An Iterative Procedure for Solving the Riccati Equation A 2 R RA 1 = A 3 + RA 4 R. M.THAMBAN NAIR (I.I.T. Madras)

An Iterative Procedure for Solving the Riccati Equation A 2 R RA 1 = A 3 + RA 4 R M.THAMBAN NAIR (I.I.T. Madras) Abstract Let X 1 and X 2 be complex Banach spaces, and let A 1 BL(X 1 ), A 2 BL(X 2 ), A 3 BL(X 1, X 2 ) and A 4 BL(X 2, X 1 ). We propose an iterative procedure which is a modified form of Newton s iterations for obtaining approximations for the solution R BL(X 1, X 2 ) of the Riccati equation A 2 R RA 1 = A 3 + RA 4 R, and show that the convergence of the method is quadratic. The advantage of the present procedure is that the conditions imposed on the operators A 1, A 2, A 3, A 4 are weaker than the corresponding conditions for Newtons iterations, considered earlier by Demmel (1987), Nair (1989) and Nair(1990) in the context of obtaining error bounds for approximate spectral elements. Also, we discuss the application of suggested procedure to spectral approximation under perturbation of the operator. AMS Subject Classification : 45B05, 47A10, 65F35, 65R20. 1 Introduction Let X 1 and X 2 be complex Banach spaces, and let A 1 : X 1 X 1, A 2 : X 2 X 2, A 3 : X 1 X 2, A 4 : X 2 X 1 be bounded linear operators. We shall specify conditions on the above operators so that the Riccati equation A 2 R RA 1 = A 3 + RA 4 R has a unique solution R BL(X 1, X 2 ), and also introduce an iterative procedure for obtaining R k BL(X 1, X 2 ), k = 1, 2,... such that and R (R 1 +... + R k ) 0, as k, R (R 1 +... + R k+1 ) α R (R 1 +... + R k ) 2 for certain α > 0. The above Riccati equation arises naturally while looking for an invariant subspace of a bounded linear operator having a specified complimentary subspace. To see this let A : X X 1

be a bounded linear operator on a complex Banach space X and P 0 (bounded linear) projection operator. Suppose : X X be a X 1 = P 0 (X), X 2 = (I P 0 )(X), and let R : X 1 X 2 be a bounded linear operator and P : X X be defined by P x = P 0 x + RP 0 x, x X. Then it can be easily seen that P is a projection operator along X 2. Moreover, P (X) is invariant under A if and only if R satisfies a Riccati equation A 2 R RA 1 = A 3 + RA 4 R, (1.1) with A 1 : X 1 X 1, A 2 : X 2 X 2, A 3 : X 1 X 2, A 4 : X 2 X 1 given by A 1 = P 0 A X1, A 2 = (I P 0 )A X2, A 3 = (I P 0 )A X1, A 4 = P 0 A X2. (For details one may see [5] or [6]). Thus the problem of finding the invariant subspace P (X) is equivalent to solving the Riccati equation (1.1). Now suppose that (S k ) is a sequence of bounded linear operators from X 1 to X 2 such that R S k 0 as k, where R : X 1 X 2 satisfies (1.1). Then the operators are projections along X 2, and we have P k = P 0 + S k P 0, k = 1, 2,..., P P k 0 as k. Recall that, for subspaces M 1 and M 2 of X, if we define the distance from M 1 to M 2 as sep(m 1, M 2 ) := sup{dist(x, M 2 ) : x M 1, x = 1}, then gap(m 1, M 2 ), the gap between M 1 and M 2 (Kato [1]), is given by gap(m 1, M 2 ) = max {sep(m 1, M 2 ), sep(m 2, M 1 )}. If we denote then it follows that M k = P k (X), M = P (X), sep(m k, M) R S k, gap(m k, M) P P k = (R S k )P 0. 2

