Hermitian matrix inequalities and a conjecture

Introduction

1.

A ≥ (>)0 if all eigenvalues of A are nonnegative (positive) or equivalently if (u, Au) ≥ (>)0 for all nonzero vectors 𝑢 ∈ 𝐶 𝑛 . 2. A≥ (>)B if A -B ≥ (>)0, or equivalently if (u,Au) ≥ (>) (u,Bu) for all nonzero vectors 𝑢 ∈ 𝐶 𝑛 . Unfortunately this ordering is only a partial one. Thus it is not true that if A is not greater than or equal to B then A must be smaller than B. Another source of trouble is the fact that matrix multiplication is not compatible with the ordering (unless all matrices involved are mutually commutative). Most inequalities involving multiplications of real numbers cease to hold when the real numbers are replaced by Hermitian matrices. Two simple examples are: 1. It is not true that A≥0 and B ≥ 0 imply that AB + BA ≥ 0. Simply take A = ( 2 0 0 1 ) and B = ( 1 1 1 1 ).

It is not true that

To illustrate the delicacy of Hermitian inequalities further, we note two more examples: 3. Let A and B be nonnegative Hermitian such that A + B > 0, and let X be an n X n Hermitian matrix that satisfies the inequality

Inthe scalar case (n = 1), it is obvious that the real number X must be positive. However, that is not always true for n;,. 2. A counterexample is

Then (A + B)X + X(A + B) =AB+ BA but X ≥ 0. However, if in addition A > 0 and instead of (1.1) X satisfies

Then X ≥ 0. In fact if we have equality in (1.2), then X is given byX = 𝐴 -1 -(A+ B) -1 ≥ 0.

pair (A, P) for which the solution X is not nonnegative definite. For each dimension n, there is a cutoff value t0 such that the result holds for t ∈ (-2, t0) but not for t > t0. It is also Loewner in his 1934 paper [12] . It turns out that the condition on/ depends on the size of the matrices in question. More It is necessary that f ∈ Pn (a, b) possess continuous derivatives up to order 2 n -3 and that 𝑓 (2𝑛-3) be convex.

Loewner went on to show that the set

Pn(a, b) is miraculously tied up with the theory of analytic functions. Indeed P(a, b) is precisely the set of real-valued functions on (a, b) that are continuable analytically to the upper half plane in which the functions take values with positive imaginary parts (the so-called Pick functions).

For an exposition of this beautiful theory, we refer the readers to the monograph by Donoghue [6] , and the article by Ando [l] . From an advanced point of view, Loewner's result is elegant, but the proof is by no means short or elementary. For instance, it is not at all obvious that f(t) = √𝑡 belongs to P(0, ∞). Perhaps this is the reason why the fact "tr ∈ P(0, ∞), 0 < r < l" has been rediscovered more than once. We shall come back to this in the next section. It is an interesting exercise to try to discover as many functions in P(a, b) as possible by entirely elementary means. In Section 2 we look at this problem. In Sections 3 and 4 we discuss a conjecture and some results that are inspired by an attempt to look for further inequalities.

2. An inequality-generating theorem. We shall establish the following general result.

Theorem 1: Let A, B, C, D be n X n Hermitian matrices. Suppose that A commutes with C and that B commutes with D. If A ≥ B ≥ 0 and C ≥ D ≥ 0, then for any positive r and s such that r + s ≤ 1 we have

The proof of Theorem 1 depends on the following very special case. (Take r = l/2, s = 0, and C = D = I, the identity matrix).

Theorem 2: If

As pointed out in the introduction, Theorem 2 is an obvious corollary of Loewner's deep result. R. Bellman [3] proved Theorem 2 analytically using ideas in the theory of differential equations. Y. H. Au-Yeung [2] gave a simple algebraic proof of the following generalization.

Special cases of Theorem 1 were used by Au-Yeung in his proof of Theorem 3 but Theorem 1 has not been stated in such generality. Theorem 3 was in fact "discovered" earlier in 1951 by E. Heinz [7] , not just for matrices but for general Hermitian operators in a Hilbert space. In the following year T. Kato [8] gave a shorter proof. These authors seem not to have been aware of Loewner's result at the time the papers were written. The various proofs are, however, different and are worth recording. In [9] the second author gave yet another proof that resulted in a stronger form of Theorem 3. Recently the operator version of Theorem 2 found an interesting application in linear neutron transport theory; see [11]. An important normequivalence result can now be given a simple elementary proof. This is an impetus for further investigation of related inequalities. Still more recently Theorem 2 was used in a study of the oscillation of second order systems of differential equations of the form X"(t) + Q(t)X(t) = 0, where Q(t) is an n X n Hermitian matrix for all t and X(t) is an n-vector, see [5] . The (ollowing proof of Theorem 2 is a simplification of a proof that appeared in Marshall and Olkin [13] . Alternative proofs by other authors can also be found in [13] .

Proof of Theorem 2: If suffices to establish the conclusion under the additional condition that A > 0, since the general case follows by continuity. Let

Theorem 1 is thus a consequence of the following well-known result due to Lyapunov. CX + XC = P Has a unique solution X. Moreover if P ≥ 0, then so is X Proof. For the sake of completeness we give the proof. That (2.1) has a unique solution X can be checked directly after we notice that we can assume without loss of generality that C is diagonal.

In particular, by taking adjoints of both sides of (2.1), we conclude that X must be Hermitian whenever P is. Suppose finally that P ≥ 0 and u is an eigenvector of X corresponding to the eigenvalue 𝛼.

