projection methods for computing moore-penrose … · projection methods for computing...

PROJECTION METHODS FOR COMPUTINGMOORE-PENROSE INVERSES OF UNBOUNDED

OPERATORS

S. H. KULKARNI AND G. RAMESH

Abstract. In this article we give a characterization of the con-vergence of projection methods which are useful for approximatingthe Moore-Penrose inverse of a closed densely defined operator be-tween Hilbert spaces. We also consider questions related to thestability of projection methods. In the process we prove someresults related to the perturbation of closed range operators andperturbations of Moore-Penrose inverses. We illustrate the mainresult with an example.

1. Introduction

Projection methods are efficient and widely used tools to approxi-mate the solution of a given operator equation. In this method theinfinite dimensional problem (operator equation) can be reduced to asequence of finite dimensional operator equations (matrix equations).Hence it can be solved with the help of the known techniques of thefinite dimensional case. So this method has advantage both theoreticalas well as computational point of view.

Our main aim is to solve the operator equation

Tx = y

where T is a densely defined closed and unbounded operator betweenHilbert spaces H1 and H2. In an earlier paper [15], we studied thisproblem with the assumption that T has a bounded inverse. In thepresent paper, we do not make this assumption. Consequently, we lookfor the least square solution of minimal norm of the equation Tx = y.

In this article we give a necessary and sufficient condition for theconvergence of projection methods to such a least square solution of

Date: May 10, 2008.2000 Mathematics Subject Classification. 47A50,47A58.Key words and phrases. Densely defined operator, closed operator, Moore-

Penrose inverse, the reduced minimum modulus, generalized projection method,weakly bounded operator, perturbation.

1

2 S. H. KULKARNI AND G. RAMESH

minimal norm. We also show that these methods are stable under aparticular class of bounded perturbations.

To approximate the Moore-Penrose inverse of an operator by projec-tion methods, first we should be able to approximate the given operatorwith the help of a pair of sequences of projections. We call such a pairto be admissible for the given operator (see Definition 3.2 ).

In case, the operator is bounded, then we can always find an ad-missible sequence of projections for the given operator. But, since un-bounded operators are defined on subspaces of Hilbert space, we haveto impose some conditions on the sequence of projections. Thus it isdifficult to find an admissible sequence in this case. In this article wehave proved that such a sequence of projections can be constructed byintroducing some new operators (see Section 2).

The projection methods discussed in this article generalize the resultsof the article [15] in three directions: First, we extend the results of[15] for the usual inverse to the Moore-Penrose inverse. The second is,in this article we consider operators between different Hilbert spaceswhere as in [15] we have considered operators on a separable Hilbertspace. The third is that we assume that the range of the operator isseparable instead of the whole space.

The results related to the perturbation of closed range operators andthe Moore-Penrose inverses are the genaralizations of the results in [21]from the case of rectangular matrices to the case of closed operatorson Hilbert spaces. Several authors studied the perturbation resultsfor the Moore-Penrose inverses of bounded operators on Hilbert spaces[2, 5, 6, 7] and Banach spaces [18]. Some of these results are generalizedin the present paper.

This paper is organized as follows: In the second section we definethe convergence of generalized projection methods and give a necessaryand sufficient condition for the convergence. This is illustrated with anexample. In the third section the existence of admissible sequenceof projections is discussed. The fourth section contains perturbationresults for closed range operators and Moore-Penrose inverses. Thefinal section deals with the stability of generalized projection methodsunder bounded perturbations.

2. Notations and Preliminaries

Throughout the paper we denote infinite dimensional complex Hilbertspaces by H, H1, H2, H3, and inner product and the correspondingnorm on a Hilbert space are denoted by 〈., .〉 and ‖ · ‖ respectively.

L(H1, H2):= The set of all linear operators between H1 and H2.

GENERALIZED PROJECTION METHODS 3

L(H) := L(H, H)

If T ∈ L(H1, H2), then the domain , null space and the range spaceof T are denoted by D(T ), N(T ) and R(T ) respectively.

For S, T ∈ L(H1, H2) and U ∈ L(H2, H3), D(S +T ) = D(S)∩D(T )and (T + S)x = Tx + Sx for all x ∈ D(T + S). The domain of theoperator UT is given by D(UT ) = {x ∈ D(T ) : Tx ∈ D(U)} and inthis case UTx = U(Tx) for all x ∈ D(T ).

B(H1, H2):= The space of all bounded linear operators from H1 intoH2.

B(H) := B(H, H).

C(H1, H2) := {T ∈ L(H1, H2) : T is closed}.

C(H) := C(H, H)

Note 2.1. By the Closed Graph Theorem [16, 21.1, Page 420 ], it fol-lows that a closed operator T ∈ C(H1, H2) with D(T ) = H1 is bounded.

If S and T are two operators, then by S ⊆ T we mean that S is therestriction of T to D(S). i.e., D(S) ⊆ D(T ) and Sx = Tx, for all x ∈D(S) and in that case, we may also write S as T |D(S).

If M is a subspace of H, then M and M⊥ denote the closure and theorthogonal complement of M in H respectively. If M is closed, thenPM denote the orthogonal projection onto M .

