefly, algorithm for optimal design of block digital filters (BDFs) scanning the recent literature, one can see their utility was proposed based on skew circulant matrix. is appreciated in the design of digital filters [1–3], image Besides, some scholars have given various algorithms for processing [4–6], communications [7], signal processing [8], the determinants and inverses of nonsingular circulant matri- and encoding [9]. They have been put on firm basis with the ces [10, 11]. Unfortunately, the computational complexity of work of Davis [10] and Jiang and Zhou [11]. these algorithms is very amazing with the order of matrix The skew circulant matrices as preconditioners for lin- ear multistep formulae- (LMF-) based ordinary differential increasing. However, some authors gave the explicit determi- equations (ODEs) codes. Hermitian and skew-Hermitian nants and inverse of circulant and skew circulant involving Toeplitz systems are considered in [12–15]. Lyness and some famous numbers. For example, Jaiswal evaluated some Sørevik employed a skew circulant matrix to construct 𝑠- determinants of circulant whose elements are the generalized dimensional lattice rules in [16]. Spectral decompositions of Fibonacci numbers [19]. Lind presented the determinants skew circulant and skew left circulant matrices were discussed of circulant and skew circulant involving the Fibonacci in [17]. Compared with cyclic convolution algorithm, the numbers [20]. Dazheng [21] gave the determinant of the skew cyclic convolution algorithm [8]isabletoperform Fibonacci-Lucas quasicyclic matrices. Shen et al. considered filtering procedure in approximate half of computational cost circulant matrices with the Fibonacci and Lucas numbers for real signals. In [2] two new normal-form realizations are and presented their explicit determinants and inverses by presented which utilize circulant and skew circulant matrices constructing the transformation matrices [22]. Gao et al. [23] as their state transition matrices. The well-known second- gave explicit determinants and inverses of skew circulant and order coupled form is a special case of the skew circulant skew left circulant matrices with the Fibonacci and Lucas form. Li et al. [18] gave the style spectral decomposition numbers. Jiang et al. [24, 25] considered the skew circulant 2 Journal of Applied Mathematics and skew left circulant matrices with the 𝑘-Fibonacci num- Definition 2 (see [17]). A skew left circulant matrix over 𝐶 bers and the 𝑘-Lucas numbers and discussed the invertibility with the first row (𝑎1,𝑎2,...,𝑎𝑛) is meant a square matrix of of the these matrices and presented their determinant and the the form inverse matrix by constructing the transformation matrices, 𝑎 𝑎 𝑎 ⋅⋅⋅ 𝑎 respectively. 1 2 3 𝑛 𝑎 𝑎 ⋅⋅⋅ 𝑎 −𝑎 Recently, there are several papers on the norms of some 2 3 𝑛 1 special matrices. Solak [26]establishedthelowerandupper ( . ) 𝑎3 cc c . , (5) boundsforthespectralnormsofcirculantmatriceswith . theclassicalFibonacciandLucasnumbersentries.˙Ipek [27] . 𝑎𝑛 −𝑎1 ⋅⋅⋅ −𝑎𝑛−2 𝑎 −𝑎 ⋅⋅⋅ −𝑎 −𝑎 investigated an improved estimation for spectral norms of ( 𝑛 1 𝑛−2 𝑛−1)𝑛×𝑛 these matrices. Shen and Cen [28] gave upper and lower bounds for the spectral norms of 𝑟-circulant matrices in the denoted by SLCirc(𝑎1,𝑎2,...,𝑎𝑛). forms of 𝐴=𝐶𝑟(𝐹0,𝐹1,...,𝐹𝑛−1), 𝐵=𝐶𝑟(𝐿0,𝐿1,...,𝐿𝑛−1), and they also obtained some bounds for the spectral norms of Lemma 3 (see [10, 17]). Let 𝐴=SCirc(𝑎1,𝑎2,...,𝑎𝑛) be skew Kronecker and Hadamard products of matrix 𝐴 and matrix circulant matrix; then 𝐵.AkbulakandBozkurt[29] found upper and lower bounds 𝐴 𝐴 for the spectral norms of Toeplitz matrices such that 𝑎𝑖𝑗 ≡ (i) isinvertibleifandonlyiftheeigenvaluesof 𝐹 𝑏 ≡𝐿 𝑖−𝑗 and 𝑖𝑗 𝑖−𝑗. The convergence in probability and in 𝑘 distribution of the spectral norm of scaled Toeplitz, circulant, 𝜆𝑘 =𝑓(𝜔 𝜂) =0,̸ (𝑘=0,1,2,...,𝑛−1) , (6) reverse circulant, symmetric circulant, and a class of 𝑘- circulant matrices is discussed in [30]. 