Mathematical Notes, vol. 72, no. 2, 2002, pp. 285–290.

Translated from Matematicheskie Zametki, vol. 72, no. 2, 2002, pp. 315–320.

Original Russian Text Copyright c©2002 by A. S. Kostenko.

On the Defect Indexof Quadratic Self-Adjoint Operator Pencils

A. S. Kostenko

Received December 5, 2001

Key words: spectrum of an operator pencil, instability index, e-dichotomy, defect index, Hilbertspace, connective motion.

Let H be a separable Hilbert space.In the book [1] by Kopachevskii, Krein, and Ngo Zui Kan, in the course of studying the problem

of convection in a vessel partially filled with a liquid, the spectral properties of pencils of the form

L(λ) = λA+1

λB + εK − I (1)

are examined; such pencils are known as Krein pencils. It is assumed in [1] that A and Bare compact nonnegative operators, K = K∗ is a compact operator in H , and ε is a positiveparameter.Recall that a point λ0 ∈ C is an eigenvalue of a pencil L if there exists a nonzero vector x0 ∈ H

such that L(λ0)x0 = 0.

In [1], it is shown that the eigenvalues of a pencil L corresponding to normal motions of theform exp(−λx) lie in the right half-plane if ε < 1/‖K‖ .If ε > 1/‖K‖ , then the operator I − εK can have a finite number κ of negative eigenvalues.

In this case, the spectrum of the problem no longer necessarily lies in the right half-plane Cr .Namely, it turns out that the left half-plane Cl can contain a finite number of eigenvalues. Thismeans that normal oscillations are unstable.Recall that the number of eigenvalues of a pencil L in the left half-plane Cl is called the

instability index of the pencil L and is denoted by κ−(L) .Kopachevskii and Pivovarchik [2] obtained a sufficient condition for the instability of normal

convective motions; namely, they specified a number ε2 determined by the operators A , B ,and K (under the assumption that KerB �= {0}) such that the instability index κ−(L) is positiveat ε > ε2 .Shkalikov [3, 4] studied operator pencils of the form

L(λ) = Fλ2 + (D + iG)λ+ T ,

where F , D , G , and T are operators in a Hilbert space H satisfying the following conditions:F = F ∗ is bounded and has bounded inverse, T = T ∗ has bounded inverse, D and G aresymmetric densely defined operators, T is bounded, and D ≥ 0. For such pencils, upper boundsfor the instability index κ−(L) under certain additional conditions on the operators F and Twere obtained in [3, 4].

0001-4346/2002/7212-0285$27.00 c©2002 Plenum Publishing Corporation 285

286 A. S. KOSTENKO

Closest to this paper is the work [5] of Sukhocheva, who found sufficient conditions on thecoefficients of matrix pencils of the form

L(λ) = λA+1

λB −C (2)

for the equality κ−(L) = 2κ to hold. Here A , B , and C are Hermitian matrices acting in thespace H = Cn and satisfying the following conditions:

A = ‖aij‖i,j=1,2,...,n , A > 0, aii >

n∑j=1j �=i

|aij |+ εAi , εAi ≥ 0, (3)

B = ‖bij‖i,j=1,2,...,n , B > 0, bii >

n∑j=1j �=i

|bij |+ εBi , εBi ≥ 0, (4)

C = Jκ = diag(In−κ , −Iκ). (5)

Theorem 1 [5]. Suppose that the Hermitian matrices A , B and C satisfy conditions (3)–(5) anddetermine a matrix pencil of the form (2). If the spectrum of the pencil Ld(λ) lies in the angle

Φ =n⋂i=1

{Ψi : | tanΨi| ≤ 2(εAi εBi )1/2}, (6)

then 2κ eigenvalues of the pencil L lie in the open left half-plane, and 2(n− κ) eigenvalues lie inthe open right half-plane.

