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Başlık: The eigenvectors of a combinatorial matrixYazar(lar):AKKUS, İlkerCilt: 60 Sayı: 1 Sayfa: 009-014 DOI: 10.1501/Commua1_0000000665 Yayın Tarihi: 2011 PDF

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Vo lu m e 6 0 , N u m b e r 1 , P a g e s 9 — 1 4 ( 2 0 1 1 ) IS S N 1 3 0 3 — 5 9 9 1

THE EIGENVECTORS OF A COMBINATORIAL MATRIX

ILKER AKKUS

A. In this paper, we derive the eigenvectors of a combinatorial matrix whose eigenvalues studied by Kilic and Stanica. We follow the method of Cooper and Melham since they considered the special case of this matrix.

1. I  

In [7], Peele and St˘anic˘a studied n×n matrices with the (i, j) entry the binomial coefficientj−1i−1, respectively,n−ji−1and derived many interesting results on powers of these matrices. In [8], one of them found that the same is true for a much larger class of what he called netted matrices, namely matrices with entries satisfying a certain type of recurrence among the entries of all 2 × 2 cells.

Let Rn be the matrix whose (i, j) entries are ai,j=

i−1

n−j



, which satisfy

ai,j−1= ai−1,j−1+ ai−1,j. (1.1)

The previous recurrence can be extended for i ≥ 0, j ≥ 0, using the boundary conditions a1,n = 1, a1,j = 0, j = n. Remark the following consequences of the

boundary conditions and recurrence (1.1): ai,j = 0 for i + j ≤ n, and ai,n+1 =

0, 1 ≤ i ≤ n.

The matrix Rn was firstly studied by Carlitz [2] who gave explicit forms for the eigenvalues of Rn. Let fn+1(x) = det (xI − Rn) be the characteristic polynomial

of Rn. Thus fn(x) = n+1  r=0 (−1)r(r+1)/2  n r  F xn−r wherenr

F denote the Fibonomial coefficient, defined (for n≥ r > 0) by

 n r  F = F1F2. . . Fn (F1F2. . . Fr) (F1F2. . . Fn−r) Received by the editors Feb. 15, 2011, Accepted: March 24, 2011. 2000 Mathematics Subject Classification. 11B39, 15A15, 15A18.

Key words and phrases. Fibonomial coefficients, binomial coefficients, eigenvector, Pascal matrix.

c

2011 A nkara U niversity

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withnn

F =

n

0



F = 1. Carlitz showed that

fn(x) = n−1 j=0  x− φj¯φn−j

where φ, ¯φ =1 ±√5/2. Thus the eigenvalues of Rnare φn, φn−1¯φ, . . . , φ¯φn−1, ¯φn.

In [7] it was proved that the entries of the power Re

n satisfy the recurrence

Fe−1a(e)i,j = Fea(e)i−1,j+ Fe+1a(e)i−1,j−1− Fea(e)i,j−1, (1.2)

where Fe is the Fibonacci sequence. Closed forms for all entries of Re

n were not

found, but several results concerning the generating functions of rows and columns were obtained (see [7, 8]). Further, the generating function for the (i, j)-th entry of the e-th power of a generalization of Rn, namely

Qn(a, b) =  ai+j−n−1bn−j  i− 1 n− j  1≤i,j≤n is

B(e)n (x, y) = (Ue−1+ Uey)(b Ue−1+ y Ue)

n−1

Ue−1+ Uey− x(Ue+ Ue+1y)

.

Regarding this generalization, in [3], the authors gave the characteristic polyno-mial of Qn(a, b) and the trace of kth power of Qn(a, b, ), that is, tr

 Qk

n(a, b)

 , by using the method of Carlitz [2]:

trQkn(a, b)= Ukn

Uk , (1.3)

whereni

U stands for the generalized Fibonomial coefficient, defined by

n r  U = U1U2. . . Un (U1U2. . . Ur) (U1U2. . . Un−r) , for n≥ i > 0, wherenn U = n 0  U= 1.

In [7], the authors proposed a conjecture on the eigenvalues of matrix Rn, which was proven independently in [1] and the unpublished manuscript [9]. Also, in [6], they found the eigenvectors of Rn.

In [7, 8], it was shown that the inverse of Rn is the matrix Rn−1=  (−1)n+i+j+1  n− i j− 1  1≤i,j≤n , and, in general, the inverse of Qn(a, b) is

Q−1n (a, b) = 

(−1)n+i+j+1an+1−i−jbi−n

 n− i j− 1  1≤i,j≤n . (1.4) Let φ = 1+ √ 5 2 , ¯φ = 1− √ 5

2 be the golden section and its conjugate. The

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Let the sequences{un}, {vn} be defined by un = aun−1+ bun−2

vn = avn−1+ bvn−2,

for n > 1, where u0 = 0, u1= 1, and v0 = 2, v1 = a, respectively. Let α, β be the

roots of the associated equation x2− ax − b = 0. The next lemma can be found in [4].

Lemma 1.1. For k≥ 1 and n > 1,

ukn = vkuk(n−1)+ (−1)k+1bkuk(n−2) (1.5)

vkn = vkvk(n−1)+ (−1)k+1bkvk(n−2).

