positivity manuscript No. (will be inserted by the editor)
Some inequalities involving determinants, eigenvalues, and Schur complements in Euclidean Jordan algebras M. Seetharama Gowda · Jiyuan Tao
February 9, 2010, Revised July 8, 2010
Abstract In this paper, using Schur complements, we prove various inequalities in Euclidean Jordan algebras. Specifically, we study analogues of the inequalities of Fischer, Hadamard, Bergstrom, Oppenheim, and other inequalities related to determinants, eigenvalues, and Schur complements. Keywords Euclidean Jordan algebras · Cauchy interlacing theorem · Schur complement · Schur interlacing inequalities Mathematics Subject Classification (2000) 15A33 · 17C20 · 17C55 1 Introduction Inequalities abound in matrix theory. A number of them are derived from classical results involving norms, eigenvalues, Schur complements, Schur products, etc. In this paper, we consider some inequalities that arise from Schur complements and study their analogues in the setting of Euclidean Jordan algebras. Given a complex matrix M in the block form AB M= (1) CD M. Seetharama Gowda Department of Mathematics and Statistics University of Maryland Baltimore County Baltimore, Maryland 21250, USA, E-mail:
[email protected] Jiyuan Tao Department of Mathematical Sciences Loyola University Maryland Baltimore, Maryland 21210, USA, E-mail:
[email protected]
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with A invertible, the Schur complement of A in M is M/A = D − CA−1 B. The Schur complement enjoys numerous properties such as the Schur determinantal formula, the Haynsworth inertia formula, the Guttman rank formula, etc., and appears in various applications [22]. While the above definition continues to hold for quaternionic matrices, for lack of associativity property, it is not defined for octonionic matrices; lack of a nontrivial multiplicative determinant for matrices over quaternions and octonions [2] is perhaps another reason why Schur complements have not been studied for matrices over such numbers. However, the concept of Schur complement can be studied for square Hermitian matrices (of suitable size) over quaternions and octonions, and more generally over Euclidean Jordan algebras. Consider a Euclidean Jordan algebra V (which is a finite dimensional real inner product space with a compatible Jordan product), an idempotent c ∈ V (that is, c2 = c), and the corresponding Peirce decomposition [5] 1 V = V (c, 1) ⊕ V (c, ) ⊕ V (c, 0), 2
(2)
where V (c, γ) = {x ∈ V : x ◦ c = γx} and γ ∈ {0, 21 , 1}. For any x ∈ V , let x = u + v + w,
(3)
where u ∈ V (c, 1), v ∈ V (c, 21 ), and w ∈ V (c, 0). When u is invertible in the Euclidean Jordan (sub)algebra V (c, 1), let u−1 ∗ denote the inverse of u in V (c, 1). In this case, the Schur complement of u in x is defined by x/u := w − Pv (u−1 ∗ ), where, for any element a ∈ V , the quadratic representation Pa is defined on V by Pa (z) = 2a ◦ (a ◦ z) − (a ◦ a) ◦ z (z ∈ V ). (4) The concept of Schur complement was introduced in [15], see also [14], where it was shown that x/u ∈ V (c, 0) and det(x) = det(u) det(x/u).
(5)
Here, the determinants det(x), det(u), and det(x/u) are taken in the algebras V , V (c, 1), and V (c, 0) respectively. In a recent article [8], Gowda and Sznajder showed that (i) In(x) = In(u) + In(x/u), (ii) x > 0 (≥ 0) in V if and only if u > 0 (≥ 0) in V (c, 1) and x/u > 0 (≥ 0) in V (c, 0), and (iii) rank(x) = rank(u) + rank(x/u),
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where In(x) and rank(x) denote the inertia and rank of an element x respectively, and x ≥ 0 means that x belongs to the symmetric cone in V . With appropriate identification, the above results reduce to the familiar results of matrix theory in the algebras Herm(Rn×n ) and Herm(C n×n ) of all n × n Hermitian matrices over real numbers and complex numbers. Items (i) − (iii) show that they continue to hold even in Herm(Hn×n ) (the algebra of all n×n Hermitian matrices over quaternions), Herm(O3×3 ) (the algebra of all 3 × 3 Hermitian matrices over octonions), and the Jordan spin algebra Ln . As we shall see, these items will allow us to extend many of the (real/complex) matrix theory results to Hermitian matrices over quaternions/octonions and to objects in Ln . In this paper, we study some inequalities that arise via Schur complements in the setting of Euclidean Jordan algebras. Specifically, we study analogues of inequalities due to Fischer, Hadamard, Bergstrom, Oppenheim, and inequalities related to determinants and eigenvalues. The organization of the paper is as follows. In Section 2, we recall various concepts, results, and notation from Euclidean Jordan algebra theory. Section 3 deals with some inequalities involving quadratic representations and determinants. In Section 4, we study analogues of inequalities of Hadamard, Fischer, and Bergstrom. An inequality involving the sum of two Schur complements is also presented in this section. In Section 5, we consider interlacing inequalities for Schur complements. Section 6 deals with the (new) concept of Schur (or Hadamard) product of two objects in a Euclidean Jordan algebra when their Peirce decompositions with respect to a Jordan frame are given. A generalization of Schur’s product theorem and related results are also studied in this section. Finally, in Section 7, we present a converse of the Schur complement interlacing theorem.