Thus the convergence of (S k ) to R implies the convergence of (M k ) to M in the sense of sep and gap. In practical situations, what one would intially have is an approximate spectral subspace M 0, in the sense that M 0 is a spectral subspace of a perturbed operator A 0. In such situation, the projection P 0 may be the spectral projection with M 0 = P 0 (X) associated with a spectral set Λ 0 of A 0. Then, what one would like to look for is a spectral subspace M associated with a spectral set Λ of A. The application of our results, with suitable conditions on M 0, does in fact give a sequence (M k ) of subspaces and a sequence (Λ k ) of closed subsets of the complex plane defined by Λ k = σ (P 0 A Mk ), k = 1, 2,..., such that and sep(m k, M) 0, gap(m k, M) 0 sep(λ k, Λ) := sup{dist(λ, Λ) : λ Λ k } 0 as k, where M is the spectral subspace of A associated with a spectral set Λ. In all the above results, the crucial role is played by the size of the quantity Here, δ := sep(a 1, A 2 ) is defined by ɛ := A 3 A 4 δ 2. { 0 if 0 σ(t ) sep(a 1, A 2 ) = 1/ T 1 otherwise, where T : BL(X 1, X 2 ) BL(X 1, X 2 ) is defined by T (B) = A 2 B BA 1, B BL(X 1, X 2 ). It is to be remarked that iterative procedures for approximately solving Riccati equation (1.1) and the corresponding problem of approximating an invariant subspace/spectral subspace have been considered by many authors (see e.g. Stewart [9], Demmel [2], Nair ([5], [6], [7]). The results in Stewart [9] and Nair [5] are based on a Piccard type iteration, namely, A 2 R k+1 R k+1 A 1 = A 3 + R k A 4 R k, R 0 = 0, (1.2) and provide linear convergence under the assumption that ɛ < 1/4. Demmel [2] and Nair [7] use Newton s method, namely, (A 2 R k A 4 ) R k+1 R k+1 (A 1 + A 4 R k ) = A 3 R k A 4 R k, R 0 = 0, (1.3) 3

and obtained quadratic convergence whenever ɛ < 1/12. Note that the quadratic convergence of the iteration in (1.3) is achieved not only by imposing stronger assumption on the matrices A 1, A 2, A 3, A 4, but by also requiring to solve the equation each time with new coefficient matrices. In [6], the author considered two iterative procedures which modify the iterations (1.2) and (1.3) and obtained improved estimates together with linear convergence if ɛ < 1/4 and quadratic convergence in the case of ɛ < ( 3 1)/4. where The new iterative procedure suggested in this paper is and R 1 as the unique solution of A (k) 2 R k+1 R k+1 A (k) 1 = R k A 4 R k, (1.4) A (k) 1 = A (k 1) 1 A 4 R k, A (k) 2 = A (k 1) 2 R k A 4, A (0) 2 R 1 R 1 A (0) 1 = A 3 with A (0) 1 = A 1, A (0) 2 = A 2. Of course, in this procedure also, the coefficient matrices have to be computed at each step of the iteration. But advantage of the present procedure over existing Newton s-type methods is that the condition to be imposed on the matrices A 1, A 2, A 3, A 4 are weaker. We, in fact, show that the iterative procedure (1.4) is valid for ɛ < 1/4, and it provides quadratic convergence whenever ɛ < 3/16. Observe that 1 12 < 3 1 4 < 3 16 < 1 4. 2 Basic Definitions and Preliminary Results Let X, X 1 and X 2 be complex Banach spaces. We denote by BL(X 1, X 2 ), the space of all bounded linear operators from X 1 into X 2, and denote BL(X, X) by BL(X). The following result is, by now, well known in the literature. The if part of the result was proved by Rosenblum [8] in the context of A 1, A 2 are in a Banach algebra, and the only if part was proved by Stewart [9] while obtaining error bounds for invariant subspaces of a closed linear operator in a Hilbert space. Stewart s result was extended by Nair [4] (quoted in [5]) for Banach space operators, with an alternate proof. PROPOSITION 2.1 For A 1 BL(X 1 ) and A 2 BL(X 2 ), consider the map T : BL(X 1, X 2 ) BL(X 1, X 2 ) defined by Then T (B) = A 2 B BA 1, B BL(X 1, X 2 ). 0 σ(t ) if and only if σ(a 1 ) σ(a 2 ) =. 4

