the cauchy interlacing theorem in simple euclidean jordan ...gowda/tech-reports/trgow08-02.pdf ·...

July 23, 2009 16:41 Linear and Multilinear Algebra cauchy-lmatrevisedf

Linear and Multilinear AlgebraVol. 00, No. 00, Month 200x, 1–23

The Cauchy interlacing theorem in simple Euclidean Jordan

algebras and some consequences

M. Seetharama Gowdaa∗ and J. Tao b

aDepartment of Mathematics and Statistics, University of Maryland, Baltimore County,Baltimore, MD 21250; bDepartment of Mathematical Sciences, Loyola College in

Maryland, Baltimore, Maryland 21210

(Received 12 August 2008)

In this article, based on the min-max theorem of Hirzebruch, we formulate and prove theCauchy interlacing theorem in simple Euclidean Jordan algebras. As a consequence , werelate the inertias of an element and its principal components and extend some well knownmatrix theory theorems and inequalities to the setting of simple Euclidean Jordan algebras.

Keywords: Euclidean Jordan algebras, quadratic representations, min-max theorem ofHirzebruch, Cauchy interlacing theorem, Schur’s theorem, Hadamard’s inequality, Fan’strace inequality

AMS Subject Classification: 15A33; 17C20; 17C55

1. Introduction

In matrix theory, the well known Cauchy’s interlacing theorem states that if A isa (complex) Hermitian matrix of size n×n and B is a leading principal submatrixof A of size k × k, then

λ↓i (A) ≥ λ↓i (B) ≥ λ↓n−k+i(A) (i = 1, 2, . . . , k),

where λ↓1(A), λ↓2(A), · · · , λ↓n(A) denote the eigenvalues of A written in the decreas-ing order (with a similar notation for B). This result has many interesting conse-quences. For example, given any two (complex) Hermitian matrices A and B, wehave the following:

(1) If B is a principal submatrix of A, then π(B) ≤ π(A) and ν(B) ≤ ν(A), whereπ(A) and ν(A) denote the number of positive and negative eigenvalues of Arespectively (and likewise for B);

(2) A is positive definite (semidefinite) if and only if all its leading principal minorsare positive (respectively, nonnegative);

(3) Schur’s theorem: diag(A) ≺ λ(A), where diag(A) and λ(A) refer to the diago-nal of A and the vector of eigenvalues of A (written in the decreasing order),respectively;

(4) Hadamard’s inequality: If A is positive semidefinite with diagonal elementsa11, a22, . . . , ann, then det(A) ≤ a11a22 · · · ann;

∗Email: [email protected]

ISSN: 0308-1087 print/ISSN 1563-5139 onlinec© 200x Taylor & FrancisDOI: 10.1080/03081080xxxxxxxxxhttp://www.informaworld.com


2 M. Seetharama Gowda and J. Tao

(5) Fan’s inequality: trace(AB) ≤∑n

1 λ↓i (A)λ↓i (B).

In this paper, we extend Cauchy’s interlacing theorem to simple Euclidean Jor-dan algebras and deduce analogs of the above statements in the setting of simpleEuclidean Jordan algebras.

Let (V, , 〈·, ·〉) be a simple Euclidean Jordan algebra of rank r (see Section 2 fordefinition). For any nonzero idempotent c in V , let

Pc(x) := 2c (c x)− c2 x

define the quadratic representation of c. Also, let

V (c, 1) := x ∈ V : c x = x.

Then V (c, 1) is a simple subalgebra of V and for any z ∈ V ,

z := Pc(z) ∈ V (c, 1).

Denoting the eigenvalues of any z in V by the decreasing sequence

λ↓1(z), λ↓2(z), · · · , λ↓r(z)

and similarly for z in V (c, 1), we state the Cauchy interlacing inequalities in simpleEuclidean Jordan algebras:

λ↓i (z) ≥ λ↓i (z) ≥ λ↓r−k+i(z) (i = 1, 2, . . . , k),

where k is the rank of V (c, 1).To recover the matrix theoretic result, one has to take V = Herm(Cn×n) (the

space of all n×n complex Hermitian matrices with trace inner product and Jordanproduct defined by X Y := 1

2(XY + Y X)) and the idempotent

C :=[Ik×k 0

0 0

],

where Ik×k is the identity matrix of size k × k.

Our proof of the interlacing inequalities/theorem is based on the min-max the-orem of Hirzebruch [13] which generalizes the well-known Courant-Fischer-Weylmin-max theorem (see [3]) to simple Euclidean Jordan algebras. As a consequenceof the above inequalities, we state analogs of items/statements (1)− (5) above forEuclidean Jordan algebras. We also state a result that describes the inertia of anelement in a Euclidean Jordan algebra when a specific Peirce decomposition of thatelement is known.

In the case of simple algebras Herm(Rn×n) (the space of all n×n real symmet-ric matrices) and Herm(Cn×n), our results reduce to the classical results. Whenspecialized to Herm(Hn×n) (the space of all n × n quaternionic Hermitian ma-trices) our interlacing inequalities reduce to a result of Tam [20] who proved theinterlacing inequalities in the setting of Lie algebras. Our results are new in the(only other) simple algebras Ln (the so called Jordan spin algebra) for n > 2 andHerm(O3×3) (the space of all 3 × 3 octonionic Hermitian matrices). (We remarkthat in the setting of Herm(O3×3) our results are for ‘spectral’ eigenvalues, asthere is a difference between such eigenvalues and real eigenvalues, see Sec. 2.2.)


Cauchy interlacing theorem 3

2. Euclidean Jordan Algebras

We assume that the reader is familiar with the basic Euclidean Jordan algebratheory which can be found, for example, in Faraut and Koranyi [7]. For briefintroductions, see Schmieta and Alizadeh [19], and Gowda, Sznajder, and Tao[10].

Let the triple (V, , 〈·, ·〉) denote a Euclidean Jordan algebra, where (V, 〈·, ·〉) isa finite dimensional inner product space over R (the field of real numbers) and(x, y) 7→ xy : V ×V → V is a bilinear mapping satisfying the following conditionsfor all x and y: x y = y x, x (x2 y) = x2 (x y), and 〈x y, z〉 = 〈y, x z〉. Wedenote the unit element of V by e. In V , the cone of squares x2 : x ∈ V (whichis a symmetric cone) is denoted by K. For an element z ∈ V , we write

z ≥ 0 (z > 0) if and only if z ∈ K (z ∈ Ko),

where Ko denotes the interior of K.