2 Preliminaries Throughout this paper, we let (V, ◦, h·, ·i) denote a Euclidean Jordan algebra of rank r [5], [9], [18]. The symmetric cone of V is the cone of squares K := {x2 : x ∈ V }. We use the notation x ≥ y (x > y) when x − y ∈ K (respectively, x − y ∈ interior(K)). As is well known, any Euclidean Jordan algebra is a product of simple Euclidean Jordan algebras and every simple algebra is isomorphic to the Jordan spin algebra Ln or to the algebra of all n × n real/complex/quaternion Hermitian matrices or to the algebra of all 3 × 3 octonion Hermitian matrices. Given a Euclidean Jordan algebra V of rank r, for every x ∈ V , there exist a Jordan frame {e1 , . . . , er } and unique real numbers λ1 , . . . , λr – called the (spectral) eigenvalues of x – such that x = λ1 e1 + · · · + λr er .
(6)
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We remark that these eigenvalues coincide with the real right eigenvalues of matrices in Herm(Rn×n ), Herm(C n×n ), and Herm(Hn×n ). However, they can be different for matrices in Herm(O3×3 ), see [3], [17]. Corresponding to the spectral decomposition of x given by (6), the determinant, inertia and rank of x are respectively defined by det(x) := λ1 λ2 . . . λr ,
(7)
In(x) := (π(x), ν(x), δ(x)), and rank(x) = π(x) + ν(x),
(8)
where π(x), ν(x), and δ(x) are, respectively, the number of eigenvalues of x which are positive, negative, and zero, counting multiplicities. We note that x > 0 (x ≥ 0) if and only if all eigenvalues of x are positive (respectively, nonnegative). When all the eigenvalues λi are nonzero, we define the inverse of x by x−1 := λ1−1 e1 + · · · + λ−1 r er .
Referring to the space V (c, 1) defined in (2), if {e1 , e2 , . . . , ek } is a Jordan frame in V (c, 1) with c = e1 + e2 + · · · + ek and u = λ1 e1 + λ2 e2 + · · · + λk ek is invertible in V (c, 1), we define the (partial) inverse of u in V (c, 1) by −1 −1 u−1 ∗ = λ1 e1 + · · · + λk ek + 0fk+1 + · · · + 0fr ,
where {fk+1 , fk+2 , . . . , fr } is any Jordan frame in V (c, 0). Given two elements a and b, we say that a and b operator commute if a ◦ (b ◦ x) = b ◦ (a ◦ x) ∀ x ∈ V. (This is equivalent to saying that a and b have their spectral decompositions with respect to a common Jordan frame.) Let {e1 , e2 , . . . , er } be a Jordan frame in V . For i, j ∈ {1, 2, . . . , r}, the Peirce eigenspaces are given by Vii := {x ∈ V : x ◦ ei = x} = Rei and when i 6= j, Vij := {x ∈ V : x ◦ ei =
1 x = x ◦ ej }. 2
It is well known that V is an orthogonal direct sum of Vij s with i ≤ j. In particular, any element x ∈ V can be written as an orthogonal sum x=
X
1≤i≤j≤r
xij =
r X
xi ei +
i=1
X
xij
(9)
1≤i 1. We write any element x in the form x0 x= (11) x′
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with x0 ∈ R and x′ ∈ Rn−1 . The inner product in Ln is the usual inner product on Rn . The Jordan product x ◦ y in Ln is defined by
x0 y0 hx, yi x◦y = ◦ ′ := . x′ y x0 y ′ + y0 x′ Then Ln is a Euclidean Jordan algebra of rank 2 and for any element x ∈ Ln , see Example 10 in [18], the eigenvalues and the determinant are given by λ1 (x) = x0 + ||x′ ||,
λ2 (x) = x0 − ||x′ ||,
(12)
and det(x) = x20 − ||x′ ||2 . Now consider any Jordan frame {e1 , e2 } in Ln . Then there exists (see e.g., Lemma 2.3.1, [20]) a unit vector u′ ∈ Rn−1 such that e1 :=
1 1 1 1 and e := . 2 2 u′ 2 −u′
With respect to this, any x ∈ Ln given by (11) has a Peirce decomposition x=
x0 x′
= x1 e1 + x2 e2 + x12 = x1
1 1 1 1 0 + x + ′ 2 z 2 u′ 2 −u′
(13)
where z ′ ∈ Rn−1 with hu′ , z ′ i = 0. (This is easy to verify, see e.g., Lemma 2.3.4, [20].) Given the above Peirce decomposition with u := x1 e1 6= 0, we have x/u = x2 e2 − Px12 (x−1 1 e1 ) =
det(x) x1 x2 − ||x12 ||2 e2 = e2 . x1 det(u)
From now onwards, our setting is that of a (general) Euclidean Jordan algebra.