In [5], using the sequence vk, they defined the n×n matrix Hnvk, bkas follows:

Hnvk, bk=  vki+j−n−1  − (−b)k n−j  i− 1 n− j  1≤i,j≤n . As in equation (1.4), they also found the inverse of the matrix Hn, namely

Hn−1(vk, bk) =  (−1)j+1(−b)−k(n−i)vn+1−i−j  n− i j− 1  i,j . It is well known that for n≥ −1,

un+1= r  r n− r  a2r−nbn−r. (1.6)

Thus the authors [5] generalized this identity as well as they gave the following results:

Lemma 1.2. For k > 0 and n ≥ −1,

uk(n+1) uk =  r  r n− r  v2r−nk  − (−b)k n−r.

Lemma 1.3. For all m > 0,

trHnmvk, bk= uknm uk . Theorem 1.4. The eigenvalues of Hnvk, bkare

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2. T      Hnvk, bk

In [6], the authors considered the matrix Qn(a, b) and gave its eigenvectors. In this section, we consider the generalization of matrix Qn(a, b) namely Hn

 vk, bk

and then determine its eigenvectors by using the method given in [6]. Let 0 ≤ p ≤ n − 1 be a fixed integer,

f (x) = (x − αk)p(x − βk)n−1−p= n−1 r=0 srxr, and s = (s0, s1, ..., sn−1)T. Theorem 2.1. For m≥ 0 f(m)(x) = m! f (x) (x − αk)m(x − βk)m m  j=0  p m− j  n− 1 − p j  (x − αk)j(x − βk)m−j. Proof. It can be proved with the use of Leibniz’s formula for the m-th derivative of a product of two functions. We recall the Leibniz’s formula: For m≥ 0

dm dxmg(x)h(x) = m  j=0  m j  g(m−j)(x)h(j)(x). (2.1)

We use the notation xn to denote the falling factorial, and hence

f(m)(x) = m  j=0  m j  pm−j(x − αk)p−m+j(n − 1 − p)j(x − βk)n−1−p−j = m! f (x) (x − αk)m(x − βk)m m  j=0  p m− j  n− 1 − p j  (x − αk)j(x − βk)m−j, as claimed. 

Lemma 2.2. Suppose that 0 ≤ m ≤ n − 1 be a fixed integer. Then,

sn−1−m= m  j=0 (−1)m  p m− j  n− 1 − p j  αk(m−j)βkj and  Hnvk, bks n−1−m= n−1  r=m (−1)m(−b)km  r m  vkr−msr.

Proof. The proof of the first equation can be followed from computing the coefficient

of xn−1−m of f (x) by multiplying x− αkp times x− βk n−1−p. The second

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Theorem 2.3. Hnvk, bks = (αk)n−1−pβkps. (2.2) Proof. Consider  Hnvk, bks n−1−m = n−1  r=m (−1)m(−b)km  r m  vr−mk sr = (−1) m(−b)km m! n−1  r=m srrmvkr−m = (−1) m(−b)km m! f (m)(v k) = (−1) m(−b)km m! (vk− αk)p(vk− βk)n−1−p (vk− αk)m(vk− βk)m × m! m  j=0  p m− j  n− 1 − p j  (vk− αk)j(vk− βk)m−j = (−1)m(αβ)kmβk(p−m)αk(n−1−p−m)× m  j=0  p m− j  n− 1 − p j  βkjαk(m−j) = αk(n−1−p)βkp m  j=0 (−1)m  p m− j  n− 1 − p j  αk(m−j)βkj = αk(n−1−p)βkpsn−1−m.

Thus the proof is complete. 

ÖZET: Bu çalı¸smada, Kilic ve Stanica tarafından özde˘gerleri ver-ilen bir kombinatoryal matrisin özvektörleri, Cooper ve Melham’ın metodu takip edilerek elde edilmi¸stir.

R    

[1] D. Callan and H. Prodinger, An involutory matrix of eigenvectors, Fibonacci Quart. 41(2) (2003), 105—107.

[2] L. Carlitz, The characteristic polynomial of a certain matrix of binomial coefficients, Fibonacci Quart. 3 (1965), 81—89.

[3] C. Cooper and R. Kennedy, Proof of a result by Jarden by generalizing a proof by Carlitz, Fibonacci Quart. 33(4) (1995), 304—310.

[4] E. Kilic and P. St˘anic˘a, Factorizations of binary polynomial recurrences by matrix methods, to be appear in Rocky Mountain J. Math.

[5] E. Kilic, G.N. St˘anic˘a, P. St˘anic˘a, Spectral properties of some combinatorial matrices, 13th International Conference on Fibonacci Numbers and Their Applications, 2008.

[6] R.S. Melham and C. Cooper, The eigenvectors of a certain matrix of binomial coefficients, Fibonacci Quart. 38 (2000), 123-126.

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[7] R. Peele and P. St˘anic˘a, Matrix powers of column-justified Pascal triangles and Fibonacci sequences, Fibonacci Quart. 40 (2002), 146—152.

[8] P. St˘anic˘a, Netted matrices, Int. J. Math. Math. Sci. 39 (2003), 2507—2518.

[9] P. St˘anic˘a and R. Peele, Spectral properties of a binomial matrix, unpublished manuscript, http://arxiv.org/abs/math/0011111, 2000.

Current address : Ilker Akkus :Ankara University, Faculty of Sciences, Dept. of Mathematics, Ankara, TURKEY

E-mail address : iakkus@science.ankara.edu.tr URL: http://communications.science.ankara.edu.tr

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