On formulae for the determinant of symmetric pentadiagonal Toeplitz matrices

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.


Introduction
We consider here the problem of finding the determinant of the m × m symmetric pentadiagonal Toeplitz matrix 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 P m have intrigued the researchers for decades. If c = 0, then P m is reduced to a tridiagonal matrix and there exists a closed form of det (P m ) 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 P m [1,2,5,8]. In [5,7], det (P m ) is explicitly computed using the kernel of the Chebyshev polynomials {T n } , {U n } , {V n } and {W n } [11] and, as a consequence, the eigenvalues M. Elouafi (B) Classes Préparatoites aux Grandes Ecoles d'Ingénieurs, Lyc ée Moulay Alhassan, Tangier, Morocco E-mail: med3elouafi@gmail.com of the matrix P m are localized by means of explicitly given rational functions. The formulae are simplified to give det(P m ) as polynomials of the parameters a, b, c [6].
In the new formula presented here, det(P m ) is given as the product of two polynomials given in a standard form. Here is our main result: and where

Proof of the main result
Since det(P m (a, b, c)) = c m det P m a c , b c , 1 , then we can assume for simplicity that c = 1. We denote by ζ j , 1 ζ j , j = 1, 2, the roots of the polynomial g(x) = x 4 + bx 3 + ax 2 + bx + 1 assumed pairwise distinct and different of ±1.
Recall that the Chebyshev polynomials {T n }, {U n }, {V n } and {W n } are orthogonal polynomials over (−1, 1) with respect to the weight 1 1−x and 1−x 1+x , respectively, and we have for We shall use the following formula for det(P m ): We have Proof See [4].
We have by the Vieta' formulae: This implies that (u Proof Using the notations from Lemma 2.1, we obtain Remark that On the other hand Similarly, we obtain that By the same method, we get Finally det (P m (a, b, 1) A straightforward computation (using the Maple software for example) shows that and this completes the proof of the Lemma.
and v + 1 = 1 2 The term 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 = a 2 + 1 and (u + 1) (v + 1) = b 2 2 . For this, we distinguish two cases: Case 1: m = 2n + 1. Using the following expression of U 2n+2 (x) [3]: On the other hand, we have for x, y : and for k ≥ 1 Applying those formulae for x = √ 1 + u and y = √ and for k ≥ 1: Case 2: m = 2n. We have [3]: μ n,k (2x) 2k+1 , μ n,k = (−1) k n + 1 + k n − k , and thus As for the odd case, we have for x, y : This implies which completes the proof of Theorem 1.1.

Numerical computation of det (P m )
In this section, we shall derive from the formulae (1) and (2) an efficient algorithm for computing det (P m ) .
We are lead to evaluate sums of the form where x = a+2c 2b , and {P r } 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 Mp = q, where M is the (N + 1) Let be the row vector such that Thus, y k are computed by putting y N +1 = y N +2 = 0 and performing the three-term recurrence It follows that For P k = T k and P k = 1 b U k−1 , with U −1 = 0, respectively, we obtain where y n+2 = y n+3 = 0 and y k = 2x y k+1 − y k+2 + γ n,k c n+1−k b k , for k = n + 1, . . . , 0.
pentadiagonal Toeplitz matrix using computer algebra systems such as MAPLE, MATHEMATICA, MATLAB and MACSYMA.
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