3 Inequalities involving quadratic representations and determinants In this section, we present a few inequalities involving quadratic representations and determinants. These results, while known, are presented here for completeness, and for the novelty/simplicity of their proofs. We begin with a lemma that will facilitate the proofs of some of our results. Lemma 1 Let x ≥ 0 and y ≥ 0. Then there exists d > 0 such that a = Pd (x) and b = Pd (y) operator commute. In particular, if x > 0, we can take a = e.
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Proof Without loss of generality, we assume that x + y 6= 0. Let the spectral decomposition of x + y be given by x+y =
k X
λi ei +
1
r X
0ei ,
k+1
where λi > 0. Now, define d=
r k X X 1 √ ei + ej . λi 1 k+1
Then Pd (x + y) =
k X
ei =: c.
1
Put a := Pd (x) and b := Pd (y). Then a ≥ 0 and b ≥ 0. Since a + b = c ∈ V (c, 1) ∩ K and V (c, 1) ∩ K is a face of K (see, [7], Theorem 3.1), we have a, b ∈ V (c, 1). As c is the unit element of V (c, 1), we see that a and b (= c − a) operator commute in V (c, 1), and hence in V . Finally, if x > 0, we may choose a d > 0 such that Pd (x) = e (which can be seen by putting y = 0 in the previous argument). Hence e = Pd (x) operator commutes with Pd (y). Corollary 1 (Cf. [6] or [5], Exercise 7, page 59) If x ≥ y > 0, then x−1 ≤ y −1 . Proof Let x ≥ y > 0. When x and y operator commute, we can compare the eigenvalues of x and y, and get the result. In the general case, by Lemma 1, there exists a d > 0 such that a = Pd (x) and b = Pd (y) operator commute. Since a ≥ b > 0, we have a−1 ≤ b−1 and a−1 ≤ b−1 ⇒ [Pd (x)]−1 ≤ [Pd (y)]−1 ⇒ Pd−1 (x−1 ) ≤ Pd−1 (y −1 ) ⇒ x−1 ≤ y −1 .
This completes the proof. Corollary 2 If x ≥ 0 and y ≥ 0, then det(x + y) ≥ det(x) + det(y). Proof For any Q two sets Q of nonnegative numbers xi and yi , i = 1, 2, . . . , r, Qr r r x + (x + y ) ≥ i i i 1 yi . This inequality proves our result when x and 1 1 y operator commute (so that they have spectral expansions with respect to a common Jordan frame). Now assume that x ≥ 0 and y ≥ 0 are arbitrary. Then by Lemma 1, there exists d > 0 such that a = Pd (x) and b = Pd (y) operator commute. Since a ≥ 0 and b ≥ 0, by the previously considered case, we have det(a + b) ≥ det(a) + det(b). Thus det Pd (x + y) ≥ det Pd (x) + det Pd (y). Since det(Pd (x)) = (det(d))2 det(x), etc., see Prop. 1, we have det(x + y) ≥ det(x) + det(y).
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Corollary 3 (Cf. [5], Exercise 4, page 59) If x ≥ y ≥ 0, then det(x) ≥ det(y). Proof When x ≥ y ≥ 0, from the previous Corollary, det(x) = det(x − y + y) ≥ det(x − y) + det(y) ≥ det(y). 4 Inequalities involving Schur complements In matrix theory, the well-known Fischer’s inequality ([12], Theorem 7.8.3) asserts that if the complex matrix M given by (1) is Hermitian and positive semidefinite, then det(M ) ≤ det(A) det(D). We state an analogue of this result in Euclidean Jordan algebras; see [10], Corollary 4.9, for a different proof. Proposition 2 (A generalized Fischer inequality) For x ≥ 0, consider the Peirce decomposition x = u + v + w of the form (3). Then det(x) ≤ det(u) det(w). Proof By continuity, we may assume that u is invertible in V (c, 1). Since x ≥ 0 in V , we have, by Item (ii) in Section 1, u > 0 in V (c, 1) and x/u ≥ 0; thus, u−1 ∗ ≥ 0. Now, by the definition of Schur complement, w − x/u = w − (w − −1 Pv (u−1 ∗ )) = Pv (u∗ ) ≥ 0. Hence w ≥ x/u ≥ 0. Applying Corollary 3 in the det(x) algebra V (c, 0), we have det(w) ≥ det(x/u). Replacing det(x/u) by det(u) (see (5)) and simplifying, we get the required inequality. The inequality in the above proposition can be slightly extended. First, we introduce some notation. Let {e1 , e2 , . . . , er } be a Jordan frame in V with subsets α and β, and their complements α′ and β ′ . Define P s(α) := ei ∈α ei and xα := Ps(α) (x) (x ∈ V ).