𝑓(𝑥) = ∑𝑛 𝑎 𝑥𝑗−1 𝜔= (2𝜋𝑖/𝑛) 𝜂= Beginning with Mirsky [31], several authors [32–38]have where 𝑗=1 𝑗 , exp ,and (𝜋𝑖/𝑛) obtained bounds for the spread of a matrix. exp ; The purpose of this paper is to obtain the explicit deter- (ii) if 𝐴 is invertible, then the inverse of 𝐴 is a skew circulant minants, explicit inverses, norm, and spread of skew circulant matrix. type matrices involving any continuous Lucas numbers. And we generalize the result [23]. In passing, the norm and spread Lemma 4 (see [17]). Let 𝐴=SLCirc(𝑎1,𝑎2,...,𝑎𝑛) be skew of skew circulant type matrices have not been researched. It left circulant matrix and let n be odd; then is hoped that this paper will help in changing this. More work 𝑛 continuing the present paper is forthcoming. (𝑗−(1/2))(𝑘−1) 𝑛−1 𝑟 𝜆𝑗 =±∑𝑎𝑘𝜔 , (𝑗 = 1, 2, . . ., ), In the following, let be a nonnegative integer. We adopt 2 00 =1 𝑘=1 the following two conventions , and, for any sequence (7) 𝑛 𝑛 {𝑎𝑛}, ∑𝑘=𝑖 𝑎𝑘 =0in the case 𝑖>𝑛. 𝑘−1 The Lucas sequences are defined by the following recur- 𝜆(𝑛+1)/2 = ∑ 𝑎𝑘(−1) , rence relations [21–23, 27–29]: 𝑘=1 𝐿 =𝐿 +𝐿 , 𝐿 =2,𝐿 =1, 𝑛+1 𝑛 𝑛−1 where 0 1 (1) where 𝜆𝑗,𝑗 = 1,2,...,(𝑛−1)/2, (𝑛 + 1)/2 are the eigenvalues 𝐴 for 𝑛≥0.Thefirstfewvaluesofthesequencesaregivenby of . the following table: Lemma 5 (see [23]). With the orthogonal skew left circulant 𝑛 0123456789 matrix . (2) 𝐿𝑛 213471118294776 10⋅⋅⋅00 The {𝐿𝑛} is given by the formula 00⋅⋅⋅0−1 𝐿 =𝛼𝑛 +𝛽𝑛, 00⋅⋅⋅−10 𝑛 (3) Θ:=( ) , (8) 2 . . . . where 𝛼 and 𝛽 are the roots of the characteristic equation 𝑥 − . . d . . 0−1⋅⋅⋅0 0 𝑥−1=0. 𝑛×𝑛
Definition 1 (see [17]). A skew circulant matrix over 𝐶 with it holds that the first row (𝑎1,𝑎2,...,𝑎𝑛) is meant a square matrix of the form SCirc (𝑎1,𝑎2,...,𝑎𝑛) =ΘSLCirc (𝑎1,𝑎2,...,𝑎𝑛) . (9) 𝑎 𝑎 ... 𝑎 𝑎 1 2 𝑛−1 𝑛 Lemma 6 −𝑎𝑛 𝑎1 𝑎2 ... 𝑎𝑛−1 (see [23]). If . . ( . −𝑎 𝑎 d . ) , −1 . 𝑛 1 . (4) [SCirc (𝑎1,𝑎2,...,𝑎𝑛)] = SCirc (𝑏1,𝑏2,...,𝑏𝑛), (10) . −𝑎3 . dd 𝑎2 −𝑎 −𝑎 ... −𝑎 𝑎 then ( 2 3 𝑛 1 )𝑛×𝑛 −1 denoted by SCirc(𝑎1,𝑎2,...,𝑎𝑛). [SLCirc (𝑎1,𝑎2,...,𝑎𝑛)] = SLCirc (𝑏1,−𝑏𝑛,...,−𝑏2). (11) Journal of Applied Mathematics 3
Lemma 7 (see [27, 28]). Let {𝐿𝑛} be the Lucas numbers; then (ii) and if 𝐴 is Hermitian, then 𝑛−1 (𝑖) ∑𝐿 =𝐿 −1, 𝑠 (𝐴) ≥2max 𝑎𝑖𝑗 . 𝑖 𝑛+1 (12) 𝑖 =𝑗̸ (20) 𝑖=0
𝑛−1 2 2. Determinant and Inverse of Skew Circulant (𝑖𝑖) ∑𝐿𝑖 =𝐿𝑛𝐿𝑛−1 +2, (13) 𝑖=0 Matrix with the Lucas Numbers
𝑛−1 In this section, let 𝐴𝑟,𝑛 = SCirc(𝐿𝑟+1,...,𝐿𝑟+𝑛) be skew (𝑖𝑖𝑖) ∑𝑖𝐿𝑖 = (𝑛−1) 𝐿𝑛+1 −𝐿𝑛+2 +4. (14) circulant matrix. Firstly, we give a determinant explicit 𝑖=0 formula for the matrix 𝐴𝑟,𝑛.Afterwards,weprovethat𝐴𝑟,𝑛 𝑛≥2 Definition 8 (see [29]). Let 𝐴=(𝑎𝑖𝑗 ) be an 𝑛×𝑛matrix. The is an invertible matrix for ,andthenwefindtheinverse 𝐴 maximum column sum matrix norm, the spectral norm, the of the matrix 𝑟,𝑛. 𝑥=−((𝐿 +𝐿 )/(𝐿 +𝐿 )) 𝑡= Euclidean (or Frobenius) norm, and the maximum row sum In the following, let 𝑟 𝑟+𝑛 𝑟+1 𝑟+𝑛+1 , 𝐿 /𝐿 𝑐=𝐿 +𝐿 𝑑=𝐿 +𝐿 𝑙 =𝐿 +𝑡𝐿 + matrix norm of matrix 𝐴 are, respectively, 𝑟+2 𝑟+1, 𝑟+1 𝑟+𝑛+1, 𝑟 𝑟+𝑛, 𝑛 𝑟+1 𝑟+𝑛 ∑𝑛−2(𝑡𝐿 −𝐿 )⋅𝑥𝑛−(𝑘+1) 𝑙 =∑𝑛−1 𝐿 ⋅𝑥𝑛−(𝑘+1) 𝑛 𝑘=1 𝑟+𝑘+1 𝑟+𝑘+2 and 𝑛 𝑘=1 𝑟+𝑘+1 . ‖𝐴‖1 = max ∑ 𝑎𝑖𝑗 , 1≤𝑗≤𝑛 Theorem 13. 𝐴 = (𝐿 ,...,𝐿 ) 𝑖=1 Let 𝑟,𝑛 SCirc 𝑟+1 𝑟+𝑛 be skew circu- lant matrix; then 1/2 ∗ ‖𝐴‖2 =(max 𝜆𝑖 (𝐴 𝐴)) , 1≤𝑖≤𝑛 det 𝐴𝑟,𝑛 =𝐿𝑟+1 [𝐿𝑟+1 +𝑡𝐿𝑟+𝑛 1/2 𝑛 (15) 2 ‖𝐴‖𝐹 =(∑ 𝑎𝑖𝑗 ) , 𝑛−2 𝑖,𝑗=1 𝑛−(𝑖+1) 𝑛−2 +∑ (𝑡𝐿𝑟+𝑖+1 −𝐿𝑟+𝑖+2)𝑥 ]⋅𝑐 , 𝑖=1 𝑛 (21) ‖𝐴‖∞ = max ∑ 𝑎𝑖𝑗 , 1≤𝑖≤𝑛 𝑗=1 where 𝐿𝑟+𝑛 is the (𝑟+𝑛)th Lucas number. Specially, when 𝑟=0, ∗ where 𝐴 denotes the conjugate transpose of 𝐴. one gets the result of [23].
