The -Fibonacci Matrix and the Pascal Matrix

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 F kF F ; n ≥ ; k;n+1 = k;n + k;n−1 1 (1) F F with the initial conditions k;0 = 0, k;1 = 1. Following [1, 6, 8, 9, 11], we define the n × n, k-Fibonacci matrix as the unipotent lower triangular Toeplitz matrix F k f f F i ≥ j n( ) = [ i;j ]i;j=1;:::;n defined with entries i;j = k;i−j+1 if , 0 otherwise. That is,   1  F   k;2 1   F F   k;3 k;2 1  Fn k  F F F  : ( ) =  k;4 k;3 k;2 1     . . . . .   ......  F F F F ::: k;n k;n−1 k;n−2 k;n−3 1

∗ E-mail: [email protected]

1403 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, T and the eigenvalues of the symmetric Fibonacci matrix FnFn . It is not diﬃcult to prove that the inverse matrix of the k F−1 f0 -Fibonacci matrix introduced above, is given by the lower triangular matrix n = [ i;j ]i;j=1;:::;n where  1 if j = i;   0 −k if j = i − 1; fi;j = − j i − ;  1 if = 2  0 otherwise.

That is,   1  −k   1   − −k  −  1 1  F 1 k   : n ( ) =  0 −1 −k 1     . . . . .   ......  0 0 0 −1 −k 1

− −1 − Let Fn 1 be the matrix obtained from Fn by deleting the first row [10]. The negative matrix of the matrix Fn 1 is [3] the matrix   k −1  k −   1 1    H k; n  0 1 k −1  ; ( ) =    . . . . .   ......  0 0 0 1 k −1 ∞ k {Fk;n} and the sequence of principal minors is precisely the -Fibonacci sequence n≥2. i−1 n ×n Pn pi;j i;j ;:::;n pi;j i ≥ j On the other hand, the well-known Pascal matrix = [ ] =1 is defined by = j−1 if , 0 otherwise. 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 n−1 −1 0 0 i+j i−1 Pn pi;j i;j ;:::;n pi;j − i ≥ j The inverse of the Pascal matrix is = [ ] =1 with = ( 1) j−1 if , 0 otherwise, see [1]. In this paper, following [7, 12], we will show two factorizations of the Pascal matrix involving the k-Fibonacci matrix. Let us define the L k l matrix n( ) = [ i;j ]i;j=1;:::;n by i − 1 i − 2 i − 3 li;j = − k − : (2) j − 1 j − 1 j − 1

For instance,   1  − k   1 1  L k  −k − k  : 5( ) =  2 1     −k 2 − 2k 3 − k 1  −k 2 − 3k 5 − 3k 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 j ≥ l l l above in the preceding row; that is, for 1, i;j = i−1;j−1 + i−1;j , which is the same rule as used in the Pascal matrix.

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2. First factorization of the Pascal matrix

Theorem 2.1. For every n ∈ N, the matrix Pn may be written in the form Pn = FnLn.

−1 Proof. It is enough to prove that Fn Pn = Ln. Since the left-hand matrices are lower triangular, the product is a −1 lower triangular matrix as well. In addition, since the elements of the main diagonal of Pn and Fn 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 Pn 0 li;j f ph;j relation (1), = h=1 i;h occurs and the proof is complete.

Corollary 2.2. For every n ≥ 3 and n ≥ j the following holds:

n−j X j r − j r − j r − n − + 1 − k + 2 − + 3 F 1 : j − j − j − k;n−j+1−r = j − r=0 1 1 1 1

Proof. Note that the entry pn;j in matrix Pn = FnLn is of the form

n−j X p F l ; n;j = k;n−j+1−r j+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 { ; − k; −k; −k; : : :} p F F − k − k F F ··· F 1 1 , n;1 = 1, we have 1 = k;n + k;n−1(1 ) ( k;n−2 + k;n−3 + + k;1) and so we arrive at the next result.

Corollary 2.3.

n X F 1 F F − : k;r = k ( k;n + k;n+1 1) r=0

This corollary appears as [5, Proposition 8].

