Selçuk J. Appl. Math. Selçuk Journal of Vol. 5. No.1. pp. 11-21, 2004 Applied Mathematics
On the Bounds for the Spectral Norms of Some Special Matrices D. Bozkurt, N. Ta¸skara and H. Köse
Department of Mathematics, Art and Science Faculty of Selçuk University, Selçuklu 42079, Konya, Turkey;
e-mail: [email protected]; [email protected]; [email protected]
Received : April 8, 2004
Summary. In this paper, we have obtained lower and upper bounds for the spectral norms of Cauchy-Toeplitz and Cauchy-Henkel Matrices
Key words: Cauchy-Toeplitz Matrices, Cauchy-Henkel Matrices.
1. Introduction
Let C = [1=(xi yj)]ni;j=1 (xi 6= yj) be a Cauchy matrix and Tn = [tj i]n 1i;j=0
be a Toeplitz matrix. In general, Cauchy-Toeplitz and Cauchy-Hankel matrices are de…ned as (1) Tn= 1 g + (i j)h n 1 i;j=0 and (2) Hn= 1 g + (i + j)h n 1 i;j=0
respectively where, h 6= 0; g and h are any numbers such that g=h are not integers. Toeplitz matrices are precisely those matrices that one constant along all diagonals parallel to the main diagonal, and thus, a Toeplitz matrix are also determined by its …rst row and column and a Hankel matrix are determined by its …rst row and last column. Furthermore, a Hankel matrix is always the symmetrical matrix.
Recently, there have been several papers on the norms of Cauchy-Toeplitz matrix and Cauchy-Hankel matrix [2,3,4,5,6,8]. We have obtained the bounds for the `p norms of Cauchy-Toeplitz matrices with diferent values of g and h s
mentioned above [2,4]. In [4], we have established an upper and lower bounds for the `pnorms and given a conjecture related with the spectral norm of Hadamard
product of the matrices Hn and Tn where, g = 1=2 and h = 1:[5], [6] and [8]
are related to spectral norm of Cauchy-Toeplitz matrices. In [5], a lower bound for the spectral norm of Cauchy-Toeplitz matrix was obtained by Tyrtyshnikov taking g = 1=2 and h = 1 in (1). Parter proved that the singular values could be related to eigenvalues of certain Hermitian Toeplitz matrices corresponding to Laurent-Fourier series [6]. Türkmen and Bozkurt have established bounds for the spectral norms of Cauchy-Toeplitz and Cauchy-Hankel matrices in the form (1) and (2) by taking h = 1 and g = 1=k where, k 2 and k integer [8].
In Section 2 of this paper, we have studied on the spectral norms of the Tn
matrix that has given in (1) and obtained lower and upper bounds for this norm. In Section 3, we have established lower and upper bounds for the spectral norm of the Hn matrix that has given in (2).
Now we give some preliminaries related to our study. It is well known the Euclidean (Frobenius) norm of the matrix A is
(3) kAkF = 0 @ m X i=1 n X j=1 jaijj2 1 A 1=2
and the spectral norm of the matrix A is also
(4) kAk2=
r max
1 i n i(A HA)
where, A is m n matrix and AH is the conjugate transpose of the matrix A.
The following inequality holds:
(5) p1
nkAkF kAk2 kAkF:
The maximum column length norm c1(:) and the maximum row length norm
r1(:) of any matrix A are de…ned by
(6) c1(A) = max j sX i jaijj2 and (7) r1(A) = max i sX j jaijj2 respectively [9].
Let A = (aij) and B = (bij) be m n matrices. The Hadamard product of A
and B is de…ned by A B = (aijbij). If C = A B then
A function is called a psi (or digamma) function if (x) = d dxfln[ (x)]g where, (x) = 1 Z 0 e ttx 1dt:
The nth derivative of a psi function is called a polygamma function. i.e. (n; x) = d dxn [ (x)] = d dxn d dx(ln[ (x)]) [5]: If n = 0 then (0; x) = (x) = d
dxfln[ (x)]g: On the other hand, if a > 0 and b
is any number and n is a positive integer, then lim
n!1 (a; n + b) = 0:
2. The Norms of Cauchy-Toeplitz Matrices
Theorem 1 Let us take g = 1=k and h = 1=(k + 1) in (1) where, 2 k 2 Z+
and n 8. Then (k + 1) sin k kTnk2; kTnk2 8 > > > > > > > > < > > > > > > > > : k + k(k + 1) + (k + 1) n + 1 2 1 k + n + 1 2 + 1 k 1 1 k 2 + 1 k ; n odd; 2k3+ 6k2+ 3k 2k + 1 + (k + 1) n 2 1 k 1 1 k ; n even is valid.
