On the generating matrices of the k-fibonacci numbers

Transcription

1 Proyecciones Journal of Mathematics Vol. 3, N o 4, pp , December 013. Universidad Católica del Norte Antofagasta - Chile On the generating matrices of the k-fibonacci numbers Sergio Falcon Universidad de las Palmas, España Received : February 013. Accepted : September 013 Abstract In this paper we define some tridiagonal matrices depending of a parameter from which we will find the k-fibonacci numbers. And from the cofactor matrix of one of these matrices we will prove some formulas for the k-fibonacci numbers differently to the traditional form. Finally, we will study the eigenvalues of these tridiagonal matrices. Keyword : k-fibonacci numbers, Cofactor matrix, Eigenvalues. AMS : 11B39, 34L16.

2 348 Sergio Falcon 1. Introduction The generalization of the Fibonacci sequence has been treated by some authors as e.g. Hoggat V.E. [6] and Horadam A.F. [7]. One of these generalizations has been found by Falcon S. and Plaza A. to study the method of triangulation 4TLE [1] and that we define below. We define the k-fibonacci numbers [1,, 3] by mean of the recurrence relation F k,n+1 = kf k,n + F k,n 1 for n 1 with the initial conditions F k,0 =0and F k,1 =1. The recurrence equation of this formula is r k r 1=0whose solutions are σ 1 = k+ k +4 and σ = k k +4. The Binet formula for these numbers is F k,n = σn 1 σn σ 1 σ From the definition of the k-fibonacci numbers, the first of them are presented in Table??. First k-fibonacci numbers F k,0 =0 F k,1 =1 F k, = k F k,3 = k +1 F k,4 = k 3 +k F k,5 = k 4 +3k +1 F k,6 = k 5 +4k 3 +3k For k = 1, the classical Fibonacci sequence {0, 1, 1,, 3, 5, 8,...} is obtained and for k = it is the Pell sequence {0, 1,, 5, 1, 9,...}. Tridiagonal matrices and the k-fibonacci numbers In this section we extend the matrices definedin[4]andappliedthemto the k-fibonacci numbers in order to prove some formulas differently to the traditional form.

3 On the generating matrices of the k-fibonacci numbers The determinant of a special kind of tridiagonal matrices Let us consider the n-by-n tridiagonal matrices M n, a b c d e c d e M n = c d e ċ d e c d Solving the sequence of determinants, we find M 1 = a M = d M 1 bċ M 3 = d M cė M 1 M 4 = d M 3 cė M... And, in general, (.1) M n+1 = d M n cė M n 1.. Some tridiagonal matrices and the k Fibonacci numbers If a = d = k, b = e =1,andc = 1, the matrices M n are transformed in the tridiagonal matrices H n (k) = k 1 1 k 1 1 k k 1 1 k In this case, and taking into account Table??, the above formulas are transformed in H 1 (k) = k = F k, H (k) = k k 1( 1) = k +1=F k,3 H 3 (k) = k(k +1) ( 1)1k = k 3 +k = F k,4 and Formula (.1) is H n (k) = F k,n+1 for n 1.

4 350 Sergio Falcon The k-fibonacci numbers can also be obtained from the symmetric tridiagonal matrices k i i k i i k i Hn(k) 0 = i k i i k where i is the imaginary unit, i.e. i = 1. If a = k +1,b = c = e =1,d = k +, the tridiagonal matrices obtained are k k k + 1 O n (k) = k k + In this case, it is O n (k) = F k,n+1 for n 1. So, with O 0 (k) = F k,1 = 1, the sequence of these determinants is the sequence of odd k Fibonacci numbers {1,k +1,k 4 +3k +1}. Finally, if a = k, b =0,c =1,d = k +, ande =1,forn 1weobtain the even k Fibonacci numbers from the determinant of the matrices E n (k) = k 0 1 k k k k + because E n (k) = F k,n for n 1with E 0 (k) = F k,0 =0. 3. Cofactor matrices of the generating matrices of the k- Fibonacci numbers The following definitions are well-known: [9] If A is a square matrix, then the minor of its entry a ij,alsoknownasthe (i, j) minor of A, is denoted by M ij and is defined to be the determinant of the submatrix obtained by removing from A its i th row and j th column. It follows C ij =( 1) i+j M ij and C ij called the cofactor of a ij, also referred to as the (i, j) cofactor of A.

