Fibonacci matrix and the Pascal matrix - Springer Link

Cent. Eur. J. Math. • 9(6) • 2011 • 1403-1410 DOI: 10.2478/s11533-011-0089-9

Central European Journal of Mathematics

The k-Fibonacci matrix and the Pascal matrix Research Article Sergio Falcon1∗

1 Department of Mathematics and Institute for Applied Microelectronics (IUMA), University of Las Palmas de Gran Canaria, 35017-Las Palmas de Gran Canaria, Spain

Received 21 March 2011; accepted 8 August 2011 Abstract: We define the k-Fibonacci matrix as an extension of the classical Fibonacci matrix and relationed with the k-Fibonacci numbers. Then we give two factorizations of the Pascal matrix involving the k-Fibonacci matrix and two new matrices, L and R. As a consequence we find some combinatorial formulas involving the k-Fibonacci numbers. MSC:

15A36, 11C20, 11B39

Keywords: Pascal matrix • k-Fibonacci numbers • Factorization of a matrix

© Versita Sp. z o.o.

1.

Introduction

Many generalizations of the Fibonacci sequence have been introduced and studied [6, 9]. Here we use the k-Fibonacci numbers defined as follows [4, 5]: for any positive real number k, the k-Fibonacci sequence, say {Fk,n }n∈N , is defined recurrently by Fk,n+1 = kFk,n + Fk,n−1 , n ≥ 1, (1) with the initial conditions Fk,0 = 0, Fk,1 = 1. Following [1, 6, 8, 9, 11], we define the n × n, k-Fibonacci matrix as the unipotent lower triangular Toeplitz matrix Fn (k) = [fi,j ]i,j=1,...,n defined with entries fi,j = Fk,i−j+1 if i ≥ j, 0 otherwise. That is,      Fn (k) =     

∗

 1 Fk,2 Fk,3 Fk,4 .. . Fk,n

1 Fk,2 Fk,3 .. . Fk,n−1

1 Fk,2 .. . Fk,n−2

1 .. . Fk,n−3

..

. ... 1

    .    

E-mail: [email protected]

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The k-Fibonacci matrix and the Pascal matrix

From now on, we will designate the matrix Fn (k) as Fn . Lee et al. [8] discussed the factorizations of the Fibonacci matrix corresponding to the classical Fibonacci sequence, and the eigenvalues of the symmetric Fibonacci matrix Fn FnT . It is not difficult to prove that the inverse matrix of the 0 k-Fibonacci matrix introduced above, is given by the lower triangular matrix Fn−1 = [fi,j ]i,j=1,...,n where   1   −k 0 fi,j = −1     0 That is,

if j = i, if j = i − 1, if j = i − 2, otherwise.

     −1 Fn (k) =     

 1 −k −1 0 .. . 0

Let Fn−1 be the matrix obtained from Fn−1 by deleting matrix  k 1   0 H(k, n) =   .  .  . 0

1 −k −1 .. . 0

1 −k .. . 0

1 .. .. . . −1 −k

    .     1

the first row [10]. The negative matrix of the matrix Fn−1 is [3] the −1 k −1 1 k −1 .. .. .. . . . 0 0 1



.. k

   ,   

. −1

and the sequence of principal minors is precisely the k-Fibonacci sequence {Fk,n }∞ n≥2 . On the other hand, the well-known n × n Pascal matrix Pn = [pi,j ]i,j=1,...,n is defined by pi,j = That is,   1  1  1    1  2 1    . Pn =  1 3 3 1     ..  .. .. .. ..  .  . . . .




n−1 n−1 n−1 n−1 n−1 0

1

2

3

i−1 j−1




if i ≥ j, 0 otherwise.

n−1


0 0 i+j i−1 if i ≥ j, 0 otherwise, see [1]. In this paper, The inverse of the Pascal matrix is P−1 n = [pi,j ]i,j=1,...,n with pi,j = (−1) j−1 following [7, 12], we will show two factorizations of the Pascal matrix involving the k-Fibonacci matrix. Let us define the matrix Ln (k) = [li,j ]i,j=1,...,n by       i−1 i−2 i−3 li,j = −k − . (2) j −1 j −1 j −1 For instance,



    L5 (k) =   

1 1−k −k −k −k

1 2−k 2 − 2k 2 − 3k

1 3−k 5 − 3k

   .  

