Selçuk J. Appl. Math. Selçuk Journal of Vol. 11. No.1. pp. 107-116 , 2010 Applied Mathematics
On The Norms of Circulant Matrices with Generalized Fibonacci Numbers
Ay¸se Nalli, Murat ¸Sen
Department of Mathematics, Faculty of Sciences, Selcuk University, Campus, Konya 42075, Türkiye
e-mail: aysenalli@ gm ail.com
Received Date: May 11, 2009 Accepted Date: February 16, 2010
Abstract. In this paper, we obtain a generalization of [6]. We …rst construct the so-called circulant matrix with the generalized Fibonacci numbers and then present lower and upper bounds for the Euclidean and spectral norms of this matrix.
Key words: Circulant matrix; generalized Fibonacci number; matrix norm. 2000 Mathematics Subject Classi…cation: 15A99, 15A60, 11B39.
1. Introduction
A Toeplitz matrix is an n n matrix Tn = [tk;j ; k; j = 0; 1; :::; n 1], where
tk;j = tk j. This structure is rather interesting for the rich theoretical properties
and applications.
For any given b0; b1; :::; bn 12 C, the circulant matrix B = (bi;j)n n is de…ned
by bi;j= bj i(mod n), that is,
B = 2 6 6 6 6 6 4 b0 b1 b2 bn 1 bn 1 b0 b1 bn 2 bn 2 bn 1 b0 bn 3 .. . ... ... . .. ... b1 b2 b3 b0 3 7 7 7 7 7 5
A circulant matrix is special Toeplitz matrix where each row vector is rotated one element to the right relative to the preceding row vector. In numerical analysis circulant matrices are important because they can be quickly solved using the discrete Fourier transform.
Solak and Bozkurt [5] de…ned almost circulant matrix as follows : Dn=
a; i = j
1
k(j i) k(mod n); otherwise
where a 2 R f0gand k = 1; 2; :::; n 1 and then established upper bound for the lp matrix and lp operator norms of the matrix Dn .
The Fibonacci numbers are integer sequence fFng1n=0 de…ned by the linear
recurrence equation, with F0= 0, F1= 1 and
(1) Fn+2= Fn+1+ Fn :
The sequence is given by
N 0 1 2 3 4 5 6 7 : : :
Fn 0 1 1 2 3 5 8 13 : : :
The Lucas numbers are integer sequence fLng1n=0 de…ned by the linear
recur-rence equation, with L0= 2, L1= 1, and
(2) Ln+2= Ln+1+ Ln :
The sequence is given by
N 0 1 2 3 4 5 6 7 : : :
Ln 2 1 3 4 7 11 18 29 : : :
The generalized Fibonacci numbers are integer sequence fGng1n=0 de…ned by
the linear recurrence equation, with G0= a, G1= b (a; b 2 R) and
(3) Gn+2= Gn+1+ Gn :
The sequence is given by
n 0 1 2 3 4 5 6 7 : : :
Gn a b (a + b) (a + 2b) (2a + 3b) (3a + 5b) (5a + 8b) (8a + 13b) : : :
(3) is generalized of (1) and (2) Furthermore there is relationship
between Fibonacci and generalized Fibonacci numbers.
Fibonacci, Lucas numbers and their generalization have many interesting prop-erties and applications to almost every …eld of science and art. For the beauty and rich applications of these numbers and their relatives one may see Vajda’s book [9] and Koshy’s book [3].
Solak [6] de…ned the circulant matrix with Fibonacci number as
(5) A = Fmod(j i;n)
n i;j=1
where Fn is the n th Fibonacci number, the circulant matrix with the Lucas
number as
(6) B = Lmod(j i;n)
n i;j=1
where Ln is the n th Lucas number and then showed lower and upper bounds
with the Fibonacci number for the Euclidean and spectral norms of the matrices A and B .
Civciv and Türkmen [1] constructed the circulant matrix with the Lucas number and then presented lower and upper bounds for the Euclidean and spectral norms of this matrix as a function of n and Ln is n th Lucas number. In [1]
is studied the norm bounds for the Hadamard inverse of this matrix.
In this study, we …rst construct the so-called circulant matrix with the gen-eralized Fibonacci numbers and then present lower and upper bounds for the Euclidean and spectral norms of this matrix.
We begin with some background necessary to understand this study. For more comprehensive treatments on matrices, we refer readers to [2].
Let A be any n n matrix. The Euclidean norm of the matrix A is de…ned as
kAkE= 0 @ n X i=1 n X j=1 jaijj2 1 A 1 2
The spectral norm of the matrix A is
kAk2=
r max
1 i n i(A H:A)
where i is an eigenvalue of AH A and AHis conjugate transpose of the matrix
(7) p1
nkAkE kAk2 kAkE
De…ne the maximum column lenght norm c1(:) and the maximum row lenght
norm r1(:) of any matrix A respectively by
c1(A) = max j v u u t n X i=1 jaijj2 and r1(A) = max i v u u t n X j=1 jaijj2
Let A = [aij] and B = [bij] be matrices of the same size, not necessarily square.