5 On the generating matrices of the k-fibonacci numbers 351 Define the cofactor matrix of A, asthen n matrix C whose (i, j) entryis the (i, j) cofactor of A. Finally, the inverse matrix of A is A 1 = 1 A CT,where A is the determinant of the matrix A (assuming non zero) and C T is the transpose of the cofactor matrix C or adjugate matrix of A. On the other hand, let us consider the n-by-n nonsingular tridiagonal matrix (3.1) T = a 1 b 1 c 1 a b. c bn 1 c n 1 a n In [8], Usmani gave an elegant and concise formula for the inverse of the tridiagonal matrix T 1 =(t i,j ): (3.) t i,j = ( 1) i+j 1 θ n b i b j 1 θ i 1 φ j+1 if i j ( 1) i+j 1 θ n c j c i 1 θ j 1 φ i+1 if i > j where θ i verify the recurrence relation θ i = a i θ i 1 b i 1 c i 1 θ i for i =,...n with the initial conditions θ 0 =1andθ 1 = a 1. Formula (.1) is one special case of this one. φ i verify the recurrence relation φ i = a i φ i+1 b i c i φ i+ for i = n 1,...1 with the initial conditions φ n+1 =1andφ n = a n Observe that θ n = det(t ) Cofactor matrix of H n (k) For the matrix H n (k), it is a i = k, b i =1,c i = 1, θ i = F k,i+1 and φ j = F k,n j+. Consequently, ( 1) i+j 1 F k,i F k,n j+1 if i j F ((H n (k)) 1 k,n+1 ) i,j = 1 F k,j F k,n i+1 if i > j F k,n+1

6 35 Sergio Falcon We will work with the cofactor matrix whose entries are ( ( 1) i+j F k,j F k,n i+1 if i j c i,j (H n (k)) = F k,i F k,n j+1 if i < j So, c j,i (H n (k)) = ( 1) i+j c i,j (H n (k)) if i>j. In this form, the cofactor matrix of H n (k) forn is C n 1 (k) = F k,n F k,n 1 F k,n F k,n 3 F k, F k,1 F k,n 1 F k, F k,n 1 F k, F k,n F k, F k,n 3 F k, F k, F k, F k,n F k, F k,n F k,3 F k,n F k,3 F k,n 3 F k,3 F k, F k,3 F k,n 3 F k, F k,n 3 F k,3 F k,n 3 F k,4 F k,n 3 F k,4 F k, F k, F k, F k, F k, F k,3 F k, F k,4 F k, F k,n 1 F k, F k,n 1 F k,1 F k, F k,3 F k,4 F k,n 1 F k,n On the other hand, taking into account the inverse matrix A 1 = 1 A Adj(A), it is easy to prove Adj(A) = A n 1. So, C n 1 (k) = F n 1 k,n+1. In this form, for n =, 3, 4,...,itis C 1 (k) = C (k) = F k, F k,1 F k,1 F = Fk,3 F k, k, + Fk,1 = F k,3 F k,3 F k, F k,1 F k, F k, F k, F k, F k,1 F k, F k,3 = F k,4 µ Fk,(F k,3 +F k,3 +1)=Fk,4 Fk,3 F k,1 k (F k,3 + F k,1) = F k,4 Fk,3 Fk,1 = kf k,4 F k,4 F k,3 F k, F k,1 C 3 (k) = F k,3 F k, F k,3 F k, F k, F k, F k, F k, F k, F k,3 F k, F k,3 F k,1 F k, F k,3 F = F k,5 3 k,4 (F k, + F k,3)f k,5 = F 3 k,5 F k, + F k,3 = F k,5 C 4 (k) = F 4 k,6 F k,3(f k, + F k,4 ) F k,6 = F 4 k,6 µ Fk,4 F k, k (F k,4 + F k,) = F k,6 F k,4 F k, = kf k,6 Generalizing these results, and taking into account F k,n = F k,n+1 F k,n 1 k,we find the following two formulas for the k-fibonacci numbers according to that n is odd or even []: Fk,n+1 + Fk,n = F k,n+1 and Fk,n+1 Fk,n 1 = k F k,n

