On formulae for the determinant of symmetric pentadiagonal Toeplitz matrices

Arabian Journal of Mathematics, Dec 2017

We show that the characteristic polynomial of a symmetric pentadiagonal Toeplitz matrix is the product of two polynomials given explicitly in terms of the Chebyshev polynomials.

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On formulae for the determinant of symmetric pentadiagonal Toeplitz matrices

On formulae for the determinant of symmetric pentadiagonal Toeplitz matrices Mohamed Elouafi Mathematics Subject Classification ⎜ b a ⎜ ⎜⎜ c b Pm = Pm ( We show that the characteristic polynomial of a symmetric pentadiagonal Toeplitz matrix is the product of two polynomials given explicitly in terms of the Chebyshev polynomials. We consider here the problem of finding the determinant of the m × m symmetric pentadiagonal Toeplitz matrix b - 1 Introduction ⎛ a 0 . . . . . . . . . b c · · · 0 ⎞ . . . ... ⎟ . . ⎟⎟ . 0 ⎟⎟ . b c ⎟⎟ ⎟ a b ⎠⎟ b a This class of matrices arises naturally in many applications, such as signal processing, trigonometric moment problems, integral equations and elliptic partial differential equations with boundary conditions [ 9 ]. Computing the determinant of the matrix Pm have intrigued the researchers for decades. If c = 0, then Pm is reduced to a tridiagonal matrix and there exists a closed form of det (Pm ) from which the eigenvalues of the matrix are explicitly given. It is becoming a challenge to find similar formulae for the general case and so far, little is known about the eigenvlaues of Pm [ 1,2,5,8 ]. In [ 5,7 ], det (Pm ) is explicitly computed using the kernel of the Chebyshev polynomials {Tn} , {Un} , {Vn} and {Wn} [ 11 ] and, as a consequence, the eigenvalues of the matrix Pm are localized by means of explicitly given rational functions. The formulae are simplified to give det(Pm ) as polynomials of the parameters a, b, c [ 6 ]. In the new formula presented here, det(Pm ) is given as the product of two polynomials given in a standard form. Here is our main result: Theorem 1.1 We have and where det(P2n+1) = 2 γn,k cn+1−k bk Tk det(P2n) = μn,k cn−k bk Vk n+1 k=0 n k=0 , γn,k = (−1)k n + 1 + k , μn,k = (−1)k n + 1 + k . n + 1 − k n − k 2 Proof of the main result Since det(Pm (a, b, c)) = cm det Pm ac , bc , 1 , then we can assume for simplicity that c = 1. We denote by ζ j , ζ1j , j = 1, 2, the roots of the polynomial g(x) = x4 + bx3 + ax2 + bx + 1 assumed pairwise distinct and different of ±1. Recall that the Chebyshev polynomials {Tn}, {Un}, {Vn} and {Wn} are orthogonal polynomials over (−1, 1) with respect to the weight √11−x2 , √1 − x2, 11+−xx and 11−+xx , respectively, and we have for ζ ∈ C∗ ⎪⎧⎨ Tn 21 ζ + ζ 1 ⎪⎩ Vn 21 ζ + ζ 1 = 21 ζ n + ζ −n , Un 21 ζ + ζ 1 = ζ n+1/2+ζ −n−1/2 ζ 1/2+ζ −1/2 , Wn 21 ζ + ζ 1 = ζ n+ζ1−−ζζ−−1n−1 , = ζ n+1/2−ζ −n−1/2 ζ 1/2−ζ −1/2 . We shall use the following formula for det(Pm): Lemma 2.1 For J ⊂ {1, 2}, let IJ (k) = 1 i f k ∈ J −1 i f k ∈/ J , and (1) (2) (3) ωJ = ζkIJ (k), γJ = (ζkIJ (k) − ζ jIJ ( j)). 