Here Ld is the diagonal of the pencil L , i.e.,

Ld(λ) = λdiagA+1

λdiagB − Jκ ,

where diagA is the matrix obtained from A by “killing” its nondiagonal elements.We use the following notation: Cl (Cr) is the open left (respectively, right) half-plane, and

κ−(A) (κ−(L)) is the number of the eigenvalues of the operator A (respectively, of the pencil L)lying in Cl .In this paper, we obtain simple sufficient conditions ensuring the equality κ−(L) = 2κ and the

e-dichotomy of pencils of the form (2). Our approach differs from those used in the works [1, 2, 4, 5]mentioned above.The proof of the main result uses the following generalization of the Sylvester criterion (see,

e.g., [6]).

Lemma 1. Let A = A∗ be a bounded operator in H = H1 ⊕H2 having the block representation

A =

(A11 A12A21 A22

),

where Aij = A∗ji . Suppose that the operators Aij : Hj → Hi satisfy the following conditions:

(1) the operator A11 invertible and κ−(A11) = m ;(2) κ−(A22 −A21(A11)−1A12) = n .

Then κ−(A) = m+ n .

Recall (see [7]) that an operator (pencil) is called e-dichotomic if its spectrum is disjoint fromthe imaginary axis.The following theorem is the main result of this paper.

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DEFECT INDEX OF QUADRATIC SELF-ADJOINT OPERATOR PENCILS 287

Theorem 2. Let A , B , and C be self-adjoint operators in a Hilbert space H , and let L be aquadratic pencil of the form

L(λ) = λ2A− λC +B. (7)

Suppose that the operators A and B are positive definite, C is invertible, and κ−(C) = κ . If theoperator

R0 := CA−1C −B (8)

is positive definite, then the pencil L is e-dichotomic and κ−(L) = 2κ .

Proof. Since the operator A is positive definite, we have σ(L) = σ(L) , where L(λ) is a pencil ofthe form

L(λ) = λ2I − λCA−1 +BA−1. (9)

It is easy to see that the linearizer of the pencil L(λ) has the form

T =

(0 −I

BA−1 CA−1

), (10)

and σ(L) = σ(T ) .By the Lyapunov Theorem [7], to prove the equality κ−(L) = 2κ and the e-dichotomy of the

pencil L , it suffices to specify an operator W = W ∗ such that κ−(W ) = 2κ and Re(WT ) ispositive definite.Let us show that the operator

W =

(12A−1CA−1 1

2A−1

12A−1 C−1

)(11)

has the required properties.It is easy to see that

κ−

(1

2A−1CA−1

)= κ−

(1

2C

)= κ, κ−

(C−1− 1

2A−1(1

2A−1CA−1

)−11

2A−1)= κ−(C) = κ.

By Lemma 1, κ−(W ) = 2κ .Next, (10) and (11) imply

2Re(WT ) =

(12A−1BA−1 1

2A−1BC−1

12C−1BA−1 1

2A−1

). (12)

Let us show that the operator Re(WT ) is positive definite. We have A−1BA−1 � 0; hence, by

the Sylvester criterion, it suffices to show that the operator R0 := A−1 − C−1BC−1 is positive

definite. But this is equivalent to the operator R0 of the form (8) being positive definite.

Thus, by the Lyapunov theorem, κ−(T ) = 2κ , and so κ−(L) = κ−(L) = κ−(T ) = 2κ . �The following theorem shows that, in Theorem 2, the assumption that the operator A is positive

definite can be omitted, provided that we appropriately modify the positive definiteness require-ment on the operator R0 of the form (8) is appropriately altered.

Theorem 3. Let A , B , and C be self-adjoint operators determining a pencil of the form (7)and satisfying the following conditions: A is nonnegative (A > 0), B is positive definite, C hasbounded inverse, and κ−(C) = κ . If the operators

Rε := C(A+ εI)−1C −B, ε > 0, (13)

are positive definite for all ε > 0 , then the pencil L is e-dichotomic and κ−(L) = 2κ .

MATHEMATICAL NOTES Vol. 72 No. 2 2002

288 A. S. KOSTENKO

Proof. (i) First, let us prove the e-dichotomy of L .