Note that xα , which belongs to V ( s(α), 1), is nothing but a component of x in the Peirce decomposition (3) corresponding to the idempotent s(α). With this notation, the inequality in the above proposition yields: For α ∩ β = ∅ and z ≥ 0 in V , det(zα∪β ) ≤ det(zα ) det(zβ ). (14) Also, when x and xα are invertible, we get the following from Theorem 1(a) and (5): det(xα ) . (15) (x−1 )α′ = (x/xα )−1 and det (x−1 )α′ = det(x) We now state the promised extension: For any α, β ⊆ {e1 , e2 , . . . , er } and x ≥ 0 in V , det(xα∪β ) det(xα∩β ) ≤ det(xα ) det(xβ ). (16) To see this, assume without loss of generality, α ∪ β = {e1 , e2 , . . . , er } and x > 0 in V . Then α′ ∩ β ′ = ∅. Applying (14) to z = x−1 we get det((x−1 )α′ ∪β ′ ) ≤ det((x−1 )α′ ) det((x−1 )β ′ ).
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This leads, upon using determinantal relations in (15) to (16). For complex Hermitian matrices, the inequality (16) is often known as the Hadamard-Fischer inequality, see [1]. An immediate consequence of Proposition 2 is the following: Pr P Proposition 3 (A generalized Hadamard inequality) If x = 1 xi ei + i 0 in V (c, 1) and so −1 u−1 ∗ ≥ 0. From Theorem 1, we have x − x/u = Px (u∗ ) ≥ 0. Now assume that V is simple and rank of c is k < r. Since u is invertible in V (c, 1), the rank of u in V (which is the number of nonzero eigenvalues of u in V ) is k. The same is true for u−1 ∗ . Now, from Theorem 11 in [11], −1 π(Px (u−1 ∗ )) ≤ π(u∗ ) = k,
where π of an element is the number of positive eigenvalues of that element. −1 Since Px (u−1 ∗ ) cannot have any negative eigenvalues (as Px (u∗ ) ≥ 0), it ↓ ↓ −1 follows that λk+1 (x − x/u) = λk+1 (Px (u∗ )) = 0. This implies that λ↓i (x − x/u) = 0 for all i ≥ k + 1, completing the proof. Theorem 4 For x ≥ 0 and y ≥ 0, consider Peirce decompositions x = u + v + w and y = p + q + r of the form (3) with u and p invertible in V (c, 1). Then (x + y)/(u + p) ≥ x/u + y/p. −1 −1 Proof Noting that Pv (u−1 ∗ ), Pq (p∗ ), and Pv+q ((u + p)∗ ) belong to V (c, 0) (see Theorem 1), we claim that
a := u + v + Pv (u∗−1 ) ≥ 0
and b := p + q + Pq (p−1 ∗ ) ≥ 0.
As u > 0 and v > 0 in V (c, 1), we have a/u = 0 and b/p = 0. By Item (ii) in Section 1, a ≥ 0 and b ≥ 0 in V . Hence the claim. Now −1 a + b = (u + p) + (v + q) + [Pv (u−1 ∗ ) + Pq (p∗ )] ≥ 0
with u + p > 0 in V (c, 1). Hence by Item (ii) in Section 1, (a + b)/(u + p) ≥ 0, that is, [Pv (u∗−1 ) + Pq (p∗−1 )] − Pv+q ((u + p)−1 (17) ∗ ) ≥ 0.
10 −1 As x/u = w − Pv (u−1 ∗ ), y/p = r − Pq (p∗ ), and (x + y)/(u + p) = w + r − −1 Pv+q ((u + p)∗ ) ), we have −1 −1 (x + y)/(u + p) − [x/u + y/p] = [Pv (u−1 ∗ ) + Pq (p∗ )] − Pv+q ((u + p)∗ ) ≥ 0.
This completes the proof. −1 −1 Remarks. As x − Px (u−1 ∗ ) = x/u = w − Pv (u∗ ), y − Py (p∗ ) = y/p = −1 −1 r − Pq (p∗ ), and x + y − Px+y ((u + p)∗ ) = w + r − Pv+q ((u + p)−1 ∗ ), the inequality (17) can be rewritten as −1 −1 Px (u−1 ∗ ) + Py (p∗ ) ≥ Px+y ((u + p)∗ ).
