A Note on the Determinant of Five-Diagonal Matrices with Fibonacci Numbers

Transcription

1 Int. J. Contemp. Math. Sciences, Vol. 3, 2008, no. 9, A Note on the Determinant of Five-Diagonal Matrices with Fibonacci Numbers Hacı Civciv Department of Mathematics Faculty of Art and Science Selcuk University, Konya, Turkey hacicivciv@gmail.com, hacicivciv@selcuk.edu.tr Abstract It is always fascinating to see what results when seemingly different areas of mathematics overlap. This article reveals one such result; number theory and Linear algebra (with the help of orthogonal polynomials) are interwined to yield the determinant of five-diagonal matrix with Fibonacci numbers. Keywords: Determinants; Fibonacci Numbers; Five-diagonal matrices 1 Introduction There is a long tradition of using matrices and determinants to study Fibonacci numbers. For example, Bicknell-Johnson and Spears [1] use elementary matrix operations and determinants to generate classes of identities for generalized Fibonacci numbers, and Cahill and Narayan [2] show how Fibonacci and Lucas numbers arise as determinants of some tridiagonal matrices. In another paper, in an attempt to solve a recently published problem [4], the authoe needs to compute L 4n+8 +1 L 4n L 4n L 4n+4 +1 L 4n L 4n 4 L 4n +1 L 4n L 4n 8 where L n is the nth Lucas number defined recursively by, L 0 =2,L 1 =1, and L n = L n 1 + L n 2,n 2.

2 420 H. Civciv To study its generalization Kwong [3] first defined, for any real numbers a, b, c, d, e and f with a, c, e 0, any integers i, j, k 1, and any integer n, Δ(L) = al n+i+j+k+2 + b cl n+i+j+k + d el n+i+j + f al n+i+k+2 + b cl n+i+k + d el n+i + f al n+k+2 + b cl n+k + d el n + f, and analogously Δ(F )= af n+i+j+k+2 + b cf n+i+j+k + d ef n+i+j + f af n+i+k+2 + b cf n+i+k + d ef n+i + f af n+k+2 + b cf n+k + d ef n + f, where F n is the nth Fibonacci number defined recursively by F 0 =0,F 1 =1, and F n = F n 1 + F n 2,n 2, and then he find that the values of these two determinants can be expressed in a rather neat manner, and they only differ by a constant. For example, Strang [6, 7] presents a family of tridiagonal matrices given by: A (n) = , where A (n) isn n. The determinants A (k) are the Fibonacci numbers F 2k+2. Webb and Parberry [8] have showed the following complex factorization: n 1 F n = k=1 ( 1 2i cos πk ), n 2 n,where F n is nth Fibonacci number, by considering the roots of Fibonacci polynomials. In this short note, we study the determinant of a five-diagonal matrix with Fibonacci numbers.

3 Matrices with Fibonacci numbers The determinant of five-diagonal matrix with Fibonacci numbers Let A is the following k k (k 3) five-diagonal matrix: A k = 1 F n F n 1 F n+1 F n F n 1 F n+1 1 2F n F n 1.. F n F n 1 F n F n+1 F n F n F n F n 1 F n+1 F n F n 1 F n+1 1 F n F n 1 k k, where F n (n 2) is the nth Fibonacci number. In order to derive the determinant of the matrix A, we introduce the real sequences S k and T k such that and S 1 = 1, S 2 = 1+a 2, S k = S k 1 + a 2 S k 2,k 3, T 1 = 1, T 2 = 1+b 2, T k = T k 1 + b 2 T k 2,k 3, where a = 1 2 (F n+1 + F n 2 ) and b = 1 2 (F n+1 F n 2 ). Then, we have det A k = S k T k,k 3. (2.1) In order to derive S k,k =1, 2,..., we define the k k tridiagonal matrix of the form: 0 ia ia 0 ia M k =. ia 0.., with i = ia ia 0 Note that S k = det (I + M k ). Here I is the k k identity matrix. We know that the determinant of a square matrix can be found by taking the product of its eigenvalues. Therefore, we will compute the spectrum of M k in order to find

4 422 H. Civciv an alternate formulation for M k. It is not diffucult to see that λ j =1+μ j, j =1, 2,..., k,where λ j and μ j,, 2,..., k, are the eigenvalues of I + M k and M k, respectively. Therefore, S k = (1 + μ j ),n 1. (2.2) In order to determine the μ j s, we know that each μ j is a zero of the characteristic polynomial p k (μ) = M k μi. Note that M k μi =(ia) k G k, where G k = μ 1 ia 1 μ 1 ia 1 μ... ia μ ia. Therefore, since ia 0 the determinant M k μi = 0 if and only if G k =0. In [2], it was obtained that G k = 0 if and only if μ j = 2ia cos Combining (2.2) and (2.3), we get S k =,j =1, 2,..., k. (2.3) ( ) 1 2ia cos,k 1. (2.4) Similarly, we obtain T k = ( ) 1 2ib cos,k 1. (2.5)

5 Matrices with Fibonacci numbers 423 Taking (2.1), (2.4) and (2.5) into account we compute det A k = 1 i [(F n+1 + F n 2 )+(F n+1 F n 2 )] cos = = = [(F n+1 + F n 2 ). (F n+1 F n 2 )] cos 2 1 i [(F n 1 + L n 1 )+(L n F n )] cos [(F n 1 + L n 1 )(L n F n )] cos 2 Consequently, we have det A k = 1 2i (F n F 1 n ) cos ) 1 2i (F n +( 1) n+1 F n 1 cos +4F nf 1 n cos 2 +4( 1)n F n F n 1 cos 2 1 2iFn+2 cos k+1 4F nf n 1 cos 2 k+1, n odd 1 2iFn 2 cos +4F. k+1 nf n 1 cos 2 k+1, n even By choosing values for the entries of A k we can obtain matrices where the sequence det A k,k 1 follows a pattern of recurrence found in subsequences of the Fibonacci sequence. Here, we also present yet another problem where the Fibonacci sequence surprisingly appears. Problem 1 References k k = Fk+1 2,k 3. [1] M. Bicknell-Johnson and C. Spears, Classes of identities for the generalized Fibonacci numbers G n = G n 1 + G n c from matrices with constant valued determinants, Fibonacci Quart. 34 (1996),

6 424 H. Civciv [2] N. Cahill and D. Narayan, Fibonacci and Lucas numbers as tridiagonal matrix determinants, Fibonacci Quart. 42 (2004), [3] H. Kwong, Two determinants with Fibonacci and Lucas entries, Appl. Math. and Comp., (2007), doi: /j.amc [4] Br. J. Mahon, Elementary Problem B-1016, Fibonacci Quart. (2006), 44; 182. [5] T. Koshy, Fibonacci and Lucas Numbers with Applications, Wiley, [6] G. Strang, Introduction to linear algebra, 2nd edition, Welleslay MA, Wellesley-Cambridge, [7] G. Strang and K. Borre, Linear algebra, Geodesy and GPS, Welleslay MA, Wellesley-Cambridge, [8] W. A. Webb and E. A. Parberry, Divisibility properties of Fibonacci polynomials, The Fibonacci Quart. (1969), 7.5, Received: October 9, 2007