It is easy to see that σ(L) = σ(L) , where

L(λ) = λ2B−1/2AB−1/2 − λB−1/2CB−1/2 + I. (14)

Thus we can consider the quadratic pencil

L(λ) = λ2A− λC + I , (15)

where A = B−1/2AB−1/2 , C = B−1/2CB−1/2 .Let us show that the spectrum of this pencil is disjoint from the imaginary axis. Let λ be an

eigenvalue pencil L , and let f be the corresponding eigenvector.It is easy to see that a sufficiently small neighborhood of zero contains no points of the spectrum

of the pencil L(λ) .Therefore,

(L(λ)f , f) = λ2(A2f , f)− λ(Cf , f) + (f , f) = 0, (16)

(L(λ)f , Af) = λ2(A2f , f)− λ(Cf , Af) + (f , Af) = 0,(Cf , L(λ)f) = λ2(Cf , Af)− λ(C2f , f) + (Cf , f) = 0.

Now, suppose that λ = ia , where a ∈ R . Then the first equation in (16) implies that (Cf , f) = 0and a2 = (f , f)/(Af , f) .Next, multiplying the second equation by λ and adding the third to the result, we obtain

a2(A2f , f) + (C2f , f) = (Af , f). (17)

Substituting the value found for a2 , we obtain

(f , f)(A2f , f)− (Af , f)2(Af , f)

+ (C2f , f) = 0. (18)

Since the operator A is nonnegative, the first term in (18) is also nonnegative by the Cauchy–Bunyakovskii inequality. The operator C2 is positive definite, because C is invertible. Therefore,(18) can hold only if f = 0.

It is easy to show that the continuous spectrum of the pencil L(λ) is disjoint from the imaginaryaxis. Indeed, suppose that λ0 = ia with a ∈ R belongs to the continuous spectrum of L . Thenthere exists a sequence {fn}∞n=1 such that ‖fn‖ = 1 and ‖(L(ia)fn)‖ < 1/n .Replacing f with fn in (16) and repeating the preceding evaluations, we obtain the inequality

∣∣∣∣ (fn , fn)(A2fn , fn)− (Afn , fn)2(Afn , fn) + (C2fn , fn)

∣∣∣∣ < Mn , (19)

where M is a positive constant independent of n . Since A > 0, the first term is nonnegative.

Next, since the operator C is invertible, C2 is positive definite, i.e., there exists a δ > 0 such that(C2f , f) > δ(f , f) for all f ∈ H . Therefore, (19) cannot hold for large n . Thus the spectrum ofthe pencil is disjoint from the imaginary axis.(ii) Consider the pencil

Lε(λ) = λ2(A+ εI)− λC +B. (20)

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DEFECT INDEX OF QUADRATIC SELF-ADJOINT OPERATOR PENCILS 289

The linearizer of the pencil Lε(λ) = λ2(A+ εI)B−1 − λCB−1 + I has the form

Tε =

(CB−1 −(A+ εI)B−1I 0

). (21)

Since B is invertible, we have σ(Lε) = σ(Lε) = σ(Tε) .Since the operator Rε is positive definite, Theorem 2 implies the equality κ−(Tε) = 2κ and the

e-dichotomy of the pencils Lε for all ε > 0.The operators Tε converge uniformly to T0 as ε approaches 0 , and they are e-dichotomic; hence

the theorem about the continuity of a finite system of isolated eigenvalues (see [8, p. 270]) implies

the equality κ−(L) = κ−(L0) = κ−(T0) = 2κ , which proves the second assertion of Theorem 3. �Corollary 1. Suppose that A , B , and C are operators in a Hilbert space H , C∗ = C hasbounded inverse, A = A∗ > 0 , and B = B∗ > δI , where δ ∈ R , δ > 0 . If κ−(C) = κ and

‖A‖ · ‖B‖ · ‖C−1‖2 < 1, (22)

then κ−(L) = 2κ .

