On Some Pentadiagonal Matrices: Their Determinants and Inverses

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On Some Pentadiagonal Matrices: Their Determinants and Inverses Annales Univ. Sci. Budapest., Sect. Comp. 51 (2020) 3950 ON SOME PENTADIAGONAL MATRICES: THEIR DETERMINANTS AND INVERSES Carlos M. da Fonseca (Safat, Kuwait) László Losonczi (Debrecen, Hungary) Dedicated to the 70th birthday of Professor Antal Járai Communicated by László Szili (Received February 7, 2020; accepted June 2, 2020) Abstract. In this paper we consider pentadiagonal (n+1)×(n+1) matri- ces with two super-diagonals and two sub-diagonals at distances k and 2k from the main diagonal. We give an explicit formula for their determinants provided that (n + 1)=3 ≤ k ≤ n=2: We consider the Toeplitz and imper- fect Toeplitz versions of such matrices, and give explicit formulas for their determinants. We also show that the inverse matrix can be obtained as the product of an upper triangle matrix with two super-diagonals, a diagonal and a lower triangle matrix with two sub-diagonals. 1. Introduction Let n; k; ` be given positive integers with 1 ≤ k < ` ≤ n and denote by Mn the set of n × n complex matrices. Consider the pentadiagonal matrix A0 = (aij) 2 Mn+1 where for i; j = 0; 1; : : : ; n, 8 Lj if j − i = −`; > > lj if j − i = −k; > < di if j − i = 0; aij = if > ri j − i = k; > if > Ri j − i = `; :> 0 otherwise: Key words and phrases: Toeplitz matrix, pentadiagonal, tridiagonal matrices. 2010 Mathematics Subject Classication: Primary 15A15, 15B9, Secondary 65F40, 15A42. 40 C. M. da Fonseca and L. Losonczi Notice that the numbering of entries starts with zero. We also denote the matrix A0 by An+1;k;`(L; l; d; r; R) and denote by D0 = Dn+1;k;`(L; l; d; r; R) its determinant where the diagonal vectors L; l; d; r; R are dened by L = (L0;:::;Ln−`); l = (l0; : : : ; ln−k); d = (d0; : : : ; dn); R = (R0;:::;Rn−`); r = (r0; : : : ; rn−k): We will refer to the diagonal vectors L; l; d; r; R as `th sub-diagonal, kth sub- diagonal, (main) diagonal, kth super-diagonal, `th super-diagonal, respectively. If L; R are zero vectors then our matrix becomes a tridiagonal one denoted by An+1;k(l; d; r) and its determinant by Dn+1;k(l; d; r). If 1 ≤ k = ` ≤ n, then the two super-diagonals and the two sub-diagonals slip together and our matrix becomes a tridiagonal matrix An+1;k;k(L; l; d; r; R) = An+1;k(L + l; d; r + R) and its determinant D0 = Dn+1;k(L + l; d; r + R): We shall call An+1;k;`(L; l; d; r; R) a (general) k; `-pentadiagonal matrix while An+1;k(l; d; r) will be termed as k-tridiagonal. In [6] we developed a method to reduce the determinant of k; `-pentadiagonal matrices to tridiagonal determinants provided that k + ` ≥ n + 1. If k ≥ ≥ (n + 1)=3, then by suitable matrix multiplications we can reduce a k; 2k- pentadiagonal matrix to a tridiagonal one. A similar reduction of tridagonals leads to diagonal matrices. In this way we get an explicit formula for the determinant of a general k; 2k-pentadiagonal matrix. In case of Toeplitz pentadiagonal matrices the diagonal vectors are con- stant vectors, i.e., L = (L; : : : ; L), R = (R; : : : ; R) 2 Rn+1−`, l = (l; : : : ; l), r = (r; : : : ; r) 2 Rn+1−k, d = (d; : : : ; d) 2 Rn+1 and for the matrix and its determinant the notations An+1;k;`(L; l; d; r; R) and Dn+1;k;`(L; l; d; r; R) will be used. For imperfect Toeplitz matrices (the term used by Marr and Vineyard [9]) the main diagonal is n+1 (d − α; : : : ; d − α; d; : : : ; d; d − β; : : : ; d − β) 2 R | {z } | {z } | {z } k n+1−2k k while the other diagonals are the same as above. To indicate the imperfectness we add a superscript (α,β) to the notation of Toeplitz pentadiagonal matrices. Thus an imperfect -pentadiagonal matrix (α,β) has the k; 2k An+1;k;2k(L; l; d; r; R) form On some pentadiagonal matrices 41 0 d − α r R 1 . B .. C B d − α r C B .. .. C B . R C B C B d − α C B C B .. C B l d . C B C B l d C B . C B .. .. .. C B C B d r C B C B .. C B . d − β C B C B L d − β C B . C @ .. .. .. A L l d − β As far as we can tell, the rst particular