Selçuk J. Appl. Math. Selçuk Journal of Vol. 5. No.1. pp. 23-30, 2004 Applied Mathematics
Lower Bounds For The Norms Of Hilbert Matrix and Hadamard Product of Cauchy-Toeplitz and Cauchy-Hankel Matrices
A. Dilek Güngör, Ali Sinan
Department of Mathematics Art and Science Faculty Selçuk University, 42031 Konya, Turkey;
e-mail: agungor@selcuk.edu.tr; asinan@selcuk.edu.tr
Received : April 15, 2004
Summary. In this study, we have found lower bounds for the spectral norms and Euclidean norms of Hilbert matrix H = (1= (i + j 1) )ni;j=1 and its Hadamard square root H = (1= (i + j 1)1=2)n
i;j=1.
In addition, we have established lower bounds for the spectral norms and Euclid-ean norms of the Hadamard product of Cauchy-Toeplitz and Cauchy-Hankel matrices.
Key words:Hilbert matrix, Hadamard square root Hilbert matrix, Hadamard product Cauchy-Toeplitz and Cauchy-Hankel matrices, Norm, Lower bound.
1. Introduction and Preliminaries
Let A = (aij) be an n n symmetric matrix with all positive entries. Then the
Hadamard inverse of A, given by A ( 1)= (1=a
ij)ni;j=1 is positive semide…nite,
and the Hadamard square root by A 1=2= (a1=2
ij )ni;j=1 [10].
The matrix
(1) H = (1=(i + j 1))ni;j=1
is well known as Hilbert matrix. Hence the Hadamard square root of Hilbert matrix, denoted by
Let C = [1=(xi yj)]ni;j=1(xi 6= yj) be a Cauchy matrix and Tn = [tj i]ni;j=0
be a Toeplitz matrix. In general Cauchy-Toeplitz matrix is being de…ned as
(3) Tn=
1 g + (i j)h
n
i;j=1
where h 6= 0; g ve h are any numbers and g=h is not integer. Toeplitz matrices are precisely those matrices that one constant along all diagonals parallel to the main diagonal, and thus a Toeplitz matrix is determined by its …rst row and column.
On the other hand, let Hn = [hi+j]ni;j=0 be a Hankel matrix. Every n n
Cauchy-Hankel matrix is of the form
(4) Hn=
1 g + (i + j)h
n
i;j=1
where h 6= 0; g ve h are any numbers and g=h is not integer. Hankel matrices are the symmetrical.
Recently, there have been several papers on the norms of Cauchy-Toeplitz matrix and Cauchy-Hankel matrix [1,2,4,5]. Refs. [4,5] are interested in the spectral norm of Cauchy-Toeplitz matrix. 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 (3). Parter proved that singular values could be related to eigenvalues of certain Hermitian Toeplitz matrices corresponding to Laurent-Fourier series [4]. Solak, Turkmen and Bozkurt [7] obtained upper bounds for the lp norm of the Hilbert matrix and its Hadamard square root. Turkmen
and Bozkurt have established a lower and an upper bounds for the Euclidean norm of the Hadamard product Cauchy-Toeplitz and Cauchy-Hankel matrices [6]. Güngör [9] obtained lower bounds for the spectral norm and Euclidean norm of Cauchy-Toeplitz and Cauchy-Hankel matrices in the forms (3) and (4) by taking g = 1=2 and h = 1 using B matrix de…ned in [8].
In Section 2 of this paper, we have obtained lower bounds for the Spectral and Euclidean norms of the Hilbert matrix and its Hadamard square root. In Section 3, we have established lower bounds for the spectral norm and the Euclidean norm of the Hadamard product Cauchy-Toeplitz and Cauchy-Hankel matrices in the forms (3) and (4) by taking g = 1=2 and h = 1 using B matrix de…ned in [8].
Now, we give some preliminaries related to our study. Let A be an n n complex matrix. The well known Euclidean norm of matrix A is
(5) k A kE= 0 @ n X i=1 n X j=1 j aij j2 1 A 1=2
and also the spectral norm of the matrix A is (6) k A k2= r max 1 i n i(A HA);
where i are eigenvalues of AHA and AH is conjugate transpose of the matrix
A. A function is called polygamma function if (x) = d d(x)flog[ (x)]g ; where (x) = 1 Z 0 e ttx 1dt:
The function (m; x) have the property: lim
n!1 (a; n + b) = 0;
where a > 0 and b are any numbers and n is a positive integer [3].