Lemma 2.4. For n ≥ , 2 n X n − n − n − 2 − k 2 − 2 F : j − j − j k;j−1 = 1 (3) j=2 2 1

Proof. By a suitable change of indices, the proposed identity is equivalent to

n n X n X n n F k F : j k;j+1 = 1 + j + j k;j+1 j=0 j=0 + 1 + 2

This equation is trivial for n = 0 and n = 1 because 1 F 1 F k; LHS = k;1 + k;2 = 1 + 0 1 k 1 1 F k: RHS = 1 + + k;1 = 1 + 1 2

1405 The k-Fibonacci matrix and the Pascal matrix

Let us suppose n > 1. Then

n− n− n X1 n n X1 n F F F kF F kF F LHS = k;1 + j k;j+1 + n k;n+1 = 1 + j ( k;j + k;j−1) + k;n + k;n−1 0 j=1 j=1 n n n− X n X n X1 n n k F F k F : = 1 + j k;j + j k;j−1 = 1 + j + j k;j+1 = RHS j=1 j=1 j=0 + 1 + 2

Lemma 2.5. For n ≥ , 3 n X n − n − n − n − 2 − k 2 − 2 F n − k2 3 k : j − j − j k;j = ( 2) + + 1 (4) j=3 2 1 2

1 1 Proof. an;j Fk;j n − k − By induction. Let us denote as the coeﬃcient of in (4). For = 3, (LHS)3 = 1 2 1 Fk; Fk; k2 m ≤ n 3 3 = 3 = + 1 = (RHS)3 and the relation holds. Let us suppose the relation is true for any . Since m m−1 m−1 i = i + i−1 , then

n n n X+1 X X+1 a F a F a F (LHS)n+1 = n+1;j k;j = n;j k;j + n;j−1 k;j j=3 j=3 j=3 n n X X a F a kF F = (RHS)n + n;j k;j+1 = (RHS)n + n;j k;j + k;j−1 j=2 j=2 n n − n − X − k n − − ( 2)( 3) kF k a F = (RHS)n + 1 ( 2) k;2 + (RHS)n + n;j k;j−1 2 j=2 n − n − n − k − k n − − ( 2)( 3) k2 n − k2 2 k = (1 + )(RHS)n + 1 ( 2) + 1 = ( 1) + + 1 = (RHS)n+1 2 2

having used the formula (3).

Theorem 2.6. For every n ∈ , L−1 l0 with N n = [ i;j ]i;j=1;:::;n

i X i − l0 − i+r 1 F : i;j = ( 1) r − k;r−j+1 r=j 1

−1 −1 Proof. It is enough to use the fact that Pn Fn = Ln .

3. 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 R k r deﬁne the matrix n( ) = [ i;j ]i;j=1;:::;n by

i − 1 i − 1 i − 1 ri;j = − k − : j − 1 j j + 1

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For instance,   1  − k   1 1   − k − k  R k  2 2 1  : 6( ) =    −2 − 3k 2 − 3k 3 − k 1     −5 − 4k −6k 5 − 4k 4 − k 1  −9 − 5k −5 − 10k 5 − 10k 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 = RnFn.

−1 Proof. It is enough prove that PnFn = Rn.

Corollary 3.2. For every n ≥ j, we have

n X n − n − n − n − 1 − k 1 − 1 F 1 : r − r r k;r−j+1 = j − r=j 1 + 1 1

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

Corollary 3.3.

n X n − n − n − 1 − k 1 − 1 F : r − r r k;j = 1 r=1 1 + 1

Theorem 3.4. The inverse matrix of R is R−1 r0 with entries n n = [ i;j ]i;j=1;2;:::;n

i X i j − r − r0 − i+r + 1 F : i;j = ( 1) j − k;r−j+1 r=j 1

−1 −1 Proof. It is enough to use the fact that FnPn = Rn .

4. 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. T Pc Sn sr;l r;l ;:::;n Sn FnF Sn sr;l Fk;r−j Fk;l−j Let = [ ] =1 be the product matrix = n . Then, the entries of this matrix are = j=0 , 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

l X sr;l = Fk;r−j Fk;l−j : (5) j=0

1407 The k-Fibonacci matrix and the Pascal matrix

This formula also can be expressed as l X s F F : r;l = k;j k;r−l+j (6) j=0 k s s F The entries of the ﬁrst line (row or column) of this matrix are the -Fibonacci numbers, r;1 = 1;r = k;r, while the s s F P R F entries of the second line are r;2 = 2;r = k;r+1. Moreover, taking into account that n = n n, we can deduce −1 T −1 T Sn = Rn PnFn or Sn = PnLn Fn .