1 4−k 1

We will designate the matrix Ln (k) as Ln . In this note, we use the k-Fibonacci sequence to extend the results of [7, 12]. Note that if k = 1 in (2) then li,j is the matrix defined in [12]. The recursive formula for matrix Ln is that each entry is the sum of the elements to the left and above in the preceding row; that is, for j ≥ 1, li,j = li−1,j−1 + li−1,j , which is the same rule as used in the Pascal matrix.

1404

S. Falcon

2.

First factorization of the Pascal matrix

Theorem 2.1. For every n ∈ N, the matrix Pn may be written in the form Pn = Fn Ln . It is enough to prove that Fn−1 Pn = Ln . Since the left-hand matrices are lower triangular, the product is a lower triangular matrix as well. In addition, since the elements of the main diagonal of Pn and Fn−1 are 1, their product has also 1 there. That is, li,j = 1 if j = i, 0 if j > i. Moreover, for i ≥ 2, and by relation (2) and the recurrence P 0 relation (1), li,j = nh=1 fi,h ph,j occurs and the proof is complete.

Proof.

Corollary 2.2. For every n ≥ 3 and n ≥ j the following holds:  n−j  X j +r−1 r=0

Proof.

j −1

 −k

     j +r−2 j +r−3 n−1 − Fk,n−j+1−r = . j −1 j −1 j −1

Note that the entry pn,j in matrix Pn = Fn Ln is of the form pn,j =

n−j X

Fk,n−j+1−r lj+r,j ,

r=0

from which the relation follows. In particular, for j = 1, taking into account that the terms of the first column of matrix Ln are of the form {1, 1 − k, −k, −k, . . .}, pn,1 = 1, we have 1 = Fk,n + Fk,n−1 (1 − k) − k (Fk,n−2 + Fk,n−3 + · · · + Fk,1 ) and so we arrive at the next result.

Corollary 2.3. n X

Fk,r =

r=0

1 (Fk,n + Fk,n+1 − 1). k

This corollary appears as [5, Proposition 8].

Lemma 2.4. For n ≥ 2,

 n  X n−2 j −2

j=2

Proof.

 −k

   n−2 n−2 − Fk,j−1 = 1. j −1 j

(3)

By a suitable change of indices, the proposed identity is equivalent to n   X n j=0

j

Fk,j+1

   n   X n n = 1+ k + Fk,j+1 . j +1 j +2 j=0

This equation is trivial for n = 0 and n = 1 because     1 1 Fk,1 + Fk,2 = 1 + k, 0 1      1 1 RHS = 1 + k + Fk,1 = 1 + k. 1 2 LHS =

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The k-Fibonacci matrix and the Pascal matrix

Let us suppose n > 1. Then

LHS =

    n−1   n−1   X X n n n n Fk,1 + Fk,j+1 + Fk,n+1 = 1 + (kFk,j + Fk,j−1 ) + kFk,n + Fk,n−1 0 j n j j=1 j=1

=1+

     n n   n−1   X X X n n n n k Fk,j + Fk,j−1 = 1 + k + Fk,j+1 = RHS. j j j +1 j +2 j=1 j=0 j=1

Lemma 2.5. For n ≥ 3,

 n  X n−2 j −2

j=3

 −k

     n−2 n−2 n−3 − Fk,j = (n − 2) k 2 + k +1 . j −1 j 2

(4)



By induction. Let us denote as an,j the coefficient of Fk,j in (4). For n = 3, (LHS)3 = 11 − k 12 − 1 F = Fk,3 = k 2 + 1 = (RHS)3 and the relation holds. Let us suppose the relation is true for any m ≤ n. Since 3
k,3

m = m−1 + m−1 , then i i i−1

Proof.


n+1 X

(LHS)n+1 =

an+1,j Fk,j =

j=3

n X

an,j Fk,j +

n+1 X

j=3 n X

= (RHS)n +

an,j−1 Fk,j

j=3

an,j Fk,j+1 = (RHS)n +

j=2

n X

an,j kFk,j + Fk,j−1




j=2

 n X (n − 2)(n − 3) an,j Fk,j−1 kFk,2 + k (RHS)n + = (RHS)n + 1 − k (n − 2) − 2 j=2     (n − 2)(n − 3) n−2 = (1 + k)(RHS)n + 1 − k (n − 2) − k 2 + 1 = (n − 1) k 2 + k + 1 = (RHS)n+1 2 2 

having used the formula (3).

