M athematical I nequalities & A pplications
Volume 16, Number 1 (2013), 93–99
doi:10.7153/mia-16-06
COMPLETE INTERPOLATION OF MATRIX VERSIONS OF HERON AND HEINZ MEANS
RUPINDERJIT K AUR
AND
M ANDEEP S INGH
(Communicated by Josip Peˇcari´c) Abstract. of a matrix version of Heron mean, Fα (a,b) = √ The interpolation and comparison + (1 − α ) ab + α a+b 2 , 0 6 α 6 1, a,b ∈ R is considered by R. Bhatia in [1]. We shall discuss the complete interpolation and comparison of matrix version of such means by extending the range of α from [0,1] to R+ . We shall also discuss some more results involving Heinz means.
1. Introduction In what follows, the capital letters A, B,C, . . . denote the n × n ( n arbitrary but fixed) matrices over the algebra of complex numbers, i.e. elements of Mn . By Pn and Sn , we denote the set of positive definite and the set of positive semidefinite matrices respectively. The Schur product of two matrices A = (ai j )i, j and B = (bi j )i, j in Mn is defined to be the matrix A ◦ B whose i, j− entry is ai j bi j . For any matrix A ∈ Mn , σ (A) denotes the set of singular values of A i.e. eigenvalues of (A∗ A)1/2 . The symbol |||.||| denotes unitarily invariant norms throughout this paper. Heron and Heinz means are two families of means defined respectively as, √ a+b (1.1) Fα (a, b) = (1 − α ) ab + α 2 a1−α bα + aα b1−α Hα (a, b) = (1.2) 2 for 0 6 α 6 1 and a, b ∈ R+ . The first family is clearly the linear interpolant between arithmetic and geometric mean and satisfies Fα 6 Fβ whenever α 6 β and α , β ∈ R+ . Using simple arguments it is proved in [1] that Hν (a, b) 6 Fα (ν ) (a, b)
(1.3)
for α (ν ) = (2ν − 1)2 and 0 6 ν 6 1. There is yet, another mean of interest in several branches of science like geometry, statistics and thermodynamics, the logarithmic mean, defined as L(a, b) =
a−b = loga − logb
Z1
at b1−t dt.
(1.4)
0
Mathematics subject classification (2010): 15A60, 47A30, 47A63, 47A64. Keywords and phrases: Positive definite matrices, matrix means, norm inequalities. c D l , Zagreb
Paper MIA-16-06
93
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In [1], R. Bhatia proved that f (α ) 6 f (1/2) for α ∈ [0, 1/2], while f (α ) is an increasing function of α on [1/2, 1] with f (α ) as one of the possible matrix version of (1.1) and is defined as follows AX + XB 1/2 1/2 f (α ) = (1 − α )A XB + α (1.5) , 2 for A, B ∈ Pn and X ∈ Mn . For the matrix version of (1.2) and (1.4), and more about these, the reader may refer to [2],[4] and [5]. In this note, we shall prove f (α ) 6 f (1/2) for 0 6 α 6 1/2 and f (α ) is an increasing function for α ∈ [1/2, ∞). This is in fact the generalization of monotonic property of matrix version of (1.1) as in the case of α ∈ R+ of Fα (a, b) for a, b ∈ R+ . As an outcome of above result we shall also present a possible generalized matrix analogue of (1.3) i.e. AX + XB 1 1−ν 1−ν 1/2 1/2 ν ν + A XB ||| 6 (1 − α )A XB + α |||A XB (1.6) 2 2 for 1/4 6 ν 6 3/4 and α ∈ [1/2, ∞). A comparison of possible matrix version of (1.4) i.e. Z1 At XB1−t dt 0
for A, B ∈ Pn , X ∈ Mn and (1.5) for α ∈ [1/2, ∞) is established. Some more general results are indicated. We shall further conclude similar result for another linear interpolant matrix version of the Heinz and the arithmetic mean, i.e. AX + XB α g(α ) = (1 − )(A2/3 XB1/3 + A1/3XB2/3 ) + α . 2 2 2. Main results
We shall use the following Theorem (2.1) and Lemma (2.2) in the sequel. T HEOREM 2.1. For A, B ∈ Mn with A as positive semidefinite, we have |||A ◦ B||| 6 max aii |||B||| where aii ’s for i = 1, 2, · · · , n are the diagonal entries of matrix A. L EMMA 2.2. Let σ1 , σ2 , · · · , σn be positive numbers, −1 6 r 6 1, and −2 < t 6 2. Then the n × n matrix ! σir + σ rj W= σi2 + t σi σ j + σ 2j is positive semidefinite.