Proof. By the de…nition of the matrix Tn and the equality (3), we have
kTnk2F = nk 2+ n 1 X s=1 (n s) 2 6 4 1 1 k s k+1 2 + 1 1 k+ s k+1 2 3 7 5 = nk2 +(k + 1) 3 nk(k + 1)2 k 1; n k + 1 k (k + 1) 2 n +k + 1 k (k + 1)3 nk(k + 1)2 k 1; n + k + 1 k + (k + 1) 2 1 k + 1 k +(k + 1)2 1 + k + 1 k (k + 1)3 nk(k + 1)2 k 1; 1 k + 1 k +(k + 1) 3+ nk(k + 1)2 k 1; 1 + k + 1 k (k + 1) 2 n k + 1 k
where, is a polygamma function (see [7]). If we divide both sides by n of this equality and take limit as n ! 1; then we obtain
lim n!1 1 nkTnk 2 F = nlim!1 8 > < > :k 2+ n 1X s=1 (1 s n) 2 6 4 1 1 k s k+1 2+ 1 1 k + s k+1 2 3 7 5 9 > = > ; = 2(k + 1)2 sin2 k : Thus, lim n!1 1 p nkTnkF = (k + 1) sin k : Then we have (k + 1) sin k kTnk2:
Now, let us establish the upper bounds for kTnk2: Take Tn= A B such that
A = B = 2 4q 1 1 k+ i j k+1 3 5 n n :
kTnk2 r1(A)c1(B) from (8) where, r1(A) and c1(B) are the maximum row
length norm and column length norm of the matrices A and B; respectively. r1(A) and c1(B) are ((n 1)=2)th row length norm and column length norm of
the matrices A and B for n odd. Then
r1(A) = c1(B) = v u u u tk(k + 1) 0 @1 + (n 1)=2 X s=1 1 sk 1 + (n 1)=2 X s=2 1 sk + 1 1 A + k = k(k + 1) + (k + 1) n + 1 2 1 k + n + 1 2 + 1 k 1 1 k 2 + 1 k + k 1=2 :
r1(A) and c1(B) are second row length norm and (n 1)th column length norm
of the matrices A and B for n even, repectively. Then
r1(A) = c1(B) = v u u tk(k + 1) 1 + 1 k + 1+ 1 2k + 1+ n 3X s=1 1 sk 1 ! = 2k 3+ 6k2+ 3k 2k + 1 + (k + 1) n 2 1 k 1 1 k 1=2 :
Conjecture 2 Let us take g = 1=k and h = 1=(k +1) in (1) where, 3 k 2 Z+. Then
kTnk2 k(k + 1) + 3:
Conjecture 3 Let us take g = 1=(k +1) and h = 1=k in (1) where, 2 k 2 Z+.
Then k sin k+1 kTnk2 k sin k+1 + "; where, 10 9 " 2:10 9:
3. The Norms of Cauchy-Hankel Matrices
Let mbe entries of the matrix Hn: Then we can …nd the function f (x) such
that (9) m= 1 2 Z f (x)e imxdx (m = 0; 1; 2; :::):
If we can separate f (x) = f1(x)f2(x) such that f1(x) and f2(x) are real and
complex valued functions, respectively and jf2(x)j = 1 (see [6]), then
(10) kHnk2 sup jf1(x)j :
Theorem 4 f g = 1=k and h = 1=(k + 1) in (2), then for the spectral norm of the matrix Hn; (11) kHnk2 (k + 1)2 kn 1; n + 1 + 1 k 1; 1 + 1 k +(k + 1) 2 n 2 n + 1 + 1 k 1 + 1 k 2n +1 k + (k + 1)2(2nk + 1) kn 1; n + 1 + 1 k 1; 2n +1 k 1=2 kHnk2 k sin k(k + 1) 9 > > > > > > > > > > > > > > > > > > = > > > > > > > > > > > > > > > > > > ;
Proof. We have the following equality from the de…nition of Euclidean norms: 1 nkHnk 2 F = 1 n n X s=1 2 6 6 6 4 s 1 k + s 1 k + 1 2 3 7 7 7 5+ 1 n n 1X s=1 2 6 6 6 4 n s 1 k+ n + s 1 k + 1 2 3 7 7 7 5 = (k + 1) 2 kn 1; n + 1 + 1 k 1; 1 + 1 k +(k + 1) 2 n 2 n + 1 + 1 k 1 + 1 k 2n + 1 k +(k + 1) 2(2nk + 1) kn 1; n + 1 + 1 k 1; 2n + 1 k :
