Selçuk J. Appl. Math. Selçuk Journal of Vol. 7. No.2. pp. 3-11, 2006 Applied Mathematics
On The Applications Of Hadamard Square Root And Hadamard In-verse
Mehmet Akbulak, Durmu¸s Bozkurt, Ramazan Türkmen, Kenan Kaygısız
Department of Mathematics, Art and Science Faculty, Selcuk University, 42031, Konya, Turkey
e-mail:m akbulak@ selcuk.edu.tr, db ozkurt@ selcuk.edu.tr, rturkm en@ selcuk.edu.tr,kenan2727@ yaho o.com
Summary.In this study, we have established upper and lower bounds for the spectral norm of Hadamard square root and Hadamard inverse of the matrix 121where the matrix 121is Cauchy-Toeplitz matrix. We have also obtained nonzero eigenvalues of the Hadamard product of 121 and 121 where the matrix 121 is Cauchy-Hankel matrix. Finally, we have defined Hadamard condition number for Cauchy-Toeplitz matrix and calculated an upper bound for this condition number.
Key words: Hadamard square root, Hadamard inverse, Hadamard product, Cauchy-Toeplitz matrix, Cauchy-Hankel matrix, condition number.
1. Introduction
Let = () be × matrix with nonnegative entries. Then the Hadamard square root of is given by ◦12 = (12
). Let = () be × matrix with all nonzero entries. Then the Hadamard inverse of is given by ◦−1 = (1)[1].
The well known Frobenius norm of the matrix is
(1) kk = ⎛ ⎝ X =1 X =1 ||2 ⎞ ⎠ 12
Also the spectral norm of the matrix is kk2=
r max 166(
where is × and is the conjugate transpose of the matrix . The following inequality holds:
(2) √1
kk 6 kk26 kk
where k · k is Frobenius norm of the matrix (see [2]). The maximum column norm 1(·) and the maximum row norm 1(·) of any matrix are defined by
(3) 1() = max sX ||2 and (4) 1() = max sX ||2 respectively [3, 4].
Let = () be × matrix. The 2-minors of the matrix are defined by µ ¶ = ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ where and and 16 6 . If
µ ¶ = ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯
then the minors are called principal 2-minors of the matrix , where 16 6 .
Let = () and = () be × matrices. The Hadamard product of and is defined by ◦ = (). If = ◦ then
kk26 1()1()[4]. (1.5) A function Ψ is called a psi (or digamma) function if
Ψ() = {ln[Γ()]} where Γ() = ∞ Z 0 −−1
Ψ( ) = [Ψ()] = ½ (ln[Γ()]) ¾
If = 0 then Ψ(0 ) = Ψ() ={ln[Γ()]}. On the other hand, if 0, is any number and is positive integer, then
lim
→∞Ψ( + ) = 0[5] The condition number () = kk2
° °−1°°
2of a nonsingular matrix plays an important role in the numerical solution of linear systems since it measures the sensitivity of the solution of linear systems = to the perturbations on and where kk2 donetes the spectral norm.
Let = [1(− )]=1 (6= ), = [−]=0−1 and = [+]=0−1 be a Cauchy, a Toeplitz and a Hankel matrices, respectively. In general, Cauchy-Toeplitz and Cauchy-Hankel matrices are defined as in the following forms:
= ∙ 1 + ( − ) ¸−1 =0 = ∙ 1 + ( + ) ¸−1 =0
where 6= 0, and are some numbers and is not integer. If we take = 12 and = 1, then we get
121= ∙ 1 1 2+ − ¸−1 =0 121= ∙ 1 1 2+ + ¸−1 =0 Then the Hadamard inverses of the matrices 121and 121 are
121◦−1 = ∙ 1 + 2 ( − ) 2 ¸−1 =0 121◦−1 = ∙ 1 + 2 ( + ) 2 ¸−1 =0 and the Hadamard square roots of the matrices¯¯121
¯ ¯ and 121are ¯ ¯121 ¯ ¯◦12 = ⎡ ⎣q 1 1 2+ | − | ⎤ ⎦ −1 =0 121◦12 = ⎡ ⎣q 1 1 2+ + ⎤ ⎦ −1 =0
2. Norms Of Hadamard Square Roots And Hadamard Inverses Theorem 1. Let ¯¯121
¯ ¯◦12
be the Hadamard square root of the matrix ¯ ¯121 ¯ ¯. Then ½µ 2 + 1 ¶ ∙ Ψ µ +3 2 ¶ − 2 + + 2 ln 2 ¸ +2 ¾12 6°°°¯¯121 ¯ ¯◦12°° ° 2 and ° ° °¯¯121 ¯ ¯◦12°° ° 26 ⎧ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎩ 12£Ψ¡3−2 2 ¢ − Ψ¡2−2 ¢ −Ψ¡4− 2 ¢ +8 3− − 2 ln 2 ¤12 odd 12£Ψ¡1− 2 ¢ + Ψ¡+32 ¢ −2 + ln¡16 2 ¢¤12 even is valid.