A Euclidean Jordan algebra is said to be simple if it is not the direct productof two (non-trivial) Euclidean Jordan algebras. The classification theorem (Chap-ter V, Faraut and Koranyi [7]) says that every simple Euclidean Jordan alge-bra is isomorphic to the (Jordan spin) algebra Ln or to the algebra of all n × nreal/complex/quaternion Hermitian matrices with (real) trace inner product andX Y = 1

2(XY +Y X) or the algebra of all 3×3 octonion Hermitian matrices with(real) trace inner product and X Y = 1

2(XY + Y X).Furthermore, the structure theorem, see (Chapters III and V, Faraut and Koranyi

[7]) says that any Euclidean Jordan algebra is a (Cartesian) product of simpleEuclidean Jordan algebras.

Let V be a Euclidean Jordan algebra of rank r. An element c ∈ V is called anidempotent if c2 = c; it is a primitive idempotent if it is nonzero and cannot bewritten as the sum of two nonzero idempotents. A finite set e1, . . . , er is said tobe a Jordan frame in V if each ei is a primitive idempotent in V , eiej = δijei forall i, j = 1, 2, . . . , r and e1 + e2 + · · ·+ er = e. (These conditions imply 〈ei, ej〉 = 0for all i 6= j = 1, 2, . . . , r.)

The spectral theorem says that for each element x in V , there exists a Jordanframe e1, . . . , er and real numbers λi (i = 1, 2, . . . , r) such that

x = λ1e1 + · · ·+ λrer. (1)

The above expression is the spectral decomposition (or the spectral expan-sion) of x. The real numbers λi (also written as λi(x)) are the (spectral) eigenval-ues of x; these are uniquely defined, even though the Jordan frame correspondingto x need not be unique. For any x ∈ V given by (1), we define the trace anddeterminant of x by

trace(x) := λ1 + λ2 + · · ·+ λr

and

det(x) := λ1λ2 · · ·λr.



We also define the inertia of x by

In(x) = (π(x), ν(x), δ(x)),

where π(x), ν(x), and δ(x) are, respectively, the number of eigenvalues of x whichare positive, negative, and zero, counting multiplicities. Clearly,

π(x) + ν(x) + δ(x) = r

for all x.We note that 〈u, v〉t := trace(u v) defines another inner product on V so that

(V, , 〈·, ·〉t) is also an Euclidean Jordan algebra.

Fix a Jordan frame e1, e2, . . . , er in a Euclidean Jordan algebra V . For i, j ∈1, 2, . . . , r, define the Peirce spaces

Vii := x ∈ V : x ei = x = Rei

and when i 6= j,

Vij := x ∈ V : x ei =12x = x ej.

Then we have the following.

Theorem 2.1 : (Theorem IV.2.1, Faraut and Koranyi [7]) The space V is theorthogonal direct sum of spaces Vij (i ≤ j). Furthermore,

Vij Vij ⊂ Vii + Vjj ,Vij Vjk ⊂ Vik if i 6= k, andVij Vkl = 0 if i, j ∩ k, l = ∅.

Thus, given a Jordan frame e1, e2, . . . , er, we can write any element x ∈ V as

x =r∑

i=1

xiei +∑i<j

xij (2)

where xi ∈ R and xij ∈ Vij . This expression is the Peirce decomposition of xwith respect to e1, e2, . . . , er.

Given a Euclidean Jordan algebra V , we may write it as a product of sim-ple Jordan algebras: V = V1 ⊕ V2 ⊕ . . . ⊕ VN . Then any primitive idempotentin V is of the form (0, 0, · · · , ci, 0, · · · , 0) for some primitive idempotent ci inVi. Consequently, for any x = (x1, x2, . . . , xN ) in the direct product, In(x) =(∑N

i π(xi),∑N

i ν(xi),∑N

i δ(xi)).

For any primitive idempotent c, let

V (c,12) := x : x c =

12x.

Proposition 2.2: The following are equivalent:

(i) V is simple.



(ii) For any two orthogonal primitive idempotents c1 and c2,

V (c1,12) ∩ V (c2,

12)

is nonzero.

Proof : The implication (i) ⇒ (ii) is given in Corollary IV.2.4 of [7]. We onlyprove the converse. Assume (ii) and suppose (for simplicity) that V is the directproduct of simple algebras V1 and V2. If ei (i = 1, 2) is a primitive idempotentin Vi, then c1 = (e1, 0) and c2 = (0, e2) are primitive idempotents in V that areorthogonal. It is clear that zero is the only element x = (x1, x2) ∈ V1⊕V2 with theproperty x c1 = 1

2x = x c2. This contradicts (ii). Hence V is simple.

For a given a ∈ V , we define the corresponding Lyapunov transformation La :V → V by

La(x) = a x

and the quadratic representation Pa by

Pa(x) := 2a (a x)− a2 x.

We say that elements a and b operator commute if La and Lb commute, i.e.,

LaLb = LbLa.

It is known that a and b operator commute if and only if a and b have their spectraldecompositions with respect to a common Jordan frame (Lemma X.2.2, Faraut andKoranyi [7] or Theorem 27, Schmieta and Alizadeh [19]). In particular, if a and boperate commute, then PaPb = Pab.

We recall the following from Gowda, Sznajder and Tao [10]:

Proposition 2.3: For x, y ∈ V , the following conditions are equivalent:

(i) x ≥ 0, y ≥ 0, and 〈x, y〉 = 0.(ii) x ≥ 0, y ≥ 0, and x y = 0.

In each case, elements x and y operator commute.

An easy consequence is the following.

Proposition 2.4: Let the Peirce decomposition of an element x with respect toa Jordan frame e1, e2, . . . , er be given by

x =r∑

i=1

xiei +∑i<j

xij .

If x ≥ 0 and xk = 0 for some k, then xlk = xkj = 0 for all l, j with l < k < j.

This can be easily seen by noting that when x ≥ 0, 0 = 〈x, ek〉 ⇒ x ek = 0 and0 = x ek = xkek + 1

2(∑

l<k xlk +∑

k<j xkj) implies xlk = 0 = xkj for all l < k < jdue to the orthogonality of the spaces Vij .

The following result describes the effect of La and Pa on any element x.



Proposition 2.5: Suppose that e1, e2, . . . , er is a Jordan frame,

a = a1e1 + a2e2 + . . . + arer,

and

x =r∑

i=1

xiei +∑i<j

xij

(with xi ∈ R and xij ∈ Vij) be the Peirce decomposition of x with respect to thisJordan frame. Then

La(x) =r∑

i=1

aixiei +∑i<j

ai + aj

2xij

and

Pa(x) =r∑

i=1

ai2xiei +

∑i<j

aiajxij .

Proof : From Theorem 2.1, we have a ei = aiei and a xij = ai+aj

2 xij . Fromthese, the stated expressions for La and Pa follow.