2 k=1 1 We have 1 1 . where d = ζ2 + ζ2 − ζ1 − ζ1 Proof See [4]. det (Pm) = d2 2 k=1(ζk − ζk−1) J Let us put α = ζ1ζ2 , β = ζ1ζ2−1 and u = 21 α + α−1 , v = 21 β + β−1 . We have by the Vieta’ formulae: 1≤ j<k≤2 m+1 (−1)|J| γJ ωJ 2 × J 2 b a = − Proof Using the notations from Lemma 2.1, we obtain det(Pm) = and Remark that and hence Similarly, we obtain that αm2+1+1 α− 2 m+1 − α 1 − = α1/2 m+2 α 2 α− 2 − m+2 − 1 α−1/2 1/2 = Um+1 α 2 + α−1/2 = Um+1 1 u + 2 (F1) By the same method, we get J γJ ωnJ+1 = (ζ2−1 − ζ1)(β − 1) β , A straightforward computation (using the Maple software for example) shows that and this completes the proof of the Lemma. tRheemsaerckon2d.3-orWdeerheaqvueautio+nvx +2−2 =a2 a+2+1 1xa+nd (b2u +2=1)0(.v T+h1is)g=ivesb2 fo2r. eTxhaemn,pule+ 1 and v + 1 are the zeros of J 2 C = and The term Um2 +1 1+2u2(u−−Uvm)2 +1 1+2v is a symmetric polynomial of u + 1 and v + 1 and, consequently, it can be expressed in terms of the elementary symmetric polynomials u + 1 + v + 1 = a2 + 1 and (u + 1) (v + 1) = b2 2 . For this, we distinguish two cases: Case 1: m = 2n + 1. Using the following expression of U2n+2 (x )[ 3 ]: U2n+2 (x ) = (−1)k n+1 k=0 2n + 2 − k k (2x )2n+2−2k n+1 k=0 = (−1)n+1 γn,k (2x )2k , γn,k = (−1)k n + 1 + k , n + 1 − k U2n+2 1 +2 v = (−1)n+1 n+1γn,k2k (1 + u)k + (1 + v)k , On the other hand, we have for x, y : we obtain and for k ≥ 1 and gives and for k ≥ 1: and thus Applying those formulae for x = √1 + u and y = √1 + v where x2k + y2k = (xy)k x k x −k y + y = 2(xy)k Tk 2y + 2yx , x x2k − y2k = (xy)k x k x −k y − y = (xy)k xy − xy Uk−1 2y + 2x x y = x2 − y2 (xy)k−1 Uk−1 2y + 2yx . x xy = b2, x2 − y2 = u − v, 2xy + 2yx = x22+xyy2 = 2 + u + v = a2+b2, b (1 + u)k + (1 + v)k = 2 b2 k Tk a2+b2 , Case 2: m = 2n. We have [ 3 ]: (1 + u)k − (1 + v)k = (u − v) b2 k−1 Uk−1 a2+b2 k=0 n U2n+1 (x) = (−1)k 2n +k1 − k (2x)2n+1−2k n = (−1)n μn,k (2x)2k+1 , μn,k = (−1)k n +n −1 +k k , k=0 U2n+1 As for the odd case, we have for x , y : x 2k+1 + y2k+1 = (x y)k+1/2 x 2k+1 − y2k+1 = (x y)k+1/2 = (x y)k+1/2 = (x y)k+1/2 and and This implies x y x y x y x y √ √ n , , (1 + u)k+1/2 + (1 + v)k+1/2 = 1 + u + 1 + v (1 + u)k+1/2 − (1 + v)k+1/2 = 1 + u − 1 + v which completes the proof of Theorem 1.1. 3 Numerical computation of det (Pm) In this section, we shall derive from the formulae (1) and (2) an efficient algorithm for computing det (Pm ) . We are lead to evaluate sums of the form N k=0 SN = αk Pk (x ) where x = a +2b2c , and { Pr } are polynomials that satisfy the three-term recurrence Such sums can be computed efficiently through the following method described in [ 11 ]: Equation (4) may be written in matrix notation as M p = q, where M is the (N + 1) × (N + 1) matrix Pr (x ) − 2x Pr −1(x ) + Pr −2(x ) = 0. (4) 0 . . . 