For A 1 BL(X 1 ) and A 2 BL(X 2 ), define { 0 if 0 σ(t ) sep(a 1, A 2 ) = 1/ T 1 otherwise, where T : BL(X 1, X 2 ) BL(X 1, X 2 ) is defined by T (B) = A 2 B BA 1, B BL(X 1, X 2 ). Using the standard perturbation theory arguments (cf. Kato [1]), we obtain the following result ( see e.g. Nair[4], [5]). PROPOSITION 2.2 Let A 1, V 1 BL(X 1 ) and A 2, V 2 BL(X 2 ). Then In particular, if then the operator is invertible. sep(a 1 + V 1, A 2 + V 2 ) sep(a 1, A 2 ) V 1 V 2. sep(a 1, A 2 ) > 0 and V 1 + V 2 < sep(a 1, A 2 ), B (A 2 + V 2 )B B(A 1 + V 1 ), B BL(X 1, X 2 ), For closed subsets Λ 1 and Λ 2 of of the complex plane, define We note that for ω > 0, sep(λ 1, Λ 2 ) := sup{dist(λ, Λ 2 ) : λ Λ 1 }. sep(λ 1, Λ 2 ) < ω Λ 1 {z : dist(z, Λ 2 ) < ω}. The following proposition essentially states that the spectrum of a bounded operator is upper semi-continuous, that is, the spectrum does not expand too much by small perturbation of the operator, and its proof is based on standard arguments in perturbation theory (cf. Kato [1]). For the sake of completeness of the exposition, we give the proof in detail. PROPOSITION 2.3 Let B BL(X) and B k BL(X), k = 1, 2,.... If B B k 0 as k, then sep (σ(b k ), σ(b)) 0 as k. 5

Proof. Suppose B BL(X) and B k BL(X), k = 1, 2,..., such that B B k 0 as k. Let c > 0 be such that B k c for all k = 1, 2,..., and let c 0 = 1+max{c, B }. Then we have σ(b k ) σ(b) {λ : λ < c 0 }, k = 1, 2,.... For ω > 0, let ω = {z : dist (z, σ(b)) < ω}. Note that ω is an open subset of the complex plane containing σ(b). Assume that ω is small enough such that ω {λ : λ < c 0 }. Let C ω = {λ : λ c 0 }\ ω. Clearly C ω is a compact subset of the complex plane and C ω σ(b) =. Therefore, it follows that, for every z C ω, z σ(b k ) whenever B B k max µ C ω (B µi) 1 < 1. Since B B k 0 as k, there exists a positive integer N such that for all k N, Since it follows that Thus, sep (σ(b k ), σ(b)) 0 as k. λ σ(b k ) = λ C ω. σ(b k ) {λ : λ c 0 } = ω C ω, σ(b k ) ω, k N. Let A BL(X). A subset Λ of the complex plane is called a spectral set of A if both Λ and σ(a)\λ are closed subsets of the complex plane. If Λ is a spectral set of A, then (c.f. Taylor[10], Nair[4]) there is an open set Ω such that its boundary Γ consists of a finite number of simple closed contours and σ(a) (Ω Γ) = Λ. The range of the spectral projection (c.f. Kato[1], Limaye[3]) P Λ = 1 (A zi) 1 dz 2πi Γ is called the spectral subspace of A associated with the spectral set Λ. Here is a characterization of a spectral subspace which is found to be useful in spectral approximation ( cf. Nair[4], [5]). 6