Proof. It is easy to see that the assumption that the operator R0 of the form (8) is positivedefinite is equivalent to the inequality

‖B1/2C−1A1/2‖ < 1, (23)

which follows from the estimate (22). �Remark 1. If the operator A is positive definite and B is nonnegative, Theorem 1 remainsvalid, except the pencil L is no longer e-dichotomic (because L(0) = B , and this operator is notinvertible). To show this, consider the pencil

L(λ) := λ2B − λC +A. (24)

If λ0 ∈ σ(L) , then 1/λ0 ∈ σ(L) . Let us apply Theorem 3 to the pencil L . Since C is invertibleand R0 is positive definite, we have

A−1 − C−1(B + εI)C−1 � 0 (25)

for sufficiently small ε > 0; hence

C(B + εI)−1C −A� 0. (26)

Thus κ−(L) = 2κ . If λ belongs to Cl (C+) , then 1/λ also belongs to C− (respectively, to C+).According to the aforesaid, κ−(L) = 2κ .

Example 1. Suppose that

H = C2 , B = diag

(1

2,1

3

), J = diag(1, −1), A =

(1 1/21/2 1

).

It is easy to see that ‖A‖ = 1.5 and ‖B‖ = 0.5; therefore, ‖A‖ ‖B‖ < 1. Applying Corollary 1,we obtain κ−(L) = 2.Let us verify that the assumptions of Theorem 1 hold. We have

σ(Ld) = {(1 ± i)/2 ; (−1± i√3)/2} and Φ = {ϕ : | tanϕ| <

√2/3}.

It is easy to see that Theorem 1 from [5] does not apply, because (1 + i)/2 /∈ Φ.

MATHEMATICAL NOTES Vol. 72 No. 2 2002

290 A. S. KOSTENKO

Example 2. Suppose that

H = C2 , B = I , J = diag(1, −1), A =

(3 0.10.1 3

).

It is easy to see that det(A) = 8.9 and

R0 =

( −5.98.9

189

189

−5.98.9

).

Thus the operator R0 is not positive definite, and Theorem 2 does not apply.Let us verify that the assumptions of Theorem 1 hold. We have

σ(Ld) = {(1± i√11)/6 ; (−1± i√11)/6} and Φ = {ϕ : | tanϕ| < √11.6}.

It is easy to see that σ(Ld) ⊂ Φ; hence, by Theorem 1, we obtain κ−(L) = 2.These examples show that Theorem 2 does not cover Theorem 1. At the same time, it applies

to operator (rather than only to matrix) pencils, and its assumptions are unitary invariant, asopposed to those of Theorem 1.Note also that the assumptions of Theorem 1 mean that the matrices A and B are “close” to

diagonal matrices, and they are not unitary invariant.

ACKNOWLEDGMENTS

The author wishes to express his gratitude to his teacher M. M. Malamud for advice andto C. M. Malamud and L. L. Oridoroga for many useful discussions.

REFERENCES

1. N. D. Kopachevskii, S. G. Krein, and Zui Kan Ngo, Operator Methods in Linear Hydrodynamics [inRussian], Nauka, Moscow, 1989.

2. N. D. Kopachevskii and V. N. Pivovarchik, Zh. Vychisl. Mat. i Mat. Fiz. [Comput. Math. and Math.Phys.], 33 (1993), no. 1, 101–118.

3. A. A. Shkalikov, Uspekhi Mat. Nauk [Russian Math. Surveys], 51 (1996), no. 5, 195–196.4. A. A. Shkalikov, in: Operator Theory Advances Appl., vol. 87, Bikhauser Verlag, 1996, pp. 258–285.5. L. I. Sukhocheva, Mat. Zametki [Math. Notes], 61 (1997), no. 3, 381–390.6. M. M. Malamud, Ukrain. Mat. Zh. [Ukrainian Math. J.], 44 (1992), no. 2, 215–233.7. Yu. L. Daletskii and M. G. Krein, Stability of Solutions of Differential Equations in a Banach Space[in Russian], Nauka, Moscow, 1970, p. 536.

8. T. Kato, Perturbation Theory for Linear Operators, Springer-Verlag, Heidelberg, 1966.

Donetsk National UniversityE-mail : karabashi@yahoo.com

MATHEMATICAL NOTES Vol. 72 No. 2 2002