Theorem 4.1. Elements sr;l of the matrix Sn fulﬁl the relations

s s ; 1;0 = 0;1 = 0 (7) s 1 F F ; r;r = k k;r k;r+1 (8) s k s s ; r < l; r;l = r;l−1 + r;l−2 (9) s k s s ; r > l: r;l = r−1;l + r−2;l (10)

Pr Proof. r l Sn sr;r F 2 1 Fk;rFk;r Equation (7): For = , the entries of main diagonal of are obtained and = j=1 k;r = k +1 by applying a formula found in [2] for the sum of the squares of k-Fibonacci numbers. Equation (9): Let l = r + h. Then,

r r X X s s F F F k F F r;l = r;r+h = k;r−j k;r+h−j = k;r−j k;r+h−j−1 + k;r+h−j−2 j=0 j=0 r r X X k F F F F k s s k s s : = k;r−j k;r+h−j−1 + k;r−j k;r+h−j−2 = r;r+h−1 + r;r+h−2 = r;l−1 + r;l−2 j=0 j=0

Finally, taking into account that matrix Sn is symmetric the formula (10) can be obtained.

|F | |S | n k Q Given that n = 1, we have n = 1. For = 10 and = 1, the matrix 10 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 k s 1 F F s s are -Fibonacci numbers, the entries of the main diagonal are r;l = k k;r k;r+1 and r;l = l;r. In the following theorem, Pm Sn Fk;lFk;r− m−l we will give a very simple expression for these last entries of the matrix , which have the form l=1 ( ).

Theorem 4.2. For h ≥ , we have 0 n X − n F F 1 F F − ( 1) + 1 F : k;j k;j+h = k k;n+1 k;n+h k;h (11) j=0 2

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

X1 F F F F F ; 1 F F 1 k F F : (LHS) = k;j k;j+h = k;1 k;1+h = k;1+h (RHS) = k k;2 k;1+h = k k;1+h = k;1+h j=0

Let us suppose this formula is true until n − 1, that is

n− X1 − n−1 F F 1 F F − ( 1) + 1 F : k;j k;j+h = k k;n k;n−1+h k;h j=0 2

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Then

n n− X X1 − n−1 F F F F F F 1 F F − ( 1) + 1 F F F k;j k;j+h = k;j k;j+h + k;n k;n+h = k k;n k;n−1+h k;h + k;n k;n+h j=0 j=0 2 − n−1 1 F F − ( 1) + 1 F = k;n k;n+1+h k;h k 2 − n − n 1 F F − nF − ( 1) + 1 F 1 F F − ( 1) + 1 F : = k;n k;n+1+h + ( 1) k;h k;h = k;n+h k;n+1 k;h k 2 k 2 after applying to the penultimate expression the D’Ocagne identity [4, 5]:

F F − F F − nF : k;m k;n+1 k;m+1 k;n = ( 1) k;m−n Theorem 4.3 (main result). For r ≥ l,

 1  Fk;rFk;l if l is odd;  k +1 sr;l = (12)  1 F F − F if l is even. k k;r k;l+1 k;r−l

Proof. 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):

l X − l s F F 1 F F − ( 1) + 1 F : r;l = k;j k;j+r−l = k k;r k;l+1 k;r−l j=0 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  1 l  Fk;rFk;l if l is odd, X  k +1 Fk;r−j Fk;l−j =  j=0  1 F F − F if l is even. k k;r k;l+1 k;r−l

These formulas have been proved in [2].

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

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

1409 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.

References

[1] Call G.S., Velleman D.J., Pascal’s matrices, Amer. Math. Monthly, 1993, 100(4), 372–376 [2] Falcón S., On sequences of products of two k-Fibonacci numbers, Int. J. Contemp. Math. Sci. (in press) [3] Falcón S., Some tridiagonal matrices and the k-Fibonacci numbers. Appl. Math. Comput. (in press) [4] Falcón S., Plaza Á., On the Fibonacci k-numbers, Chaos Solitons Fractals, 2007, 32(5), 1615–1624 [5] Falcón S., Plaza Á., The k-Fibonacci sequence and the Pascal 2-triangle, Chaos Solitons Fractals, 2007, 33(1), 38–49 [6] Lee G.-Y., Kim J.-S., The linear algebra of the k-Fibonacci matrix, Linear Algebra Appl., 2003, 373, 75–87 [7] Lee G.-Y., Kim J.-S., Cho S.-H., Some combinatorial identities via Fibonacci numbers, Discrete Appl. Math., 2003, 130(3), 527–534 [8] Lee G.-Y., Kim J.-S., Lee S.-G., Factorizations and eigenvalues of Fibonacci and symmetric Fibonacci matrices, Fibonacci Quart., 2002, 40(3), 203–211 [9] Lee G.-Y., Lee S.-G., Shin H.-G., On the k-generalized Fibonacci matrix Qk , Linear Algebra Appl., 1997, 251, 73–88 [10] Peart P., Woodson L., Triple factorization of some Riordan matrices, Fibonacci Quart., 1993, 31(2), 121–128 [11] Zhang Z., The linear algebra of the generalized Pascal matrix, Linear Algebra Appl., 1997, 250, 51–60 [12] Zhang Z., Wang X., A factorization of the symmetric Pascal matrix involving the Fibonacci matrix, Discrete Appl. Math., 2007, 155(17), 2371–2376

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