亚洲免费av电影一区二区三区,日韩爱爱视频,51精品视频一区二区三区,91视频爱爱,日韩欧美在线播放视频,中文字幕少妇AV,亚洲电影中文字幕,久久久久亚洲av成人网址,久久综合视频网站,国产在线不卡免费播放

        ?

        LIE-TROTTER FORMULA FOR THE HADAMARD PRODUCT *

        2020-08-02 05:11:48JingWANG王靜
        關(guān)鍵詞:王靜

        Jing WANG (王靜) ?

        School of Information, Beijing Wuzi University, Beijing 101149, China E-mail: wangjingzzumath@163.com

        Yonggang LI (李永剛)

        College of Science, Zhengzhou University of Aeronautics, Zhengzhou 450015, China E-mail: liyonggang914@126.com

        Huafei SUN (孫華飛)

        School of Mathematics and Statistics, Beijing Institute of Technology, Beijing 100081, China Beijing Key Laboratory on MCAACI, Beijing 100081, China E-mail: huafeisun@bit.edu.cn

        Abstract Suppose that A and B are two positive-definite matrices, then, the limit of (Ap/2BpAp/2)1/p as p tends to 0 can be obtained by the well known Lie-Trotter formula. In this article, we generalize the usual product of matrices to the Hadamard product denoted as ? which is commutative, and obtain the explicit formula of the limit (Ap? Bp)1/p as p tends to 0. Furthermore, the existence of the limit of (Ap? Bp)1/p as p tends to +∞ is proved.

        Key words Lie-Trotter formula; reciprocal Lie-Trotter formula; Hadamard product; positive-definite matrix

        1 Introduction

        Let M(n,C) denote the space of all n×n matrices with complex entries, H(n) denote the vector space of Hermitian n×n matrices, and H+(n) denote the set of n×n positive-definite matrices. For X, Y ∈ M(n,C), the well-known Lie-Trotter formula, as originally established in [1, 2], is

        The Lie-Trotter formula can easily be modified to symmetric form ([3]), especially, when restricted to matrices A, B ∈H+(n), the formula (1.1) can be rephrased as

        A similar formula holds for the limit of(Ap?Bp)2/pas p tends to 0,where A?B is the geometric mean of A and B ([4]). In [5, 6], the authors considered the explicit formula of the limit of(Ap/2BpAp/2)1/pas p tends to+∞, which is called the reciprocal Lie-Trotter formula,and this formula can be obtained by the log-majorization relation ([7]).

        It is known that for any two positive-definite matrices A, B, the Hadamard product of A and B denoted by A ?B is also a positive-definite matrix ([8]). For A, B ∈H+(n), T. Ando settled affirmatively the conjecture of Johnson and Bapat on the Hadamard product ([9])

        where λi(A) are the eigenvalues of A, and λ1(A) ≥λ2(A) ≥···≥λn(A). In [10], by studying the eigenvalues of(Ap?Bp)1/pfor p ∈(0,1],G.Visick presented a number of intervening terms for inequality (1.3). Besides, to progress further on the Lie-Trotter formula (1.2), we find that the limit of (Ap?Bp)1/pas p tends to 0 is related to formula (1.2). That is, according to the latter case for the Hadamard product, the Lie-Trotter formula can be derived. Within this motivation, we investigate the Lie-Trotter problem that the limit of (Ap?Bp)1/pas p →0 for positive-definite matrices A and B, as well as the reciprocal Lie-Trotter problem. It is interesting that the explicit Lie-Trotter formula for the Hadamard product is obtained,and the existence of the reciprocal Lie-Trotter formula for the Hadamard product is proved.

        The remainder of the article is organized as follows. In Section 2,we review the fundamental notions and definitions, and show some important conclusions of operator-monotone function.The Lie-Trotter formula for the Hadamard product is obtained in Section 3. In Section 4, we prove the existence of the reciprocal Lie-Trotter formula for the Hadamard product.

        2 Preliminaries

        In this section, we recall some notions and definitions from matrix analysis, and introduce some important results of the operator-monotone function, which are used through the article(refer to [11–15]).

        Let Cnbe the n-dimensional complex vector space with the inner product

        where x,y ∈Cn, and superscript xHdenotes the conjugated transpose of x.

        We say A ∈H(n) is positive-semidefinite, that is, A ≥0, if A satisfies

        and positive-definite, that is, A>0, if

        We denote that A ≥B, B ∈H(n), if

        2.1 The Hadamard product and the tensor product

        For any two matrices A=(aij), B =(bij) in M(n,C), the Hadamard product (also known as the Schur product or the entrywise product) A ?B is defined as ([16, 17])

        It is noticed that the Hadamard product is different from the usual matrix product, and the most important is the commutativity of Hadamard multiplication, that is, A ?B =B ?A.