To minimize the numerical round-o¤ errors in solving the system Ax = b; it is normally convenient that the rows of A be properly scaled before the solution procedure begins. One way is to premultiply by the diagonal matrix
(7) D=diag 1 r1(A) ; 2 r2(A) ; :::; n rn(A) ;
where ri(A) is the Euclidean norm of the i th row of the matrix A and 1; 2; :::; n are positive real numbers such that
(8) 21+ 22+ ::: + 2n= n:
Clearly, the Euclidean norm of the coe¢ cient matrix B = DA of the scaled system is equal topn and if 1 = 2 = ::: = n = 1 then each row of B is a
unit vector in the Euclidean norm. Also, we can de…ne B = AD,
(9) D=diag 1 c1(A) ; 2 c2(A) ; :::; n cn(A) ;
where ci(A) is the Euclidean norm of the i th columm of the matrix A: Again,
k BkE =
p
n and if 1 = 2 = ::: = n = 1 then each column of B is a unit
vector in the Euclidean norm. We know for the matrix B that
B:
2. Lower Bounds for Norms of Hilbert Matrix and its Hadamard Square Root
Theorem 1 Let the matrix H and i’s (i = 0; :::; n 1 and 20+ 21+
::: + 2
n 1= n ) be as in (1) and (8) ; respectively. Then
(11) 2 4n n 1X i=0 2 i (1; n + i + 1) + (1; i + 1) ! 13 5 1=2 k H kE and hence (12) n 1 X i=0 2 i (1; n + i + 1) + (1; i + 1) ! 1=2 k H k2 Proof. We have k D kE= n 1 X i=0 2 i (1; n + i + 1) + (1; i + 1) !1=2
from the de…nition (7) : Since k B kE=pn for B = DA we obtain (11) from
the inequality (10) : Hence, since 1 p
n k H kE k H k2; we obtain (12) :
Theorem 2 Let the matrix H 1=2 and
i’s (i = 0; :::; n 1 and 20+ 2
1+ ::: + 2n 1= n ) be as in (2) and (8) ; respectively. Then
(13) 2 4n n 1 X i=0 2 i (n + i + 1) (i + 1) ! 13 5 1=2 k H 1=2kE and hence (14) n 1 X i=0 2 i (n + i + 1) (i + 1) ! 1=2 k H 1=2k2:
Proof. We have k D kE= n 1 X i=0 2 i (n + i + 1) (i + 1) !1=2
from the de…nition (7) : Since k B kE=pn for B = DA , we obtain (13) from
the inequality (10) : Hence, since 1 p n k H 1=2 kE k H 1=2k2; we obtain (14) :
3. Lower Bounds for the Norms of Hadamard Product of Cauchy-Toeplitz and Cauchy-Hankel Matrices
Theorem 3 Let Tn and Hn be Cauchy-Toeplitz and Cauchy-Hankel matrices
as in (3) and (4) ;respectively, where g = 1=2 and h = 1. Let i’s (i = 1; :::; n
and 2
1+ 22+ ::: + 2n = n ) be as in (8). Let the operation " " be a
Hadamard product i.e., if A = (aij) and B = (bij) are n n matrices,
then A B = (aijbij) :Then (15) 2 6 4n 0 @ n X j=1 4 2 j a b 1 A 13 7 5 1=2 k Tn HnkE and hence (16) 0 @ n X j=1 4 2 j a b 1 A 1=2 k Tn Hnk2 where (17) a = 1 j3 ( 3 2+ n + j) + ( 3 2 j) + 1 j2 (1; 3 2 + j) + (1; 3 2 j) and (18) b = 1 j3 ( 3 2 + n j) + ( 3 2 + j) + 1 j2 (1; 3 2+ n + j) + (1; 3 2+ n j) :
Proof. We can formulate the Hadamard product of Tn and Hn matrices
where g = 1=2 and h = 1 as in the following form:
(19) Tn Hn= 4 2 6 6 6 6 4 1 5 1 7 ::: 1 [1+2(n+1)][1 2(n 1)] 1 21 1 9 ::: 1 [1+2(n+2)][1 2(n 2)] : : ::: : : : ::: : 1 [1+2(n+1)][1+2(n 1)] : ::: 1 1+4n 3 7 7 7 7 5: We have k D kE= 0 @ n X j=1 4 2 j a b 1 A 1=2
from de…nition (9) where a and b as in (17) and (18) , respectively. Since k B kE=pn for B = AD , we obtain (15) from the inequality (10) : Hence,
since
1 p
n k Tn HnkE k Tn Hn k2; we obtain (16) :
Example 1 Let 0= 1= ::: = n 1= 1 and
c = 2 4n n 1X i=0 2 i (1; n + i + 1) + (1; i + 1) ! 13 5 1=2 :
For Euclidean norm of the matrix H in (1), we obtain the following values:
n c k H kE 1 pn k H kE k H k2 1 1 1 1 1 5 0:4861024026 1:580906263 :7070027742 1:567050691 10 0:3456005779 1:785527123 :5646332532 1:751919670 20 0:2447829834 1:969813453 :4404636783 1:907134720 50 0:1548996009 2:190011373 :3097143786 2:076296683 Example 2 Let 0= 1= ::: = n 1= 1 and
d = 2 4n n 1X i=0 2 i (n + i + 1) (i + 1) ! 13 5 1=2 :
For Euclidean norm of the matrix H 1=2in (2), we obtain the following values: n d k H 1=2k E 1 pn k H 1=2k E k H 1=2k2 1 1 1 1 1 5 1:052417645 2:540934711 1:136340548 2:533602599 10 1:056539472 3:657243232 1:156521857 3:638302962 20 1:058039521 5:218441843 1:166879070 5:180584414 50 1:058671811 8:295614384 1:173177036 8:219123246
Example 3 Let 1= 2= ::: = n = 1 ande =
2 4n Pn j=1 4 2 j a b ! 13 5 1=2
where a and b as in (17) and (18) :For Euclidean norm of the matrix Tn
Hn in (19), we obtain the following values:
n e k Tn Hn kE 1 p n k Tn Hn kE k Tn Hn k2 1 0:8 0:8 0:8 0:8 5 0:44836704 1:31519951 0:5881751040 0:999474704 10 0:24737993 1:39612151 0:4414923887 1:000163696 20 0:12939636 1:43769129 0:3214775476 1:000261912 50 0:05321964 1:46282405 0:2068745608 1:000276124 We have seen that the bounds for 0 = 1 = ::: = n 1= 1 are better
than those for i’s (i = 0; :::; n 1) such that 20+ 21+ ::: + 2n 1= n: For
example, let 0= 0:5; 1= 0:9; 2= 1:1; 3= 1:3 and 4=
p
1:04. For these values, a and b bounds are 0:2173916033 k H kE, 0:4706554789 k Hn kE
and 0:412621844 k Tn Hn kE, respectively.
References
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