k; `-pentadiagonal Toeplitz matrix was studied in 1928 by Jen® Egerváry and Otto Szász [3]. One of the matrices they ¯ considered was An+1;k;`(δ; ,¯ −λ, , δ) with k+` = n+1 and jj = jδj = 1. Their result has been largely ignored in the pure and applied matrix theory as we can see e.g. in [2, 10]. For some other papers on the topic, the reader is refereed to [1,6, 11, 12]. An interesting graph theoretical approach can be found in [7,5]. An exhaustive list of recent references is given in the survey [4]. Our paper was inspired by Marr and Vineyard [9] who have shown that the product of two 1-tridiagonal Toeplitz matrices is an imperfect Toeplitz matrix (α,β) which is related to the corresponding Toeplitz matrix by a two-step An+1;1;2 recursion. In this way they succeeded to express the determinant Dn+1;1;2 of 1; 2-pentadiagonal Toeplitz matrix in terms of Chebyshev polynomials of second kind. A similar approach was used in [13] to nd a formula for the inverse of a 1; 2-pentadiagonal Toeplitz matrix. 2. Reduction of general k; 2k-pentadiagonal determinants to tridiagonal ones Our starting point is the matrix A0 = An+1;k;2k(L; l; d; r; R) where 1 ≤ k < < 2k ≤ n and k + 2k ≤ n + 1, i.e., (n + 1)=3 ≤ k ≤ n=2 < (n + 1)=2. Let us dene the matrices B = (bij);C = (cij),F = (fij);G = (gij) of Mn+1 by 42 C. M. da Fonseca and L. Losonczi 8 1 if i=j; > <> lj b = − if i=j+k; j = 0; : : : ; k−1; ij d > j > : 0 otherwise; 8 1 if i=j; > > r < − i if i=0; : : : ; k−1; j =i+k; cij = d > i > : 0 otherwise; 8 1 if i=j; > <> Lj f = − if i=j+2k; j =0; : : : ; n−2k; ij d > j > : 0 otherwise; 8 1 if i=j; > <> Ri g = − if i=0; : : : ; n−2k; j =i+2k; ij d > i > : 0 otherwise: Performing four matrix multiplications (for some details see [6], Theo- rem 3, (i)) we get the matrix L ∗ where A1 = F BA0CG = E1 A1 E1 2 Mk is a diagonal matrix with diagonal elements d0; : : : ; dk−1 and ∗ (1) (1) (1) (1) (1) (1) (1) (1) (1) (1) A1 =An+1−k;k;k(L ; l ; d ; r ;R )=An+1−k;k(L +l ; d ; r +R ) is tridiagonal, where the iterated diagonal vectors are L r L r L(1) = − 0 0 ;:::; − n−2k n−2k ; d0 dn−2k (1) l = (lk; : : : ; ln−k) ; (1) l0r0 lk−1rk−1 L0R0 Ln−2kRn−2k d = dk − ; : : : ; d2k−1 − ; d2k − ; : : : ; dn − ; d0 dk−1 d0 dn−2k (1) r = (rk; : : : ; rn−k) ; R l R l R(1) = − 0 0 ;:::; − n−2k n−2k : d0 dn−2k On some pentadiagonal matrices 43 3. Further reduction to diagonal matrix The matrix A1 can be reduced to a diagonal matrix with two matrix mul- tiplications. Dene the matrices U = (uij);V = (vij) 2 Mn+1 by 8 1 if i = j; > > > < lj − Lj−krj−k=dj−k uij = − if i = j + k; j = k; : : : ; n − k; > dj − lj−krj−k=dj−k > > :> 0 otherwise; 8 1 if i = j; > > > < ri − Ri−kli−k=di−k vij = − if i = k; : : : ; n − k; j = i + k; > dj − li−kri−k=di−k > > :> 0 otherwise. Multiplying A1 from the left by U, then from the right by V we get a diagonal matrix A2 = UA1V with diagonal (3.1) d0; : : : ; dk−1; l0r0 lk−1rk−1 dk − ; : : : ; d2k−1 − ; d0 dk−1 L0r0 R0l0 lk − rk − L0R0 d0 d0 d2k − − ; l0r0 d0 dk − d0 Ln−2krn−2k Rn−2kln−2k ln−k − rn−k − Ln−2kRn−2k dn−2k dn−2k : : : ; dn − − ; ln−2krn−2k dn−2k dn−k − dn−2k where in the above lines k; k; n + 1 − 2k elements are listed. Because the determinants of the matrices B; C; F; G; U; V all are 10s, the determinant of our general pentadiagonal matrix An+1;k;2k is the product of the diagonal elements of A2: To calculate this product we multiply the elements of the rst two groups pairwise, to obtain d0dk − l0r0; : : : ; dk−1d2k−1 − lk−1rk−1: 44 C. M. da Fonseca and L. Losonczi Simplifying the general term of the third group we get, for j = 0; : : : ; n − 2k, Lj rj Rjlj lk+j − d rk+j − LjRj j dj d2k+j − − = dj ljrj dk+j − dj d d d −d l r −L R d −l r d +l R l +r L r = j j+k j+2k j j+k j+k j j j+k j j j+2k j j j+k j j j+k : djdj+k−ljrj Thus we have proved Theorem 3.1. Assuming (n+1)=3≤ k ≤n=2 the determinant Dn+1;k;2k(L; l; d; r; R) of the general k; 2k-pentadiagonal matrix is (3.2) n−2k Q (djdj+kdj+2k −djlj+krj+k −LjRjdj+k −ljrjdj+2k +ljRjlj+k +rjLjrj+k)· j=0 k−1 Q · (djdj+k−ljrj) j=n+1−2k Remark 3.1.
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