aa r X i v : . [ m a t h . F A ] M a y SOME INEQUALITIES FOR THE MATRIX HERON MEAN
DINH TRUNG HOA
Abstract.
Let
A, B be positive definite matrices, p = 1 , r ≥
0. It is shown that || A + B + r ( A♯ t B + A♯ − t B ) || p ≤ || A + B + r ( A t B − t + A − t B t ) || p . We also prove that for positive definite matrices A and B det( P t ( A, B )) ≤ det( Q t ( A, B )) , where Q t ( A, B ) = (cid:0) A t + B t (cid:1) /t and P t ( A, B ) is the t -power mean of A and B . As aconsequence, we obtain the determinant inequality for the matrix Heron mean: for anypositive definite matrices A and B, det( A + B + 2( A♯B )) ≤ det( A + B + A / B / + A / B / )) . These results complement those obtained by Bhatia, Lim and Yamazaki (LAA, (2016) 112-122). Introduction
Let M n be the space of n × n complex matrices and M + n the positive part of M n . Denoteby I the identity element of M n . For self-adjoint matrices A, B ∈ M n the notation A ≤ B means that B − A ∈ M + n . For a real-valued function f of a real variable and a self-adjointmatrix A ∈ M n , the value f ( A ) is understood by means of the functional calculus.For 0 ≤ t ≤ t -geometric mean of A and B is defined as A♯ t B = A / ( A − / BA − / ) t A / . The geometric mean
A♯B := A♯ / B is the midpoint of the unique geodesic A♯ t B con-necting two points A and B in the Riemannian manifold of positive matrices.Recently, Bhatia et al. [1] proved that for any positive definite matrices A and B andfor p = 1 , , || A + B + 2 rA♯B || p ≤ || A + B + r ( A / B / + B / A / ) || p . (1)When r = 1, inequality holds for p = ∞ .For the case p = 2 the proof of (1) is based on the following fact: for any positivedefinite A and B , λ ( A / ( A♯B ) A / ) ≺ log λ ( A / B / A / ) , (2) Mathematics Subject Classification.
Key words and phrases. operator ( r, s )-convex functions, operator Jensen type inequality, operatorHansen-Pedersen type inequality, operator Popoviciu inequality. where the notation λ is used for the n -tuple of eigenvalues of a matrix A in decent orderand λ ( A ) ≺ log λ ( B ) means that k Y j =1 λ i ( A ) ≤ k Y j =1 λ i ( B ) , ≤ k ≤ n and inequality holds when k = n .For p = ∞ inequality (1) was proved by using a result of Lim and Yamazaki [3, Theorem4.1] || P t ( A , A , · · · , A n ) || ∞ ≤ || Q t ( A , A , · · · , A n ) || ∞ , (3)where the power mean P t ( A , A , · · · , A n ) of A , A , · · · , A n [2] and is the unique solutionof the matrix equation X = 1 m m X i =1 X♯ t A i and Q t ( A , A , · · · , A n ) = (cid:0) m m X i =1 A ti (cid:1) /t . For m = 2 Lim and P´ a flia [2, Remark 3.10] show that P t ( A, B ) = A♯ /t (cid:0)
12 ( A + A♯ t B ) (cid:1) = A / (cid:0) I + ( A − / BA − / ) t (cid:1) /t A / . Hopefully, for t = 1 / P / ( A, B ) = 14 ( A + B + A♯B ) , Q / ( A, B ) = 14 ( A + B + A / B / + B / A / ) . And so, the inequality (1) for p = ∞ is obtained from (3) choosing m = 2 and t = 1 / a, b is defined as H t ( a, b ) = (1 − t )( a + b t √ ab, ≤ t ≤ . (4)The Kubo-Ando extension of this to matrices is(1 − t ) A + B t ( A♯B )that connects the arithmetic mean and the geometric mean, and a naive extension is(1 − t ) A + B t A / B / + B / A / A s B − s + B s A − s . So, inequality (1) is a special case of the following || (1 − t ) A + B t ( A♯B ) || p ≤ || (1 − t ) A + B t A / B / + B / A / || p with t = 1 / . Notice that another naive extension of the Heron mean for positive definite matrices A and B is defined as (1 − t ) A + B t A t B − t + A − t B t . OME INEQUALITIES FOR THE MATRIX HERON MEAN 3
In this paper, we extend the inequality (2) to t -geometric means. More precisely, weprove that for any positive definite matrices A, B and for any t ∈ [0 , λ ( A / ( A♯ t B ) A / ) ≺ log λ ( A − t B t A − t ) . Using this extension, we prove the following result:
Theorem 1.1.