An algebra automorphism on a Euclidean Jordan algebra is an invertiblelinear transformation that preserves the Jordan product. Such an automorphismmaps a Jordan frame onto a Jordan frame; hence for any element x, the eigenval-ues λi(x) and diagonal numbers x1, x2, . . . , xr in the Peirce decomposition (2) areinvariant under algebra automorphisms. In addition, it is known that in a simpleEuclidean Jordan algebra, any Jordan frame can be mapped onto any other by analgebra automorphism, see Prop. IV.2.5 in [7].

2.1. The algebra Ln

For the Jordan spin algebra Ln the underlying space is Rn, n > 1. We write anyelement x in the form

x =[x0

x′

](3)

with x0 ∈ R and x′ ∈ Rn−1. The inner product in Ln is the usual inner producton Rn. The Jordan product x y in Ln is defined by

x y =[x0

x′

]

[y0

y′

]:=

[〈x, y〉

x0y′ + y0x

′

].

Then Ln is a Euclidean Jordan algebra of rank 2 and for any element x ∈ Ln, seeExample 10 in [19], the eigenvalues and the determinant are given by

λ1(x) = x0 + ||x′||, λ2(x) = x0 − ||x′||, (4)



and

det(x) = x20 − ||x′||2.

Now consider any Jordan frame e1, e2 in Ln. Then there exists (see e.g., Lemma2.3.1, [21]) a unit vector u ∈ Rn−1 such that

e1 :=12

[1u

]and e2 :=

12

[1−u

].

With respect to this, any x ∈ Ln given by (3) has a Peirce decomposition[x0

x′

]= x1e1 + x2e2 + x12 = x1

12

[1u

]+ x2

12

[1−u

]+

[0v

](5)

where v ∈ Rn−1 with 〈u, v〉 = 0. (This is easy to verify, see e.g., Lemma 2.3.4,[21].)

2.2. Hermitian matrices over quaternions and octonions

Throughout this paper, R and C, denote, respectively, the set of all real numbersand complex numbers. The linear space of quaternions - denoted by H - is a 4-dimensional linear space over R with a basis 1, i, j, k. The space H is made intoan algebra by means of the conditions

i2 = j2 = k2 = −1 and ijk = −1.

For any x = x0 1 + x1 i + x2 j + x3 k ∈ H, we define the conjugate by

x := x0 1− x1 i− x2 j − x3 k.

The linear space of octonions over R - denoted by O - is an 8-dimensional linearspace with basis 1, e1, e2, e3, e4, e5, e6, e7. The space O becomes an algebra viathe following multiplication table on the non-unit basis elements [23]:

e1 e2 e3 e4 e5 e6 e7

e1 −1 e3 −e2 e5 −e4 −e7 e6

e2 −e3 −1 e1 e6 e7 −e4 −e5

e3 e2 −e1 −1 e7 −e6 e5 −e4

e4 −e5 −e6 −e7 −1 e1 e2 e3

e5 e4 −e7 e6 −e1 −1 −e3 e2

e6 e7 e4 −e5 −e2 e3 −1 −e1

e7 −e6 e5 e4 −e3 −e2 e1 −1

For an element

x = x0 1 + x1e1 + x2e2 + x3e3 + x4e4 + x5e5 + x6e6 + x7e7



in O, we define the conjugate by

x = x0 − x1e1 − x2e2 − x3e3 − x4e4 − x5e5 − x6e6 − x7e7.

Let F denote the set of all reals/complex numbers/quaternions/octonions. Wewrite Fn for the space of all n×1 vectors over F and Fn×n for the space of all n×nmatrices over F . For a matrix A ∈ Fn×n, we define the conjugate A and transposeAT in the usual way. We say that a square matrix A ∈ Fn×n is Hermitian if Acoincides with its conjugate transpose, that is, if A = A∗ := (A)T . We let

Herm(Fn×n):=set of all n× n Hermitian matrices with entries from F .

In Herm(Fn×n), the Jordan product and the inner product are respectively givenby

X Y =12(XY + Y X) and 〈X, Y 〉 = Re trace(XY ).

As noted previously, Herm(Rn×n), Herm(Cn×n), Herm(Hn×n), and Herm(O3×3)are examples of simple Euclidean Jordan algebras.

For a matrix A ∈ Fn×n, an element λ ∈ F is a left (right) eigenvalue of A if thereis a nonzero x ∈ Fn such that Ax = λx (respectively, Ax = xλ); when λ ∈ R, wecall λ a real eigenvalue.

In the context of Hermitian matrices over reals/complex numbers/quaternions,real eigenvalues coincide with the spectral eigenvalues. However, as noted below,in the context of octonions, these numbers can be different. For discussions onleft/right/real eigenvalues of quaternionic/octonionic matrices, we refer to [5], [17],[23] and the references therein.

Let A ∈ Herm(O3×3) be given by

A :=

p a ba q cb c r

,

where p, q, r ∈ R and a, b, c ∈ O. Then, see [17],

det(A) = pqr + 2Re(b(ac))− r|a|2 − q|b|2 − p|c|2. (6)

For objects a, b, c ∈ O and for the matrix A given above, we let

[a, b] := ab− ba, [a, b, c] := (ab)c− a(bc),

and Φ(a, b, c) := 12 Re ([a, b]c). Also, let

s(A) := pq + qr + rp− |a|2 − |b|2 − |c|2.

(Recall that tr(A) = p + q + r.)Then as noted in Remark 1 of [17], the spectral eigenvalues of the above A are

the roots of

det(λI −A) = λ3 − (trA)λ2 + s(A)λ− det(A) = 0.



Furthermore, as noted in Lemma 1O3 , [5], the real eigenvalues of A are the rootsof

det(λI −A) = λ3 − (trA)λ2 + s(A)λ− det(A) = r

where r is either of the two roots of

r2 + 4Φ(a, b, c)r − |[a, b, c]|2 = 0. (7)

The real eigenvalues of A can also be studied via a 24× 24 real symmetric matrixconstructed in the following way [23]. Regard any octonion x ∈ O as a columnvector ~x ∈ R8. Then any u ∈ O induces an 8 × 8 real matrix defined by ω(u)~x =−→ux. Now for the given matrix A = [akl] ∈ Herm(O3×3), we let ω(A) denote the(block) matrix with blocks ω(akl). In [23], Tian shows that ω(A) is a real symmetricmatrix whose (real) eigenvalues coincide with the real eigenvalues of A. In fact, ifµ1, µ2, . . . , µ6 are the real eigenvalues of A coming from (7), then these are theeigenvalues of ω(A), each with multiplicity four. It was made explicit in Remark 4of [17] that the real eigenvalues and the spectral eigenvalues of A can be different.