Lemma 9 (see [30]). If 𝐴 is an 𝑛×𝑛real symmetric or normal Proof. Obviously, det 𝐴𝑟,1 =𝐿𝑟+1 satisfies the equation. In the matrix, then one has case 𝑛>1,let ‖𝐴‖2 = max 𝜆𝑖 , 1≤𝑖≤𝑛 (16) 1 𝑡1 where 𝜆𝑖 (𝑖=1,2,...,𝑛)aretheeigenvaluesof𝐴. (11−1) (001−1−1) 𝐴=(𝑎) 𝑛×𝑛 ( ) Definition 10 (see [31, 32]). Let 𝑖𝑗 be an matrix Σ=( ) , 𝜆 𝑖 = 1, 2, . . .𝑛 𝐴 (. ) with eigenvalues 𝑖, .Thespreadof is defined (. ccc ) as 01cc 𝑠 (𝐴) = 𝜆 −𝜆 . 01−1c 0 max 𝑖 𝑗 (17) 𝑖,𝑗 (0 1 −1 −1 ) (22) Beginning with Mirsky [31], several authors [32–38]have 100⋅⋅⋅00 obtained bounds for the spread of a matrix. 0𝑥𝑛−2 0⋅⋅⋅01 0𝑥𝑛−3 0⋅⋅⋅10 Lemma 11. Let 𝐴=(𝑎𝑖𝑗 ) be an 𝑛×𝑛matrix. An upper bound ( ) Ω1 = (. . . . .) for the spread due to Mirsky [31] states that . . . d . . 2 0𝑥1⋅⋅⋅00 𝑠 (𝐴) ⩽ √2‖𝐴‖2 − | 𝐴|2, (18) 010⋅⋅⋅00 𝐹 𝑛 tr ( ) where ‖𝐴‖𝐹 denotes the Frobenius norm of 𝐴 and tr 𝐴 is trace be two 𝑛×𝑛matrices; then we have of 𝐴. 𝐿𝑟+1 𝑙𝑛 𝑐13 ⋅⋅⋅ 𝑐1,𝑛−1 𝑐1𝑛 Lemma 12 𝐴=(𝑎 ) 𝑛×𝑛 (see [38]). Let 𝑖𝑗 be an matrix; then 0𝑙𝑛 𝑐23 ⋅⋅⋅ 𝑐2,𝑛−1 𝑐2𝑛 00𝑐 (i) if 𝐴 is real and normal, then ( ) Σ𝐴 Ω = ( 00𝑑d ) , 𝑟,𝑛 1 (23) 1 . . 𝑠 (𝐴) ≥ ∑𝑎 , . . d 𝑐 𝑛−1 𝑖𝑗 (19) 𝑖 =𝑗̸ ( 00 𝑑) 𝑐 4 Journal of Applied Mathematics
𝑙 where that 𝜔 𝜂 = −((𝐿𝑟+1 +𝐿𝑟+𝑛+1)/(𝐿𝑟 +𝐿𝑟+𝑛)) is a real number. Since 𝑐1𝑗 =𝐿𝑟+𝑛+2−𝑗,𝑐2𝑗 =𝑡𝐿𝑟+𝑛+2−𝑗 −𝐿𝑟+𝑛+3−𝑗, (24) (2𝑙+1) 𝜋𝑖 (𝑗=3,4,...,𝑛). 𝜔𝑙𝜂= exp 𝑛 (28) (2𝑙 + 1) 𝜋 (2𝑙+1) 𝜋 So it holds that = +𝑖 , cos 𝑛 sin 𝑛 det Σ det 𝐴𝑟,𝑛 det Ω1 𝑙 it yields that sin((2𝑙 + 1)𝜋/𝑛),sowehave =0 𝜔 𝜂=−1for =𝐿𝑟+1 [𝐿𝑟+1 +𝑡𝐿𝑟+𝑛 0 < (2𝑙 + 1)𝜋/𝑛.Since <2𝜋 𝑥=−1is not the root of the equation 𝐿𝑟+1 +𝐿𝑟+𝑛+1 +(𝐿𝑟 +𝐿𝑟+𝑛)𝑥 = 0 (𝑛 >2).Weobtain (25) 𝑘 𝑘 𝑛−2 𝑓(𝜔 𝜂) =0̸ ,forany𝜔 𝜂 (𝑘=1,2,...,𝑛−1), while 𝑛−(𝑖+1) + ∑ (𝑡𝐿𝑟+𝑘+1 −𝐿𝑟+𝑘+2)𝑥 ] 𝑛 𝑘=1 𝑗−1 𝑓(𝜂)= ∑𝐿𝑗𝜂 ⋅(𝐿 +𝐿 )𝑛−2. 𝑗=1 𝑟+1 𝑟+𝑛+1 (29) 𝐿 +𝐿 +(𝐿 +𝐿 )𝜂 Σ= Ω =(−1)(𝑛−1)(𝑛−2)/2 = 𝑟+1 𝑟+𝑛+1 𝑟 𝑟+𝑛 =0.̸ While taking det det 1 ,wehave 1−𝜂−𝜂2
