HADAMARD PRODUCT VERSIONS OF THE CHEBYSHEV AND KANTOROVICH INEQUALITIES

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Hadamard Product Versions Jagjit Singh Matharu and

Jaspal Singh Aujla vol. 10, iss. 2, art. 51, 2009

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HADAMARD PRODUCT VERSIONS OF THE CHEBYSHEV AND KANTOROVICH INEQUALITIES

JAGJIT SINGH MATHARU AND JASPAL SINGH AUJLA

Department of Mathematics National Institute of Technology Jalandhar 144011, Punjab, INDIA

EMail:matharujs@yahoo.com aujlajs@nitj.ac.in

Received: 10 February, 2009

Accepted: 15 April, 2009

Communicated by: S. Puntanen

2000 AMS Sub. Class.: Primary 15A48; Secondary 15A18, 15A45.

Key words: Chebyshev inequality, Kantorovich inequality, Hadamard product

Abstract: The purpose of this note is to prove Hadamard product versions of the Chebyshev and the Kantorovich inequalities for positive real numbers. We also prove a generalization of Fiedler’s inequality.

Acknowledgements: The authors thank a referee for useful suggestions.

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1. Introduction

In what follows, the capital letters A, B, C, . . . denote m × m complex matrices, whereas the small letters a, b, c, . . . denote real numbers, unless mentioned other- wise. By X ≥ Y we mean that X − Y is positive semidefinite (X > Y mean X −Y is positive definite). For A = (aij)and B = (bij), A◦B = (aijbij) denotes the Hadamard product of A and B. According to Schur’s theorem [4, Page 23] the Hadamard product is monotone in the sense thatA ≥ B, C ≥ D implies A◦C ≥B◦D. The tensor productA⊗B is them²×m² matrix

(1.1)





a₁₁B · · · a_1mB ... ... a_m1B · · · a_mmB



.

Marcus and Khan in [10] made the simple but important observation that the Hadamard product is a principal submatrix of the tensor product. The inequality

(1.2)

n

X

i=1

w_ia_i

! _n X

i=1

w_ib_i

!

≤

n

X

i=1

w_ia_ib_i

holds for all a₁ ≥ a₂ ≥ · · · ≥ a_n ≥ 0, b₁ ≥ b₂ ≥ · · · ≥ b_n ≥ 0 and weights w_i ≥ 0, i = 1, . . . , n. Hardy, Littlewood and Polya [6, page 43] attribute this inequality to Chebyshev. For0< a≤a_i ≤b, w_i ≥0, i= 1,2, . . . , n, Kantorovich’s inequality states that

(1.3)

n

X

i=1

w_ia_i

! _n X

i=1

wi

a_i

!

≤ (a+b)² 4ab

n

X

i=1

w_i

!2

.

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In Section2, we state and prove matrix versions of inequalities (1.2) and (1.3) involving the Hadamard product. A generalization of Fiedler’s inequality is also proved in this section. There are several generalizations of Kantorovich and Fiedler’s inequality; see [2,3,8,9].

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2. The Chebyshev and Kantorovich Inequalities:

Matrix Versions

We begin with a Hadamard product version of inequality (1.2).

Theorem 2.1. LetA1 ≥ · · · ≥An≥0andB1 ≥ · · · ≥Bn ≥0. Then

(2.1)

n

X

i=1

wiAi

!

◦

n

X

i=1

wiBi

!

≤

n

X

i=1

wi

! _n X

i=1

wi(Ai◦Bi)

! ,

wherew_i ≥0, i= 1, . . . , n,are weights.

Proof. We have

n

X

i=1

w_i

! _n X

i=1

w_i(A_i◦B_i)

!

−

n

X

i=1

w_iA_i

!

◦

n

X

i=1

w_iB_i

! (2.2)

=

n

X

i,j=1

(w_iw_j(A_j ◦B_j)−w_iw_j(A_i◦B_j))

= 1 2

n

X

i,j=1

w_iw_j(A_j ◦B_j)−w_iw_j(A_i◦B_j)

+w_jw_i(A_i◦B_i)−w_jw_i(A_j◦B_i)

= 1 2

n

X

i,j=1

w_iw_j(A_i−A_j)◦(B_i−B_j).

