Several Theorems for the Trace of Self-Conjugate Quaternion Matrix

Modern Applied Science September, 2008

Several Theorems for the Trace of Self-conjugate Quaternion Matrix

Qinglong Hu Department of Engineering Technology Xichang College Xichang, Sichuan, 615013, China E-mail: [email protected]

Limin Zou(Corresponding author) College of Mathematics and Physics Chongqing University Chongqing, 400044, China E-mail: [email protected]

The research is Supported by Chongqing University postgraduates’ Science and Innovation Fund (200801A 1 A0070266). Abstract The purpose of this paper is to discuss the inequalities for the trace of self-conjugate quaternion matrix. We present the inequality for eigenvalues and trace of self-conjugate quaternion matrices. Based on the inequality above, we obtain several inequalities for the trace of quaternion positive definite matrix. Keywords: Quaternion matrix, Trace, Inequalities, Eigenvalues 1. Introduction Quaternion was introduced by the Irish mathematician Hamilton (1805-1865) in 1843. The literature on quaternion matrices, though dating back to 1936 [1], is fragmentary. Quaternion is mostly used in computer vision because of their ability to represent rotations in 4D spaces. It is also used in programming video games and controlling spacecrafts [2, 3] and so forth. The research on mathematical objects associated with quaternion has been dynamic for years. There are many research papers published in a variety of journals each year and different approaches have been taken for different purposes, and the study of quaternion matrices is still in development. As is expected, the main obstacle in the study of quaternion matrices is the non-commutative multiplication of quaternion. The theory on eigenvalues, determinant, singular values and trace of real and complex matrices has been well established. On the contrary, little is known for the trace of quaternion matrices. As usual, R and C are the set of the real and complex numbers. We denote by H (in honor of the inventor, Hamilton) the set of real quaternion: . HaaaiajakaaaaR==++{ 01 2 + 30123,,,, ∈}

* For aa=++01 aiajakH 2 + 3 ∈ , the conjugate of a is aa==−− a01 aiajak 2 − 3 and the norm of a is 22221/2 nn× n×1 N() a= aa = aa = ( a01 +++ a a 23 a ) . Let H and H be respectively the collections of all nn× matrices with entries in H and n -column vectors. Let In be the collections of all nn× unit matrices with entries in H . −−−− For XH∈ n×1 , X T is the transpose of X . If T , then T is the conjugate of X XXXX= (,12K n ) XXXX= (,12 ,,K n ) −− − and * is the conjugate transpose of X . The norm of X is defined to be * . For an XXXX= (,12 ,,K n ) N ()XXX=

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− matrix , the conjugate transpose of is the matrix **T . nn× A =∈()aaHij n× n ( ij ) A nn× AA==() aij n× n The research of matrices is continuously an important aspect of algebra problems over quaternion division algebra, the subject, such as eigenvalues, singular values, congruent and positive definite of self- conjugate matrices as well as sub-determinant of self- conjugate matrices and so on, has been extensively explored [4-15], while little is known for the trace of quaternion matrices. For linear algebraists and matrix theories, some basic questions on the trace of quaternion matrix are different from real or complex matrix. For instance, if A and B are the nn× matrices, then n Tr AB= Tr BA and are not always right. In this section, we introduce the notation and terminology. In () () Tr() A = ∑λi i section 2, we define some definitions and recall several lemmas. In section 3, we discuss the inequality about eigenvalues and trace of self-conjugate quaternion matrices. In section 4, we conclude the paper with several inequalities for the trace of quaternion positive definite matrix obtained by the result in section 3 and the Holder’s inequality over quaternion division algebra. 2. Definitions and Lemmas We begin this section with some basic definitions and lemmas. Definition 2.1 Let A∈ H nn× . A is said to be the self-conjugate quaternion matrix if A* = A . H()n,* is the set of self-conjugate quaternion matrices, Hn( ,>) is the set of quaternion positive definite matrices.