det 𝐴𝑟,𝑛 It follows from Lemma 3 that the conclusion holds. =𝐿 [𝐿 +𝑡𝐿 𝑛−2 𝑟+1 𝑟+1 𝑟+𝑛 Lemma 15. Let the matrix H =[ℎ𝑖𝑗 ]𝑖,𝑗=1 be of the form (26) 𝑛−2 {𝐿𝑟+1 +𝐿𝑟+𝑛+1 =𝑐, 𝑖=𝑗, +∑ (𝑡𝐿 −𝐿 )𝑥𝑛−(𝑖+1)] { 𝑟+𝑘+1 𝑟+𝑘+2 ℎ = 𝐿 +𝐿 =𝑑, 𝑖=𝑗+1, 𝑘=1 𝑖𝑗 { 𝑟 𝑟+𝑛 (30) 0, 𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒. 𝑛−2 { ⋅(𝐿𝑟+1 +𝐿𝑟+𝑛+1) . −1 𝑛−2 Then the inverse H =[ℎ ] of the matrix H is equal to This completes the proof. 𝑖,𝑗 𝑖,𝑗=1
𝑖−𝑗 Theorem 14. Let 𝐴𝑟,𝑛 = SCirc(𝐿𝑟+1,...,𝐿𝑟+𝑛) be skew circu- (−𝑑) { ,𝑖≥𝑗, 𝐴 𝑖−𝑗+1 lant matrix; then 𝑟,𝑛 is an invertible matrix. Specially, when ℎ𝑖𝑗 = { 𝑐 (31) 𝑟=0,onegetstheresultof[23]. {0, 𝑖 < 𝑗. 𝑛=2 𝐴 =𝐿2 + Proof. Taking in, Theorem 13, we have det 𝑟,2 𝑟+1 𝑟=0 2 Specially, when , one gets the result of [23]. 𝐿𝑟+2 =0̸ .Hence𝐴𝑟,2 is invertible. In the case 𝑛>2,since 𝐿 =𝛼𝑟+𝑛 +𝛽𝑟+𝑛 𝛼+𝛽=1 𝛼𝛽 = −1 𝑛−2 𝑟+𝑛 ,where , ,wehave Proof. Let 𝑒𝑖𝑗 = ∑𝑘=1 ℎ𝑖𝑘ℎ𝑘𝑗 .Obviously,𝑒𝑖𝑗 =0for 𝑖<𝑗.Inthe 𝑛 case 𝑖=𝑗,weobtain𝑒𝑖𝑖 =ℎ𝑖𝑖ℎ𝑖𝑖 =(𝐿𝑟+1 +𝐿𝑟+𝑛+1) ⋅ (1/(𝐿𝑟+1 + 𝑘 𝑘 𝑗−1 𝐿 )) = 1 𝑖≥𝑗+1 𝑓(𝜔 𝜂) = ∑𝐿𝑟+𝑗(𝜔 𝜂) 𝑟+𝑛+1 .For ,weobtain 𝑗=1 𝑛−2 𝑛 𝑗−1 𝑒 = ∑ℎ ℎ =ℎ ℎ +ℎ ℎ = ∑ (𝛼𝑟+𝑗 +𝛽𝑟+𝑗)(𝜔𝑘𝜂) 𝑖𝑗 𝑖𝑘 𝑘𝑗 𝑖,𝑖−1 𝑖−1,𝑗 𝑖𝑖 𝑖𝑗 𝑘=1 𝑗=1 (32) (−𝑑)𝑖−𝑗−1 (−𝑑)𝑖−𝑗 𝛼𝑟+1 (1 + 𝛼𝑛) 𝛽𝑟+1 (1 + 𝛽𝑛) (27) =𝑑⋅ +𝑐⋅ =0. = + 𝑐𝑖−𝑗 𝑐𝑖−𝑗+1 1−𝛼𝜔𝑘𝜂 1−𝛽𝜔𝑘𝜂 −1 𝑘 Hence, we get HH =𝐼𝑛−2,where𝐼𝑛−2 is (𝑛 − 2) × (𝑛 −2) 𝐿𝑟+1 +𝐿𝑟+𝑛+1 +(𝐿𝑟 +𝐿𝑟+𝑛)𝜔 𝜂 −1 = identity matrix. Similarly, we can verify that H H =𝐼𝑛−2. 1−𝜔𝑘𝜂−𝜔2𝑘𝜂2 Thus, the proof is completed. (𝑘=1,2,...,𝑛−1) , Theorem 16. Let 𝐴𝑟,𝑛 = SCirc(𝐿𝑟+1,...,𝐿𝑟+𝑛) be skew circu- 𝑙 lant matrix; then where 𝜔=exp(2𝜋𝑖/𝑛), 𝜂=exp(𝜋𝑖/𝑛). If there exists 𝜔 𝜂(𝑙= 1,2,...,𝑛−1) 𝑓(𝜔𝑙𝜂) = 0 𝐿 +𝐿 + 1 such that ,weobtain 𝑟+1 𝑟+𝑛+1 (𝐴 )−1 = ⋅ (𝑦 ,𝑦 ,...,𝑦 ), (𝐿 +𝐿 )𝜔𝑙𝜂=0 1−𝜔𝑙𝜂−𝜔2𝑙𝜂2 =0̸ 𝑟,𝑛 SCirc 1 2 𝑛 (33) 𝑟 𝑟+𝑛 ,for , and hence it follows 𝑙𝑛 Journal of Applied Mathematics 5 where Since the last row elements of the matrix Ω are (0, 1, 𝜔23,𝜔24,...,𝜔2,𝑛−1,𝜔2𝑛), then the last row elements of 𝑛−3 −1 −1 (−𝑑) the matrix Ω(D ⊕H ) are (0, 1/𝑙𝑛,𝑇23,𝑇24,...,𝑇2𝑛 ), where 𝑦 =1−[(𝐿 −𝑡𝐿 )⋅ 1 𝑟+3 𝑟+2 𝑐𝑛−2 𝑛−2 (−𝑑)𝑖−1 𝑛−3 (−𝑑)𝑖−1 𝑇 = ∑𝜔 ⋅ , +∑ (𝐿 −𝑡𝐿 )⋅ ], 23 2,2+𝑖 𝑖 𝑟+𝑛+2−𝑖 𝑟+𝑛+1−𝑖 𝑖 𝑖=1 𝑐 𝑖=1 𝑐 (38) 𝑛−2 (−𝑑)𝑖−1 𝑛+1−𝑘 (−𝑑)𝑖−1 𝑦 =−𝑡−∑ (𝐿 −𝑡𝐿 )⋅ , 𝑇 = ∑ 𝜔 ⋅ (𝑘=3,4,...,𝑛) . 2 𝑟+𝑛+1−𝑖 𝑟+𝑛−𝑖 𝑖 2𝑘 2,𝑘−1+𝑖 𝑐𝑖 𝑖=1 𝑐 𝑖=1