Proof. By the definition of the matrix ¯¯121 ¯
¯◦12 and the equality (1), we have ° ° °¯¯121 ¯ ¯◦12°° °2 = 2 + 4 X−1 =1 − 2 + 1 = (2 + 1) ∙ Ψ +3 2− 2 + + 2 ln 2 ¸ + 2 Then we obtain 1 √ ° ° °121◦12 ° ° ° = ½µ 2 + 1 ¶ ∙ Ψ µ +1 2 ¶ − 2 ¸ + µ 2 + 1 ¶ ( + 2 ln 2) + 2 ¾12 ≤ °°°121◦12 ° ° ° Now, let us establish the upper bound for
° ° °¯¯121 ¯ ¯◦12°° ° 2. If we take ¯ ¯121 ¯ ¯◦12 = ◦ such that = []×=£√2¤× and =
h 1p1 + 2 | − |i ×, then we get°°°¯¯121 ¯ ¯◦12°° °
2 6 1()1()from (1.5) where 1() and 1() are the maximum row norm and column norm of the matrices and , respectively.
(5) 1() =
√ 2 and
1() = v u u u t 2 X =1 1 (1 + − 2)+ X = 2+1 1 (1 + 2 − ) (6) = √1 2 ∙ Ψ µ 1 − 2 ¶ + Ψ µ + 3 2 ¶ − 2 − ln µ 16 2 ¶¸12
for even and
1() = ⎡ ⎣ (X−1)2 =1 1 (−1 + 2 − 2)+ X =+1 2 1 (3 + 2 − 2) ⎤ ⎦ 12 (7) = √1 2 ∙ Ψ µ 3 − 2 2 ¶ − Ψ µ 2 − 2 ¶ − Ψ µ 4 − 2 ¶ +8 3− − 2 ln 2 ¸12
for odd. Hence, we establish the upper bound by (5), (6) and (7) for°°°¯¯121 ¯ ¯◦12°°
°2. Thus, we get the results.
Theorem 2. Let 121◦−1 be the Hadamard inverse of the matrix 121. Then r 1 6 3+ 1 126 ° ° °121◦−1 ° ° ° 26 1 6 p 124− 32 holds.
Proof. Firstly, we establish lower bound and secondly, upper bound for the matrices 121◦−1 . By (1), we have (8) °°°121◦−1 °°° 2 = 2 − 1 4 + 1 4 X−1 =1 £ (2 − 2 − 1)(2 + 1)2¤
If we evaluate the sum in the right side of (8), then we get ° ° °121◦−1 ° ° °2 = 1 6 4+ 1 12 2 By the inequality (2), we have
1 √ ° ° °121◦−1 ° ° ° = r 1 6 3+ 1 126 ° ° °121◦−1 ° ° °2
Secondly, let us establish upper bound. If we take 121◦−1 = ◦ such that = [12]×and = [1 + 2( − )]◦, then we get
1() = 1 2 √ and 1() = " X =1 (2 − 1)2 #12 = µ 4 3 3 −13 ¶12 By the inequality (5), we have
° ° °121◦−1 ° ° ° 26 s 1 4 µ4 3 3−1 3 ¶ =1 6 p 124− 32 Thus, the proof is completed.