Let
A, B be positive definite matrices and p = 1 , and r ≥ . Then || A + B + r ( A♯ t B + A♯ − t B ) || p ≤ || A + B + r ( A t B − t + A − t B t ) || p . (5)Also, using the approach in [4] we show that for positive definite matrices A, B and forany z in the strips S / = { z ∈ C : Re( z ) ∈ [1 / , / } , | Tr( A / B z A / B − z ) | ≤ Tr( AB ) . Inequalities
Proposition 2.1.
Let
A, B be positive definite matrices. Then for any t ∈ [0 , λ ( A / ( A♯ t B ) A / ) ≺ log λ ( A − t/ B t A − t/ ) . (6) Proof.
Firstly, let’s prove λ ( A / ( A♯ t B ) A / ) ≤ λ ( A − t/ B t A − t/ ) . (7)This inequality is equivalent to the statement A − t/ B t A − t/ ≤ I = ⇒ A / ( A♯ t B ) A / ≤ I, which in turn is equivalent to B t ≤ A t − = ⇒ ( A − / BA − / ) t ≤ A − . (8)That can be proved by using the Furuta inequality which states that if 0 ≤ Y ≤ X , thenfor all p ≥ r ≥ X r Y p X r ) / ≤ ( X p +2 r ) /p . (9)Let apply (9) to X = A t − , Y = B t , p = t and r = − t − , we get (8), and hence (7).Denote by C k ( X ) the k -th compound of X ∈ M n , k = 1 , . . . , n . Note that for anypositive definite matrices X, Y , C k ( A / ( A♯ t B ) A / ) = C k ( A ( A − / BA − / ) t A )= C k ( A ) C k (( A − / BA − / ) t ) C k ( A )= C / k ( A )( C / k ( A )( C − / k ( A ) C k ( B ) C − / k ( A )) t C / k ( A )) C / k ( A )= C / k ( A )( C k ( A ) ♯ t C k ( B )) C / k ( A ) . (10) DINH TRUNG HOA
In the other hand, λ ( C k ( A / ( A♯ t B ) A / )) = k Y i =1 λ i ( A / ( A♯ t B ) A / ) , k = 1 , . . . , n − , . (11)On account of (7) and (11) for 1 ≤ k ≤ n we have k Y i =1 λ i ( A / ( A♯ t B ) A / ) = λ ( C k ( A / ( A♯ t B ) A / )) ≤ λ ( C k ( A ) − t C k ( B ) t C k ( A ) − t )= k Y i =1 λ i ( A − t B t A − t ) . The equality holds for k = n , since det( A / ( A♯ t B ) A / = det( A − t B t A − t ) . Thus, we have proved (6). (cid:3)
The following special case of Proposition will be used in the proof of the main result.
Corollary 2.2.
For any positive definite matrices A and B, Tr( A ( A♯ t B )) ≤ Tr( A − t B t ) , t ∈ [0 , . In order to prove the next result, let’s recall the generalized Hølder inequality for trace[5, Theorem 2.8]: let p + q + r = 1 for p, q, r ≥ X, Y, Z be matrices in M n , thenTr( XY Z ) ≤ || XY Z || ≤ || X || p || Y || q || Z || r . We also need the famous Lieb-Thirring inequality: Tr(( AB ) m ) ≤ Tr( A m B m ). Theorem 2.3.
Let
X, Y be positive definite matrices and z ∈ S / = { z ∈ C : Re ( z ) ∈ [ , ] } . Then | Tr( X / Y z Y / Y − z ) | ≤ Tr( XY ) . (12) Proof.