1 (34) 𝑦3 =−(𝐿𝑟+3 −𝑡𝐿𝑟+2)⋅ , −1 𝑐 Hence, it follows from Lemma 15 that letting 𝐴𝑟,𝑛 = SCirc(𝑦1,𝑦2,...,𝑦𝑛),thenitslastrowelementsare(−𝑦2, 2 (−𝑑)𝑖−1 −𝑦 ,...,−𝑦 ,𝑦 𝑦 =−∑ (𝐿 −𝑡𝐿 )⋅ , 3 𝑛 1)whicharegivenbythefollowingequations: 4 𝑟+1+𝑖 𝑟+𝑖 𝑖 𝑖=1 𝑐 𝑡 2 (−𝑑)𝑘−5+𝑖 −𝑦 = +𝑇 𝑦 =−∑ (𝐿 −𝑡𝐿 )⋅ 2 𝑙 23 𝑘 𝑟+1+𝑖 𝑟+𝑖 𝑘−4+𝑖 𝑛 𝑖=1 𝑐 𝑡 1 𝑛−2 (−𝑑)𝑖−1 (𝑘=5,6,...,𝑛) . = + ∑ (𝐿 −𝑡𝐿 )⋅ , 𝑟+𝑛+1−𝑖 𝑟+𝑛−𝑖 𝑖 𝑙𝑛 𝑙𝑛 𝑖=1 𝑐 𝑟=0 Specially, when ,onegetstheresultof[23]. 1 1 −𝑦3 =𝑇2,𝑛 = (𝐿𝑟+3 −𝑡𝐿𝑟+2)⋅ , Proof. Let 𝑙𝑛 𝑐 −𝑦 =𝑇 −𝑇 𝑙 4 2,𝑛−1 2𝑛 1− 𝑛 𝜔 𝜔 ⋅⋅⋅ 𝜔 𝐿 13 14 1𝑛 𝑟+1 1 2 (−𝑑)𝑖−1 (01𝜔23 𝜔24 ⋅⋅⋅ 𝜔2𝑛) = ∑ (𝐿 −𝑡𝐿 )⋅ , ( ) 𝑟+1+𝑖 𝑟+𝑖 𝑖 0 0 1 0 ⋅⋅⋅ 0 𝑙𝑛 𝑖=1 𝑐 Ω2 = ( ) , (35) (0 0 0 1 ⋅⋅⋅ 0 )
. . . . . −𝑦5 =𝑇2,𝑛−2 −𝑇2𝑛−1 −𝑇2𝑛 . . . . d . (0 0 0 0 ⋅⋅⋅ 1 ) 1 2 (−𝑑)𝑖 = ∑ (𝐿 −𝑡𝐿 )⋅ , 𝑙 𝑟+1+𝑖 𝑟+𝑖 𝑐𝑖+1 where 𝑛 𝑖=1 −𝑦 =𝑇 −𝑇 −𝑇 1 𝑙 𝑘 2,𝑛−𝑘+3 2,𝑛−𝑘+4 2,𝑛−𝑘+5 𝜔 = [ 𝑛 (𝑡𝐿 −𝐿 )−𝐿 ] 1𝑗 𝐿 𝑙 𝑟+𝑛+2−𝑗 𝑟+𝑛+3−𝑗 𝑟+𝑛+2−𝑗 𝑟+1 𝑛 1 2 (−𝑑)𝑘−5+𝑖 (36) = ∑ (𝐿 −𝑡𝐿 )⋅ , 1 𝑟+1+𝑖 𝑟+𝑖 𝑘−4+𝑖 𝑙𝑛 𝑖=1 𝑐 𝜔2𝑗 = ⋅(𝐿𝑟+𝑛+3−𝑗 −𝑡𝐿𝑟+𝑛+2−𝑗) (𝑗=3,4,...,𝑛). 𝑙𝑛 . . Then, we have
−𝑦𝑛 =𝑇23 −𝑇24 −𝑇25 𝐿𝑟+1 0 0 0 ⋅⋅⋅ 0 0𝑙0 0 ⋅⋅⋅ 0 𝑛 𝑛−2 (−𝑑)𝑖−1 𝑛−3 (−𝑑)𝑖−1 0 0 𝑐 0 ⋅⋅⋅ 0 = ∑𝜔 ⋅ − ∑𝜔 ⋅ ( ) 2,2+𝑖 𝑖 2,3+𝑖 𝑖 Σ𝐴𝑟,𝑛Ω1Ω2 = ( 0 0 𝑑 𝑐 ⋅⋅⋅) 0 , (37) 𝑖=1 𝑐 𝑖=1 𝑐 ...... d ...... 𝑛−4 (−𝑑)𝑖−1 0 0 0 0 ⋅⋅⋅ 𝑐 − ∑𝜔 ⋅ ( ) 2,4+𝑖 𝑖 𝑖=1 𝑐 so Σ𝐴𝑟,𝑛Ω1Ω2 = D ⊕ H,where𝐷=diag(𝐿𝑟+1,𝑙𝑛) is a 1 2 (−𝑑)𝑛−5+𝑖 diagonal matrix and D ⊕ H is the direct sum of D and H.If = ∑ (𝐿 −𝑡𝐿 )⋅ , −1 −1 −1 𝑟+1+𝑖 𝑟+𝑖 𝑛−4+𝑖 we denote Ω=Ω1Ω2,thenweobtain𝐴𝑟,𝑛 =Ω(D ⊕H )Σ. 𝑙𝑛 𝑖=1 𝑐 6 Journal of Applied Mathematics
1 2 (−𝑑)𝑖−1 𝑦1 = −𝑇23 −𝑇24 𝑦 =−∑ (𝐿 −𝑡𝐿 )⋅ , 𝑙𝑛 4 𝑟+1+𝑖 𝑟+𝑖 𝑖 𝑖=1 𝑐 1 1 (−𝑑)𝑛−3 = − [(𝐿 −𝑡𝐿 )⋅ 2 (−𝑑)𝑘−5+𝑖 𝑙 𝑙 𝑟+3 𝑟+2 𝑐𝑛−2 𝑦 =−∑ (𝐿 −𝑡𝐿 )⋅ , (𝑘=5,6,...,𝑛) . 𝑛 𝑛 𝑘 𝑟+1+𝑖 𝑟+𝑖 𝑘−4+𝑖 𝑖=1 𝑐 𝑛−3 (−𝑑)𝑖−1 +∑ (𝐿 −𝑡𝐿 )⋅ ]. (41) 𝑟+𝑛+2−𝑖 𝑟+𝑛+1−𝑖 𝑖 𝑖=1 𝑐 This completes the proof. (39) 3. Norm and Spread of Skew Circulant Matrix Hence, we obtain with the Lucas Numbers Theorem 17. 