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For a proof of Theorem (2.1) reader may refer to Horn and Johnson ([6], p. 343) and for Lemma (2.2), Bhatia and Parthasarthy [3] and Zhan ([8], p. 75–76). T HEOREM 2.3. For A, B ∈ Pn and X ∈ Mn and |||.||| any unitarily invariant norm, the function AX + XB 1/2 1/2 f (α ) = (1 − α )A XB + α 2
is increasing for α ∈ [1/2, ∞) and f (α ) 6 f (1/2) for α ∈ [0, 1/2].
Proof. We first prove the result for α > 0 and A = B, i.e., AX + XA α f (α ) = (1 − α )A1/2XA1/2 + α = 2 p(α ), 2 where, p(α ) = ||| h(α )A1/2 XA1/2 + AX + XA ||| and h(α ) = 2(α −1 − 1). We may assume without loss of generality, A = diag(λ1 , λ2 · · · , λn ), λi > 0, (due to unitarily invariant property of |||.|||). Then
where
h(α )A1/2 XA1/2 + AX + XA 1/2 1/2 = h(α )λi λ j + λi + λ j xi j i, j 1/2 1/2 h(α )λi λ j + λi + λ j ◦ h(β )A1/2 XA1/2 + AX + XA = 1/2 1/2 h(β )λi λ j + λi + λ j i, j 1/2 1/2 = Z ◦ h(β )A XA + AX + XA , Z
1/2 1/2 λ j + λi + λ j = 1/2 1/2 h(β )λi λ j + λi + λ j
h(α )λi
. i, j
Now the matrix Z can be written as 1/2 1/2 (h(α )−h(β ))λi λ j h( α )−h( β ) 1+ = (1) + λ 1/2 λ 1/2 , i j i, j 1/2 1/2 1/2 1/2 h(β )λi λ j +λi +λ j h(β )λi λ j +λi + λ j i, j
i, j
which will be positive semidefinite if the matrix, h(α ) − h(β )
1/2 1/2 λ j + λi + λ j
h(β )λi
, i, j
is positive semidefinite. By Lemma (2.2), the later matrix is positive semidefinite if and only if h(α ) > h(β ) and 2 > h(β ) > −2. Since h(α ) = 2(α −1 − 1), so continuously
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decreasing function on positive half line from [1/2, ∞)→ (−2,2]. Hence h(α ) > h(β )
α )+2 for all β > α . Using Theorem (2.1), we get p(α ) 6 h( h(β )+2 p(β ). This proves the result for A = B and α ∈ [1/2, ∞). For α ∈ (0, 1/2], we have 2 6 h(α ) < ∞ and h(α ) > h(1/2) = 2 and so the matrix Z with β = 1/2 is positive semidefinite using Lemma (2.2). The case α = 0 trivially holds, since by Lemma (2.2) the matrix
1/2 1/2 λj 1/2 1/2 λi λ j + λi + λ j i, j
λi
1/2 = λi
1 1/2 1/2 λ j + λi + λ j
λi
λ 1/2 j
i, j
is positive semidefinite. This gives us the desired result for this case, i.e. α p(α ) 6 1 for all α ∈ [0, 1/2]. 2 p(1/2). Equivalently saying that f (α ) 6 f (1/2) A0 0X The general case follows on replacing A by and X by . 0B 0 0 C OROLLARY 2.4. Let A, B, X ∈ Mn with A, B positive definite. Then for any unitarily invariant norm |||.||| and a matrix monotone increasing function f : (0, ∞) → (0, ∞) with f ⊥ (x) = x( f (x))−1 , 1 |||A1/4 ( f (A1/2 )X f ⊥ (B1/2 ) + f ⊥ (A1/2 )X f (B1/2 ))B1/4 ||| 2 AX + XB 1/2 1/2 6 (1 − α )A XB + α 2
(2.1)
holds for α ∈ [1/2, ∞).