Now, let us show that there exista second equality in (11). If we consider the equality in (9), we have hs = 1 1 k+ s k + 1 = 1 2 Z f (x)e isxdx = 1 2 Z t(s)e(1k+ s k+1)ixe isxdx = 1 2 Z t(s)e(k1+ s k+1 s)ixdx = t(s) 2 i 1 k+ s k + 1 s e(k1+ s k+1 s)ix = t(s) sin 1 k+ s k + 1 s 1 k + s k + 1 s where, s = 0; 2; :::; 2n 2 and t(s) is a function of s: If t(s) is solved from the last equality, then
t(s) = (k sk 2+ 1) (k + sk + 1) sin k sk 2+ 1 k(k + 1) and f (x) = (k sk 2+ 1) (k + sk + 1) sin k sk 2+ 1 k(k + 1) e(k1+k+1s )ix:
We can seperate f (x) = f1(x)f2(x) such that f1(x) is real valued. Then
jf(x)j = jf1(x)f2(x)j = jf1(x)j : If we take jf1(x)j = (k sk2+ 1) (k + sk + 1) sin k sk 2+ 1 k(k + 1) ;
then since k sk2+ 1 (k + sk + 1) k and sin k sk2+ 1 k(k + 1) sin k(k + 1) ; we have jf1(x)j = k sin k(k + 1) :
Thus, the proof of the theorem is completed.
Theorem 5 Let g = 1=(k + 1) and h = 1=k as in (2). Then k2 n 2 n + 1 1 k + 1 1 1 k + 1 2n 1 k + 1 +2 1 2 n(k + 1) k 2 1; n + 1 1 k + 1 + k2 n(k + 1) 1; 1 1 k + 1 2 1 1 n(k + 1) k 2 1; 2n 1 k + 1 1=2 kHnk2 9 > > > > > > = > > > > > > ; ; kHnk2 (k 1) sin k(k + 1) where, 3 k 2 Z+ and n 2:
Proof. We can take the following equality from the de…nition of Euclidean norms: 1 nkHnk 2 F = 1 n n X s=1 2 6 6 6 4 s 1 k + 1+ s 1 k 2 3 7 7 7 5+ 1 n n 1X s=1 2 6 6 6 4 n s 1 k + 1+ n + s 1 k 2 3 7 7 7 5 = k 2 n 2 n + 1 1 k + 1 1 1 k + 1 2n 1 k + 1 +2 1 2 n(k + 1) k 2 1; n + 1 1 k + 1 + k2 n(k + 1) 1; 1 1 k + 1 2 1 1 n(k + 1) k 2 1; 2n 1 k + 1 9 > > > > > > > > > > > > > > > = > > > > > > > > > > > > > > > ; :
Let us consider equation (9) for the upper bound of the spectral norm of the matrix Hn:Then, (12) hs= 1 1 k+1+ s k = 1 2 Z f (x)e isxdx = g(s) 2 Z e(k+11 + s k)ixe isxdx
where, s = 0; 2; :::; 2n 2 and g(s) is a function of s. If we solve g(s) in the equation (12), we have g(s) = (k sk 2+ s) (k + sk + s) sin k sk 2+ s k(k + 1) : Then we obtain f (x) = (k sk 2+ s) e(k+11 +ks)ix (k + sk + s) sin k sk 2+ s k(k + 1) :
The real valued function f1(x) is
f1(x) = (k sk2+ s) (k + sk + s) sin k sk 2+ s k(k + 1) : Since k sk2+ s k + sk + s k 1 and sin k sk 2+ s k(k + 1) sin k(k + 1) ; we obtain jf1(x)j = (k sk2+ s) (k + sk + s) sin k sk 2+ s k(k + 1) (k 1) sin k(k + 1) : Since kHnk2 jf(x)j = sup jf1(x)j ; we have kHnk2 (k 1) sin k(k + 1) :
Thus, the proof of the theoerem is completed. 4. Numerical Results Let a = k + k(k + 1) + (k + 1) n + 1 2 1 k + n + 1 2 + 1 k 1 1 k 2 + 1 k ;
b = 2k 3+ 6k2+ 3k 2k + 2 + (k + 1) n 2 1 k 1 1 k ; c = (k + 1) 2 kn 1; n + 1 + 1 k 1; 1 + 1 k +(k + 1) 2 n 2 n + 1 + 1 k 1 + 1 k 2n + 1 k + (k + 1)2(2nk + 1) kn 1; n + 1 + 1 k 1; 2n +1 k 1=2 and d = k 2 n 2 n + 1 1 k + 1 1 1 k + 1 2n 1 k + 1 +2 1 2 n(k + 1) k 2 1; n + 1 1 k + 1 + k2 n(k + 1) 1; 1 1 k + 1 2 1 1 n(k + 1) k 2 1; 2n 1 k + 1 1=2
Let k = 2 and n odd.