𝐴 = (𝐿 ,...,𝐿 ) 1 1 (−𝑑)𝑛−3 Let 𝑟,𝑛 SCirc 𝑟+1 𝑟+𝑛 be skew 𝑦 = − [(𝐿 −𝑡𝐿 )⋅ circulant matrix; then three kinds of norms of 𝐴𝑟,𝑛 are given 1 𝑙 𝑙 𝑟+3 𝑟+2 𝑐𝑛−2 𝑛 𝑛 by 𝑛−3 𝑖−1 (−𝑑) 𝐴 = 𝐴 =𝐿 −𝐿 , +∑ (𝐿 −𝑡𝐿 ) ], 𝑟,𝑛1 𝑟,𝑛∞ 𝑟+𝑛+2 𝑟+2 (42) 𝑟+𝑛+2−𝑖 𝑟+𝑛+1−𝑖 𝑐𝑖 𝑖=1 𝐴 = √𝑛(𝐿 𝐿 −𝐿 𝐿 ). 𝑟,𝑛𝐹 𝑟+𝑛 𝑟+𝑛+1 𝑟 𝑟+1 (43) 𝑡 1 𝑛−2 (−𝑑)𝑖−1 𝑦 =− − ∑ (𝐿 −𝑡𝐿 )⋅ , 2 𝑟+𝑛+1−𝑖 𝑟+𝑛−𝑖 𝑖 𝑙𝑛 𝑙𝑛 𝑖=1 𝑐 Proof. By Definition 8 and (12), we have 1 1 𝑛 𝑦 =− (𝐿 −𝑡𝐿 )⋅ , 𝐴 = 𝐴 = ∑𝐿 =𝐿 −𝐿 . 3 𝑙 𝑟+3 𝑟+2 𝑐 𝑟,𝑛1 𝑟,𝑛∞ 𝑟+𝑖 𝑟+𝑛+2 𝑟+2 (44) 𝑛 𝑖=1 1 2 (−𝑑)𝑖−1 𝑦 =− ∑ (𝐿 −𝑡𝐿 )⋅ , By Definition 8 and (13), we have 4 𝑟+1+𝑖 𝑟+𝑖 𝑖 𝑙𝑛 𝑖=1 𝑐 (40) 𝑛 𝑛 2 2 2 𝑖 (𝐴 ) = ∑∑𝑎 1 (−𝑑) 𝑟,𝑛𝐹 𝑖𝑗 𝑦 =− ∑ (𝐿 −𝑡𝐿 )⋅ , 𝑖=1𝑗=1 5 𝑟+1+𝑖 𝑟+𝑖 𝑖+1 𝑙𝑛 𝑖=1 𝑐 𝑛 2 2 𝑘−5+𝑖 =𝑛∑𝐿 1 (−𝑑) 𝑟+𝑖 𝑦 =− ∑ (𝐿 −𝑡𝐿 )⋅ , 𝑖=1 𝑘 𝑟+1+𝑖 𝑟+𝑖 𝑘−4+𝑖 (45) 𝑙𝑛 𝑖=1 𝑐 𝑟+𝑛 𝑟 2 2 . =𝑛(∑𝐿𝑖 − ∑𝐿𝑖 ) . 𝑖=0 𝑖=0
2 𝑛−5+𝑖 1 (−𝑑) =𝑛(𝐿𝑟+𝑛𝐿𝑟+𝑛+1 −𝐿𝑟𝐿𝑟+1). 𝑦 = ∑ (𝐿 −𝑡𝐿 )⋅ , 𝑛 𝑙 𝑟+1+𝑖 𝑟+𝑖 𝑐𝑛−4+𝑖 𝑛 𝑖=1 Thus −1 1 𝐴 = ⋅ (𝑦 ,𝑦 ,...,𝑦 ), 𝐴 = √𝑛(𝐿 𝐿 −𝐿 𝐿 ). 𝑟,𝑛 SCirc 1 2 𝑛 𝑟,𝑛𝐹 𝑟+𝑛 𝑟+𝑛+1 𝑟 𝑟+1 (46) 𝑙𝑛
Theorem 18. where Let 𝐴𝑟,𝑛 = SCirc (𝐿𝑟+1,−𝐿𝑟+2,...,−𝐿𝑟+𝑛−1,𝐿𝑟+𝑛) (47) (−𝑑)𝑛−3 𝑦 =1−[(𝐿 −𝑡𝐿 )⋅ 1 𝑟+3 𝑟+2 𝑐𝑛−2 be an odd-order alternative skew circulant matrix and let 𝑛 be odd. Then 𝑛−3 (−𝑑)𝑖−1 +∑ (𝐿 −𝑡𝐿 )⋅ ], 𝑛 𝑟+𝑛+2−𝑖 𝑟+𝑛+1−𝑖 𝑐𝑖 𝐴 = ∑𝐿 =𝐿 −𝐿 . 𝑖=1 𝑟,𝑛2 𝑟+𝑖 𝑟+𝑛+2 𝑟+2 (48) 𝑖=1 𝑛−2 (−𝑑)𝑖−1 𝑦 =−𝑡−∑ (𝐿 −𝑡𝐿 )⋅ , 2 𝑟+𝑛+1−𝑖 𝑟+𝑛−𝑖 𝑖 Proof. By Lemma 3,wehave 𝑖=1 𝑐 𝑛 1 𝑖−1 𝑗 𝑖−1 𝜆𝑗 (𝐴𝑟,𝑛)=∑(−1) 𝐿𝑟+𝑖(𝜔 𝜂) . (49) 𝑦 =−(𝐿𝑟+3 −𝑡𝐿𝑟+2)⋅ , 3 𝑐 𝑖=1 Journal of Applied Mathematics 7
So = (𝑛+2) (𝐿𝑟+𝑛+2 −𝐿𝑟+3) 𝑛 𝑖−1 𝑛 𝑛 𝜆 (𝐴 ) ≤ ∑ (−1)𝑖−1𝐿 ⋅ (𝜔𝑗𝜂) 𝑗 𝑟,𝑛 𝑟+𝑖 −2[∑ (𝑟+𝑘) 𝐿𝑟+𝑘 − ∑𝑟𝐿 𝑟+𝑘], 𝑖=1 𝑘=2 𝑘=2 (50) 𝑛 (56) = ∑𝐿𝑟+𝑖, 𝑖=1 by (12)and(14), for all 𝑗=0,1,...,𝑛−1. ∑𝑎 =2𝐿 − (𝑛−2) 𝐿 −𝑛𝐿 −2𝐿 . 𝑛 𝑖𝑗 𝑟+𝑛+3 𝑟+𝑛+2 𝑟+3 𝑟+4 (57) Since 𝑛 is odd, ∑𝑖=1 𝐿𝑟+𝑖 is an eigenvalue of 𝐴𝑟,𝑛;thatis, 𝑖 =𝑗̸ 1 . By (19), we have 𝐿𝑟+1 −𝐿𝑟+2 . 𝐿𝑟+𝑛 −1 −𝐿 𝐿 −𝐿 1 𝑟+𝑛 𝑟+1 𝑟+𝑛−1 ( 1 ) 𝑠(𝐴 )≥ 2𝐿 − (𝑛−2) 𝐿 (𝐿 −𝐿 𝐿 ) ( ) 𝑟,𝑛 𝑛−1 𝑟+𝑛+3 𝑟+𝑛+2 𝑟+𝑛−1 𝑟+𝑛 𝑟+𝑛−2 −1 (58) . . . . . . . . . −𝑛𝐿 −2𝐿 ...... 