Combining Theorem 4 with Corollaries 2 and 3, and using (5), we get the following. Corollary 4 (A Bergstrom type inequality) Let x, y, u and p be as in the previous theorem. Then det(x + y) det(x) det(y) ≥ + . det(u + v) det(u) det(p) It has been observed in [10] that in a simple algebra of rank r, the following Minkowski type inequality holds for all x, y ≥ 0: 1
1
1
[det(x + y)] r ≥ [det(x)] r + [det(y)] r . Now suppose that the conditions of Theorem 4 are in place and further suppose that V is simple, has rank r, and c has rank k < r. Then applying the above inequality in V (c, 0), which is a simple algebra of rank r − k, we get
det(x + y) det(u + v)
1 r−k
det(x) ≥ det(u)
1 r−k
det(y) + det(p)
1 r−k
.
This inequality can be regarded as a generalization of Ky Fan’s inequality, see [4]. Proposition 4 Suppose that x, y > 0. Then Pa (x−1 ) + Pb (y −1 ) ≥
1 Pa+b ((x + y)−1 ), 2
for all a, b ∈ V . Proof Fix a, b ∈ V . First we show that Pa (u)+ Pb (u) ≥ 12 Pa+b (u) for all u > 0. Fix u > 0. Then there exists d > 0 such that Pd (e) = u (see Lemma 1). Since r2 + s2 ≥ 12 (r + s)2 for all r, s ∈ V , we have Pr (e) + Ps (e) ≥ 21 Pr+s (e). Now, taking r = Pd (a) and s = Pd (b), we have PPd (a) (e) + PPd (b) (e) ≥
1 PP (a+b) (e). 2 d
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This leads to Pd Pa Pd (e) + Pd Pb Pd (e) ≥
1 1 Pd Pa+b Pd (e) and Pa (u) + Pb (u) ≥ Pa+b (u). 2 2
Now, put u = (x + y)−1 . Then Pa ((x + y)−1 ) + Pb ((x + y)−1 ) ≥
1 Pa+b ((x + y)−1 ). 2
Since x + y ≥ x ⇒ (x + y)−1 ≤ x−1 (see Corollary 1), we have Pa ((x + y)−1 ) ≤ Pa (x−1 ). Similarly, Pb ((x + y)−1 ) ≤ Pb (y −1 ). Therefore, Pa (x−1 ) + Pb (y −1 ) ≥ 1 −1 ). 2 Pa+b ((x + y) Remarks. For Hermitian matrices over reals/complex numbers, the factor 12 in the above inequality is not needed, see [21], Theorem 6.23. It is not clear if a similar result holds in the general case also.
5 Interlacing inequalities for Schur complements In this section, we state some interlacing inequalities involving Schur complements. First, we recall the following generalization of Weyl’s theorem [16]. Theorem 5 Let V be a simple Euclidean Jordan algebra of rank r, and a, b ∈ V . Then λ↓j (a) + λ↓r (b) ≤ λ↓j (a + b) ≤ λ↓j (a) + λ↓1 (b),
∀ j ∈ {1, 2, . . . , r}
and λ↓i+j−1 (a + b) ≤ λ↓i (a) + λ↓j (b),
∀ i, j ∈ {1, 2, . . . , r}, i + j − 1 ≤ r.
Theorem 6 Let V be simple with rank r. For an x ≥ 0, corresponding to an idempotent c of rank k < r, consider the Peirce decomposition x = u + v + w of the form (3) with u invertible in V (c, 1). Then the sets of nonnegative numbers {λ↓i (x) : i = 1, 2, . . . , r} and {λ↓i (x/u) : i = 1, 2 . . . , r − k} have the same number of zeros, and λ↓i (x) ≥ λ↓i (x/u) ≥ λ↓i+k (x)
(i = 1, 2, . . . , r − k).
Proof By Item (ii) of Section 1, x ≥ 0 implies that u > 0 and x/u ≥ 0; hence eigenvalues of x and x/u are all nonnegative. By Item (i) of Section 1, In(x) = In(u) + In(x/u) and so δ(x) = δ(x/u), proving the first part of the assertion. Now combining the previous theorem with Theorem 3, we get the following inequalities: λ↓i+k (x) = λ↓i+k (x/u + x − x/u) ≤ λ↓i (x/u) + λ↓k+1 (x − x/u) = λ↓i (x/u); λ↓i (x) = λ↓i (x/u + x − x/u) ≥ λ↓i (x/u) + λ↓r (x − x/u) = λ↓i (x/u).