n×n * * Definition 2.2 Let A∈ H . A is said to be the quaternion unitary matrix if A A = AA = I n . H()n,u is the set of quaternion unitary matrices.

n n Definition 2.3 Let AH∈ nn× . is said to be the trace of matrix A , remarked by Tr(A). That is . ∑ aii Tr()A = ∑ aii i=1 i=1 Lemma 1. [4] Let A∈ H ()n,* . Then, A is unitary similar to a real diagonal matrix, that is, there exists a unitary matrix UHnu∈ ( , ), such that * , U AU = diag(λ1 ,λ2 ,L,λn ) where, are the eigenvalues of A . λ1 ,λ2 ,Lλn ∈ R Lemma 2. Let A∈ H ()n,* and B ∈ Hn( ,*) . If there existsUHnu∈ ( , ) , such that B = UAU * , then, TrA= TrB . Proof. Since A∈ H ()n,* , by Lemma 1, there exists VHnu∈ ( , ) , such that * , VAV= diag(λ12,,,λλL n ) where, are the eigenvalues of A . For any , we have λ1 ,λ2 ,Lλn ∈ R UHnu∈ ( , )

nn *2 (2.2) ∑∑uuij ij=== N() u ij 1,() j 1, 2,L , n ii==11

nn *2 . (2.3) ∑∑uuij ij=== N() u ij 1,() i 1, 2,L , n jj==11 So nnnn⎧⎫⎛⎞ nnn ⎧ ⎫ TrA== a vλδ v** = λ v δ v ∑∑∑∑ii⎨⎬⎨⎬ ik⎜⎟ k klil ∑∑∑ k ik kl il iikl====1111⎩⎭⎝⎠ ikl === 111 ⎩ ⎭ nn⎧⎫⎧⎫ n n n ==λ vv**λλ vv = ∑∑⎨⎬⎨⎬lllll ∑ l ∑ ll ll ∑ l il==11⎩⎭⎩⎭ i = 1 l = 1 i = 1 Meanwhile, we have B = UAU * , then **. B = UVdiag(λλ12,,,L λn ) V U Since UV∈ H( n, u) , therefore

n . TrA== TrB ∑ λl i=1 Thus

22 Modern Applied Science September, 2008

TrA= TrB (2.4) 3. The inequality for the trace of self-conjugate quaternion matrices It is well known that the eigenvalues and trace of any self-conjugated quaternion matrix are all real numbers. In this section, we shall discuss the inequality about eigenvalues and trace of self-conjugate quaternion matrices. Theorem 1. Let , their eigenvalues are , respectively, if A∈>∈Hn( ,,) B Hn( ,*) α1 ≥ α 2 ≥ L ≥ α n β1 ≥ β2 ≥L≥ β n A, B are commutative, then . Tr() AB ≤ ∑αiiβ i=1 Proof. Since , by lemma 1, there exists unitary matrices , such that A∈∈Hn( ,*) , B Hn( ,*) UU12,,∈ Hnu( ) * (3.1) U1 AU1 = diag(α1 ,α 2 ,L,α n ) * (3.2) U 2 BU 2 = diag(β1 , β 2 ,L, β n ) where .Therefore α i > 0()i = 1,2,Ln ⎡⎤**. Tr( AB) = Tr⎣⎦ U11212122 diag(αα,,,LL αnn) U U diag( ββ ,,, β) U By (3.1) and (3.2), we have ***. U11 ABU= diag(αα 12,,,LL αnn) U 1212 U diag( ββ ,,, β) U 21 U Let UU* == U u , it is easy to know UHnu∈ ( , ) , then 12 ( ij )nn×

nn *2 ∑∑uuij ij=== N() u ij 1,() j 1, 2,L , n ii==11

nn *2 . ∑∑uuij ij=== N() u ij 1,() i 1, 2,L , n ji==11 Since ()ABBABA* ==** , and A, B are commutative, then ()ABAB* = , so ABHn∈ ( ,*) . Hence, by Lemma 2, * , then Tr( AB) = Tr( U11 ABU ) ⎡⎤* Tr( AB) = Tr⎣⎦ diag(αα12,,,LL αnn) Udiag( ββ 12 ,,, β) U