Since the Hadamard product of two positive semidefinite matrices is positive semidefinite, therefore the summand in2.2is positive semidefinite.

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Our next result is a Hadamard product version of inequality (1.3) .

Theorem 2.2. LetA₁, . . . , A_n be such that 0 < aI_m ≤ A_i ≤ bI_m, i = 1, . . . , n (hereI_m denotes them×midentity matrix). Then

(2.3)

n

X

i=1

W_i^1/2A_iW_i^1/2

!

◦

n

X

i=1

W_i^1/2A⁻¹_i W_i^1/2

!

≤ a²+b² 2ab

n

X

i=1

W_i

!

◦

n

X

i=1

W_i

!

for allW_i ≥0, i= 1, . . . , n.

Proof. We first prove the inequality

(2.4) P^1/2AP^1/2◦Q^1/2B⁻¹Q^1/2+P^1/2A⁻¹P^1/2◦Q^1/2BQ^1/2 ≤ a² +b²

ab (P ◦Q), when0< aI_m ≤ A, B ≤bI_m andP, Q≥ 0. LetA =U DU^∗ andB =VΓV^∗ with unitaryU andV, and diagonal matricesDandΓ. Then

A⊗B⁻¹+A⁻¹⊗B = (U ⊗V)(D⊗Γ + Γ⁻¹⊗D)(U ⊗V)^∗

≤(U ⊗V)

a²+b²

ab (I_m⊗I_m)

(U ⊗V)^∗

= a²+b²

ab (I_m⊗I_m), where the inequality follows from (1.3). Thus we have

P^1/2AP^1/2⊗Q^1/2B⁻¹Q^1/2 +P^1/2A⁻¹P^1/2⊗Q^1/2BQ^1/2 (2.5)

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= (P^1/2⊗Q^1/2)(A⊗B⁻¹ +A⁻¹⊗B)(P^1/2⊗Q^1/2)

≤ a²+b²

ab (P ⊗Q).

Since the Hadamard product is a principal submatrix of the tensor product, the inequality (2.4) follows from (2.5). On takingB =AandQ=P in (2.4) we see that (2.3) holds forn = 1. Further, by (2.4) we have

W_i^1/2A_iW_i^1/2◦W_j^1/2A⁻¹_j W_j^1/2+W_i^1/2A⁻¹_i W_i^1/2◦W_j^1/2A_jW_j^1/2 ≤ a²+b²

ab (W_i◦W_j) fori, j = 1, . . . , n. Summing overi, j, we have

(2.6) 2

n

X

i,j=1

h

W_i^1/2A_iW_i^1/2

◦

W_j^1/2A⁻¹_j W_j^1/2i

≤

a²+b² ab

ⁿ X

i,j=1

(W_i◦W_j),

which implies that

n

X

i=1

W_i^1/2A_iW_i^1/2

!

◦

n

X

i=1

W_i^1/2A⁻¹_i W_i^1/2

!

≤

a²+b² 2ab

n

X

i=1

W_i

!

◦

n

X

i=1

W_i

! .

The next corollary follows on takingW_i =w_iI_m,i= 1, . . . , n.

Corollary 2.3. LetA₁, . . . , A_nbe such that0< aI_m ≤A_i ≤bI_m,andw_i ≥0, i= 1, . . . , nbe weights. Then

n

X

i=1

w_iA_i

!

◦

n

X

i=1

w_iA⁻¹_i

!

≤

a² +b² 2ab

ⁿ X

i=1

w_i

!2

I_m.

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Remark 1. The casen= 1of Corollary2.3is proved in [7]. The example A=

2 1 1 1

, a= 3−√ 5

2 , b= 3 +√ 5 2 shows that the inequality

A◦A⁻¹ ≤ (a+b)² 4ab I₂ need not be true.

For our next result we need the following lemma.