Let z = + iy, y ∈ R denote any point in the vertical line of the complex planepassing x = 1 /
2. Then we have | Tr( X / Y z X / Y − z ) | = | Tr( X / Y / Y iy X / Y / Y − iy ) |≤ Tr( | X / Y / Y iy X / Y / Y − iy | ) ≤ || X / Y / Y iy || || X / Y / Y − iy || = || X / Y / || = Tr( XY ) . The first inequality is obvious, the second one follows from the Cauchy-Schwarz inequalityfor trace, and the second equality is from the fact that Y iy and Y − iy are unitary operators. OME INEQUALITIES FOR THE MATRIX HERON MEAN 5
Now let consider z = + iy, y ∈ R , a generic point in the vertical line over x = 1 / o lder inequality with + + = 1 and the Araki-Lieb-Thirringinequality we have | Tr( X / Y z X / Y − z ) | = | Tr( X / Y / Y iy X / Y / Y − iy Y / ) | = | Tr( Y / X / X / Y / Y iy X / Y / Y − iy ) |≤ || Y / X / || / || X / Y / || ≤ || Y / X / || || X / Y / || = Tr( XY ) . Mention that the map x A z = e x ln A = P k z k (ln A ) k k ! is analytic for A > f ( z ) = Tr( X / Y z X / Y − z ) is entire. Moreover, by the similar above argument for z = x + iy it is easy to see that if 0 ≤ x ≤ M ( x ) = sup {| f ( x + iy ) | : y ∈ R } of thefunction Tr( X / Y z X / Y − z ) is log-convex, that means, for any λ ∈ [0 , ,M ( λx + (1 − λ ) x ) ≤ M ( x ) λ M ( x ) − λ ≤ Tr( XY ) λ Tr( XY ) − λ = Tr( XY ) . Therefore, the bound Tr( XY ) is valid in the vertical strip 1 / ≤ Re( z ) ≤ / . Invokingthe symmetry z − z and exchanging the roles of A and B give the desired bound onthe full strip S / = { / ≤ Re( z ) ≤ / } . (cid:3) As a consequence, we have the following inequality (see [1, Inequality (39)]):
Corollary 2.4.
For any positive definite matrices and for t ∈ [0 , . Tr(( A♯ t B )( A♯ − t B )) ≤ Tr( AB ) . Now we are ready to prove the main result in this paper.
Theorem 2.5.
Let
A, B be positive definite matrices, p = 1 , and r ≥ . Then || A + B + r ( A♯ t B + A♯ − t B ) || p ≤ || A + B + r ( A t B − t + A − t B t ) || p . (13) Proof.
Since A + B + r ( A♯ t B + A♯ − t B ) ≥
0, the left hand side of (13) is Tr( A + B + r ( A♯ t B + A♯ − t B )). It is well-known that Tr( A♯ t B ) ≤ Tr( A − t B t ) and Tr( A♯ − t B ) ≤ Tr( A t B − t ).We have Tr( A + B + r ( A♯ t B + A♯ − t B )) ≤ Tr( A + B + r ( A t B − t + A − t B t )) ≤ Tr( | A + B + r ( A t B − t + A − t B t ) | ) . So for p = 1 the inequality (13) follows. DINH TRUNG HOA
Next consider the case p = 2. Notice again that Tr(( A♯ t B ) ) ≤ Tr( B t A − t ) ) (see [1,pape 121]). Similarly, we also have Tr(( A♯ − t B ) ) ≤ Tr( A t B − t ) ). ThenTr(( A♯ t B ) + ( A♯ − t B ) ) ≤ Tr( A t B − t ) + B t A − t ) ) . (14)By Proposition 2.1 we haveTr(( A + B )( A♯ t B + A♯ − t B )) ≤ Tr( A t +1 B − t + A − t B t + A t B − t + A − t B t ) . (15)Now, squaring both sides of (13), we need to showTr(( A + B ) + r ( A♯ t B ) + r ( A♯ − t B ) + 2 r ( A + B )( A♯ t B + A♯ − t B ) + 2 r ( A♯ t B )( A♯ − t B )) ≤ Tr(( A + B ) + 2 r ( A t +1 B − t + A − t B t + A t B − t + A − t B t ) + r A t B − t ) ) + r B t A − t ) + 2 r Tr( AB ) . The last inequality follows from (14), (15) and Corollary 2.4. (cid:3)
Remark 2.6.