PROPOSITION 2.4 For A BL(X), let P BL(X) be a projection such that its range M = P (X) is invariant under A. Then M is a spectral subspace of A if and only if σ(p A P (X) ) σ((i P )A (I P )(X) ) =, and in that case the associated spectral set is σ(p A P (X) ) and σ(a)\σ(p A P (X) ) = σ((i P )A (I P )(X) ). 3 Existence of the Solution Let A 1 BL(X 1 ), A 2 BL(X 2 ), A 3 BL(X 1, X 2 ) and A 4 BL(X 2, X 1 ). Assume that δ := sep(a 1, A 2 ) > 0. Let γ = A 3, η = A 4 and ɛ = ηγ δ. 2 In the following we make use of the following function : g(t) = { 1 if t = 0 ( 1 1 4t ) /2t if 0 < t 1/4. It follows that 1 g(t) 2, 0 t 1/4, and s = g(t) satisfies the relation ts 2 s + 1 = 0. THEOREM 3.1 If ɛ < 1/4, then the Riccati equation A 2 R RA 1 = A 3 + RA 4 R has a unique solution R BL(X 1, X 2 ) which satisfies the relation R γ δ g(ɛ). Proof. By the assumption δ > 0, the operator T : BL(X 1, X 2 ) BL(X 1, X 2 ) is defined by T (B) = A 2 B BA 1, B BL(X 1, X 2 ), is invertible. Let D = {B BL(X 1, X 2 ) : B γ δ g(ɛ) }. 7

For B D, let Then we have F (B) = T 1 (A 3 + BA 4 B). F (B) T 1 ( A 3 + A 4 B 2) 1 [ ( ) ] γ 2 γ + η δ δ g(ɛ) = γ ( ) 1 + ɛ(g(ɛ)) 2 δ = γ δ g(ɛ). Thus F maps the complete metric space D into iteself. Therefore, the proof will be completed once it is shown that F : D D is a contraction. For this, let B 1, B 2 D and observe that Therefore, F (B 1 B 2 ) = T 1 (B 1 A 4 B 1 B 2 A 4 B 2 ) = T 1 [B 1 A 4 (B 1 B 2 ) + (B 1 B 2 )A 4 B 2 ]. F (B 1 B 2 ) T 1 ( B 1 A 4 + A 4 B 2 ) B 1 B 2 1 δ η( B 1 + B 2 ) B 1 B 2 2ɛg(ɛ) B 1 B 2 Since 2ɛg(ɛ) = 1 1 4ɛ < 1, it follows that F is a contraction mapping on D. 4 The Iterative Procedure Let A 1 BL(X 1 ), A 2 BL(X 2 ), A 3 BL(X 1, X 2 ) and A 4 BL(X 2, X 1 ). Let γ = A 3, η = A 4, δ := sep(a 1, A 2 ) > 0, ɛ = ηγ δ 2. Recall that, by the assumption δ > 0, the operator T : BL(X 1, X 2 ) BL(X 1, X 2 ) defined by T (B) = A 2 B BA 1, B BL(X 1, X 2 ), is invertible. We also observe that if γ = 0, then R = 0 is the unique solution of the Riccati equation A 2 R RA 1 = A 3 + RA 4 R. Therefore, hereafter, we assume that γ > 0. First we prove a technical result. 8