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The above result admits a converse, see Section 7. It is known that the Cauchy interlacing theorem does not hold for the Schur complement of a Hermitian matrix but it holds for the reciprocals of nonsingular Hermitian matrices (see, [19], [22]), i.e., if H is an n × n complex Hermitian nonsingular matrix and A is a k×k nonsingular principal submatrix of H, where 1 ≤ k < n, then λi (H −1 ) ≥ λi [(H/A)
−1
] ≥ λi+k (H −1 ) (i = 1, 2, . . . , n − k).
Combining Theorem 1(a) with Theorem 2, we get the following generalization: Theorem 7 Let V be simple of rank r, c be an idempotent of rank k < r, x = u + v + w (of the form (3)) be invertible in V with u invertible in V (c, 1). Then −1
λ↓i (x−1 ) ≥ λ↓i [(x/u)
] ≥ λ↓i+k (x−1 )
(i = 1, 2, . . . , r − k).
6 Schur product and a determinantal inequality Consider two real symmetric matrices A = [aij ] and B = [bij ] in Herm(Rr×r ). Then the Schur (or Hadamard) product of A and B is the matrix C := [aij bij ]. This product enjoys numerous properties, see e.g., [12], [21], [22]. In this section, we consider a generalization of this for Euclidean Jordan algebras. Fix a Jordan frame {e1 , e2 , . . . , er } in a Euclidean Jordan algebra V and consider the Peirce decompositions of x and y with respect to this Jordan frame: r r X X X X yij . xij and y = yi ei + x= xi ei + 1
i 0. Now the first and last inequalities come from the generalized Hadamard inequality (see Prop. 3) applied to x and x △ y, respectively; we prove only the second inequality. Our proof is by induction on the rank r. There is nothing
14
to prove when r = 1, so assume r > 1 and that the result holds in all algebras of rank r − 1. Corresponding to the Jordan frame {e1 , e2 , . . . , er }, let c = e1 and consider the Peirce decompositions of the form (3): x = u + v + w and y = p + q + r, where u = x1 e1 and p = y1 e1 . Now consider z := p + q + Pq (y1−1 e1 ). Since p > 0 in V (c, 1) and z/p = Pq (y1−1 e1 ) − Pq (y1−1 e1 ) = 0, by Item (ii) of the Introduction, z ≥ 0. By the previous theorem, x △ z = u △ p + v △ q + w △ Pq (y1−1 e1 ) 0.
Note that u △ p = x1 y1 ||e1 ||2 E1 and the right-hand side in the above expression is nothing but the decomposition of x △ z of the form (3) with respect to the idempotent E1 in Herm(Rr×r ). To simplify the notation, let µ1 := x1 y1 ||e1 ||2 so that u △ p = µ1 E1 . As x1 > 0 and y1 > 0, we can compute the Schur complement of u △ p in x △ z and get 0 (x △ z)/(u △ p) = w △ Pq (y1−1 e1 ) − Pv △ q (µ−1 1 E1 ). Replacing Pq (y1−1 e1 ) by r − y/p in the above expression and simplifying, we get w △ r − Pv △ q (µ−1 1 E1 ) w △ (y/p). Now, x △ y = µ1 E1 + v △ q + w △ r implies (x △ y)/(µ1 E1 ) = w △ r − Pv △ q (µ−1 1 E1 ) w△(y/p). Using the well-known implication in matrix theory, namely, A B 0 ⇒ det(A) ≥ det(B), we get det (x △ y/(µ1 E1 )) ≥ det(w △ (y/p)). By the induction hypothesis and (10), r Y det(w△(y/p)) ≥ (x2 · · · xr )( ||ei ||2 ) det(y/p). 2
From the determinantal formula (5) applied to x △ y/(µ1 E1 ) and y/p, along with det(p) = y1 and µ1 = x1 y1 ||e1 ||2 , we have r Y det(x △ y) ≥ x1 y1 ||e1 ||2 (x2 · · · xr ) ( ||ei ||2 ) det(y)/y1 = δ (x1 · · · xr ) det(y). 2
This completes the proof.
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Recall that for z ∈ V or for a Hermitian matrix z, λ↓i (z) (i = 1, 2, . . . , r) denote the eigenvalues of z written in the decreasing order. We use the familiar notation λ↓1 (z) = λmax (z) and λ↓r (z) = λmin (z). We also recall that in any simple algebra, the norms of all primitive idempotents are the same. Theorem 10 Let V be simple and θ := ||c||2 for any primitive idempotent c. If x ≥ 0 and y ≥ 0, then for all i = 1, 2, . . . , r, x↓i λmin (y) θ ≤ λ↓i (x △ y) ≤ x↓i λmax (y) θ. Proof Let a := λmax (y) e − y
and b := y − λmin (y) e,
where e = e1 + e2 + · · · + er is the unit element in V . It is easy to verify that a ≥ 0 and b ≥ 0. Then we have x △ a 0 ⇒ x △ y λmax (y)x △ e, and x △ b 0 ⇒ x △ y λmin (y)x △ e. We note that x△e is a diagonal matrix with entries xi θ for i = 1, . . . , r. Using the well-known fact that A B in Herm(Rr×r ) implies λ↓i (A) ≤ λ↓i (B) for all i, we have x↓i λmin (y) θ ≤ λ↓i (x △ y) ≤ x↓i λmax (y) θ. This completes the proof. For a real Hermitian positive definite matrix A, the (Schur product) inequality A ◦ A−1 I holds. This is referred to as Fiedler’s inequality in [1]. A generalization of this result is given below. Theorem 11 If x > 0, then x △ x−1
r X 1
−1
In particular, when V is simple, x △ x
||ei ||2 Ei . θI, where θ := ||ei ||2 for any i.
