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volume 7, issue 2, article 65, 2006.

Received 28 January, 2006;

accepted 16 February, 2006.

Communicated by:F. Hansen

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Journal of Inequalities in Pure and Applied Mathematics

ON AN INEQUALITY INVOLVING POWER AND CONTRACTION OF MATRICES WITH AND WITHOUT TRACE

MARCOS V. TRAVAGLIA

Universidade de Brasília

Centro Internacional de Física da Matéria Condensada Caixa Postal 04513

Brasília- DF

CEP 70904-970 Brazil.

EMail:mvtravaglia@unb.br

c

2000Victoria University ISSN (electronic): 1443-5756 027-06

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On an Inequality Involving Power and Contraction of Matrices with and without Trace

Marcos V. Travaglia

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J. Ineq. Pure and Appl. Math. 7(2) Art. 65, 2006

http://jipam.vu.edu.au

Abstract

LetAandBbe positive semidefinite matrices. Assuming that the eigenvalues ofBare less than one, we prove the following trace inequalities

Trn

(BA^αB)^1/αo

≤Trn

(BA^βB)^1/βo

and Tr

n

(BA^αB)^1/α o

≤Tr

B^α/βA^αB^α/β 1/α

, for all0< α≤β. Moreover we prove that

Tr

B^α/βA^αB^α/β 1/α

≤Tr

BA^βB 1/β

,

for all0< α≤βand0< α≤1. Furthermore we prove that (BA^αB)^1/α ≤ (BA^βB)^1/β

in the cases (a)1≤α ≤βor (b) ¹₂ ≤α≤ βandβ ≥1. Further we present counterexamples involving2×2matrices showing that the last inequality is, in general, violated in case that neither (a) nor (b) is fulfilled.

2000 Mathematics Subject Classification:15A45, 15A90, 47A63.

Key words: Trace inequalities, Operator inequalities, Positive semidefinite matrix, Operator monotony, Operator concavity.

1. Introduction

LetM_nbe the space ofn×ncomplex matrices. We say thatA∈M_nis positive ifAis Hermitian, that isA^∗ = A, and its eigenvaluesλi(A)(i = 1, . . . , n) are nonnegative. A positive matrixAis denoted by0≤ Aand we say thatA ≤B if0≤B−A. The identity matrix is denoted byI.

Our main result is the proof of the following inequalities involving the matrix (BA^αB)^1/αwithα >0:

Theorem 1.1. Let beA, B ∈M_nwith0≤Aand0≤B ≤I. Defining H(α) := (BA^αB)^1/α and h(α) := Tr{H(α)} , we prove the following operator and trace inequalities.

a) For all1≤α ≤βwe have

(1.1) H(α)≤H(β).

b’) For all1/2≤α≤1andβ = 1we have

(1.2) H(α)≤H(β= 1).

b") For all0< α <1/2andβ = 1we can find matricesA≥0and0≤B ≤ Isuch that

(1.3) H(α)H(β= 1).

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b) Combining a) with b’)

(1.4) H(α)≤H(β)

holds for all1/2≤α≤β andβ ≥1.

c) For all0< α≤β we prove that (1.5) h(α)≤h(β), that is, Tr

(BA^αB)^1/α ≤Tr

(BA^βB)^1/β .

d) For all0< α≤β we have

(1.6) Trn

(BA^αB)^1/αo

≤Trn

B^α/βA^αB^α/β^1/αo .

e) For all0< α≤β and0< α≤1we prove that

(1.7) Trn

B^α/βA^αB^α/β1/αo

≤Trn

BA^βB1/βo .

Remark 1. The item a) is the main inequality of Theorem1.1. As we will see, it is a direct consequence of the following result of F. Hansen [8]:

“If f is an operator monotone function defined on the interval [0,∞), then Kf(X)K^∗ ≤f(KXK^∗)holds for everyX ≥0and contractionK.”

See also Lemma2.1.