7 On the generating matrices of the k-fibonacci numbers Cofactor matrix of O n (k) To apply formula (3.) to the matrices O n (k), we must take into account that a 1 = k +1 a i = k +,i b i = c i =1,i 1 θ i = F k,i 1,i 1 φ j = 1 k F k,(n j+), j 1 and consequently the cofactor of the (i, j) entry of these matrices is c i,j (O n (k)) = ( 1) i+j 1 k F k,j 1F k,(n i+1) if i j c j,i (O n (k)) = c i,j (O n (k)) for j>i 3.3. Cofactor matrix of E n (k) For the matrices E n (k) itis a 1 = k = F k, a i = k +,i b 1 = 0 b i+1 = c i =1,i 1 θ i = F k,(i+1), i 1 φ j = 1 k F k,(n j+), j 1 and consequently the cofactor of the (i, j) entry of these matrices is c 1,j (E n (k)) = ( 1) j+1 1 k F k,(n j+1) c i,j (E n (k)) = ( 1) i+j 1 k F k,jf k,(n i+1), if i j, i > 1 c j,i (E n (k)) = c i,j (E n (k)), if j > i > 1 4. Eigenvalues This section is dedicated to the study of the eigenvalues of the matrices H n (k), O n (k) ande n (k).

8 354 Sergio Falcon 4.1. Eigenvalues of the matrices H n (k) The matrix (3.1) has entries in the diagonals a 1,...,a n,b 1,...,b n 1,c 1,...,c n 1. It is well-known the eigenvalues of the matrix (3.1) are λ r = a + µ rπ b c cos for r =1,,...,n. n +1 Consequently, the µ eigenvalues of the matrix H n (k)wherea = k, b =1,c= 1, rπ are λ r = k +i cos n +1 If n is odd, then the matrix H n (k) has one unique real eigenvalue corresponding to r = n+1. If n is even, no one eigenvalue is real. So, the sequence of spectra of the tridiagonal matrices H n (k) forn =1,,... is Σ 1 = {k} Σ = {k ± i} n Σ 3 = k, k ± i o ( Σ 4 = k ± 1+ 5 i, k ± 1 ) 5 i = {k ± φi, k ± ( φ) 1 i} n Σ 5 = k, k ± i, k ± i o 3... where φ = 1+ 5 is the Golden Ratio. All the roots λ r lie on the segment <(λ j )=k, < =(λ j ) <. It is verified nx λ j = nk and ny λ j = F k,n+1. Moreover, taking into account the product of all eigenvalues is the determinant of the matrix H n (k) andas H n (k) = F k,n+1,itisverified that F k,n+1 = ny µ µ πj k +i cos n Eigenvalues of the matrices O n (k) Matrices O n (k) are symmetric, so all its eigenvalues are real.