3. Hadamard Condition Number
Definition 3 Let be any matrix. The value kk2 ° °◦−1°°
2is called Hadamard condition number of the matrix ,denoted by ◦()
Theorem 4. The upper bound of Hadamard condition number of¯¯121 ¯ ¯◦12is ◦³¯¯ 121 ¯ ¯◦12´ ≤ ( £ 2 √ 2+ 2¤ £Ψ¡1− 2 ¢ + Ψ¡+32 ¢+ 2 + ln¡162 ¢¤12 £ 2 √ 2+ 5¤ £Ψ¡3−2 2 ¢ − Ψ¡2−2 ¢ − Ψ¡4−2 ¢ +83− − 2 ln 2¤12 Proof.To prove this theorem it is enough to find upper bound
° ° ° °³¯¯121 ¯ ¯◦12´◦−1°°° ° 2 . If we take³¯¯121 ¯ ¯◦12´◦−1 = ◦ such that = []×= £ 1√2¤ ×and = []×= hp 1 + 2 | − |i ×, then ° ° ° ° ³¯ ¯121 ¯ ¯◦12´◦−1°°° ° 2 6 1()1()
from (5) where 1() and 1() are the maximum row and column norm of the matrices and , respectively. We get
1() = r
2 and for even
1() = v u u u t 2 X =1 (1 + − 2) + X = 2+1 (1 + 2 − ) = r 2+ 2 2 for odd 1() = v u u u t (X−1)2 =1 (−1 + 2 − 2) + X =+1 2 (3 + 2 − 2) = r 2+ 5 2
Then the upper bound of the spectral norm of³¯¯121 ¯ ¯◦12´◦−1 is ° ° ° °³¯¯121 ¯ ¯◦12´◦−1°°° ° 2 6½ 2 √ + 2 1 2 √ 3+ 5
Hence, we can write the upper bound for the Hadamard condition number of ¯ ¯121 ¯ ¯◦12 as ◦³¯¯ 121 ¯ ¯◦12´ ≤ ( £ 2 √ 2+ 2¤ £Ψ¡1− 2 ¢ + Ψ¡+3 2 ¢ + 2 + ln¡16 2 ¢¤12 £ 2 √ 2+ 5¤ £Ψ¡3−2 2 ¢ − Ψ¡2−2 ¢ − Ψ¡4−2 ¢ +83− − 2 ln 2¤12 4. An Application Of Hadamard Product
Theorem 5. Let = 121◦−1 ◦ 121◦−1 . Then nonzero eigenvalues of the matrix are 12= 2 4 + 3 8 ± 120 p 4105 + 6300 + 1002− 36003− 12804 Proof. First of all, let us show that () = 2. Multiply first row of the matrix by −42+45 −3 and add to th row of the matrix, where = 2 3 . Then we get ∼ ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ 5 4 − 7 4 − 27 4 − 55 4 9−42 4 0 485 1285 48 1625−16 0 24 64 120 4025−40 .. . ... ... ... . .. ... 0 1 2 3 4 ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦
where 1 = 122+ − 24 5 2 = 322+ 32 − 64 5 3 = 602+ 60 − 120 5 and 4= 44+ 43− 122− 4 + 8 5
Take the multiple −1of second row of the matrix and add to third row, the multiple −2 add to fourth row, and so on the multiple −−2 and add to th row, where 1= 52 +1= + ( + 3)2 = 1 2 − 3. Then, we have ∼ ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ 5 4 − 7 4 − 27 4 − 55 4 9−42 4 0 48 5 128 5 48 162−16 5 0 0 0 0 0 .. . ... ... ... . .. ... 0 0 0 0 0 ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦
Hence () = 2. Therefore, the matrix has two nonzero eigenvalues. Then the characteristic polynomial of the matrix is
∆() = + 1−1+ 2−2 where 1 = − 2 = X µ ¶ = 1 2 ; Since 1= − = X =1 µ 1 4+ ¶ = 4 − ( + 1)2 2 + 1 2 and 2 = X µ ¶ = X−1 =1 X =+1 ¯ ¯ ¯ ¯ 1 4+ 1 4+ + 2 − 2 1 4+ 2 + − 2 14+ ¯ ¯ ¯ ¯ = 4 6 45 + 5 4 + 4 18 − 3 4 − 132 90
we have ∆() = + Ã 4− ( + 1)2 2 + 1 2 ! −1 + µ 46 45 + 5 4 + 4 18− 3 4 − 132 90 ¶ −2 If we solve the equation ∆() = 0, then we obtain
= −1= = 3= 0 and 12= 2 4 + 3 8 ± 120 p 4105 + 6300 + 1002− 36003− 12804 References
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