3. The min-max theorem of Hirzebruch

For a simple Euclidean Jordan algebra V , let

J (V ) := c : c is a primitive idempotent in V.

Then J (V ) is a compact set in V (see Exercise 5, p.78 in [7]). If 〈·, ·〉 is the innerproduct in V , then there exists a positive number α such that 〈x, y〉 = α trace(xy),see [7], Prop. III.4.1. In particular, we have

α = 〈c, e〉 = 〈c2, e〉 = 〈c, c〉 = ||c||2 ∀c ∈ J (V ).

In what follows, we use the following notation: Given any set of numbersλ1, λ2, . . . , λk, we make a rearrangement and write the set as

λ↓1, λ↓2, . . . , λ

↓k,

where λ↓1 ≥ λ↓2 ≥ · · · ≥ λ↓k. Thus for any x ∈ V , λ↓i (x) (i = 1, 2, . . . , r) denote theeigenvalues of x written in the decreasing order. We also write

λ(x) := (λ↓1(x), λ↓2(x), . . . , λ↓r(x)).

Theorem 3.1 : (Min-max theorem of Hirzebruch [13]) Let V be simple. Then forany x ∈ V we have

λ↓1(x) = maxc∈J (V )

〈x, c〉〈e, c〉

,

λ↓r(x) = minc∈J (V )

〈x, c〉〈e, c〉

,



and

λ↓k+1(x) = minf1,f2,...,fk⊂J (V )

maxc∈J (V ),c⊥f1,f2,...,fk

〈x, c〉〈e, c〉

for k = 1, . . . , r − 2.

4. The Cauchy interlacing theorem

Let e1, e2, . . . , er be a Jordan frame with corresponding Peirce spaces Vij . For1 ≤ k ≤ r, let

V (k) = Ve1,e2,...,ek := x ∈ V : x (e1 + e2 · · ·+ ek) = x.

We list below some properties of this space.

Proposition 4.1: The following hold:

(i) V (k) is a subalgebra of V .(ii) V (k) = Re1 +Re2 + · · ·+Rek +

∑1≤i<j≤k Vij .

(iii) V (k) = Pe1+e2+···+ek(V ).

(iv) The symmetric cone of V (k) is equal to V (k) ∩K and is a face of K.(v) V (k) ∩K = x ∈ K : x ⊥ ek+1, . . . , er.(vi) If V is simple, then so is V (k).(vii) When V is simple,

J (V (k)) = c ∈ J (V ) : c ⊥ ek+1, . . . , er,

where J (V ) (J (V (k)) denotes the set of primitive idempotents in V (respectively,V (k)).

Proof : For Item (i), see [7], Prop. IV.1.1. For Items (ii) and (iv), see [9], Sec. 2.2and Thm. 3.1. Item (iii) is given in [7], Page 64; it also follows from Prop. 2.5.Regarding Item (v): Clearly, because of (ii) and the orthogonality of the Vij , everyelement of V (k) is orthogonal to ek+1, . . . , er. If x ∈ K and x ⊥ ek+1, . . . , er,then by Proposition 2.4 and Item (ii), x ∈ V (k). Now to prove (vi). Suppose V issimple. Assume without loss of generality that k > 1. We verify condition (ii) inProposition 2.2. Let c1 and c2 be any two orthogonal primitive idempotents in V (k).Then there exists primitive idempotents c3, . . . , ck in V (k) such that c1, c2, . . . , ckis a Jordan frame in V (k) (which can be seen by considering the spectral decom-position of e(k) − (c1 + c2) in V (k) with e(k) := e1 + e2 + · · ·+ ek denoting the unitelement in V (k)) and c1, c2, . . . , ck, ek+1, . . . , er is a Jordan frame in V . Note that

e1 + e2 + · · ·+ ek = c1 + c2 + · · ·+ ck.

Let

V =∑

i≤j≤r

V ′ij

be the Peirce decomposition of V with respect to the Jordanframe c1, c2, . . . , ck, ek+1, . . . , er. Since V is simple, V ′

12 = V (c1,12) ∩ V (c2,

12) is



nonzero. Clearly, V ′12 is a subset of∑

1≤i≤j≤k

V ′ij = Pc1+c2+···+ck

(V ) = Pe1+e2+···+ek(V ) = V (k).

Hence by Proposition 2.2, V (k) is simple. We note that Item (vi) is also proved in[8].Now for Item (vii). Let c be a primitive idempotent in V (k). Then clearly, c is anonzero idempotent in V . Suppose if possible, c = c1 + c2 where c1 and c2 arenonzero idempotents in V . Since c belongs to the symmetric cone of V (k) and ci

belong to the symmetric cone K of V , and the symmetric cone of V (k) is a face ofK (see Theorem 3.1, [9]), it follows that both c1 and c2 belong to the symmetriccone of V (k). Since c is primitive in V (k), we reach a contradiction. This proves thatJ (V (k)) ⊆ J (V ). Now by the orthogonality of spaces Vij , any element in V (k) isorthogonal to ej for k+1 ≤ j ≤ r. Thus J (V (k)) ⊆ c ∈ J (V ) : c ⊥ ek+1, . . . , er.Now to prove the reverse inclusion, let c ∈ J (V ) and c ⊥ ek+1, . . . , er. By Item(v), c ∈ V (k). Clearly c is a primitive idempotent in V (k) and we have the equalityin (vii).

Let V (k) be as in the above Proposition. For any z ∈ V , we let

z := Pe1+e2+···+ek(z) =

k∑1

ziei +∑

1≤i<j≤k

zij .

We call z, the principal component of z corresponding to e1, e2, . . . , ek; its deter-minant in V (k) will be called a principal minor of z.

In the result below, λ↓1(z), λ↓2(z), . . . , λ↓k(z) refer to the eigenvalues of z withrespect to V (k).

Theorem 4.2 : (The Cauchy interlacing theorem in simple Euclidean Jordan al-gebras) Let V be simple and e1, e2, . . . , er be a Jordan frame. For any elementz ∈ V , let z denote the principal component of z with respect to e1, e2, . . . , ek.Then

λ↓i (z) ≥ λ↓i (z) ≥ λ↓r−k+i(z) (i = 1, 2, . . . , k).

Proof : Without loss of generality, we assume that the inner product is givenby the trace inner product. In this case, ||c|| = 1 for any primitive idempotentc. When k = r, z = z and the stated inequalities are obvious. We assume that1 ≤ k < r.

Suppose i = 1. Then by the min-max theorem of Hirzebruch,

λ↓r−k+1(z) ≤ maxc∈J (V ), c⊥ek+1,...,er

〈c, z〉 = maxc∈J (V k)

〈c, z〉 = λ↓1(z).