Proof. For α = 1/2, (2.1) has already been proved in Singh and Vasudeva ([7], Theorem 1.1, p. 622). Now, using Theorem (2.3), we get the desired result. R EMARK 2.5. We remark here that above corollary (2.4) is one of the possible generalizations of an inequality by Zhan ([8] Th.4.24, p. 76). Now we shall settle the claim to prove (1.6) as assured earlier. C OROLLARY 2.6. Let A, B, X ∈ Mn with A, B positive definite. Then for any unitarily invariant norm |||.|||, 1/4 6 ν 6 3/4 and α ∈ [1/2, ∞), 1 AX + XB |||Aν XB1−ν + A1−ν XBν ||| 6 (1 − α )A1/2XB1/2 + α . 2 2
Proof. Choosing f (x) = x2ν −1/2 in corollary (2.4), we get the desired result.
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I NTERPOLATION OF MATRIX VERSIONS OF H ERON AND H EINZ MEANS
C OROLLARY 2.7. Let A, B, X ∈ Mn with A, B positive definite. Then for any unitarily invariant norm |||.||| and a matrix monotone increasing function f : (0, ∞) → (0, ∞)
λ |||A1/4 ( f (A1/2 )X + X f (B1/2))B1/4 ||| 2 f (λ ) AX + XB 6 (1 − α )A1/2XB1/2 + α 2
(2.2)
holds, where λ = min{σ (A), σ (B)} and α ∈ [1/2, ∞).
Proof. For α = 1/2, (2.2) has already been proved in Singh and Vasudeva ([7], Corollary 2.5, p. 622). Again, using Theorem (2.3), we get the desired result. C OROLLARY 2.8. Let A, B, X ∈ Mn with A, B positive definite. Then for any unitarily invariant norm |||.||| and λ = min{σ (A), σ (B)},
λ |||A1/4 (log(I + A1/2)X + Xlog(I + B1/2))B1/4 ||| 2 log(1 + λ ) AX + XB 1/2 1/2 6 (1 − α )A XB + α 2
(2.3)
holds for α ∈ [1/2, ∞).
Proof. Taking f (x) = log(1 + x) in corollary (2.7), we get the desired result.
C OROLLARY 2.9. Let A, B, X ∈ Mn with A, B positive definite. Then for any unitarily invariant norm |||.|||, |||A
1/2
XB
1/2
||| 6 |||
Z1 0
t
A XB
1−t
AX + XB 1/2 1/2 dt||| 6 (1 − α )A XB + α 2
(2.4)
holds for α ∈ [1/2, ∞) (c.f. [2], Th. 5.4.7, p. 163).
Proof. For α = 1/2, (2.4) has already been proved in Hiai and Kosaki ([5], p. 924). Further, using Theorem (2.3), we get the desired result. T HEOREM 2.10. For A, B ∈ Pn and X ∈ Mn and |||.||| any unitarily invariant norm, the function AX + XB α g(α ) = (1 − )(A2/3 XB1/3 + A1/3XB2/3 ) + α 2 2 is increasing for α ∈ [1/2, ∞) and g(α ) 6 g(1/2) for α ∈ [0, 1/2].
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Proof. Once again following the same lines of the proof of Theorem (2.3), we shall prove the result for α > 0, A = B and A = diag(λ1 , λ2 · · · , λn ). Suppose AX + XA α α 2/3 1/3 1/3 2/3 g(α ) = (1 − )(A XA + A XA ) + α = 2 q(α ), 2 2 where q(α ) = |||h1 (α )(A2/3 XA1/3 + A1/3 XA2/3 ) + AX + XA||| and h1 (α ) = 2α −1 − 1. h1 (α )(A2/3 XA1/3 + A1/3XA2/3) + AX + XA 2/3 1/3 1/3 2/3 = h1 (α )(λi λ j + λi λ j ) + λi + λ j xi j i, j = Y ◦ h1 (β )(A2/3 XA1/3 + A1/3XA2/3) + AX + XA .