A proof of c) can be obtained combining a) with d) and e). More precisely, c) follows from a) in the caseα ≥1and from d) and e) in the case0< α ≤1.

The author would like to thank F. Hansen for indicating a simpler proof of c) which does not make use of d) and e). This simpler proof is presented below.

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We would like to state our discussion, motivation and background of Theo- rem1.1as follows.

A motivation to prove the inequality (1.5) is the application of the well- known trace inequality |Tr{X}| ≤ Tr{|X|}, X ∈ M_n for the particular case where X = AB with 0 ≤ A and 0 ≤ B ≤ I. Here we use the definition

|X| := (X^∗X)^1/2. Applying this trace inequality we obtain h(1) ≤ h(2), because:

h(1) := Tr{BAB}= Tr{AB²}= Tr{A^1/2B²A^1/2} (1.8)

≤Tr{A^1/2BA^1/2}= Tr{AB}

≤Tr{|AB|}= Tr{(BA²B)^1/2}=:h(2),

where in the first inequality of (1.8) we used that0≤B² ≤Bsince0≤B ≤I.

We also used that|Tr{AB}|= Tr{AB}becauseTr{AB}= Tr{A^1/2BA^1/2} ≥ 0since bothAandB are positive matrices. Similar to (1.8), we can also show thath ₂k+1¹

≤h ₂¹k

fork = 0,1,2, . . . .

Considering the special caseα= 1/2andβ= 1of (1.2) we can easily prove thatH(1/2)≤H(1), namely:

BAB−(BA^1/2B)² =BA^1/2A^1/2B−BA^1/2B²A^1/2B

=BA^1/2(I−B²)A^1/2B ≥0,

because0≤B ≤I and soI−B² ≥0.

Now we put the inequalities presented in Theorem 1.1 in the context of known results. More precisely, we will derive two particular cases of (1.5) and (1.1) from [1], [3], [11], [10] and [6]. However, we need to impose some

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restrictions onAandBin the hypothesis of theorem1.1. These restrictions are 1)B is a projection and 2)0≤B ≤A≤I.

1) Considering the restriction that B = P is a projection we can show the following two particular cases of (1.5):

• The first particular case of (1.5) is

(1.9) h(1/k)≤h(1) for all k = 1,2,3, . . . .

We can derive this trace inequality using the following result by Ando, Hiai and Okubo [1]:

“For semidefinite matricesA,Bthe inequaltity Tr{A^p¹B^q¹· · ·A^p^NB^q^N} ≤Tr{AB}

holds withpi,qi ≥0andPN

i=1pi =PN

i=1qi = 1.”

Applying this result toB =P andp_i =q_i = 1/kwe have h(1) = Tr{P AP}= Tr{AP} ≥Trn

A^k¹P · · ·A¹^kPo

= Trn

P A¹^kP

· · ·

P A¹^kPo

= Tr{(P A^1/kP)^k}=h(1/k),

which proves (1.9).

• The second particular case of (1.5) is

(1.10) h(α)≤h(1) for all 0< α≤1.

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This trace inequality can be derived from the Berezin-Lieb inequality ([3], [11]). To understand this, recall that the Berezin-Lieb inequality states thatTr{f(P XP)} ≤ Tr{P f(X)P}holds ifP is a projection andf is a convex function on an interval containing the spectrum of X. Now taking X = A^α and f(λ) = λ^1/α (0 < α ≤ 1) we obtain (1.10), because

h(α) = Trn

(P A^αP)^1/αo

≤Tr{P(A^α)^1/αP}=h(1).

2) Considering the restriction 0 ≤ B ≤ A ≤ I we will show the following two particular cases of (1.1):

• The first particular case of (1.1) is

(1.11) H(1) ≤H(2), that is BAB ≤(BA²B)^1/2.

Remark 2. Although we haveBA²B ≤(BA²B)^1/2andBA²B ≤BAB(since 0 ≤ A, B ≤ I) we cannot conclude from these two operator inequalities that BAB ≤(BA²B)^1/2.