Here, we used Item (vii) of the previous proposition and the fact that for anyc ∈ V (k), 〈c, z〉 = 〈c, z〉.

For i = k, we have

λ↓r−k+k(z) = minc∈J (V )

〈c, z〉 ≤ minc∈J (V ), c⊥ek+1,...,er

〈c, z〉 = minc∈J (V k)

〈c, z〉 = λ↓k(z).



We now assume that 1 < i < k. For ease of notation, let

Ω := f1, f2, . . . , fi−1.

Then by the min-max theorem of Hirzebruch,

λ↓r−k+i(z) ≤ max〈c, z〉 : c ∈ J (V ), Ω ⊆ J (V k), c ⊥ Ω ∪ ek+1, . . . , er= max〈c, z〉 : c ∈ J (V k), c ⊥ Ω ⊆ J (V k)= max〈c, z〉 : c ∈ J (V k), c ⊥ Ω ⊆ J (V k)= λ↓i (z).

Thus, λ↓i (z) ≥ λ↓r−k+i(z) for all i = 1, 2, . . . , k.

Now replacing z by −z, we have

λ↓i (−z) ≥ λ↓r−k+i(−z).

Since λ↓i (−z) = −λ↓k−i+1(z) and λ↓i+r−k(−z) = −λ↓r−(i+r−k)+1(z) we have

λ↓k−i+1(z) ≤ λ↓k−i+1(z).

Putting j = k − i + 1, we get

λ↓j (z) ≤ λ↓j (z)

for j = 1, 2, . . . , k. This concludes the proof of the theorem.

Remarks. The above theorem yields the classical results when specialized to Her-mitian matrices over reals and complex numbers, and to a result of Tam ([20])when specialized to Hermitian matrices over quaternions. When specialized toHerm(O3×3), it yields interlacing inequalities for spectral eigenvalues. As notedin Section 2.2, for an element A in Herm(O3×3), the spectral eigenvalues can bedifferent from the real eigenvalues. For real eigenvalues, the interlacing inequalitiescan be obtained by applying the classical Cauchy interlacing theorem to the 24×24real symmetric matrix ω(A), see Section 2.2. For example, if µ1, µ2, . . . , µ6 are thereal eigenvalues of A coming from (7) written in the decreasing order, and δ1, δ2

are the real eigenvalues of the 2 × 2 principal submatrix of A (see the discussionfollowing Theorem 4.17 in [23]), written in the decreasing order, then

µ1 ≥ µ2 ≥ δ1 ≥ µ3 ≥ µ4 ≥ δ2 ≥ µ5 ≥ µ6.

In the case of Ln for n > 2, given the Peirce decomposition (5) and x = x1e1,the interlacing inequalities become

λ1 ≥ x1 ≥ λ2,

where λ1 and λ2 are given by (4). Since x1 = 〈x,e1〉||e1||2 = x0 + 〈x′, u〉, the above

inequalities reduce to the Cauchy-Schwarz inequality.



Corollary 4.3: Let V be any Euclidean Jordan algebra. Let c be any idempotentin V . Then for any z in V , we have

π(Pc(z)) ≤ π(z) and ν(Pc(z)) ≤ ν(z).

Proof : Without loss of generality, let c 6= 0. Then there exists a Jordan framee1, e2, . . . , er such that c = e1 + e2 + · · · + ek for some k. Let z = Pc(z). It isenough to prove that ν(z) ≤ ν(z) as the inequality regarding positive eigenvaluesfollows from replacing z by −z.First suppose that V is simple. Then by the Cauchy interlacing theorem,

λ↓i (z) ≥ λ↓i (z) ≥ λ↓r−k+i(z)

for i = 1, 2, . . . , k. Let ν(z) = q so that λ↓k−q+1(z), . . . , λ↓k(z) are all the negativeeigenvalues of z ∈ V (k) (if any) . With i = k − q + 1, it follows from the above in-equality that λ↓r−k+i(z) = λ↓r−q+1(z) is negative. This means that in the decreasingenumeration of eigenvalues of z, the eigenvalues subsequent to λ↓r−q+1(z) are alsonegative. Thus z will have at least r− (r− q + 1) + 1 = q eigenvalues. This provesthat ν(z) ≤ ν(z). By working with −z, we get the inequality π(z) ≤ π(z).Now suppose that V is not simple. We write V as a direct product of simple al-gebras: V = V1 × V2 × . . . × VN . If c is any idempotent in V and z ∈ V , we writec = (c1, c2, . . . , cN ) and z = (z1, z2, . . . , zN ) where ci, zi ∈ Vi for all i and each ci isan idempotent in Vi. We note that

Pc(z) = (Pc1(z1), . . . , PcN(zN )).

By applying the previously derived inequalities to each Pci(zi) and noting π(z) =∑N

1 π(zi) (and a similar expression for ν(z)), we get the stated inequalities in thegeneral case.

Remarks. We note that in the above corollary, the number π(Pc(z)) (likewiseν(Pc(z))) remains the same whether Pc(z) is viewed as an object of V or V (c, 1).As a consequence of the Sylvester’s law of inertia in Euclidean Jordan algebraswhich says that

In(Pa(x)) = In(x) (x ∈ V )

whenever a is invertible, see [11], one can get a generalization of the above corollary:

π(Pa(z)) ≤ π(z) ν(Pa(z)) ≤ ν(z) (z ∈ V )

for any a in V .

While giving an equivalent formulation of the inertia theorem of Ostrowski andSchneider [18], Wimmer [24] proves the following result on Hermitian matrices:Suppose A is a n× n complex Hermitian matrix given in the block form by

A =[C ?? D

]with C positive definite of size k × k and D negative definite of size n − k. ThenIn(A) = (k, n− k, 0).



In what follows, we prove an analog of this in Euclidean Jordan algebras; see [11]for another proof.

Corollary 4.4: Corresponding to a Jordan frame e1, e2, . . . , er in a Eu-clidean Jordan algebra V , let the Peirce decomposition of z be given by z =∑r

1 ziei +∑

i<j≤r zij with∑k

1 ziei +∑

i<j≤k zij > 0 in Ve1,e2,...,ek and∑r

k+1 ziei +∑k+1≤i<j zij < 0 in Vek+1,...,er. Then In(z) = (k, r − k, 0).

Proof : Let c := e1 + e2 + · · · + ek and d := ek+1 + · · · + er. Then c and dare idempotents. By our assumption, Pc(z) > 0 in Ve1,e2,...,ek and Pd(z) < 0 inVek+1,...,er. Therefore, π(Pc(z)) = k and ν(Pd(z)) = r − k. Since π(Pc(z)) ≤ π(z)and ν(Pd(z)) ≤ ν(z), we have k + (r − k) ≤ π(z) + ν(z) ≤ r. This implies thatπ(Pc(z)) = π(z) and ν(Pd(z)) = ν(z); hence In(z) = (k, r − k, 0).