9 On the generating matrices of the k-fibonacci numbers Theorem If λ i is an eigenvalue of the matrix O n (k) forafixed value k, thenλ i +k +1is eigenvalue of the matrix O n (k +1). Proof. If λ i is an eigenvalue of the matrix O n (k), then it is k +1 λ i 1 1 k O n (k) λ i I n = + λ i 1 1 k + λ i 1 = (k +1) +1 (λ i +k +1) 1 1 (k +1) + (λ i +k +1) 1 1 (k +1) + (λ i +k +1) O n (k +1) (λ i +k +1)I n Consequently, only it is necessary to find the eigenvalues of the matrix O n (1) for n =, 3,... and then, if λ j is an eigenvalue of O n (1), then (4.1) λ 0 j = λ j + k 1 is an eigenvalue of the matrix O n (k). Now we will study the spectra of these matrices. First, we present the spectra of these matrices for k = 1 and n =, 3, 4, 5, 6, 7, 8 obtained with the help of MATHEMATICA: Σ = { , } Σ 3 = { , , } Σ 4 = { , , , } Σ 5 = { , , , , } Σ 6 = { , ,.90790, , , } Σ 7 = { , , , , , , } Σ 8 = { , , ,.45674, , , , } Evidently, = (4.) X λj =3n 1 and (4.3) Y λj = F n+1 Below are the minimum and maximum eigenvalues of these spectra: n = n =3 n =4 n =5 n =6 n =7 n =8 minima Maxima In the first case, we can see this sequence is decreasing and converge to 1, and consequently, lim n min λ(o n(k)) = k.

10 356 Sergio Falcon Inthesameform,thesequenceofMaxima is increasing and converge to 5, so we can say lim n Maxλ(O n(k)) = k +4. Finally, if k 6= 1, then, taking into account Formula (4.1), the formulas (4.) and (4.3) are transformed into nx λ j (k) = X (λ i (1) + k 1) = nk +n 1 ny λ j (k) = F k,n Eigenvalues of the matrices E n (k) Finally, we say a matrix is positive if all the entries are real and nonnegative. If a matrix is tridiagonal and positive, then all the eigenvalues are real [5]. So, taking into account matrix E n (k) is tridiagonal and positive, all its eigenvalues are real. Following the same process that for the matrices O n (k), we can prove that the first eigenvalue is k and the others verify λ i (k) =λ i (1) + k 1. nx ny Moreover, λ j (k) =(n 1)(k +)+k and λ j (k) =F k,n The sequence of spectra of the matrices E n (1) is Σ = {1, 3} Σ 3 = {1,, 4} Σ 4 = {1, , 3, } Σ 5 = {1, , , , } Σ 6 = {1, ,, 3, 4, } σ 7 = {1, , , , , , } The sequence of minima eigenvalue converges to 1 (to k in general ) and the sequienceofmaximaconvergesto5(tok + 4 in the general case). Acknowledgements. This work has been supported in part by CICYT Project number MTM C03-0/MTM from Ministerio de Educación y Ciencia of Spain. References [1] Falcon S. and Plaza A., On the Fibonacci k-numbers, Chaos, Solit. & Fract. 3 (5), pp , (007).

11 On the generating matrices of the k-fibonacci numbers 357 [] Falcon S. and Plaza A., The k-fibonacci sequence and the Pascal -triangle, Chaos, Solit. & Fract. 33 (1), pp , (007). [3] Falcon S. and Plaza A., The k-fibonacci hyperbolic functions, Chaos, Solit. & Fract. 38 (), pp , (008). [4] Feng A., Fibonacci identities via determinant of tridiagonal matrix, Applied Mathematics and Computation, 17, pp , (011). [5] Horn R. A. and Johnson C. R., Matrix Analysis, p. 506, Cambridge University Press (1991) [6] Hoggat V. E. Fibonacci and Lucas numbers, Houghton Miffin, (1969). [7] Horadam A. F. A generalized Fibonacci sequence, Mathematics Magazine, 68, pp , (1961). [8] Usmani R., Inversion of a tridiagonal Jacobi matrix, Linear Algebra Appl. 1/13, pp , (1994). [9] Wikipedia, matrix Sergio Falcon University of Las Palmas de Gran Canaria, Department of Mathematics, Las Palmas de Gran Canaria, España sfalcon@dma.ulpgc.es