• We can derive the operator inequality (1.11) based on the following result in Kamei [10] which is a variation of [5]:

(1.12) 0≤B ≤Aassures

B^s/2A^pB^s/2^1+s_p+s

≥B^s/2AB^s/2 forp≥1ands≥0.

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To understand this, takep=s = 2in (1.12), namely, we obtain BA²B³₄

≥BAB.

On the other hand,

BA²B¹₂

≥ BA²B³₄ becauseBA²B ≤BIB≤I since0≤A, B ≤I.

• The second particular case of (1.1) is a generalization of the first one. More precisely

(1.13) H(1)≤H(p), that is,BAB ≤(BA^pB)^1/p for allp≥1.

We can obtain the above operator inequality using the following result (see [6]) which is also a variant of [5] and a more precise estimation than (1.12):

The functionF_r(p) = (B^rA^pB^r)^1+2r^p+2r forp≥1, r ≥0 (1.14)

is operator increasing as a function ofpwhenever0≤B ≤A.

Now the operator inequality H(1) ≤ H(p) follows from (1.14) setting r= 1, that is,

(BA^pB)^p+2³ =F₁(p)≥F₁(1) =BAB.

On the other hand,

(BA^pB)^p¹ ≥(BA^pB)^p+2³

becauseBA^pB ≤ BIB ≤ I (0 ≤ A, B ≤ I) and 3/(p+ 2) ≥ 1/pfor p≥1.

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We shall state the following couterexamples associated with Theorem1.1.

Counterexamples. In order to show that we cannot generally drop "Tr" from the inequality (1.5) apart from the cases a)1 ≤ α ≤ β or b)1/2 ≤ α ≤ β andβ ≥ 1, consider the following concrete example of2×2matrices: Let be B := ¹_{0 1/2}⁰

andA:= 64P +QwithP := 1/2 (^{1 1}_{1 1})andQ:=I−P orthog- onal projections. Since we are working with 2×2matrices, we observe that Det [H(β) − H(α)] < 0impliesH(β) H(α). Based on this observation we calculate the following determinants:

1) Det [H(1) − H(1/3)] =−81/16≈ −5.06<0 2) Det [H(2/3) − H(1/2)] = 12− ⁶³₂₆√

26≈ −0.36<0

3) Det [H(2/3) − H(1/3)] = 9−²¹¹⁵₈₃₂√

26≈ −3.96<0

4) Det [H(1/3) − H(1/6)] =−9446625/2097152 ≈ −4.5045<0 5) Det [H(1/2) − H(1/3)] =−225/128≈ −1.76<0

6) Det [H(4/3) − H(1/3)]≈ −3.5<0 and conclude that the respective affirmatives:

1) H(β)≥H(α)holds for all0< α <1/2andβ = 1 2) H(β)≥H(α)holds for all1/2≤α < β <1 3) H(β)≥H(α)holds for all0< α <1/2< β <1 4) H(β)≥H(α)holds for all0< α < β < 1/2

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5) H(β)≥H(α)holds for all0< α < β = 1/2 6) H(β)≥H(α)holds for all0< α <1/2andβ >1 are false.

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2. Proof of Theorem 1.1

Definition 2.1. We say that a real functionf is operator concave on the interval I when for all real numbers0≤λ≤1,

f((1−λ)X+λY)≥(1−λ)f(X) +λf(Y)

for every pairX,Y ∈M_nwhose spectra lie in the intervalI. Likewise we say that f is operator monotone when f(X) ≤ f(Y) for every pairX, Y ∈ M_n withX ≤Y.

Lemma 2.1. [Operator concavity, monotony and contractions, part of Theo- rems 2.1 and 2.5 of F. Hansen and G. K. Pedersen [9]]. Letf : [0,∞)→[0,∞) be a continuous function then the following conditions are equivalent:

(i) f is operator concave on[0,∞). (ii) f is operator monotone.