PROPOSITION 4.1 Let r 1 = γ/δ and ɛ < 1/4. Then the relations 2ηr k + 2η(r 1 +... + r k ) < δ and hold iteratively. r k+1 := ηrk 2 δ 2η(r 1 +... + r k ) < r k 2 Proof. Note that so that r 2 is well defined and r 2 = r 1 ( 4ηr 1 = 4ɛδ < δ, ηr1 δ 2ηr 1 ) < r 1 2. Thus the result is proved for k = 1. Assume the result for some k = n 1, n {1, 2,...}. Then for k = n we have ( 1 2ηr k + 2η(r 1 +... + r k ) < 2ηr 1 2 + 1 + 1 k 1 2 + 1 2 +... + 1 ) 2 2 k 1 = 4ηr 1 = 4ɛδ Hence r k+1 is well defined and r k+1 = < δ. ηr 2 k δ 2η(r 1 +... + r k ) = r k ( ) ηr k δ 2η(r 1 +... + r k ) < r k 2. THEOREM 4.2 Let ɛ < 1/4, A (0) 1 = A 1, A (0) 2 = A 2 and (r k ) be the sequence of nonnegative real numbers defined in Proposition 4.1. Let R 1 be the unique element in BL(X 1, X 2 ) such that A (0) 2 R 1 R 1 A (0) 1 = A 3. If ɛ < 1/4, then the following hold iteratively. (i) If A (k) 1 = A (k 1) 1 A 4 R k and A (k) 2 = A (k 1) 2 R k A 4, then δ k := sep ( A (k) 1, A (k) ) 2 > δ 2η(r1 +... + r k ) > 0. (ii) Let R k+1 be the unique element in BL(X 1, X 2 ) such that A (k) 2 R k+1 R k+1 A (k) 1 = R k A 4 R k. 9

Then In particular, R k+1 η δ 2η(r 1 +... + r k ) R k 2 r k+1. R k+1 η δ(1 4ɛ) R k 2. Proof. Let r 1, r 2,..., be as in Proposition 4.1, i.e., r 1 = γ/δ, and Clearly, r k+1 := ηrk 2, k = 1, 2,.... δ 2η(r 1 +... + r k ) R 1 = T 1 (A 3 ) γ δ. Then, by Propositions 2.2 and 4.1, δ 1 := sep ( A (1) 1, A (1) ) ( (0) 2 sep A 1, A (0) ) 2 2ηr1 = δ 2ηr 1 > 0. Hence there exists a unique R 2 BL(X 1, X 2 ) such that and it satisfies the relation R 2 A 4 R 1 2 δ 1 A (1) 2 R 2 R 2 A (1) 1 = R 1 A 4 R 1, η δ 2ηr 1 R 1 2 ηr2 1 δ 2ηr 1 = r 2. Assume the result is for all integers up to k 1, k 2. Then for k, define A (k) 1 = A (k 1) 1 A 4 R k, A (k) 2 = A (k 1) 2 R k A 4. Then R k r k, and by Propositions 2.2 and 4.1, δ k := sep ( A (k) 1, A (k) ) 2 δk 1 2ηr k δ 2η (r 1 +... + r k ) > 0. Hence there exists a unique R k+1 BL(X 1, X 2 ) such that Then we have A (k) 2 R k+1 R k+1 A (k) 1 = R k A 4 R k. R k+1 A 4 R k 2 δ k η δ 2η (r 1 +... + r k ) R k 2 ηrk 2 δ 2η (r 1 +... + r k ) = r k+1. 10

Since r j+1 < r j /2 for j = 0, 1, 2,..., we have 2η (r 1 +... + r k ) = 4ηr 1 ( 1 1 2 k ) 4ɛδ, so that we obtain η R k+1 δ(1 4ɛ) R k 2. Next we obtain estimate for the error R (R 1 +... + R k ), and obtain additional condition under which R (R 1 +...+R k ) 0 as k. For this purpose let R (0) = R and for k {1, 2,...}, let THEOREM 4.3 Let ɛ < 1/4. Then R (k) = R (k 1) R k = R (R 1 +... + R k ). R (k) η δ(1 4ɛ) R(k 1) 2. More over, R (k) γ 1 δ g(ɛ)β2k, β = ɛ 1 4ɛ g(ɛ). If, in addition, ɛ < 3/16, then R (k) 0 as k. Proof. First we note that the Riccati equation take the form A 2 R RA 1 = A 3 + RA 4 R A (k) 2 R (k) R (k) A (k) 1 = R k A 4 R k + R (k) A 4 R (k) for k = 1, 2,..., where A (k) 1 and A (k) 2 are defined iteratively as in Theorem 4.2. Thus, denoting the map B A (k) 2 B BA (k) 1, B BL(X 1, X 2 ), by T k, it follows that so that ( R (k) = Tk 1 Rk A 4 R k + R (k) A 4 R (k)) ( = R k+1 + Tk 1 R (k) A 4 R (k)), R (k) = R (k 1) R k = [ R k + T 1 k 1 ( R (k 1) A 4 R (k 1))] R k = T 1 k 1 ( R (k 1) A 4 R (k 1)), 11