Below we state an analog of statement/item (2) of the Introduction for simpleEuclidean Jordan algebras as a consequence of the Cauchy interlacing theorem. Wenote that this appears as problem 5 in Section VI.4 in [7].

Corollary 4.5: Let V be simple and e1, e2, . . . , er be a Jordan frame. For agiven z ∈ V , let

∆k(z) := det(Pe1+e2+···+ek(z)) (1 ≤ k ≤ r).

Then z > 0 (≥ 0) if and only if ∆k(z) > 0 (respectively, ≥ 0) for all k = 1, 2, . . . , r.

Proof : Suppose that z > 0. Then all the eigenvalues of z are positive and so thedeterminant of z is also positive. In addition, for any k, Pe1+e2+···+ek

(z) is positivein V (k) and hence det(Pe1+e2+···+ek

(z)) > 0. Thus z > 0 implies that ∆k(z) > 0 forall k = 1, 2, . . . , r. The case z ≥ 0 can be handled by considering z + εe for ε > 0.(Note that the eigenvalues and determinants vary continuously on the element,see e.g., [11].) For the converse result(s), we induct on the rank of V . The resultis obviously true for r = 1. So, assume the result for any simple algebra of rankr−1. For the given Jordan frame e1, e2, . . . , er, we consider the subalgebra V (r−1)

which is simple (according to Item (vi) in Prop. 4.1) and of rank r − 1. Let

z = Pe1+e2+···+er−1(z).

For ease of notation, let αi (βi) denote the eigenvalues of z (respectively, z) givenin the decreasing order. Then by the Cauchy interlacing theorem given above, wehave

α1 ≥ β1 ≥ α2 ≥ β2 · · · ≥ αr−1 ≥ βr−1 ≥ αr.

Now suppose that ∆k(z) > 0 for all k = 1, 2, . . . , r. As e1 + e2 + · · · + el ande1 + e2 + · · ·+ er−1 operator commute,

Pe1+e2+···+elPe1+e2+···+er−1 = P(e1+e2+···+el)(e1+e2+···+er−1) = Pe1+e2+···+el

for all l = 1, 2, . . . , r − 1, and hence we have ∆l(z) = ∆l(z) > 0 for all l =1, 2, . . . , r − 1. By induction, z > 0 in V (r−1) which means that βi > 0 for alli = 1, 2, . . . , r − 1. It follows that αi > 0 for all i = 1, 2, . . . , r − 1. Since

0 < ∆r(z) = det(z) = α1α2 · · ·αr



it follows that αr > 0. Hence all the eigenvalues of z are positive. This proves thatz > 0. The case z ≥ 0 follows by considering z + εe for small positive ε and usingthe continuity of eigenvalues [11].

Given a vector x = (x1, x2, . . . , xr) in Rr, we write x↓ := (x↓1, x↓2, . . . , x

↓r) for the

vector obtained by rearranging the components of x in the decreasing order. Fortwo vectors x = (x1, x2, . . . , xr) and y = (y1, y2, . . . , yr) in Rr, we say that x ismajorized by y and write x ≺ y if

k∑1

x↓i ≤k∑1

y↓i (k = 1, 2, . . . , r − 1)

and

r∑1

x↓i =r∑1

y↓i .

In matrix theory, the well known majorization theorem of Schur says that for acomplex Hermitian matrix A,

diag(A) ≺ λ(A),

where diag(A) and λ(A) refer to the diagonal of A and the vector of eigenvalues ofA (written in the decreasing order) respectively.

We have a similar result in any simple Jordan algebra.

Corollary 4.6: Let V be simple and e1, e2, . . . , er be any Jordan frame in V .Let a ∈ V with the Peirce decomposition given by

a =r∑1

aiei +∑i<j

aij .

Then

diag(a) ≺ λ(a),

where diag(a) refers to the diagonal (a1, a2, . . . , ar) of a and λ(a) refers to thevector of eigenvalues of a written in the decreasing order.

Proof : We first note that in V , given any Peirce decomposition x =∑r

1 xiei +∑ij xij with respect to e1, e2, . . . , er, we have the following:

trace(x) = λ↓1(x) + · · ·+ λ↓r(x) = x1 + x2 + · · ·+ xr.

The first equality above is the definition of trace. The second equality can be seenas follows: Let x =

∑r1 λ↓i (x)fi be the spectral decomposition of x. Then

r∑1

λ↓i (x)||fi||2 = 〈x, f1 + f2 + · · ·+ fr〉 = 〈x, e1 + e2 + · · ·+ er〉 =r∑1

xi||ei||2

due to the orthogonality of the Peirce spaces. Since ||c|| is a constant on J (V ), weget the required equality.



To prove the majorization inequality, fix any k between 1 and r. If k < r, leta :=

∑k1 aiei +

∑i<j≤k aij . Then by the Cauchy interlacing theorem, λ↓i (a) ≤ λ↓i (a)

for all i = 1, 2, . . . , k. Therefore,

a1 + a2 + · · ·+ ak = trace(a) =k∑1

λ↓i (a) ≤k∑1

λ↓i (a).

Finally, when k = r, we get

a1 + a2 + · · ·+ ar =r∑1

λ↓i (a).

Thus, diag(a) ≺ λ(a).

Remarks. An alternate proof of the above result can be given based on the Hardy-Littlewood-Polya theorem [2] which says that x ≺ y in Rr if and only if there is adoubly stochastic matrix M (which means that the entries of M are nonnegativeand each row/column sum is one) such that x = My. To derive the result givenin the above corollary, let the assumptions of the corollary be in place. Let a =λ1(a)f1 + · · · + λr(a)fr be the spectral decomposition of a where f1, f2, . . . , fris a Jordan frame. As V is simple, we may assume that the inner product is givenby the trace and (hence) the norm of any primitive idempotent is one. Now definethe matrix M whose ijth entry is given by

Mij := 〈ei, fj〉.

Then M is a nonnegative matrix and each row/column sum is either 〈ei, e〉 =||ei||2 = 1 or 〈e, fj〉 = ||fj ||2 = 1. Moreover, for each i,

r∑j=1

〈ei, fj〉λj(a) = 〈a, ei〉 = ai.

Thus M λ(a) = diag(a) proving diag(a) ≺ λ(a).