From Theorem 2.5 for s ∈ [0 ,
1] we || (1 − s )( A + B ) + s ( A♯ t B + A♯ − t B ) || p ≤ || (1 − s )( A + B ) + s ( A t B − t + A − t B t ) || p . When t = 1 / || − s A + B ) + s ( A♯B ) || p ≤ || − s A + B ) + sA / B / || p . Remark 2.7.
By the same arguments, one can show that || A + B + A♯ t B + A♯ − t B || p ≤ || A + B + A t B − t + A − t B t || p . But is we use another version of the Heinz mean ( A t B − t + B t A − t ) / A♯ t B )( B♯ t A )) ≤ Re Tr( A t B t A − t B − t ) . (16)Notice that both sides are bounded by Tr( AB ) but it is not clear that (16) is true or not.From the proof of the main theorem, it is natural to ask the following question: Is ittrue that for 0 ≤ X, Y ≤ Z such that X ≺ log YZ / XZ / ≺ log Z / Y Z / ? (17)Unfortunately, the answer is negative. Indeed, let X = ! , Y = ! , Z = ! . Now s ( X ) = s ( Y ) = (1 ,
1) but s ( Z / XZ / ) = (1 , w ( √ , √
2) = s ( Z / Y Z / ). Soit is not even true for diagonal positive definite matrices. OME INEQUALITIES FOR THE MATRIX HERON MEAN 7 Determinant Inequality for the Heron mean
Let’s recall a recent result of Audeanert [9]: for any positive semidefinite matrices A and B det( I + A♯B ) ≤ det( I + A / B / ) . (18)The author used the well-known fact that λ ( A♯B ) ≺ log λ ( A / B / ) and the functionΦ( X ) = P ni =1 log(1 + e x i ) ( X = ( x , x , · · · , x n )) is isotone (i.e. the function preservingweak majorization: x ≺ y ⇒ Φ( x ) ≺ w Φ( y ).)In fact, for matrices A and B such that λ ( A ) ≺ log λ ( B ) we havedet( I + A ) ≤ det( I + B ) . (19)A useful characterization of isotone functions in the case m = 1 is as follows: Lemma 3.1.
A differentiable function
Φ : R n → R is isotope if and only if it satisfy (1) Φ is permutation invariant; (2) for all X ∈ R n and for all i, j : ( x i − x j ) (cid:0) ∂ Φ ∂x i ( x ) − ∂ Φ ∂x j ( x ) (cid:1) ≥ . Do the similar argument as in [9] one can prove the followingdet( I + A♯ t B ) ≤ det( I + A − t B t ) . (20)Now we can use this fact to obtain some inequality for the Heron mean. Theorem 3.2.
For any positive definite matrices A and B det( P t ( A, B )) ≤ det( Q t ( A, B )) . (21) Proof.
The inequality (21) is equivalent to the followingdet /t ( A t + B t ) = det( A ) det /t ( I + A − t/ B t A − t/ ) ≥ det( A♯ /t ( A + A♯ t B ))= det( A ) · det /t ( I + ( A − / BA − / ) t )or det( I + A − t/ B t A − t/ ) ≥ det( I + ( A − / BA − / ) t ) . (22)By the Araki-Lieb-Thirring inequality we have λ ( I + ( A − / BA − / ) t ) ≺ log λ ( I + A − t/ B t A − t/ ) . Therefore, the inequality (22) follows from the last inequality and (19). (cid:3)
As a consequence, we obtain a determinant inequality for the Heron mean.
DINH TRUNG HOA
Corollary 3.3.
For any positive definite matrices A and B, det( A + B + 2( A♯B )) ≤ det( A + B + A / B / + A / B / )) . References [1] R.Bhatia, Y.Lim, T.Yamazaki.
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