Lemma 2.4. Let0≤r≤1. ThenA^r+A^−r≤A+A⁻¹ for allA >0.

Proof. Suppose thatA=UΓU^∗with unitaryU and diagonal matrixΓ. Then A^r+A^−r =U(Γ^r+ Γ^−r)U^∗

≤U(Γ + Γ⁻¹)U^∗ =A+A⁻¹

sincex^r+x^−r ≤x+x⁻¹for any positive real numberxand0≤r ≤1.

Theorem 2.5. Let0≤α < β. Then

n

X

i=1

W_i^1/2A^α_iW_i^1/2

!

◦

n

X

i=1

W_i^1/2A^−α_i W_i^1/2

!

≤

n

X

i=1

W_i^1/2A^β_iW_i^1/2

!

◦

n

X

i=1

W_i^1/2A^−β_i W_i^1/2

!

for allA_i >0andW_i ≥0,i= 1, . . . , n.

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Proof. We first prove the inequality

(2.7)

W_i^1/2A^α_iW_i^1/2

◦

W_j^1/2A^−α_j W_j^1/2 +

W_i^1/2A^−α_i W_i^1/2

◦

W_j^1/2A^α_jW_j^1/2

≤

W_i^1/2A^β_iW_i^1/2

◦

W_j^1/2A^−β_j W_j^1/2 +

W_i^1/2A^−β_i W_i^1/2

◦

W_j^1/2A^β_jW_j^1/2

for0≤α < β. Let0≤r≤1. Then

W_i^1/2A^r_iW_i^1/2

⊗

W_j^1/2A^−r_j W_j^1/2 +

W_i^1/2A^−r_i W_i^1/2

⊗

W_j^1/2A^r_jW_j^1/2

=

W_i^1/2⊗W_j^1/2

A^r_i ⊗A^−r_j +A^−r_i ⊗A^r_j

W_i^1/2⊗W_j^1/2

=

W_i^1/2⊗W_j^1/2 A_i⊗A⁻¹_j r

+ A_i⊗A⁻¹_j −r

W_i^1/2⊗W_j^1/2

≤

W_i^1/2⊗W_j^1/2 A_i⊗A⁻¹_j

+ A_i⊗A⁻¹_j ⁻¹

W_i^1/2⊗W_j^1/2

where the inequality follows from Lemma2.4. Takingr=α/βand replacingA_i by A^β_i andA_j byA^β_j, we have

W_i^1/2A^α_iW_i^1/2

⊗

W_j^1/2A^−α_j W_j^1/2 +

W_i^1/2A^−α_i W_i^1/2

⊗

W_j^1/2A^α_jW_j^1/2

≤

W_i^1/2A^β_iW_i^1/2

⊗

W_j^1/2A^−β_j W_j^1/2

+

W_i^1/2A^−β_i W_i^1/2

⊗

W_j^1/2A^β_jW_j^1/2

. Again using the fact that the Hadamard product is a principal submatrix of the tensor

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product, the preceding inequality implies (2.7). Summing overi, j in (2.7), we have

n

X

i=1

W_i^1/2A^α_iW_i^1/2

!

◦

n

X

i=1

W_i^1/2A^−α_i W_i^1/2

!

≤

n

X

i=1

W_i^1/2A^β_iW_i^1/2

!

◦

n

X

i=1

W_i^1/2A^−β_i W_i^1/2

! .

Corollary 2.6. Let0≤α < β. Then

n

X

i=1

A^α_i

!

◦

n

X

j=1

A^−α_j

!

≤

n

X

i=1

A^β_i

!

◦

n

X

j=1

A^−β_j

!

for allA_i >0,i= 1, . . . , n.

Proof. TakingW_i =I_m in Theorem2.5we get the desired result.

Corollary 2.7. Let0≤β. Then I_m ≤

n

X

i=1

W_i^1/2A^β_iW_i^1/2

!

◦

n

X

i=1

W_i^1/2A^−β_i W_i^1/2

!

for allA_i >0andW_i ≥0,i= 1, . . . , n, wherePn

i=1W_i =I_m. Proof. Takingα = 0in Theorem2.5gives the desired inequality.