        An important way of putting matrices together is to construct their tensor product (sometimes called the Kronecker product). If A, B ∈M(n,C), then, their tensor product is defined as

        The matrix A ?B is an n2-square matrix. The following formulas for tensor product are well known ([18])

        In [19, 20], the authors showed that the Hadamard product is a principal submatrix of the tensor product, and the main result can be summarized as follows.

        Lemma 2.1For any A, B ∈M(n,C), then,

        where Q?= [E11,··· ,Enn], and the n×n matrices Eiifor i = 1,··· ,n have a 1 in position(i,i) and zeros elsewhere.

        Remark 2.2The n2×n matrix Q satisfies the property Q?Q=I,where I is an n-square identity matrix.

        2.2 The operator-monotone function

        The operator-monotone function is a generalization of the real-valued function, and the special case is the matrix monotone function. For A ∈H+(n), using the spectral theorem, we have ([21])

        where U is a unitary matrix.

        For the function f(x) (x ∈(0,+∞)), we define the matrix function as

        where the operator-monotone function satisfies that

        In 1933,K.L?wner successfully characterized the operator monotonicity in term of positive semi-definiteness of the so-called L?wner matrices. A part of the deep theory of L?wner is summarized in the following lemma. A complete account of L?wner’s theory can be found in the book [22].

        Lemma 2.3The following statements are equivalent for a real-valued continuous function f on (0,+∞):

        (i) f is operator-monotone;

        (ii) fadmits an analytic continuation to the whole domain Im z≠0 in such a way that Im f(z)·Im z ≥0;

        (iii) f admits an integral representation

        where α is a real number,β is a non-negative real number,and dμ(t)is a finite positive measure on (?∞,0].

        According to L?wner’s theory, T. Ando gave a characterization for the operator-monotone function associated with the normalized positive linear map in [23].

        Lemma 2.4If a function f is operator-monotone on (0,+∞), then, the map A →f(A)is concave on H+(n). And if Φ is a normalized positive linear map on H(n), then, for any A ∈H+(n)

        3 Lie-Trotter Formula for the Hadamard Product

        This section aims at investigating the Lie-Trotter problem for the Hadamard product, and finding the explicit expression. At the same time, the proving method is applied to derived the well-known Lie-Trotter formula.

        3.1 Limit of (Ap ?Bp)1/p as p →0

        At first, we prove the following lemma.

        Lemma 3.1If A, B ∈H+(n) and p ∈(0,+∞), then, the following result holds

        ProofAccording to (2.1) and (2.2), we have

        From now on, we will prove

        that is, for the Frobenius norm,

        As (A ?B)H=AH?BH=A ?B, we have

        From the fact that Q?Q=I, we have

        and

        Noticing that the functions tr ln(Q?(A ?B)pQ), and tr ln2(Q?(A ?B)pQ) are continuously differentiable with respect to p, using the L′H?ospital’s rule, we obtain

        and

        Let g(p) = tr ln(Q?(A ?B)pQ)· (Q?(A ?B)pQ)?1· Q?(A ?B)pln(A ?B)Q. As g(p) is continuously differentiable and

        equality (3.7) can be recast as

        where

        Furthermore, we have

        where

        Thus, according to (3.5), (3.6), and (3.9), we have

        which verifies conclusion (3.4). Therefore, (3.1) can be obtained from (3.2) and (3.3).

        The main result of this article is the next theorem showing the Lie-Trotter formula for the Hadamard product.

        Theorem 3.2If A, B ∈H+(n) and p ∈(0,+∞), then, the following result holds

        ProofAccording to Lemma 3.1,and noticing that the exponential function is continuous,we have

        This completes the proof of Theorem 3.2.

        According to Theorem 3.2, we can obtain the following corollary.

        Corollary 3.3If A1,A2,··· ,Am∈H+(n)(m ≥2)and p ∈(0,+∞),the following result holds,

        3.2 Application

        Note that A, B >0,

        The Lie-Trotter formula for the Hadamard product can be rewritten as

        The result (3.10) has been studied to present some intervening terms for the log-majorization relation (1.3) as follows (see [10])

        In the following part, we provide a link between Theorem 3.2 and the Lie-Trotter formula(1.2).

        Let A, B ∈H+(n). By using the spectral decomposition, we have

        Thus,

        In fact, by calculations, we have

        Furthermore,

        According to (3.12), (3.11) is recast as

        4 Reciprocal Lie-Trotter Formula for the Hadamard Product

        In this section, we will prove the existence of the reciprocal Lie-Trotter formula for the Hadamard product. For A,B ∈H+(n), using the spectral theorem, we have

        where U = (uij) and V = (vij) are unitary matrices. Firstly, we show that the limit of(Ap?Bp)1/pas p →+∞exists for the special case that A and B are diagonal matrices of H+(n), that is

        In fact, noticing that

        we have

        If A and B are any positive-definite matrices,we have not obtained the accurate expression of the limit (Ap?Bp)1/pas p tends to +∞. However, the following result gives the existence of the reciprocal Lie-Trotter formula for the Hadamard product.