Proof Corresponding to the idempotent c := e1 , consider the Peirce decompositions of the form (3): x=u+v+w
and y := x−1 = p + q + r.
As x, y > 0, u and r are invertible in their respective algebras. Then, as in the proof of Theorem 9, −1 u + v + Pv (u−1 ∗ ) ≥ 0 and Pq (r∗ ) + q + r ≥ 0.
By Theorem 8, u + v + Pv (u∗−1 ) △ Pq (r∗−1 ) + q + r 0.
−1 Now putting Pv (u−1 ∗ ) = w − x/u, Pq (r∗ ) = p − y/r, and simplifying, we get
u △ (p − y/r) + v △ q + (w − x/u) △ r 0
16
which yields, u △ p + v △ q + w △ r u △ (y/r) + (x/u) △ r. Note that the left-hand side of the above inequality is x △ x−1 . In view of Theorem 1(a), the right-hand side of the above inequality is −1 u △ u−1 ∗ + r∗ △ r = (x1 e1 ) △ (
1 e1 ) + r △ r∗−1 = ||e1 ||2 E1 + r △ r∗−1 . x1
Thus, x △ x−1 ||e1 ||2 E1 + r △ r∗−1 .
By repeating this argument (or by induction), we may write r △ r∗−1 This leads to r X ||ei ||2 Ei . x △ x−1
Pr
2
||ei ||2 Ei .
1
2
When V is simple, θ = ||ei || is a constant for all i and the stated result follows. 7 Miscellaneous In [13], Hu and Smith showed that converse of Theorem 6 holds for complex Hermitian matrices. It turns out that a similar result holds in any simple Euclidean Jordan algebra. Theorem 12 For 1 ≤ k < r, consider two sets of nonnegative real numbers {α1 , α2 , . . . , αr } and {β1 , β2 , . . . , βr−k }, both arranged in the decreasing order, having the same number of zeros, and αi ≥ βi ≥ αi+k ≥ 0 for all i = 1, 2, . . . , r − k. Let V be any simple Euclidean Jordan algebra of rank r and c (6= 0, e) be any idempotent of rank k. Then there exists x ≥ 0 in V with u = Pc (x) invertible in V (c, 1) and λ↓i (x) = αi and λ↓j (x/u) = βj for all i = 1, 2, . . . , r and j = 1, 2, . . . , r − k. Proof Corresponding to the given αi and βj , there exists a real symmetric positive semidefinite matrix H with k × k nonsingular submatrix A such that the eigenvalues of H and H/A are αi and βj , respectively, see [13]. Since real numbers are included in complex numbers/quaternions/octonions, we have the result in the case of matrix algebras (and also in algebras isomorphic to these). Now consider the algebra Ln (see Section 2) for n > 2. Recall that rank of Ln is 2. Let c be a nonzero, non-unit idempotent in Ln . Then there is a Jordan frame {e1 , e2 } such that (see Section 2) 1 1 1 1 and e2 = c = e1 = 2 u′ 2 −u′ for some u′ ∈ R(n−1) with ||u′ || = 1. Corresponding to this Jordan frame, the Peirce decomposition of any x ∈ Ln is: x = x1 e1 + x2 e2 + x12 . Then
17
u = x1 e1 ∈ V (c, 1), v = x12 ∈ V (c, 12 ) and w = x2 e2 ∈ V (c, 0). We are given that the sets {α1 , α2 } and {β1 } have the same number of zeros and α1 ≥ β1 ≥ α2 ≥ 0 . To complete the proof, we have to construct an element x whose Peirce decomposition and eigenvalues are given by 0 x0 = x1 e1 + x2 e2 + ′ , x= (18) x′ z α1 = λ↓1 (x) = x0 + ||x′ ||, α2 = λ↓2 (x) = x0 − ||x′ ||, and β1 = λ↓1 (x/u) = det(x) det(x/u) = det(u) = αx1 α1 2 . Case 1: β1 6=√0. Then α1 6= 0 6= α2 . As n > 2, we can take z ′ ∈ Rn−1 , such α α (β −α )(α −β )
1 2 1 2 1 1 that ||z ′ || = and hz ′ , u′ i = 0. Now define x0 = 21 (α1 +α2 ), β1 x1 = αβ1 α1 2 , x2 = 2x0 − x1 , and x′ = 12 (x1 − x2 )u′ + z ′ . Then it is easily verified that x defined by (18) will have the right properties. Case 2: β1 = 0. Because of our assumptions, α1 > 0 = α2 . Now take x1 = α1 , x2 = 0, x0 = 21 α1 , z ′ = 0, and x′ = 21 α1 u′ . Then it is easily verified that x defined by (18) will have the right properties. This completes the proof.