for k = 1, 2,.... Hence, from the estimate for δ k := sep ( A (k) 1, A (k) ) 2 from Theorem 4.2, Now let Recall from Theorem 3.1 that R (k) T 1 k 1 A 4 R (k 1) 2 α = η δ(1 4ɛ), r = γ δ g(ɛ). R (0) = R γ g(ɛ) = r. δ Then from the relation R (k) α R (k 1) 2 it follows that for k = 0, 1, 2,.... Note that This completes the proof. η δ(1 4ɛ) R(k 1) 2. R (k) rβ 2k 1, β := αr = ɛ 1 4ɛ g(ɛ), β < 1 if and only if ɛ < 3 16. Remark. Let us denote the sequence of approximations of R obtained by our procedure (1.4 ) and the Newton s iteration (1.3) by (S k ) and ( S k ) respectively. Then the error bound obtained in Theorem 4.3 based on (1.4) can be written using the order function as R S k = O ( β 2k 1 ), β = ɛ 1 4ɛ g(ɛ) whereas the result obtained by Demmel [2] for ɛ < 1/12 based on (1.3) is (cf. Nair [6], Theorem 4.2) R S k = O ( µ 2k 1 ), µ = 3 2 ɛg(ɛ). We may observe that β < µ whenever ɛ < 1/12. Thus, not only that our result requires weaker assumptions on the coefficient operators, but also the estimate obtained is better for those values of ɛ for which Demmel s result is applicable. 5 Application to Spectral Variation Let A : X X be a bounded linear operator on a complex Banach space X and M 0 be a closed subspace of X. We would like to impose conditions on M 0 such that it is close to a spectral subspace, and then obtain a sequence (M k ) of closed subspaces of X which converges to a spectral subspace in the sense of gap. 12

For M 0 to be close to a spectral subspace of A we assume the following. Assumption 1. There exists a closed subspace N 0 such that Let P 0 : X X be the projection onto M 0 along N 0 and let [ ] A11 A 12 X = M 0 N 0. (5.1) A 21 A 22 be the matrix representation of A with respect to the decomposition (5.1), i.e., taking the operators A ij are defined by Recall that Q 1 = P 0, Q 2 = I P 0, X 1 = Q 1 (X), X 2 = Q 2 (X), A ij : X j X i is defined by A ij x = Q i Ax, x X j. M 0 is invariant under A if and only if A 21 = 0, and M 0 is a spectral subspace of A if and only if A 21 = 0 and σ(a 11 ) σ(a 22 ) =. Assumption 2. so that σ(a 11 ) σ(a 22 ) =, (5.2) δ := sep(a 11, A 22 ) > 0. We shall assume further that the quantity ɛ := A 21 A 12 δ 2 is small. It is to be mentioned that the condition (5.2) is satisfied if, for example, A is an approximation of another operator A 0, and M 0 is a spectral subspace of A 0 with corresponding spectral projection P 0. To see this suppose that and [ (0) A 11 A (0) 12 A (0) 21 A (0) 22 A = A 0 + V ], [ V11 V 12 V 21 V 22 are the matrix representations (w.r.t. the decomposition (5.2)) of A 0 and V respectively. Since P 0 is a spectral projection of A 0, we have A (0) 12 = 0 = A (0) 21 13 ]