n (k+1) sin(k) kT nk2 a 9 9:424777960 9:424777961 20:78095238 11 9:424777962 9:424777961 21:99307359 21 9:424777962 9:424777961 21:99307359 31 9:424777962 9:424777961 28:22402000 51 9:424777962 9:424777961 31:21236205 101 9:424777962 9:424777961 35:31270412 151 9:424777962 9:424777961 37:72576838 Let k = 5 and n odd.
n (k+1) sin(k) kT nk2 a 9 32:06877996 32:06877997 55:58598200 11 32:06877996 32:06877997 57:98982815 21 32:06877996 32:06878000 65:74314950 31 32:06877996 32:06878000 70:41545271 51 32:06877996 32:06878000 76:38883198 101 32:06877996 32:06878000 84:58807275 151 32:06877996 32:06878000 89:41392827 Let k = 2 and n even.
n (k+1) sin(k) kT nk2 b 8 9:424777960 9:424777962 19:92380952 10 9:424777960 9:424777962 20:93080253 20 9:424777960 9:424777962 23:59060237 30 9:424777960 9:424777962 24:97821210 50 9:424777960 9:424777962 26:64102947 100 9:424777960 9:424777963 28:81467630 150 9:424777960 9:424777963 30:06183362 Let k = 5 and n even.
n (k+1) sin(k) kT nk2 b 8 32:06877996 32:06877997 53:53241058 10 32:06877996 32:06877997 55:44924628 20 32:06877996 32:06878000 60:62239797 30 32:06877996 32:06878000 63:35897971 50 32:06877996 32:06878002 66:65649764 100 32:06877996 32:06878004 70:98414447 150 32:06877996 32:06878004 73:47216360 Let k = 2: n c kHnk2 sin(k k(k+1)) 3 1:850269456 3:198490263 12:56637062 10 1:400631868 4:376771212 12:56637062 20 1:129690668 4:942773458 12:56637062 30 :9842370645 5:237037924 12:56637062 50 :8193850706 5:572655256 12:56637062 100 :6304016262 5:972458999 12:56637062 150 :5376947340 6:180238268 12:56637062 Let k = 5: n c kHnk2 sin(k k(k+1)) 3 4:314158923 7:450845732 150:2745070 10 3:103833450 9:662775280 150:2745070 20 2:456858703 10:69001475 150:2745070 30 2:122376678 11:21932286 150:2745070 50 1:751441860 11:82059614 150:2745070 100 1:334996425 12:53501886 150:2745070 150 1:133672702 12:90603226 150:2745070
Let k = 2: n d kHnk2 (k 1) sin(k(k+1)) 3 2:142284570 3:766142619 6:283185308 10 1:393039419 4:314088303 6:283185308 20 1:057610707 4:544314266 6:283185308 30 :8955075289 4:659065842 6:283185308 50 :7232424134 4:786913752 6:283185308 100 :5382033919 4:936028037 6:283185308 150 :4516741567 5:012503057 6:283185308 Let k = 5: n d kHnk2 sin(k 1)( k(k+1)) 3 4:582160341 8:005571367 120:2196056 10 3:087392863 9:577013472 120:2196056 20 2:379326388 10:26606678 120:2196056 30 2:029447396 10:61506086 120:2196056 50 1:652224842 11:00792748 120:2196056 100 1:240741975 11:47120490 120:2196056 150 1:046014648 11:71079182 120:2196056 6. References
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