Then their Hadamard product ( also called Schur product ) A B is de…ned by entrywise multiplication :
A B = [aijbij] :
If k:k is any norm on m n matrices, then [10]
(8) kA Bk kAk kBk :
Let A; B and C be m n matrices. If A = B C then [4]
(9) kAk2 r1(B) c1(C) :
Let A; B be m n matrices. Kronecker product of this matrices is de…ned as
A B = 2 6 6 6 6 6 6 4 a11B a12B ::: a1nB a21B a22B ::: a2nB : : : : : : : : : : : : am1B am2B ::: amnB 3 7 7 7 7 7 7 5
2. Main Result De…nition 1. Let
v = (G0;G2; :::; Gn 1)
be a row vector in Cn and de…ne the shift operator : Cn! Cn by
(G0;G1; G2; :::; Gn 1) = (Gn 1;G0; :::; Gn 2)
where Gi , i = 1; :::; n 1 with generalized Fibonacci number. The circulant
matrix A with generalized Fibonacci number is the matrix whose kth row is given by k 1v , k = 1; 2; :::; n i.e., a matrix of the form
(10) A = 2 6 6 6 6 6 4 G0 G1 G2 : : : Gn 1 Gn 1 G0 G1 : : : Gn 2 Gn 2 Gn 1 G0 : : : Gn 3 .. . ... ... . .. ... G1 G2 G3 : : : G0 3 7 7 7 7 7 5
Theorem 1. Let A be the matrix de…ned in (10). Then
kAkE = q n(Gn 1Gn G0G1+ G20) and q (Gn 1Gn G0G1+G20) kAk2 q (Gn 1Gn G0G1+G02) (Gn 1Gn G0G1+1)
where k:k2is the spectral norm and Gnis the nth generalized Fibonacci number.
Proof: Let matrices B and C be de…ned as
B = (bij) = bbij= G(mod(j i;n)); i j ij = 1; i < j
C = (cij) = cij = G(mod(j i;n)); i < j
cij = 1; i j
such that A = B C . Then
(11) r1(B) = max i v u u t n X j=1 jbijj2= v u u tn 1X s=0 G2 s= q Gn 1Gn G0G1+ G20 and (12) c1(C) = max J v u u t n X i=1 jcijj2= v u u t1 +n 1X s=0 G2 s G20= p Gn 1Gn G0G1+ 1 where n 1P s=0 G2 s= Gn 1Gn G0G1+ G20 :
From (9), (11) and (12), we obtain an upper bound for the spectral norm : kAk2
q
(Gn 1Gn G0G1+ G02) (Gn 1Gn G0G1+ 1)
For the Euclidean norm of the matrix A , we have
(13) kAkE= v u u tnn 1X s=0 G2 s= q n(Gn 1Gn G0G1+ G20) :
From (7) and (13), we get immediately q
Gn 1Gn G0G1+ G20 kAk2 :
This completes the proof.
Corollary 1. Let G0= a = 0 , G1= b = 1 be in Theorem 1 . Then
kAkE= p nFn 1Fn and p Fn 1Fn kAk2 p (Fn 1Fn)(1 + Fn 1Fn) :
Corollary 2. Let G0= a = 2 , G1= b = 1 be in Theorem 1 . Then
kAkE= p n(Ln 1Ln+ 2) and p Ln 1Ln+ 2 kAk2 p (Ln 1Ln 1)(Ln 1Ln+ 2) :
Theorem 2. Let A be the matrix de…ned in (10). Then (14) kAk2 8 < : q a2(1 + F n 2Fn 1) + 2ab(F2n 1) + b 2 (Fn 1Fn) ; n odd q a2(1 + F n 2Fn 1) + 2ab(F2n 1 1) + b 2 (Fn 1Fn); n even and (15) kAk2 8 > > > > > > < > > > > > > : 8 < : q [a2(1 + Fn 2Fn 1) + 2ab(F2n 1) + b 2 (Fn 1Fn)] q [a2(Fn 2Fn 1) + 2ab(F2n 1) + b2(Fn 1Fn) + 1] ; n odd 8 < : q [a2(1 + Fn 2Fn 1) + 2ab(F2n 1 1) + b 2 (Fn 1Fn)] q [a2(Fn 2Fn 1) + 2ab(F2n 1 1) + b2(Fn 1Fn) + 1] ; n even
where k:k2 is the spectral norm and F denote Fibonacci number.