(iii) Kf(X)K^∗ ≤ f(KXK^∗)for every contraction K (i.e. kKk ≤ 1, where k · kis the operator norm) and for every matrixX ≥0.

(iv) P f(X)P ≤f(P XP)for all projectionsP and matricesX ≥0.

A functionfis called operator convex if the function−fis operator concave.

As an example of a contraction we have a matrixB ∈M_nwith0≤B ≤I.

Lemma 2.2. Let beR, S ∈ Mnwith0≤R and0≤ S ≤I then the following estimate holds for allα >0

(2.1) Trn

(SRS)^1/αo

≤Tr

R^1/α .

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In order to give a proof of Lemma2.2 and c) of Theorem1.1, we state the following Lemma2.3which is derived from the minimax principle for the sake of convenience for readers:

Lemma 2.3. [[7], Lemma 1.1]. If A and B are n × n positive semidefinite matrices such thatA ≥ B ≥ 0, then their eigenvalues ofAandB are ordered as

λ_j(A)≥λ_j(B) forj = 1,2, . . . , n.

Proof of Lemma2.2. First we observe that matricesXY andY Xhave the same eigenvalues with the same multiplicities for X, Y ∈ M_n. Let be 0 ≤ R and 0 ≤ S ≤ I and using this observation withX = SR^1/2 andY = R^1/2S we have

(2.2) λ_i(SRS) =λ_i(SR^1/2R^1/2S) =λ_i(R^1/2S²R^1/2)

fori= 1,2, . . . , n.SinceS² ≤S ≤I we have thatR^1/2S²R^1/2 ≤R^1/2R^1/2 = R. From the last operator inequality it follows from Lemma2.3that the eigenvalues ofR^1/2S²R^1/2 andRare ordered as

(2.3) 0≤λ_i(R^1/2S²R^1/2)≤λ_i(R)

fori= 1,2, . . . , n. From (2.2) and (2.3) we have for allα >0that Tr{(SRS)^1/α}=

n

X

i=1

λ_i(SRS)^1/α

=

n

X

i=1

λ_i(R^1/2S²R^1/2)^1/α ≤

n

X

i=1

λ_i(R)^1/α = Tr

R^1/α ,

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which proves the Lemma2.2.

Proof of a) of (1.1) in Theorem1.1. Defining r := α/β we have 0 < r ≤ 1 since0< α ≤ β. Since the functionf(t) = t^r,0 ≤r ≤1is operator concave (and monotone) on [0,∞) and the matrix B is a contraction we conclude by Lemma2.1, settingX =A^β andK =B that

(2.4) BA^αB =B(A^β)^rB ≤(BA^βB)^r. holds for all0< α≤β.

On the other hand using the fact that the function f(t) = t^s, 0 ≤ s ≤ 1is operator monotone and taking s = 1/α ≤ 1 (since1 ≤ α ), it follows from (2.4) that

(2.5) (BA^αB)^1/α ≤(BA^βB)^r/α = (BA^βB)^1/β, which proves a).

Proof of b’) and b” ) in Theorem1.1. In the case 1/2 ≤ α ≤ 1andβ = 1we have 1 ≤ r := 1/α ≤ 2. Based on the fact that the function f(t) = t^r is operator convex on [0,∞)if and only if1 ≤ r ≤ 2(see [4] Theorem V.2.9) it follows by Lemma2.1settingX :=A^1/r withr := 1/αandK :=B that

H(α) := (BA^αB)^1/α = BA^1/rB^r

= (KXK^∗)^r

≤KX^rK^∗ =BAB :=H(β = 1),

which proves b’).

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In the case0 < α < 1/2andβ = 1 we haver = 1/α > 2which means that the functionf(t) =t^r is not operator convex. It follows from Lemma2.1 that we can find a matrix X ≥ 0and a projection P such that (P XP)^1/α P X^1/αP. TakingA =X^1/αandB =P we have

H(α) := (BA^αB)^1/α = (P XP)^1/α

P X^1/αP =BAB =H(β = 1),

which proves b” ).