22 2 ⎛⎞Nu()11 Nu () 12L Nu () 1n ⎛⎞β1 ⎜⎟⎜⎟ 22 2β ⎜⎟Nu()21 Nu () 22L Nu () 2n ⎜⎟2 . = ()αα12,,,L αn ⎜⎟⎜⎟M ⎜⎟LLLL ⎜⎟22 2⎜⎟ ⎝⎠Nu()nn12 Nu ()L Nu () nn⎝⎠βn Let

22 2 ⎛⎞ξ11⎛⎞Nu()11 Nu () 12L Nu () 1n ⎛β ⎞ ⎜⎟⎜⎟ ⎜ ⎟ ξ Nu22() Nu () Nu 2 ()β ⎜⎟22= ⎜⎟21 22L 2n ⎜ ⎟ ⎜⎟MM⎜⎟ ⎜ ⎟ ⎜⎟LLLL ⎜⎟⎜⎟22 2 ⎜ ⎟ ⎝⎠ξnn⎝⎠Nu()nn12 Nu ()L Nu () nn ⎝β ⎠ then

nn . (3.3) ∑∑ξii= β ii==11 For any kkn()1≤< , we have

kkn kkk kn 2 ⎛⎞ 2 ∑∑∑ξβββiijjiijiijj==−−+Nu() ∑∑∑⎜⎟1 Nu () ∑∑ Nu () β iij===111 iij === 111⎝⎠ ijk ==+ 11 kkk⎛⎞ kn 22 ≤−∑∑∑ββik⎜⎟1 −N ()uNu ijk + β ∑∑ () ij iij===111⎝⎠ ijk ==+ 11 kkk⎛⎞⎛⎞ k 22 =−∑∑∑ββik⎜⎟⎜⎟11 −N ()uNu ijk +− β ∑ () ij iij===111⎝⎠⎝⎠ j = 1

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k . (3.4) = ∑ βi i=1 By (3.3), (3.4) ,and , then α i > 0()i = 1,2,Ln

nn . Tr() AB =≤∑∑αiiξαβ i i ii==11 4. The inequality for the trace of quaternion positive definite matrix In this section, we first obtain an inequality for the trace of two quaternion positive definite matrices based on Theorem 1.Then by Theorem 1 and the Holder’s inequality over quaternion division algebra, the inequality for trace of the sum and multiplication of quaternion positive definite matrices is obtained. Theorem 2. Let , their eigenvalues are , respectively, if AHn∈>∈>( ,,) BHn( ,) α1 ≥ α 2 ≥ L ≥ α n β1 ≥ β2 ≥ L ≥ β A, B are commutative, then Tr() A++ Tr () B Tr22() A Tr () B Tr() AB≤≤ Tr() A Tr () B ≤ . 22

nn nn Proof. Since and , by Theorem 1,we have Tr() A==>∑∑ aiiα i 0 Tr() B= ∑∑ bii=>β i 0 ii==11 ii==11

n . Tr() AB ≤ ∑αiiβ i=1 Because of A∈>Hn( , ) , then there exists UHnu∈ ( , ), such that * UAU= diag(α12,,,ααL n ) where .So, α i > 0()i = 1,2,Ln *** *. (4.1) UABUUAUUBU== diag(αα12,,,L αn ) UBU