It is well known that majorization plays an important role in proving inequalities.For example, if x = (x1, x2, . . . , xr) and y = (y1, y2, . . . , yr) belong to Rr withx ≺ y, then for any convex function φ : R→ R, we have the inequality

r∑1

φ(xi) ≤r∑1

φ(yi),

see e.g., Lemma 3.9.3 in [2]. Applying this to the convex function φ(t) := −log(t),we get:

0 ≤ x, y ∈ Rr, x ≺ y ⇒ x1x2 · · ·xr ≥ y1y2 · · · yr,

see Corollary 3.9.4 in [2]. As a consequence of this and the previous corollary, wehave the following inequality that generalizes the well-known Hadamard’s inequal-ity to simple Euclidean Jordan algebras.



Corollary 4.7: (Cf. Exercise 4, Section VI.4, [7]) In a simple Euclidean Jordanalgebra, consider an element a ≥ 0 with a Peirce decomposition a =

∑r1 aiei +∑

i<j aij . Then

det(a) ≤ a1a2 · · · ar.

Note. In view of the structure theorem, a similar statement holds in any EuclideanJordan algebra.

Based on the above Schur majorization theorem, one can derive a number ofinequalities. For example, following the arguments in [3], Sec II.3, one can showthe following:

Corollary 4.8: Let V be a simple Euclidean Jordan algebra V of rank r and c beany primitive idempotent. Then for all a > 0 and x, y ≥ 0, we have

min〈a, u〉

r: u ≥ 0, det(u) = 1

= ||c||2 [det(a)]

1r

and

[det(x + y)]1r ≥ [det(x)]

1r + [det(y)]

1r .

Suppose A is a n× n complex Hermitian matrix given in the block form by

A =[C ?? D

].

The so-called Fischer’s inequality says that when A is positive semidefinite,

det(A) ≤ det(C) det(D).

In the result below, we prove an analog of this for simple Euclidean Jordanalgebras using Hadamard’s inequality.

Corollary 4.9: Corresponding to a Jordan frame e1, e2, . . . , er in a simpleEuclidean Jordan algebra V , let the Peirce decomposition of a ≥ 0 be given bya =

∑r1 aiei +

∑i<j≤r aij . Let

u :=k∑1

aiei +∑

i<j≤k

aij ∈ Ve1,e2,...,ek

and

v :=r∑

k+1

aiei +∑

k+1≤i<j

aij ∈ Vek+1,...,er.

Then

det(a) ≤ det(u) det(v).



Proof : We may assume without loss of generality that the inner product is givenby the trace inner product. Let w := a− u− v so that a = u + v + w. Let

u = α1f1 + α2f2 + · · ·+ αkfk

be the spectral decomposition of u in Ve1,e2,...,ek and

v = αk+1fk+1 + · · ·+ αrfr

be the spectral decomposition of v in Vek+1,...,er. Note that α1, . . . , αk are theeigenvalues of u and αk+1, . . . , αr are the eigenvalues of v. Thus,

det(u) = α1α2 · · ·αk and det(v) = αk+1 · · ·αr.

Now, it is easily seen that f1, f2, . . . , fr is a Jordan frame in V . As V is simple,there exists an algebra automorphism Λ on V such that Λ(ei) = fi for all i (seeProp. IV.2.5 in [7]).

Let b := Λ−1(a) and consider the Peirce decomposition of b with respect to theJordan frame e1, e2, . . . , er:

b =r∑1

biei +∑i<j

bij .

We observe that Λ−1 is also an algebra automorphism of V and hence b ≥ 0. ByCorollary 4.7,

det(a) = det(b) ≤ b1b2 · · · br.

Now as V is simple and carries the trace inner product, Λ is orthogonal transfor-mation (see Page 56, [7]) and so Λ−1 = ΛT . Thus,

bi = 〈b, ei〉 = 〈Λ−1(a), ei〉 = 〈a,Λ(ei)〉 = 〈a, fi〉 = αi

for all i. Thus, det(a) ≤ (α1 . . . αk)(αk+1 · · ·αr) = det(u) det(v). This completesthe proof.

Our next result is an analog of Fan’s trace inequality in simple Euclidean Jordanalgebras. For matrices, Fan’s inequality (which is related to von Neumann’s traceinequality [16]) says that for Hermitian matrices

〈A,B〉 ≤n∑1

λ↓i (A)λ↓i (B).

In [22], Theobald shows that the equality holds in Fan’s inequality if and only ifA and B have simultaneous (orthogonal) diagonalization with diagonals given byλ(A) and λ(B) respectively.

This inequality (and the corresponding equality characterization) has been ex-tended to simple Euclidean Jordan algebras by Lim, Kim, and Faybusovich [15]based on Lie algebraic ideas, and by Baes [1] based on Birkhoff’s theorem on dou-bly stochastic matrices. See also [4] for a proof based on hyperbolic polynomials.In what follows, we provide an alternative proof of this generalization based on



Corollary 4.6. Our proof of the equality case is inspired by a similar proof in [4],Theorem 6.6.

Corollary 4.10: Let V be a simple Euclidean Jordan algebra of rank r. Let a, b ∈V . Then

〈a, b〉 ≤ ||c||2r∑1

λ↓i (a)λ↓i (b), (8)

where c is any primitive idempotent in V . Moreover, the equality holds in the aboveexpression if and only if there exists a Jordan frame f1, f2, . . . , fr in V such that

a = λ↓1(a)f1 + λ↓2(a)f2 + · · ·+ λ↓r(a)fr and b = λ↓1(b)f1 + λ↓2(b)f2 + · · ·+ λ↓r(b)fr.

Proof : Since V is simple, we may assume that the inner product is given bythe trace and that the norm of any primitive idempotent is one. Let the spectraldecomposition of a be given by

a = a1e1 + a2e2 + · · ·+ arer,

where ai = λ↓i (a) for all i. Corresponding to this Jordan frame, let the Peircedecomposition of b be given by

b =r∑1

biei +∑

i<j≤r

bij .

Then, by the orthogonality of the Peirce spaces,

〈a, b〉 =r∑1

aibi.

From the well-known Hardy-Littlewood-Polya (rearrangement) inequality [12], wehave

r∑1

aibi ≤r∑1

a↓i b↓i .

By Corollary 4.6, diag(b) is majorized by λ(b). Thus (see e.g., Lemma 3.9.1 in [2]),

r∑1

a↓i b↓i =

r∑1

λ↓i (a)b↓i ≤r∑1

λ↓i (a)λ↓i (b),

where the equality is due to diag(a) = λ(a). The inequality (8) follows.

Now for the equality. As the ‘if’ case is obvious, we assume that the equalityholds in (8), that is,

〈a, b〉 = 〈λ(a), λ(b)〉,



where the inner product on the right is the usual inner product in R∇. Now cor-responding to a + b, there is a Jordan frame f1, f2, . . . , fr such that

a + b = λ↓1(a + b)f1 + · · ·+ λ↓r(a + b)fr.