𝑟+3 𝑟+4 𝐿 −𝐿 𝐿 ( 𝑟+2 𝑟+3 𝑟+1 ) ( 1 ) (51) 1 −1 4. Determinant and Inverse of Skew Left 𝑛 Circulant Matrix with the Lucas Numbers ( 1 ) = ∑𝐿𝑟+𝑖 ⋅ (−1) . In this section, let 𝐴 = SLCirc(𝐿𝑟+1,...,𝐿𝑟+𝑛) be skew 𝑖=1 . 𝑟,𝑛 . left circulant matrix. By using the obtained conclusions in 1 Section 2, we give a determinant explicit formula for the ( ) matrix 𝐴𝑟,𝑛.Afterwards,weprovethat𝐴𝑟,𝑛 is an invertible To sum up, we have matrix for any positive interger 𝑛.Theinverseofthematrix 𝐴 𝑛 𝑟,𝑛 is also presented. According to Lemmas 5 and 6 and Theorems 13, 14,and max 𝜆𝑗 (𝐴 ) = ∑𝐿𝑟+𝑖. 0≤𝑗≤𝑛−1 𝑟,𝑛 (52) 𝑖=1 16, we can obtain the following theorems. Since all skew circulant matrices are normal, by Lemma 9 Theorem 20. Let 𝐴𝑟,𝑛 = SLCirc(𝐿𝑟+1,...,𝐿𝑟+𝑛) be skew left and (12), and (52), we have circulant matrix; then 𝑛 𝐴 = (−1)𝑛(𝑛−1)/2𝐿 𝐴 = ∑𝐿 =𝐿 −𝐿 , det 𝑟,𝑛 𝑟+1 𝑟,𝑛2 𝑟+𝑖 𝑟+𝑛+2 𝑟+2 (53) 𝑖=1 𝑛−2 𝑛−1−𝑖 ×[𝐿𝑟+1 +𝑡𝐿𝑟+𝑛 + ∑ (𝑡𝐿𝑟+1+𝑖 −𝐿𝑟+2+𝑖)𝑥 ] which completes the proof. 𝑘=1
𝑛−2 Theorem 19. Let 𝐴𝑟,𝑛 = SCirc(𝐿𝑟+1,...,𝐿𝑟+𝑛) be skew circu- ⋅𝑐 , 𝐴 lant matrix; then the bounds for the spread of 𝑟,𝑛 are (59)
𝑠(𝐴𝑟,𝑛)⩽√2𝑛 (𝐿𝑟+𝑛𝐿𝑟+𝑛+1 −𝐿𝑟+1𝐿𝑟+2), where 𝐿𝑟+𝑛 is the (𝑟 + 𝑛)th Lucas number. 1 Theorem 21. Let 𝐴𝑟,𝑛 = SLCirc(𝐿𝑟+1,...,𝐿𝑟+𝑛) be skew left 𝑠(𝐴𝑟,𝑛)≥ 2𝐿𝑟+𝑛+3 − (𝑛−2) 𝐿𝑟+𝑛+2 −𝑛𝐿𝑟+3 −2𝐿𝑟+4 . 𝑛−1 circulant matrix; then 𝐴𝑟,𝑛 is an invertible matrix. (54) Theorem 22. Let 𝐴𝑟,𝑛 = SLCirc(𝐿𝑟+1,...,𝐿𝑟+𝑛) be skew left Proof. The trace of 𝐴𝑟,𝑛,tr𝐴𝑟,𝑛 =𝑛𝐿𝑟+1.By(18)and(43), we circulant matrix; then have −1 1 (𝐴 ) = (𝑦,𝑦,...,𝑦), 𝑠 (𝐴𝑟,𝑛) ⩽ √2𝑛 (𝐿𝑟+𝑛𝐿𝑟+𝑛+1 −𝐿𝑟+1𝐿𝑟+2). (55) 𝑟,𝑛 SLCirc 1 2 𝑛 (60) 𝑙𝑛 Since where 𝑛 𝑛 𝑛−3 (−𝑑) ∑𝑎𝑖𝑗 = ∑ (𝑛−(𝑘−1)) 𝐿𝑟+𝑘 − ∑ (𝑘−1) 𝐿𝑟+𝑘 𝑦 =1−[(𝐿 −𝑡𝐿 ) 1 𝑟+3 𝑟+2 𝑛−2 𝑖 =𝑗̸ 𝑘=2 𝑘=2 𝑐
𝑛 𝑛 𝑛−3 (−𝑑)𝑖−1 = (𝑛+2) ∑𝐿 −2∑𝑘𝐿 + ∑ (𝐿 −𝑡𝐿 )⋅ ], 𝑟+𝑘 𝑟+𝑘 𝑟+𝑛+2−𝑖 𝑟+𝑛+1−𝑖 𝑖 𝑘=2 𝑘=2 𝑖=1 𝑐 8 Journal of Applied Mathematics
𝑦𝑘 =−𝑦𝑛−𝑘+2 Since all skew left circulant matrices are symmetrical, by Lemma 9 and (12)and(68), we obtain 2 (−𝑑)𝑛−𝑘−3+𝑖 = ∑ (𝐿 −𝑡𝐿 )⋅ , 𝑟+1+𝑖 𝑟+𝑖 𝑛−𝑘−2+𝑖 𝐴 =𝐿 −𝐿 . 𝑖=1 𝑐 𝑟,𝑛2 𝑟+𝑛+2 𝑟+2 (69) (𝑘=2,3,...,𝑛−2) .