In matrix theory, for any two real symmetric matrices, the rank of a Schur product satisfies the inequality [21] rank(A ◦ B) ≤ rank(A) rank(B). The following example shows this result need not hold in the setting of Euclidean Jordan algebras. Example. Let V = L4 . Let e1 = 12 [ 1 0 0 1 ]T , e2 = 12 [ 1 0 0 − 1 ]T , x12 = [ 0 1 0 0 ]T and y12 = [ 0 0 1 0 ]T . Then, it is easy to verify that x := e1 + e2 + x12 ≥ 0,
y := e1 + e2 + y12 ≥ 0,
while rank(x) = rank(y) = 1, rank(x △ y) = 2. References 1. Bapat, R.B., Raghavan, T.E.S.: Nonnegative Matrices and Applications. Cambridge University Press, Cambridge (1997) 2. Cohen, N., De Leo, S.: The quaternionic determinant. Elect. J. Linear Algebra. 7, 100– 111 (2000) 3. Dray, T., Manogue, C.A.: The octonionic eigenvalue problem. Adv. Appl. Clifford Algebr. 8, 341–364 (1998) 4. Fan, K.: Generalization of Bergstrom’s Inequality. The Amer. Math. Monthly. 66, 153– 154 (1959) 5. Faraut, J., Kor´ anyi, A.: Analysis on Symmetric Cones. Clarendon Press, Oxford (1994) 6. Koufany, K.: Contractions of angles in symmetric spaces. Publ. RIMS, Kyoto University. 38, 227–243 (2002) 7. Gowda, M.S., Sznajder, R.: Automorphism invariance of P and GUS properties of linear transformations in Euclidean Jordan algebras. Math. Oper. Res. 31, 109–123 (2006)
18 8. Gowda, M.S., Sznajder, R.: Schur complements, Schur determinantal and Haynsworth inertia formulas in Euclidean Jordan algebras. Linear Algebra Appl. 432, 1553–1559 (2010) 9. Gowda, M.S., Sznajder, R., Tao, J.: Some P-properties for linear transformations on Euclidean Jordan algebras. Linear Algebra Appl. 393, 203–232 (2004) 10. Gowda, M.S., Tao, J.: The Cauchy interlacing theorem in simple Euclidean Jordan algebras and some consequences. Linear Multilinear Algebra. To appear. 11. Gowda, M.S., Tao, J., Moldovan, M.M.: Some inertia theorems in Euclidean Jordan algebras. Linear Algebra Appl. 430, 1992–2011 (2009) 12. Horn, R.A., Johnson, C.R.: Matrix Analysis. Cambridge University Press, Cambridge (1985) 13. Hu, S.-A., Smith. R.L.: The Schur complement interlacing theorem. SIAM J. Matrix Anal. 16, 1013–1023 (1995) 14. Loos, O.: Finiteness condition in Jordan pairs. Math. Z. 206, 577–587 (1991) 15. Massam, H., Neher, E.: On transformations and determinants of Wishart variables on symmetric cones. J. Theoret. Probab. 10, 867–902 (1997) 16. Moldovan, M.M.: A Ger˘sgorin type theorem, spectral inequalities, and simultaneous stability in Euclidean Jordan algebras. PhD Thesis, University of Maryland Baltimore County (2009) 17. Moldovan, M.M., Gowda, M.S.: Strict diagonal dominance and a Ger˘sgorin type theorem in Euclidean Jordan algebras. Linear Algebra Appl. 431, 148–161 (2009) 18. Schmieta, S.H., Alizadeh, F.: Extension of primal-dual interior point algorithms to symmetric cones. Math. Program, Series A. 96, 409–438 (2003) 19. Smith, R.L.: Some interlacing properties of the Schur complement of a Hermitian matrix. Linear Algebra Appl. 177, 137–144 (1992) 20. Tao, J.: Some P-Properties for Linear Transformations on the Lorentz Cone. PhD Thesis, University of Maryland Baltimore County (2004) 21. Zhang, F.: Matrix Theory. Springer-Verlag, New York (1999) 22. Zhang, F.: The Schur Complement and its Applications. Springer, New York (2005)