Now the matrix Y can be written as 2/3 1/3 1/3 2/3 h1 (α )(λi λ j + λi λ j ) + λi + λ j 2/3 1/3 1/3 2/3 h1 (β )(λi λ j + λi λ j ) + λi + λ j i, j 1/3 1/3 (h1 (α ) − h1 (β ))λi λ j = 1 + 1/3 1/3 2/3 2/3 (h1 (β ) − 1)λi λ j + λi + λ j i, j α ) − h ( β ) h ( 1/3 1 1 λ 1/3 . = (1)i, j + λi j 1/3 1/3 2/3 2/3 (h1 (β ) − 1)λi λ j + λi + λ j i, j
Again by Lemma (2.2), the later matrix is positive semidefinite if and only if h1 (α ) > h1 (β ) and 2 > h1 (β ) − 1 > −2. Since h(α ) = h1 (α ) − 1 = (2α −1 − 2), so continuously decreasing function on positive half line and from [1/2, ∞) → (−2, 2]. h(β ) andso h1 (α ) > h1 (β ) for all β > α . Hence as in Theorem (2.3) we have h(α ) > )+1 Using Theorem (2.1), we get q(α ) 6 hh1 ((αβ )+1 q(β ). This proves the result for 1 A = B and α ∈ [1/2, ∞). For α ∈ (0, 1/2], we have 3 = h1 (1/2) 6 h1 (α ) < ∞ and so the matrix Y with β = 1/2 is positive semidefinite using Lemma (2.2). Thecase α = 0 can easily be seen
by the positive semidefiniteness of the matrix
1/3
λi
1
1/3 1/3 2/3 2/3 λ j +λ i +λ j
2λi
1/3
λj
i, j
which is so, using Lemma (2.2). This gives us the desired result for this case, i.e. α q(α ) 6 21 q(1/2). Equivalently saying that g(α )6 g(1/2) for all α∈ [0, 1/2]. A0 0X The general case follows on replacing A by and X by . 0B 0 0 C OROLLARY 2.11. Let A, B, X ∈ Mn with A, B positive definite and f (α ) and g(α ) are same as taken in Theorem (2.3) and (2.10) respectively. Then 1 f (0) 6 g(0) 6 f (α ) 2
(2.5)
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for α ∈ [1/2, ∞), or equivalently, for any unitarily invariant norm |||.||| and −2 < t 6 2, 1 1 |||A1/2 XB1/2 ||| 6 |||A2/3 XB1/3 + A1/3XB2/3 ||| 6 |||AX + XB + tA1/2XB1/2 |||. 2 2+t Proof. For the first inequality in (2.5) see [1] and for the second inequality take 2 ν = 2/3 in corollary (2.6) and α = 2+t . REFERENCES [1] R. B HATIA, Interpolating the arithmetic-geometric mean inequality and its operator version, Linear Algebra Appl. 413 (2006) 355–363. [2] R. B HATIA, Positive definite Matrices, Princeton University Press, Princeton, (2007). [3] R. B HATIA AND K. R. PARTHASARTHY, Positive definite functions and operator inequalities, Bull. London Math. Soc., 32 (2000) 214–228. [4] F. H IAI AND H. K OSAKI , Means of Hilbert space operators, Lecture notes in Mathematics 1820, Springer, New York, (2003). [5] F. H IAI AND H. K OSAKI , Means of matrices and comparison of their norms, Indiana Univ. Math. J., 48 (1999) 899–936. [6] R. A. H ORN AND C. R. J OHNSON, Topics in Matrix Analysis, Cambridge Univ. Press, (1990). [7] M. S INGH AND H. L. VASUDEVA, Norm inequalities involving matrix monotone functions, Math. Ineq. Appl., 7 (2004) 621–627. [8] X. Z HAN, Matrix Inequalities, Lecture notes in Mathematics 1790, Springer, New York, (2002).
(Received March 23, 2011)
Rupinderjit Kaur Department of Mathematics Sant Longowal Institute of Engineering and Technology Longowal-148106 Punjab, India e-mail: rupinder grewal
[email protected] Mandeep Singh Department of Mathematics Sant Longowal Institute of Engineering and Technology Longowal-148106 Punjab, India e-mail:
[email protected]
Mathematical Inequalities & Applications
www.ele-math.com
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