0 0 be the row vector such that It follows that yT = (y0, y1, . . . yN ) yT M = uT = (α0, α1, . . . αN ) . yk = 2x yk+1 − yk+2 + αk , for k = N , . . . , 0. SN = uT p = yT Mp = yT q =y0 P0 (x) + (P1(x) − 2x P0(x)) y1. Thus, yk are computed by putting yN+1 = yN+2 = 0 and performing the three-term recurrence Let and and where yn+1 = yn+2 = 0 and yk = 2x yk+1 − yk+2 + μn,k cn−k bk , for k = n, . . . , 0. For Pk = Tk and Pk = b1 Uk−1, with U−1 = 0, respectively, we obtain where yn+2 = yn+3 = 0 and For Pk = Vk and Pk = Wk , respectively, we obtain yk = 2x yk+1 − yk+2 + γn,k cn+1−k bk , for k = n + 1, . . . , 0. n+1 k=0 n+1 k=1 n k=0 n k=0 γn,k cn+1−k bk Tk (x) = y0 − x y1 1 γn,k cn+1−k bk−1Uk−1 (x) = b y1, μn,k cn−k bk Vk (x) = y0 − y1 μn,k cn−k bk Wk (x) = y0 + y1, Here is the implementation of the algorithm in Maple (To accelerate the algorithm, the terms γn,k cn+1−k bk and μn,k cn−k bk are computed recursively at the same time as yk . Implementation details are omitted): ## Computing det(P_2n+1) ## detP1:=proc(n,a,b,c) local i,j,r,s,x,k,t,z; i := 0; j := 0; r := (-1)ˆ(n+1)*bˆ(n+1); x:=(a+2*c)/b; t:=2*n; z:=-c/b; for k from 0 to n+1 do s:=i; i:=r+x*i-j; ## i:=simplify(r+x*i-j); if the purpose ## is to compute the characteristic ## polynomial with variable a j:=s; r:=r*z*((t+2 )*(t+1 ))/((t+k+2 )*(k+1 )); t:=t-2; od; return 2*j*(i-(j*x/2)/b; ## return simplify(2*j*(i-(j*x/2)/b); ## if the purpose is to compute the ## characteristic polynomial with ## variable a end; ## Computing det(P_2n) ## detP2:=proc(n,a,b,c) local i,j,r,s,x,k; i := 0; j := 0; r := (-1)ˆn*bˆn; x:=(a+2*c)/b; t:=2*n; z:=-c/b; for k from 0 to n do s:=i; i:=r+x*i-j; ## i:=simplify(r+x*i-j); if the purpose ## is to compute the characteristic ## polynomial with variable a j:=s; r:=r*z*((t+1 )*t))/(t+k+1 )*(k+1 )); t:=t-2; od; return iˆ(2)-jˆ(2); ## return simplify(iˆ(2)-jˆ(2)); ## if the purpose is to compute ## the characteristic polynomial ## with variable a end; One can easily check that the complexity of the algorithm is about 7N , where N is the size of the matrix. Thus, the algorithm is the fastest among many other recently proposed (we exclude those based on the roots of certain polynomials which are approximative) [ 10 ]. Moreover, subject to minor modifications as explained in Algorithm 1, the algorithm is suitable for computing the characteristic polynomial of a symmetric pentadiagonal Toeplitz matrix using computer algebra systems such as MAPLE, MATHEMATICA, MATLAB and MACSYMA. Open Access This article is distributed under the terms of the Creative Commons Attribution 4.0 International License (http:// creativecommons.org/licenses/by/4.0/), which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made. Springer Nature remains neutral with regard to jurisdictional claims in published maps and institutional affiliations. 1. 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Mohamed Elouafi. On formulae for the determinant of symmetric pentadiagonal Toeplitz matrices, Arabian Journal of Mathematics, 2017, 1-9, DOI: 10.1007/s40065-017-0194-0