Proof: From the Euclidean norm and (4), we have
kAk2E= n n 1 X s=0 G2s= n n 1 X s=0 (aFs 1+ bFs)2 kAk2E = n a 2 n 1X s=0 Fs 12 + 2ab n 1X s=0 Fs 1Fs+ b2 n 1 X s=0 Fs2 ! kAk2E= n a2(1 + F n 2Fn 1) + 2ab(Fn 12 ) + b2(Fn 1Fn) ; n odd a2(1 + F n 2Fn 1) + 2ab(Fn 12 1) + b2(Fn 1Fn) ; n even where (16) n 1X s=0 Fs 12 = Fn 2Fn 1+ 1; n 1 X s=0 Fs 1Fs= F 2 n 1 ; n odd Fn 12 1 ; n even ; n 1 X s=0 Fs2= Fn 1Fn: Hence from (7), we …nd kAk2 8 < : q a2(1 + F n 2Fn 1) + 2ab(Fn 12 ) + b2(Fn 1Fn) ; n odd q a2(1 + F n 2Fn 1) + 2ab(Fn 12 1) + b2(Fn 1Fn) ; n even :
On the other hand, let matrices B and C be de…ned as B = (bij) = bij= G(mod(j i;n)); i j bij = 1; i < j and C = (cij) = cij = G(mod(j i;n)); i < j cij = 1; i j
such that A = B C .Then
r1(B) = max i v u u t n X j=1 jbijj2= v u u tn 1X s=0 G2 s= v u u tn 1X s=0 (aFs 1+ bFs)2 r1(B) = v u u ta2 n 1 X s=0 F2 s 1+ 2ab n 1X s=0 Fs 1Fs+ b2 n 1X s=0 F2 s and c1(C) = max j v u u tXn i=1 jcijj2= v u u t1 +n 1X s=1 G2 s= v u u tn 1X s=0 (aFs 1+ bFs)2 G20+ 1 c1(C) = v u u ta2 n 1 X s=0 F2 s 1+ 2ab n 1 X s=0 Fs 1Fs+ b2 n 1X s=0 F2 s a2+ 1 : From (16), we have r1(B) = 8 < : q a2(1 + F n 2Fn 1) + 2ab(Fn 12 ) + b2(Fn 1Fn) ; n odd q a2(1 + F n 2Fn 1) + 2ab(Fn 12 1) + b2(Fn 1Fn) ; n even and c1(C) = 8 < : q a2(F n 2Fn 1) + 2ab(Fn 12 ) + b2(Fn 1Fn) + 1 ; n odd q a2(F n 2Fn 1) + 2ab(Fn 12 1) + b2(Fn 1Fn) + 1 ; n even : Thus from (9), kAk2 8 > > > > > > < > > > > > > : 8 < : q [a2(1 + F n 2Fn 1) + 2ab(F 2 n 1) + b 2 (Fn 1Fn)] q [a2(F n 2Fn 1) + 2ab(F2n 1) + b 2 (Fn 1Fn) + 1] ; n odd 8 < : q [a2(1 + F n 2Fn 1) + 2ab(F 2 n 1 1) + b 2 (Fn 1Fn)] q [a2(F n 2Fn 1) + 2ab(F2n 1 1) + b 2 (Fn 1Fn) + 1] ; n even :
Corollary 3. Let G0= a = 2 , G1= b = 1 be in Theorem 2 . Then kAk2 8 < : q Fn 1Fn+ 4(Fn 12 + Fn 2Fn 1+ 1) ; n odd q Fn 1Fn+ 4(Fn 12 + Fn 2Fn 1) ; n even and kAk2 8 > > > > > > < > > > > > > : 8 < : q [Fn 1Fn+4(F2n 1+ Fn 2Fn 1+ 1)] q [Fn 1Fn+ 4(F2n 1+ Fn 2Fn 1) + 1] ; n odd 8 < : q [Fn 1Fn+4(F2n 1+ Fn 2Fn 1)] q [Fn 1Fn+ 4(F2n 1+ Fn 2Fn 1) 3] ; n even
Corollary 4. For the spectral norm of Hadamard product of matrices A in Theorem 1 and A in Theorem 2, we have
kA A k2 u 8 > > > > > > < > > > > > > : 8 < : q [a2(1 + Fn 2Fn 1) + 2ab(F2n 1) + b 2 (Fn 1Fn)] q [a2(Fn 2Fn 1) + 2ab(F2n 1) + b2(Fn 1Fn) + 1] ; n odd 8 < : q [a2(1 + Fn 2Fn 1) + 2ab(F2n 1 1) + b 2 (Fn 1Fn)] q [a2(Fn 2Fn 1) + 2ab(F2n 1 1) + b2(Fn 1Fn) + 1] ; n even where u =np(Gn 1Gn G0G1+ G20)(Gn 1Gn G0G1+ 1) o : Proof: From (8), the proof is trivial fromTheorems 1 and 2.
Corollary 5. For the Euclidean norm of Kronecker product of matrices A in Theorem 1 and A in Theorem 2, we get
kA A k = n 8 > > > > > > < > > > > > > : 8 < : q [Gn 1Gn ab + a2] q [a2(1 + Fn 2Fn 1) + 2ab(F2n 1) + b 2 (Fn 1Fn)] ; n odd 8 < : q [Gn 1Gn ab + a2] q [a2(1 + Fn 2Fn 1) + 2ab(F2n 1 1) + b 2 (Fn 1Fn)] ; n even
Proof: Since kA A kE = kAkEkA kE , the proof is trivial fromTheorems 1 and 2.
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