Theorem 2.6. 0 For every n ∈ N, L−1 n = [li,j ]i,j=1,...,n with

l0i,j =

i X

(−1)i+r

r=j

Proof.

3.



 i−1 Fk,r−j+1 . r−1

−1 It is enough to use the fact that P−1 n Fn = Ln .

Second factorization of the Pascal matrix

Now, following [7, 12], we will show another factorization of the Pascal matrix involving the k-Fibonacci matrix. Let us define the matrix Rn (k) = [ri,j ]i,j=1,...,n by  ri,j =

1406

     i−1 i−1 i−1 −k − . j −1 j j +1

S. Falcon

For instance,

     R6 (k) =    

 1 1−k −2k −2 − 3k −5 − 4k −9 − 5k

1 2−k 2 − 3k −6k −5 − 10k

1 3−k 5 − 4k 5 − 10k

1 4−k 1 9 − 5k 5 − k 1

    .   

From now on, we will write Rn for the matrix Rn (k). The following theorem holds:

Theorem 3.1. For every n ∈ N, Pn = Rn Fn .

Proof.

It is enough prove that Pn Fn−1 = Rn .

Corollary 3.2. For every n ≥ j, we have  n  X n−1 r=j

r−1



     n−1 n−1 n−1 −k − Fk,r−j+1 = . r r+1 j −1

In particular, for j = 1, we have the following corollary which coincides with Lemma 2.4.

Corollary 3.3.  n  X n−1 r=1

r−1

 −k

   n−1 n−1 − Fk,j = 1. r r+1

Theorem 3.4. 0 The inverse matrix of Rn is R−1 n = [ri,j ]i,j=1,2,...,n with entries

0 ri,j =

i X

(−1)i+r



r=j

Proof.

4.

 i+j −r−1 Fk,r−j+1 . j −1

−1 It is enough to use the fact that Fn P−1 n = Rn .

Product of the k-Fibonacci matrix and its transpose

In this section, we will study the matrix obtained by means of the product of the k-Fibonacci matrix and its transpose, its relation with the preceding matrices Fn , Pn , Ln , and Rn , and some of its properties. P Let Sn = [sr,l ]r,l=1,...,n be the product matrix Sn = Fn FnT . Then, the entries of this matrix Sn are sr,l = cj=0 Fk,r−j Fk,l−j , where c = min (r, l). Taking into account that the product of a matrix and its transpose is a symmetric matrix, we have sr,l = sl,r , and we can suppose r ≥ l. Consequently, r ≥ j, and so

sr,l =

l X

Fk,r−j Fk,l−j .

(5)

j=0

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The k-Fibonacci matrix and the Pascal matrix

This formula also can be expressed as sr,l =

l X

Fk,j Fk,r−l+j .

(6)

j=0

The entries of the first line (row or column) of this matrix are the k-Fibonacci numbers, sr,1 = s1,r = Fk,r , while the entries of the second line are sr,2 = s2,r = Fk,r+1 . Moreover, taking into account that Pn = Rn Fn , we can deduce T −1 T Sn = R−1 n Pn Fn or Sn = Pn Ln Fn .

Theorem 4.1. Elements sr,l of the matrix Sn fulfil the relations s1,0 = s0,1 = 0,

(7)

1 = Fk,r Fk,r+1 , k = k sr,l−1 + sr,l−2 ,

r < l,

(9)

sr,l = k sr−1,l + sr−2,l ,

r > l.

(10)

sr,r sr,l

(8)

Equation (7): For r = l, the entries of main diagonal of Sn are obtained and sr,r = applying a formula found in [2] for the sum of the squares of k-Fibonacci numbers.

Proof.