Remark 2. Corollary2.7is another generalization of Fiedler’s inequality [5]

A◦A⁻¹ ≥I_m.

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Next we prove a convexity theorem involving the Hadamard product.

Theorem 2.8. The function

f(t) =A^1+t◦B^1−t+A^1−t◦B^1+t

is convex on the interval[−1,1]and attains its minimum att= 0for allA, B >0.

Proof. Sincef is continuous we need to prove only thatf is mid-point convex. Note that forA, B >0ands, tin[−1,1]the matrices

A^1+s+t A^1+s A^1+s A^1+(s−t)

,

A^1−(s+t) A^1−s A^1−s A^1−(s−t)

, B^1+s+t B^1+s

B^1+s B^1+(s−t)

,

B^1−(s+t) B^1−s B^1−s B^1−(s−t)

are positive semidefinite. Hence the matrix X=

A^1+s+t◦B^1−(s+t)+A^1−(s+t)◦B^1+s+t A^1+s◦B^1−s+A^1−s◦B^1+s A^1+s◦B^1−s+A^1−s◦B^1+s A^1+(s−t)◦B^1−(s−t)+A^1−(s−t)◦B^1+(s−t)

is positive semidefinite. Similarly, the matrix Y=

A^1+(s−t)◦B^1−(s−t)+A^1−(s−t)◦B^1+(s−t) A^1+s◦B^1−s+A^1−s◦B^1+s A^1+s◦B^1−s+A^1−s◦B^1+s A^1+(s+t)◦B^1−(s+t)+A^1−(s+t)◦B^1+s+t

is positive semidefinite. Hence

(2.8) X+Y =

f(s+t) +f(s−t) 2f(s) 2f(s) f(s+t) +f(s−t)

is positive semidefinite, which implies that f(s)≤ 1

2[f(s+t) +f(s−t)].

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This proves the convexity off. Further, note that f(t) = f(−t). This together with the convexity off implies thatf attains its minimum at 0.

Corollary 2.9. The function

g(t) = A^t◦B^1−t+A^1−t◦B^t

is decreasing on [0,1/2], increasing on [1/2,1], and attains its minimum at t = ¹₂ for allA, B >0.

Proof. The proof follows on replacingA, B byA^1/2, B^1/2 andtby ^1+t₂ in Theorem 2.8.

A norm|||·|||onm×mcomplex matrices is called unitarily invariant if|||U XV|||=

|||X|||for all unitary matricesU, V. IfAis positive semidefinite andXis any matrix, then

|||A◦X||| ≤max a_ii|||X|||

for all unitarily invariant norms||| · |||[1]. Thus the proof of the following corollary follows from Corollary2.9using the fact thatg(1/2)≤g(t)≤g(1) =g(0).

Corollary 2.10. Let0≤t≤1. Then,

2|||A^1/2◦B^1/2||| ≤ |||A^t◦B^1−t+A^1−t◦B^t||| ≤ |||A+B|||

for all unitarily invariant norms||| · |||and allA, B >0.

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References

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[2] J.K. BAKSALARYAND S. PUNTANEN, Generalized matrix versions of the Cauchy-Schwarz and Kantorovich inequalities, Aequationes Math., 41 (1991), 103–110.

[3] R.B. BAPAT AND M.K. KWONG, A generalisation of A◦A⁻¹ ≥ I, Linear Algebra Appl., 93 (1987), 107–112.

[4] R. BHATIA, Matrix Analysis, Springer Verlag, New York, 1997.

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[8] A.W. MARSHALL AND I. OLKIN, Matrix versions of the Cauchy and Kan- torovich inequalities, Aequationes Math., 40 (1990), 89–93.

[9] M. SINGH, J.S. AUJLA AND H.L. VASUDEVA, Inequalities for Hadamard product and unitarily invariant norms of matrices, Linear Multilinear Algebra, 48 (2000), 247–262.

[10] M. MARCUSANDN.A. KHAN, A note on Hadamard product, Canad. Math.

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