and δ 0 := sep(a (0) 11, A (0) 22 ) > 0. Thus [ A11 A 12 A 21 A 22 By Proposition 2.2, it follows that ] = [ (0) A 11 + V 11 V 12 V 21 A (0) 22 + V 22 δ δ 0 V 11 V 22. Hence the Assumption 2, viz., δ > 0, is satisfied if V 11 + V 22 < δ 0. Note that if the last inequality is satisfied, then the quantity ɛ satisfies the relation If we use the notation ɛ V 21 V 12 (δ 0 V 11 V 22 ) 2. A 1 = A 11, A 2 = A 22, A 3 = A 21, A 4 = A 12, then the definitions of δ and ɛ in this section agrees with those in the previous section. Suppose ɛ < 1/4, and R and R k are as in the last section. Define ]. P = P 0 + RP 0, M = P (X) and for k = 1, 2,..., let P k = P 0 + (R 1 +... + R k ) P 0, M k = P k (X). Then, by Theorem 4.3, taking M k = P k (X), we have sep(m k, M) γ δ g(ɛ)β2k 1 and gap(m k, M) P 0 γ δ g(ɛ)β2k 1, for k = 1, 2,.... If, in addition, ɛ < 3/16, then sep(m k, M) 0, gap(m k, M) 0 as k. It can be seen that P A P (X) = P (A 11 + A 12 R) P (X) 14

and (I P )A (I P )(X) = (I P ) (A 22 + RA 12 ) (I P )(X), so that σ ( P A P (X) ) = σ (A11 + A 12 R), σ ( (I P )A (I P )(X) ) = σ (A22 + RA 12 ). Hence, by Proposition 2.4, M = P (X) is a spectral subspace of A with the associated spectral set Λ := σ (A 11 + A 12 R) provided But, this is true because σ (A 11 + A 12 R) σ (A 22 + RA 12 ) =. sep (A 11 + A 12 R, A 22 + RA 12 ) δ 2η R > 0. δ (1 2ɛg(ɛ)) Next let B = A 11 + A 12 R, B k = A 11 + A 12 S k, where S k = R 1 +... R k, k = 1, 2,.... Then we have B B k R (R 1 +... + R k ) 0 as k. Thus, by Proposition 2.3, we have sep (σ(b k ), σ(b)) 0 as k. References [1] T. KATO, Perturbation Theory for Linear Operators, Springer verlag, @nd edition, 1976. [2] J.W. DEMMEL, Three methods for refining invariant subspaces, Computing 38 (1987), 43 57. [3] B.V. LIMAYE, Spectral Perturbation and Approximation with Numerical Experiments, Proc. Centre for Mathematical Analysis, Australian National Univiersity, Vol. 13, 1986. [4] M.T. NAIR, Approximation and Localization of Eigenelements, Ph.D. Thesis, I.I.T.Bombay, 1984. [5] M.T. NAIR, Approximation of spectral sets and spectral subspaces in Banach spaces, J. Indian Math. Soc., 54 (1989) 1 14. 15

[6] M.T. NAIR, On iterative refinements for spectral sets and spectral subspaces, Numer. Funct. Anal. Optim., 10 (1989) 1019 1037. [7] M.T. NAIR, Computable error estimates for Newton s iterations for refining invariant subspaces, Indian J. Pure &Appl. Math., 21 (1990) 1049 1054. [8] M. ROSENBLUM, On the operator equation BX XA = Q, Duke Math. J., 23 (1956) 263 270. [9] G.W. STEWART, Error bounds for approximate invariant subspaces of closed operarators, SIAM J. Numer. Anal., 8 (1971) 796 808. [10] A.E. TAYLOR, Spectral theory of closed distributive operators, Acta Math., 84 (1950) 189 224. Department of Mathematics Indian Institute of Technology Madras Chennai 600 036, INDIA E-Mail : mtnair@iitm.ac.in 16