Proof of c) in Theorem1.1. First we recall that the operator inequality (2.4),

(2.6) BA^αB ≤ BA^βBα/β

,

holds for all 0 < α ≤ β. From (2.6) it follows from Lemma 2.3 that the eigenvalues ofBA^αB and BA^βBα/β

are ordered as (2.7) λ_i(BA^αB)≤λ_i((BA^βB)^α/β) fori= 1, . . . , n.

From (2.6), (2.7) and since the functionf(t) = t^1/αis increasing we obtain Trn

(BA^αB)^1/αo

=

n

X

i=1

λ_i(BA^αB)^1/α

≤

n

X

i=1

λ_i

BA^βB^α/β1/α

= Trn

BA^βB1/βo , which proves c).

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Proof of d) in Theorem1.1. Setting r = α/β, S := B^1−r and R := B^rA^αB^r we have0 ≤ S ≤I (because0 < r ≤ 1) and0≤ R. Applying the inequality (2.1) for this choice ofRandSwe obtain

Trn

(BA^αB)^1/αo

= Tr

B^1−r(B^rA^αB^r)B^1−r1/α

≤Tr

B^rA^αB^r1/α , (2.8)

which proves d).

Proof of e) in Theorem1.1. First we note that we can expressTr

B^rA^αB^r1/α

as a norm, namely:

(2.9) Tr

B^rA^αB^r1/α

=kB^rA^αB^rk^1/α_1/α=

B^r(A^β)^rB^r

1/α 1/α , wherek · k_1/αis the1/α-trace norm which is an unitarily invariant norm (note that1/α≥1since in our hypothesis0< α≤1).

On the other hand a result from [4] (Theorem IX.2.10) states that for every unitarily invariant norm||| · |||we have

(2.10) |||B^rA^rB^r||| ≤ |||(BAB)^r|||

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for all0≤r≤1ifAandB are positive matrices. It follows from (2.10) that B^r(A^β)^rB^r

1/α 1/α≤

BA^βBr

1/α 1/α

= Tr

BA^βB r

1/α

= Tr

BA^βBr/α

= Tr{(BA^βB)^1/β} (2.11)

where we could drop the | · | within the trace in the above estimate because BA^βBis a positive matrix. Now the proof of e) in Theorem1.1follows directly from (2.9) and (2.11).

Acknowledgments: The author was first motivated to work on the inequalities presented in this paper during investigations on the Hubbard model together with V. Bach from the Institute of Mathematics at the University of Mainz, Germany.

The Hubbard model describes electrons in solids and poses several mathematical challenges (e.g. [12]). In this model, the extreme points of the set {B ∈Mn|0≤B ≤I}correspond to many-electron functions (Slater Determi- nants). We searched for extreme points that minimize the functional energy of the model. Applying the trace inequality (1.5) for the caseβ = 1V. Bach and the author could reproduce a special result from V. Bach and J. Poelchau [2] for the Hubbard model. Details on the Hubbard model and this application of the trace inequality on it go beyond the scope of this study and will be published separately.

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I would like to thank V. Bach for his participation in discussions, incen- tivation and proof-reading and acknowledge the hospitality of the Institute of Mathematics at the University of Mainz, where part of the current results were obtained. The author would further like to thank R. Bhatia from the Indian Statistical Institute in New Delhi for taking a look at the first version of my manuscript and indicating some references. I would also like to thank F. Hansen from the Institute of Economics at the University of Copenhagen for indicating some original references and a simpler proof than the original for trace inequality (1.5). Finally the author thanks for the financial support from Convênio Fundação Universidade de Brasília – Instituto Brasileiro de Energia e Materi- ais and Ministério da Ciência e Tecnologia. I would like to thank the careful (including the verification of counterexamples) and helpful work of the referee.

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[10] E. KAMEI, A satellite to Furuta’s inequality, Math. Japon., 33 (1988), 883–886.

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