We show that a = λ↓1(a)f1 + · · ·+ λ↓r(a)fr and b = λ↓1(b)f1 + · · ·+ λ↓r(b)fr. Let

u :=r∑

i=1

λ↓i (a)fi.

Then λ(u) = λ(a) and ||u||2 =∑r

1 |λ↓i (a)|2 = ||a||2. An application of (8) yields

〈a + b, b〉 ≤ 〈λ(a + b), λ(b)〉 and 〈b, u〉 ≤ 〈λ(b), λ(u)〉 = 〈λ(b), λ(a)〉.

Now,

||a− u||2 = ||a||2 + ||u||2 − 2〈a, u〉

= 2||a||2 − 2〈a + b, u〉+ 2〈b, u〉

= 2||a||2 − 2〈r∑

i=1

λ↓i (a + b)fi,

r∑i=1

λ↓i (a)fi〉+ 2〈b, u〉

= 2||a||2 − 2〈λ(a + b), λ(a)〉+ 2〈b, u〉

≤ 2||a||2 − 2〈a + b, a〉+ 2〈λ(b), λ(a)〉

= 2||a||2 − 2〈a + b, a〉+ 2〈a, b〉

= 0.

Thus we have a = u =∑r

i=1 λ↓i (a)fi. Similarly, b =∑r

i=1 λ↓i (b)fi. This completesthe proof.

Remarks. To illustrate the inequality (8), consider the Jordan spin algebra Ln,n > 2. Then using the notation of Section 2.1, for any

x =[x0

x′

](9)

with x0 ∈ R and x′ ∈ Rn−1, we have

λ↓1(x) = x0 + ||x′|| and λ↓2(x) = x0 − ||x′||.

Also, any primitive idempotent is given by c = 12

[1u

]with u ∈ Rn−1, ||u|| = 1;

hence ||c||2 = 12 . Thus, for elements x, y ∈ Ln, the inequality (8) reads:

x0y0 + 〈x′, y′〉 ≤ 12

(x0 + ||x′||)(y0 + ||y′||) + (x0 − ||x′||)(y0 − ||y′||)

which reduces to the Cauchy-Schwarz inequality in Rn−1.



We note that the inequality (8) and the corresponding equality characterizationcontinues to hold in a general Euclidean Jordan algebra when the inner productis given by the trace. This can be seen by noting ||c||2 = 1 for any primitiveidempotent and then using the structure theorem.

5. Some converse results

Motivated by questions raised by a referee, in this section, we prove results con-verse to the Cauchy interlacing theorem (Theorem 4.2) and the Schur majorizationtheorem (Corollary 4.6).

Theorem 5.1 : Let V be simple and e1, e2, . . . , er be a Jordan frame. Considertwo decreasing finite sequences of real numbers αi and βi such that αi ≥ βi ≥αr−k+i, (i = 1, 2, . . . , k). Then there exists x ∈ V such that λ(x) = (α1, α2, . . . , αr)and λ(x) = (β1, β2, . . . , βk).

Proof : Since algebra automorphisms preserve eigenvalues, we may assume thatV is either a matrix algebra or Ln, n > 2. When V is a matrix algebra, we mayassume (by using an algebra automorphism, if necessary) that the Jordan frameis the canonical one given by E1, E2, . . . , En, where Ei is the matrix with onein the (i, i) slot and zeros elsewhere. Then corresponding to αi and βi, there ex-ists a real symmetric matrix with eigenvalues αi and an appropriate submatrixwith eigenvalues βi, see [6]. Since real numbers are included in complex num-bers/quaternions/octonions, we have the result in the case of matrix algebras. Nowconsider the algebra Ln. Let e1, e2 be a Jordan frame in Ln and α1 ≥ β1 ≥ α2. Tocomplete the proof, we have to construct an element x whose Peirce decompositionand eigenvalues are given by (see Section 2.1)

x =[x0

x′

]= β1e1 + x2e2 +

[0v

], (10)

and α1 = x0 + ||x′||, α2 = x0 − ||x′||.

Now, as in Section 2.1, let e1 = 12

[1u

], e2 = 1

2

[1−u

], ||u|| = 1. As n > 2, we can

take v ∈ Rn−1, such that ||v|| =√

(β1 − α2)(α1 − β1) and 〈v, u〉 = 0. Now definex0 = 1

2(α1 + α2), x2 = 2x0 − β1, and x′ = 12(β1 − x2)u + v. Then it can be easily

verified that x defined by (10) will have the right properties.

Now we prove an analog of the converse of Schur’s majorization theorem forsimple Euclidean Jordan algebras.

First we recall that in the Peirce decomposition

x =r∑1

xiei +∑

xij

with respect to e1, e2, . . . , er, the diagonal of x is given by diag(x) =(x1, x2, . . . , xr). We speak of x1, x2, . . . , xr as the diagonal numbers of x with re-spect to e1, e2, . . . , er.

Theorem 5.2 : Let V be simple and e1, e2, . . . , er be a Jordan frame. If a =(a1, a2, . . . , ar) ≺ b = (b1, b2, . . . , br), then there exists x ∈ V such that diag(x) =(a1, a2, . . . , ar) and λ(x) = (b↓1, b

↓2, . . . , b

↓r).


22 REFERENCES

Proof : In the case of Herm(Rn×n) with the canonical Jordan frame, the resultfollows from Theorem 4.3.32 in [14]. Since real numbers are included in complexnumbers/quaternions/octonions, we have the result in the case of any matrix al-gebra when the Jordan frame is the canonical one. Since eigenvalues and diagonalnumbers of an element remain the same under an algebra automorphism, we havethe result in any matrix algebra (and with any Jordan frame). The result continuesto hold for any simple algebra isomorphic to a matrix algebra. Now consider thealgebra Ln, n > 2. In Ln, let e1, e2 be a Jordan frame, d = (d1, d2), λ = (λ1, λ2)with λ1 ≥ λ2 and d ≺ λ. We construct an element

x =[x0

x′

]= d1e1 + d2e2 +

[0v

], (11)

such that λ1 = x0 + ||x′|| and λ2 = x0 − ||x′||.

Let e1 = 12

[1u

], e2 = 1

2

[1−u

], ||u|| = 1. Take v ∈ Rn−1, such that ||v|| =√

(λ1 − d2)(λ1 − d1) and 〈v, u〉 = 0. Put x0 = 12(λ1 + λ2) = 1

2(d1 + d2), andx′ = 1

2(d1−d2)u+v. Then it is easy to verify that λ1 = x0+||x′|| and λ2 = x0−||x′||.Then x, defined by (11), will have the right properties.

Acknowledgments: We wish to thank the referee for pointing out reference [20]and raising (converse) questions which resulted in Section 5.

References

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