• Then 0 ≤ (u,Pu) = (u,CXu) +(u,XCu) = 2𝛼(u,Cu) It follows that 𝛼 ≥ 0 and hence X is nonnegative. An alternative proof is to make use of the explicit formula X =∫ 𝑒 -𝐶𝑡 𝑃𝑒 -𝐶𝑡 ∞ 0 dt which can be found in Bellman [4, p. 179]. The following lemmas can be found in [2,9] . Lemma 2: Suppose C≥ 0. Then A ≥ B implies that CAC ≥ CBC. If furthermore C > 0, then the converse also holds.

Proof: It is not difficult to prove the lemma using the definition of ordering via the inner product.

Lemma 3:

Proof: Let I denote the identity matrix. First notice that X ≥ I implies that x-1 ≤ I. This can be verified either directly from the definition of x-1 or from Lemma 2 with A = X, B = I, and C = 𝑋 -1 2 Now Lemma 2 gives 𝐵 -1 2 𝐴𝐵 -1 2 ≥ 𝐼 from the hypothesis that A ≥ B. Taking the inverse we get 𝐵

Lemma 2 then gives the conclusion. We are now ready to give a proof of Theorem 1. Proof of Theorem l. As in the proof of Theorem 2, we may assume without loss of generality that B > 0. By Lemma 3, B -1 ≥ A -1 . Using Lemma 4 with X= (BD) 1/2 , Y= (AC) 1/2 , F = A -1 and G = B - 1 , we obtain the conclusion of the Theorem for the special case r = l/2 and s = 1 -r. Replacing A and B by (AC)1/2 and (BD) 1/2 , respectively, in the special case r = l/2 of the Theorem, we obtain the special case r = l/4, s = 3/4. Replacing C and D by (AC)1/2 and (BD)1/2, respectively, in the special case r = l/2, we obtain the special case r = 3/4, s = 1/4. Repeating this process, we see that the Theorem holds when r is any dyadic fraction, namely, when r is of the form k/2 n (n = l, 2,... ; k = l, 2,..., 2 n -1) and s = 1 -r. Since such fractions are dense in [0,1], the Theorem holds for all general r ∈ (0, 1) and s = 1 -r by continuity. For s < 1 -r, the required inequality can be written as

Where r' = r/(r + s), s' = s/(r +s) and r' + s' = 1

We can thus complete the proof of Theorem 1 by using the result established above and Theorem 3, which as has been pointed out is a special case of Theorem 1 when C = D = I, r + s = 1. An immediate consequence of Theorem 1 is the following property of the Pick functions. Let P0 (0, ∞) be the subfamily {ƒ∈ P(0,∞): ƒ: (0,∞) → [0,∞)} of P(0,∞). Theorem 4: For any r, s > 0 such that r + s ≤ l and any f, g ∈ P0 (0,∞), we have f r , g s ∈ P0 (0,∞), More generally, let r 1 > 0 (i = 1,2,...,n) be such that ∑ r 1 m k=1 ≤ 1. Then

∈ 𝑃 0 (0, ∞) for all ƒi ∈ P0 (0, ∞) (i = 1,2,….,n). In other words, the set of functions {Inƒ: ƒ ∈ P0 (0, ∞)} is convex. Let us show how further examples of functions in P0 (0, oo) can be generated. We assume throughout the hypotheses A ≥ B ≥ 0; in a few cases the stronger condition B ≥ 0 is needed. Notice that in any case we need only establish the result under the stronger condition B > 0 as the general case follows by continuity. Also notice that Theorem 4 can be used in the obvious way to yield generalizations. Let r, s > 0, r + s ≤ l and 𝛼, 𝛽 > 0. Then a simple consequence of Theorem 4 is (1) (𝛼𝐼 + 𝐴) 𝑟 (𝛽𝐼 + 𝐴) 𝑠 ≥ (𝛼𝐼 + 𝐵) 𝑟 (𝛽𝐼 + 𝐵) 𝑠 1. For all a ≥ 𝛾 > 0, 0 ≤ 𝑟 ≤ 1.

(2) (𝛾𝐼 + 𝐴)(𝛼𝐼 + 𝐴) -𝑟 ≥ (𝛾𝐼 + 𝐵) 𝑟 (𝛼𝐼 + 𝐵) -𝑟 2. or equivalently for all 𝛽 > 0, 𝛽 -1 ≥ 𝛾

(3) (𝛾𝐼 + 𝐴)(𝐼 + 𝛽𝐴) -𝑟 ≥ (𝛾𝐼 + 𝐵)(𝐼 + 𝛽𝐵) -𝑟 Indeed by Lemma 3, A -1 ≤ B -1 . Thus (I + 𝛼𝐴 -1 ) ≤ (I + 𝐵 -1 ). Lemma 3 again yields (I + 𝛼𝐴 -1 ) -1 ≥ (I + 𝐵 -1 ) - with r = l. The general case then follows upon applying Theorem 1.

In (2) we let 𝛾 = 0 and r = l and integrate the resulting inequality with respect to a over [0,1] and finally invoke Theorem 1 to get (5) 𝐴[𝑙𝑛 (𝐼 + 𝐴 -1 )] 𝑟 ≥ 𝐵[𝑙𝑛 (𝐼 + 𝐵 -1 )] 𝑟 On the other hand, if in (3) we let r = l, restrict 𝛾 to be in [0,l], and integrate with respect to 𝛽 over [0,1], we obtain (6) (𝐼 + 𝛾𝐴 -1 ) ln(I + A) ≥ (𝐼 + 𝛾𝐵 -1 ) 𝑙𝑛(𝐼 + 𝐵).

In particular we have Here f(t) = ln tis an example of a function belonging to P(0, ∞) but not to P0 (0, ∞). Inequality (8) Can also be derived from Theorem 3 via differentiation. By Theorem 3, 𝐴 𝑟 -I ≥ 𝐵 𝑟 -I.

Dividing by r and taking limits as r → 0, we have

References

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