Selçuk J. Appl. Math. Selçuk Journal of Vol. 12. No. 1. pp. 71-76, 2011 Applied Mathematics
On The Spectral Norms of Hankel Matrices with Fibonacci and Lucas Numbers
Süleyman Solak1, Mustafa Bahsi2
1Selcuk University Education Faculty 42090 Meram, Konya, Turkiye
e-mail:ssolak@ selcuk.edu.tr
2Cumra Imam Hatip Lisesi, Cumra Konya, Turkiye
e-mail:m hvbahsi@ yaho o.com
Received Date: March 2, 2010 Accepted Date: November 30, 2010
Abstract. This paper is concerned with the work of the authors’ [M.Akbulak and D. Bozkurt, on the norms of Hankel matrices involving Fibonacci and Lucas numbers, Selçuk J. Appl. Math. 9 (2), (2008), 45-52] on the spectral norms of the matrices = [+]=0−1 and = [+]=0−1 where and denote
the Fibonacci and Lucas numbers, respectively[1]. Akbulak and Bozkurt have found the inequalities for bounds of spectral norms of matrices and . As for us, we have found the equalities for the spectral norms of matrices and . Key words: Spectral norm; Hankel matrix; Fibonacci number; Lucas number. 2000 Mathematics Subject Classification: 15A60,15A15,15B05,11B39.
1. Introduction and Preliminaries
The matrix = []=0−1 , where = + is called the Hankel matrix. In
section 2, we calculate the spectral norms of Hankel matrices
(1) = [+]=0−1
and
(2) = [+]=0−1
where and denote th Fibonacci and Lucas numbers, respectively.
Now we start with some preliminariesLet be any x matrix. The spectral norm of the matrix is defined as kk2=
q max
1≤≤(
conjugate transpose of matrix . For a square matrix , the square roots of the eigenvalues of are called the singular values of . Genarally, we denote
the singular values as =
np
is eigenvalue of matrix
o
Moreover, the spectral norm of matrix is the maximum singular value of matrix . If is symetric matrix, then the spectral norm of matrix is maximum eigenvalue of matrix The equation det( − )=0 is known as the characteristic equation of and the left-hand side known as the characteristic poynomial of . The solutions of characteristic equation are known as the eigenvalues of matrix.
Fibonacci and Lucas numbers are the numbers in the following sequences, re-spectively:
0 1 1 2 3 5 8 and 2 1 3 4 7 11 18 in addition, these numbers are defined backwards by
0 1 −1 2 −3 5 −8 and 2 −1 3 −4 7 −11 18 .
The sequence of Fibonacci numbers is defined by recurrence relation
=−1+ −2 with initial values 0= 0 and 1= 1The sequence of Lucas
numbers is defined by recurrence relation =−1+ −2 with initial values
0= 2 and 1= 1
2. Main Results
Theorem 1. Let the matrix A be as in (1). Then the spectral norm of matrix A is kk2= ⎧ ⎪ ⎨ ⎪ ⎩ 2−1−1+√2−12 −22−1+42+1 2 even 2−1−1+√2−12 −22−1+42−3 2 odd
Proof. Since matrix A is symetric, the spectral norm of A is maximum eigen-value of matrix A. For this, we must find the roots of the characteristic polyno-mial (3) det( − ) = ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ − 0 −1 −−1 −1 − 2 − .. . ... . .. ... −−2 −−1 −2−3 −−1 − − 2−2 ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯
Then there are two position. Position I. If n is even:
If we apply the row and column operations to (3), then we have
det( − ) = ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ − 0 0 0 0 0 − 0 0 0 .. . ... ... ... . .. ... ... ... 0 0 0 0 − 0 0 0 0 0 − −2 −−1 0 0 0 0 0 −−1 − 2 ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ Thus | − | = −2[( − −2)( − 2) − 22−1] = −2[2− (2−1− 1) − 2] .
Hence, the roots of the equation | − | = 0 are
12= 2−1− 1 ± q 2 2−1− 22−1+ 42+ 1 2 = 0 = 3 4 Hereby kk2= 2−1− 1 + q 2 2−1− 22−1+ 42+ 1 2
Position II. If n is odd:
If we apply the row and column operations to (3), then we have
det(−) = ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ − 0 0 0 0 0 − 0 0 0 .. . ... ... ... . .. ... ... ... 0 0 0 0 − 0 0 0 0 0 − −1−3 −2+ −1−2 0 0 0 0 0 −−1−2 − 2−1 ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯
The value of the det( − ) is
| − | = −2[2− (2−1+ −1−3) + 22−1−3
−2−1−2+ 2−12−2]
Hence, the roots of the equation | − | = 0 are 12= 2−1− 1 ± q 2 2−1− 22−1+ 42− 3 2 = 0 = 3 4
Since matrix A is symetric, then
kk2= 2−1− 1 + q 2 2−1− 22−1+ 42− 3 2
Theorem 2. Let the matrix B be as in (2). Then the spectral norm of matrix B is kk2= ⎧ ⎪ ⎨ ⎪ ⎩ 2−1+1+(2−1−1)√5 2 n even 2−1+1+√5(2−1−1)2+4 2 n odd
Proof. Since matrix B is symetric, the spectral norm of B is maximum eigen-value of matrix B. For this, we must find the roots of the characteristic polyno-mial (4) det( − ) = ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ − 0 −1 −−1 −1 − 2 − .. . ... . .. ... −−2 −−1 −2−3 −−1 − − 2−2 ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯
Then there are two position. Position I. If n is even:
If we apply the row and column operations to (4), then we have
det( − ) = ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ − 0 0 0 0 0 − 0 0 0 .. . ... ... ... . .. ... ... ... 0 0 0 0 − 0 0 0 0 0 − 2−2− 1 −2−1+ 1 0 0 0 0 0 −2−1+ 1 − 2 ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯
Thus
| − | = −2[( − 2−2− 1)( − 2) − (2−1− 1)2]
= −2[2− (2−1+ 1) + 2− 2]
Hence, the roots of the equation | − | = 0 are 12= 2−1+ 1 ± (2−1− 1) √ 5 2 = 0 = 3 4 Hereby kk2= 2−1+ 1 + (2−1− 1) √ 5 2
Position II. If n is odd:
If we apply the row and column operations to (4), then we have
det( − ) = ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ − 0 0 0 0 0 − 0 0 0 .. . ... ... ... . .. ... ... ... 0 0 0 0 − 0 0 0 0 0 − 2−4− 1 −22−2+ 1 0 0 0 0 0 −2−3+ 1 − 22−1 ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯
The value of the det( − ) is
| − | = −2[2− (2−4+ 22−1+ 1) + 22−12−4+ 22−1
−22−22−3+ 22−2+ 2−3− 1]
= −2[2− (2−1+ 1) + 2− 3]
Hence, the roots of the equation | − | = 0 are
12= 2−1+ 1 ± p 5(2−1− 1)2+ 4 2 = 0 = 3 4
Since matrix A is symetric, then kk2=
2−1+ 1 +
p
5(2−1− 1)2+ 4
References
1. Akbulak, M. and Bozkurt, D. (2008): On the norms of Hankel matrices involving Fibonacci and Lucas numbers, Selçuk J. Appl. Math., 9, no.2, 45-52.