1 𝑦 =−𝑦 =(𝐿 −𝑡𝐿 )⋅ , Theorem 25. Let 𝐴𝑟,𝑛 = SLCirc(𝐿𝑟+1,...,𝐿𝑟+𝑛) be skew left 𝑛−1 3 𝑟+3 𝑟+2 𝑐 circulant matrix; the bounds for the spread of 𝐴𝑟,𝑛 are 𝑦 =−𝑦 𝑛 2 2 { 2 𝑛−2 𝑖−1 √𝑀− 𝑁 , 𝑖𝑓 𝑛 𝑖𝑠 𝑜𝑑𝑑, (−𝑑) 2𝐿𝑟+𝑛 ≤𝑠(𝐴 )≤ 𝑛 (70) =𝑡+∑ (𝐿 −𝑡𝐿 )⋅ . 𝑟,𝑛 { 𝑟+𝑛+1−𝑖 𝑟+𝑛−𝑖 𝑖 {√𝑀, 𝑖𝑓 𝑛 𝑖𝑠𝑒V𝑒𝑛, 𝑖=1 𝑐 (61) where 𝑀=2𝑛(𝐿 𝐿 −𝐿 𝐿 ), 5. Norm and Spread of Skew Left Circulant 𝑟+𝑛 𝑟+𝑛+1 𝑟+1 𝑟 (71) Matrix with the Lucas Numbers 𝑁=𝐿𝑟+𝑛−1 +𝐿𝑟−1. Theorem 23. Let 𝐴𝑟,𝑛 = SLCirc(𝐿𝑟+1,...,𝐿𝑟+𝑛) be skew left Proof. Since 𝐴𝑟,𝑛 is a symmetric matrix, by (20), circulant matrix. Then three kinds of norms of 𝐴𝑟,𝑛 are given by 𝑠(𝐴 )≥2max 𝑎𝑖𝑗 =2𝐿𝑟+𝑛. 𝑟,𝑛 𝑖 =𝑗̸ (72) 𝐴 = 𝐴 =𝐿 −𝐿 , 𝑟,𝑛1 𝑟,𝑛∞ 𝑟+𝑛+2 𝑟+2 (62) 𝐴 𝑛 𝐴 = √𝑛(𝐿 𝐿 −𝐿 𝐿 ). The trace of 𝑟,𝑛 is, if is odd, 𝑟,𝑛𝐹 𝑟+𝑛 𝑟+𝑛+1 𝑟 𝑟+1 (𝐴 ) Proof. Using the method in Theorem 17 similarly, the conclu- tr 𝑟,𝑛 sion is obtained. =𝐿𝑟+1 −𝐿𝑟+2 +𝐿𝑟+3 −⋅⋅⋅+𝐿𝑟+𝑛 Theorem 24. Let =𝐿𝑟+1 +𝐿𝑟+1 +𝐿𝑟+3 +⋅⋅⋅+𝐿𝑟+𝑛−2 (73) 𝐴𝑟,𝑛 = SLCirc (𝐿𝑟+1,−𝐿𝑟+2,...,−𝐿𝑟+𝑛−1,𝐿𝑟+𝑛) (63) =2𝐿𝑟+1 +𝐿𝑟+1 +𝐿𝑟+2 +⋅⋅⋅+𝐿𝑟+𝑛−3 be an odd-order alternative skew left circulant matrix; then 𝑛−3 𝑛 =2𝐿 + ∑𝐿 . 𝐴 = ∑𝐿 =𝐿 −𝐿 . 𝑟+1 𝑟+𝑖 𝑟,𝑛2 𝑟+𝑖 𝑟+𝑛+2 𝑟+2 (64) 𝑖=1 𝑖=1 Proof. According to Lemma 4, By (12), we have 𝑛 (𝐴 )=𝐿 +𝐿 =𝑁. 𝜆 (𝐴 )=±∑(−1)𝑖−1𝐿 𝜔(𝑗−(1/2))(𝑘−1) , tr 𝑟,𝑛 𝑟+𝑛−1 𝑟−1 (74) 𝑗 𝑟,𝑛 𝑟+𝑖 (65) 𝑖=1 Let 𝑀=2𝑛(𝐿𝑟+𝑛𝐿𝑟+𝑛+1−𝐿𝑟+1𝐿𝑟);then,by(18), (62), and (74), for 𝑗=1,2,...,(𝑛−1)/2,and we obtain 𝑛 𝜆 (𝐴 )=∑𝐿 . 2 (𝑛+1)/2 𝑟,𝑛 𝑟+𝑖 (66) 𝑠(𝐴 )⩽√𝑀− 𝑁2. (75) 𝑖=1 𝑟,𝑛 𝑛
So If 𝑛 is even, then 𝑛 𝑖−1 𝑖−1 𝜆𝑗 (𝐴𝑟,𝑛) ≤ ∑ (−1) 𝐿𝑟+𝑖(−1) tr (𝐴𝑟,𝑛)=𝐿𝑟+1 −𝐿𝑟+1 +𝐿𝑟+3 𝑖=1 (67) (76) 𝑛 𝑛+1 −𝐿𝑟+3 ⋅⋅⋅−𝐿𝑟+𝑛−1 =0. = ∑𝐿 , (𝑗 = 1, 2, . . ., ). 𝑟+𝑖 2 𝑖=1 By (18), (62), and (76), we have By (66)and(67), we have √ 𝑠(𝐴𝑟,𝑛)⩽ 𝑀. (77) 𝑛 max 𝜆𝑖 (𝐴𝑟,𝑛) = ∑𝐿𝑟+𝑖. (68) 0≤𝑖≤(𝑛+1)/2 𝑖=1 So the result follows. Journal of Applied Mathematics 9
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