* * Since BHn∈>( , ) , then B and In are self-congruent, hence UBU and In are self-congruent, that is , UBUis quaternion positive definite matrix. For any main sub-matrix L of UABU* , by (4.1), we know that L can be obtained by the main sub-matrix G of UBU* .Then row＝ row , ||LL ||==> | |ααijLL | G | αα ij || G || 0 where ||⋅ || is the determinant of quaternion matrix defined by Chen L X [10]. So, UABU* is quaternion positive definite matrix, hence Tr( AB) > 0 .Since

nn n

Tr() A Tr () B−≥ Tr() AB ∑∑αβii −≥ ∑ αβ ii 0 ii==11 i = 1 so Tr() AB≤ Tr() A Tr () B . (4.2) Because of Tr22( A) +≥ Tr( B) 2 Tr( A) Tr( B) then 22TrATrBTrATrATrBTrB222( ) +≥+( ) ( ) 2( ) ( ) + 2( ) namely

22 2 Tr() A++ Tr () B⎛⎞ Tr() A Tr () B ≥ ⎜⎟ 22⎝⎠ so Tr() A++ Tr () B Tr22() A Tr () B ≤ . (4.3) 22

24 Modern Applied Science September, 2008

By (4.2) and (4.3), the conclusion holds. Theorem 3. Let . Their eigenvalues are , respectively, if AHn∈>∈>( ,,) BHn( ,) α1 ≥ α 2 ≥ L ≥ α n β1 ≥ β2 ≥L≥ β 11 p >+=1, 1 ,and A, B are commutative, then pq

11 Tr() AB≤ () Tr() Apqp () Tr() B q . Proof. By Theorem 1 and the Holder’s inequality, we have

11 nnnpq 11 ⎛⎞⎛⎞pq Pp qq . Tr() AB≤≤∑∑∑αβii⎜⎟⎜⎟ α i β i =() Tr() A() Tr () B iii===111⎝⎠⎝⎠ Specially, when pq==2 , we have

Tr() AB≤ ( Tr() A22) ( Tr () B ) . Theorem 4. Let . Their eigenvalues are , respectively, if AHn∈>∈>( ,,) BHn( ,) α1 ≥ α 2 ≥ L ≥ α n β1 ≥ β2 ≥L≥ β p >1 ,and A, B are commutative, then

1 11 (Tr()() A+≤ Bp ) p () Tr() Appp +() Tr() B p . p 11 Proof. Let q = , then pq>>1, 0, += 1 .Since p −1 pq ()()AB+=p AAB +pp−11 + BAB () +− then, by Theorem 3 and the Holder’s inequality, we have TrAB⎡⎤⎡+=ppp TrAAB +++−−11 BAB ⎤ ⎣⎦⎣() () () ⎦

11 11 pp−−11qqqq PPpp⎡⎤⎡⎤⎛⎞ ⎛⎞ ≤+++()TrA Tr⎜⎟() A B() TrB Tr ⎜⎟() A B ⎣⎦⎣⎦⎢⎥⎢⎥⎝⎠() ⎝⎠()

1 11 p−1 q q ⎡ ⎤ ⎡⎤⎛⎞PPp p =+⎢⎥Tr⎜⎟()() A B⎢() TrA +() TrB ⎥ ⎣⎦⎝⎠⎣ ⎦

p−1 p p 11 ⎡⎤⎛⎞p−1 ⎡ ⎤ =+⎢⎥Tr A Bp−1 TrAPPp + TrB p ⎜⎟()() ⎢()()⎥ ⎣⎦⎢⎥⎝⎠⎣ ⎦

p−1 11 p ⎡ ⎤ ⎡⎤p PPp p . =+Tr()() A B⎢() TrA +() TrB ⎥ ⎣⎦⎣ ⎦ That is

p−1 11 pp⎡ ⎤ ⎡⎤p PPp p . Tr⎡⎤() A+≤ B Tr() () A + B⎢() TrA +() TrB ⎥ ⎣⎦⎣⎦⎣ ⎦ So

1 11 (Tr()() A+≤ Bp ) p () Tr() Appp +() Tr() B p .

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