        Theorem 4.1If A,B ∈H+(n) and p ∈(0,+∞), then, the following limit exists

        ProofLetBecause the mapis a normalized positive linear map, we have

        which shows that Q?is operator-monotone. By Lemma 2.4, for the operator-monotone map f :, p ∈(1,+∞), we have

        We can verify that (Q?Q)1/pis monotonically increasing with respect to p ∈(1,+∞). In fact, let(s ∈(1,+∞)), then, (4.3) can be rewritten as

        Furthermore, we have

        As

        and Ap≤(A)I, Bp≤(B)I for any p ∈(0,+∞), according to (4.2) and (4.5), we can obtain

        and

        where λ1(A) and λ1(B) are the maximum eigenvalues of A and B, respectively.

        Consequently, for p ∈(1,+∞), from (4.6), we have

        From now on, we will prove the existence of the following limit

        In fact, we denote the following Schatten 1-norm as

        where si(A) (i = 1,··· ,n) are the singular values of A, and s1(A) ≥s2(A) ≥··· ≥sn(A),especially, for A ∈H+(n),

        where λi(A) are the eigenvalues of A, and λ1(A)≥λ2(A)≥···≥λn(A).

        Because (Q?(A ?B)pQ)1/pfor p ∈(1,+∞) is monotonically increasing, that is, for 1

        Notice that if A ≤B, then, tr A ≤tr B. Thus, according to (4.9), we have

        that is, ?ε>0, ?N >0, when P1,P2>N, then,

        Therefore, we have

        where the first equality in (4.11)holds because (Q?(A ?B)pQ)1/pis an operator-monotone for p ∈(1,+∞). This verifies conclusion (4.8). From formula (3.2), we finishe the proof of the existence of (4.1).

        From the proof of Theorem 4.1, we can obtain the following corollaries.

        Corollary 4.2If A1,A2,··· ,Am∈H+(n) (m ≥2) and p ∈(0,+∞), then, the following limit exists

        Corollary 4.3If A1,A2,··· ,Am∈H+(n) (m ≥2), then, the following inequality of the H?lder type holds

        In this article,we can not give the explicit expression of the reciprocal Lie-Trotter formula,but we can obtain the first eigenvalue of the reciprocal Lie-Trotter formula for the Hadamard product (refer to [6]).

        Proposition 4.4If A, B ∈H+(n), then, the first eigenvalue of the following limit

        ProofLetBy the definition of matrix function, we have

        On the one hand, we have

        thus,

        On the other hand, noting that

        we have

        According to (4.12) and (4.13), we have

        猜你喜歡
        王靜
        Reciprocal transformations of the space–time shifted nonlocal short pulse equations
        Fusionable and fissionable waves of(2+1)-dimensional shallow water wave equation
        The Management Methods And Thinking Of Personnel Files
        客聯(lián)(2021年9期)2021-11-07 19:21:33
        The Development of Contemporary Oil Painting Art
        青年生活(2019年16期)2019-10-21 01:46:49
        王靜博士簡介
        Income Inequality in Developing Countries
        商情(2017年17期)2017-06-10 12:27:58
        Let it Go隨它吧
        Rumor Spreading Model with Immunization Strategy and Delay Time on Homogeneous Networks?
        鳳崗鬼事
        RIGIDITY OF COMPACT SURFACES IN HOMOGENEOUS 3-MANIFOLDS WITH CONSTANT MEAN CURVATURE?
        国产国产人免费人成免费视频| 亚洲素人av在线观看| 91精品国产九色综合久久香蕉 | 人妻精品久久久久中文字幕| 精品久久久中文字幕人妻| 精品综合久久久久久8888 | 中文字幕一区在线直播| 国产一区二区三区免费观看在线| 中文在线天堂网www| 日韩有码中文字幕第一页| 国产精品久色婷婷不卡| 摸进她的内裤里疯狂揉她动图视频| 人人狠狠综合久久亚洲婷婷| 极品美女尤物嫩模啪啪| 久久精品蜜桃亚洲av高清| 久久香蕉国产线看观看精品yw| 亚洲综合色一区二区三区小说| 久久久精品国产亚洲av网| 日本一区二区三级在线观看 | 偷拍偷窥女厕一区二区视频| 国产一区二区精品久久| 国产精品国产三级国产专区5o| 国产一区二区三区在线观看黄 | 男女调情视频在线观看| 国产又色又爽又黄刺激在线视频 | av免费网站免费久久网| 国产午夜伦鲁鲁| 欧美性福利| 亚洲国产av午夜福利精品一区| 无码国产成人午夜电影在线观看| 国产免费无码一区二区三区 | 日本人妻三级在线观看| 特黄 做受又硬又粗又大视频| 越南女子杂交内射bbwbbw| 2021年性爱喷水视频| 亚洲精品一区二区高清| 亚洲日韩一区二区三区| 91久久福利国产成人精品| 免费人成在线观看播放视频| 国产偷国产偷精品高清尤物| 欧美人妻日韩精品|