Pr

j=1

2 Fk,r = k1 Fk,r Fk,r+1 by

Equation (9): Let l = r + h. Then, sr,l = sr,r+h =

r X

Fk,r−j Fk,r+h−j =

j=0

=k

r X

r X

Fk,r−j k Fk,r+h−j−1 + Fk,r+h−j−2




j=0

Fk,r−j Fk,r+h−j−1 +

j=0

r X

Fk,r−j Fk,r+h−j−2 = k sr,r+h−1 + sr,r+h−2 = k sr,l−1 + sr,l−2 .

j=0

Finally, taking into account that matrix Sn is symmetric the formula (10) can be obtained. Given that |Fn | = 1, we have |Sn | = 1. For n = 10 and k = 1, the matrix Q10 defined in [8] is found. In the sequel, we will indicate the general expression of terms of the matrix Sn . We have seen the first two lines of Sn are k-Fibonacci numbers, the entries of the main diagonal are sr,l = k1 Fk,r Fk,r+1 and sr,l = sl,r . In the following theorem, P we will give a very simple expression for these last entries of the matrix Sn , which have the form m l=1 Fk,l Fk,r−(m−l) .

Theorem 4.2. For h ≥ 0, we have

n X

 Fk,j Fk,j+h =

j=0

Proof.

1 k

Fk,n+1 Fk,n+h −

 (−1)n + 1 Fk,h . 2

We will prove this theorem by induction. For n = 1, in this formula is 1 X

(LHS) =

Fk,j Fk,j+h = Fk,1 Fk,1+h = Fk,1+h ,

(RHS) =

j=0

1 1 Fk,2 Fk,1+h = k Fk,1+h = Fk,1+h . k k

Let us suppose this formula is true until n − 1, that is n−1 X j=0

1408

 Fk,j Fk,j+h =

1 k

Fk,n Fk,n−1+h −

 (−1)n−1 + 1 Fk,h . 2

(11)

S. Falcon

Then n X

Fk,j Fk,j+h =

j=0

n−1 X

 Fk,j Fk,j+h + Fk,n Fk,n+h =

j=0

1 k

Fk,n Fk,n−1+h −

(−1)n−1 + 1 Fk,h 2

 + Fk,n Fk,n+h

  (−1)n−1 + 1 1 Fk,n Fk,n+1+h − Fk,h k 2     (−1)n + 1 (−1)n + 1 1 1 Fk,n Fk,n+1+h + (−1)n Fk,h − Fk,h = Fk,n+h Fk,n+1 − Fk,h . = k 2 k 2 =

after applying to the penultimate expression the D’Ocagne identity [4, 5]: Fk,m Fk,n+1 − Fk,m+1 Fk,n = (−1)n Fk,m−n .

Theorem 4.3 (main result). For r ≥ l,  1   Fk,r Fk,l+1 k =
 1 F F k,r k,l+1 − Fk,r−l k

sr,l

Proof.

if l is odd, (12) if l is even.

If in (11) we change n by l and h by r − l, we obtain the entries sr,l of the matrix Sn , according to (6): sr,l =

l X

 Fk,j Fk,j+r−l =

j=0

1 k

Fk,r Fk,l+1 −

 (−1)l + 1 Fk,r−l . 2

Then (12) follows. Taking into account formulas (5) and (12), the following two corollaries can be deduced.

Corollary 4.4. If r ≥ l, then l X

Fk,r−j Fk,l−j

j=0

 1   Fk,r Fk,l+1 k =
 1 F F k,r k,l+1 − Fk,r−l k

if l is odd, if l is even.

These formulas have been proved in [2].

Corollary 4.5. The inverse matrix of Sn is the product (Fn−1 )T Fn−1 , so, 2 + k2 0 −1  0 2 + k2 0   −1 0 2 + k2   .. . .. .. = . .   0 0 0   0 0 0 0 0 0 

S−1 n

··· 0 0 ··· 0 0 ··· 0 0 .. .. .. . . . · · · 2 + k2 0 ··· 0 1 + k2 · · · −1 −k

 0 0   0   ..  . .   −1  −k  1

For k = 1, the matrix Q−1 10 already defined in [8] is found. The decomposition of the defined positive symmetric matrix S by means of the product of a lower triangular matrix Fn and its transpose matrix is a Cholesky decomposition, so it is unique.

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The k-Fibonacci matrix and the Pascal matrix

Acknowledgements The author sincerely thanks the anonymous referee for valuable comments and suggestions, which significantly improved the quality of this paper. This work has been supported in part by CICYT Project number MTM2008-05866-C03-02/MTM from Ministerio de Educación y Ciencia of Spain.

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