Suppose X1 and X2 are subspaces of a Hilbert space with X1∩X2 ={0}. Then we use the notation X1 ⊕ X2 to denote the direct sum ofX1 and X2, and X1 ⊕⊥ X2 to denote the orthogonal direct sum of X1

and X2 whenever 〈x, y〉 = 0 for every x ∈ X1 and y ∈ X2.

Definition 2.2. [3, 08,310] An operator T ∈ L(H1, H2) with

domain D(T ) is said to be densely defined if D(T ) = H1. The subspaceC(T ) := D(T ) ∩N(T )⊥ is called the carrier of T .

Note 2.3. If T ∈ C(H1, H2), then D(T ) = N(T ) ⊕⊥ C(T ) [3, page340].

Definition 2.4. [8, 20] For T ∈ L(H), the resolvent of T is denotedby ρ(T ) and is defined as

ρ(T ) = {λ ∈ C : T − λI is bijective and (T − λI)−1 ∈ B(H)}.

The spectrum σ(T ), is defined by

σ(T ) = C \ ρ(T ).


Definition 2.5. [Moore-Penrose Inverse] [3] Let T ∈ C(H1, H2) bedensely defined. Then there exists a unique densely defined operatorT † ∈ C(H2, H1) with domain D(T †) = R(T ) ⊕⊥ R(T )⊥ and has thefollowing properties;

(1) TT †y = PR(T ) y, for all y ∈ D(T †).

(2) T †Tx = PN(T )⊥ x, for all x ∈ D(T ).

(3) N(T †) = R(T )⊥.

This operator T † is called the Moore-Penrose inverse of T .The following property of T † is also well known. For every y ∈ D(T †),let

L(y) := {x ∈ D(T ) : ||Tx− y|| ≤ ||Tu− y|| ∀ u ∈ D(T )}.Here any u ∈ L(y) is called a least square solution (lss) of theoperator equation Tx = y. The vector x = T †y ∈ L(y) and satisfies,||T †y|| ≤ ||x|| ∀ x ∈ L(y) and is called the least square solutionof minimal norm . A different treatment of T † is described in [3,Pages 336, 339, 341], where the authors call it “the Maximal Tsenggeneralized Inverse”.

Definition 2.6. [3, 13] Let T ∈ C(H1, H2). The reduced minimummodulus of T is defined by γ(T ) := inf {||Tx|| : x ∈ C(T ), ||x|| = 1}.

Some known properties of γ(T ) are listed in the following Proposi-tion. These are extensively used in the due course.

Proposition 2.7. [3, 13] For a densely defined T ∈ C(H1, H2), thefollowing statements are equivalent.

(1) R(T ) is closed.(2) R(T ∗)is closed.(3) T0 := T |C(T ) has a bounded inverse.(4) γ(T ) > 0.(5) T † is bounded.(6) γ(T ) = γ(T ∗)(7) R(T ∗T ) is closed.(8) R(TT ∗) is closed.

Proposition 2.8. [17] Let T ∈ C(H1, H2) be densely defined. Then wehave the following.

(1) γ(T ∗T ) = γ(T )2

(2) If R(T ) is closed, then γ(T ) =1

||T †||(3) If T = T ∗, then γ(T ) = inf {|λ| : λ ∈ σ(T ) \ {0}}.

In the following proposition we list some well known facts.


Proposition 2.9. [3] Let T ∈ C(H1, H2) be a densely defined operator.Then

(1) N(T ) = R(T ∗)⊥

(2) N(T ∗) = R(T )⊥

(3) N(T ∗T ) = N(T ) and

(4) R(T ∗T ) = R(T ∗).

Lemma 2.10. [9, Theorem 8.1, Page 69] Let T ∈ B(H) be such that||T || < 1. Then (I − T ) is invertible,(I − T )−1 =

∑∞k=0 T k and

||(I − T )−1|| ≤ 1

1− ||T ||.

Theorem 2.11. [1, 4] Let {Hk}, k = 1, 2, 3, . . . be closed subspacesof H and let Pk = PHk

. Suppose {Pk} is a monotone(Hk ⊆ HK+1 orHk+1 ⊆ Hk) sequence of orthogonal projections. Then the strong limitP = lim

k→∞PHk

exists and P is the projection onto ∩kHk in case Pk is

non-increasing and onto ∪kHk if {Pk} is non-decreasing.

Theorem 2.12 (Uniform boundedness principle). [9, Theorem 14.3,Page 83] Let X be a Banach space and Y be a normed linear space.Suppose F is a subset of B(X, Y ) with the property that for each x ∈ X,supA∈F

||Ax|| < ∞. Then supA∈F

||A|| < ∞.

3. Generalized Projection Methods

3.1. The Method and a Characterization. Let T ∈ C(H1, H2) bedensely defined with closed and separable range. Let {Yn} be an in-creasing sequence of subspaces of R(T ) such that ∪∞n=1Yn = R(T )and {Xn}, an increasing sequence of subspaces of N(T )⊥ such that∪∞n=1Xn = N(T )⊥. Let Pn : H2 → H2 and Qn : H1 → H1 be sequencesof bounded projections with R(Pn) = Yn and R(Qn) = Xn for each n.

Let Tn := PnTQn and Tn = Tn|Xn . Our aim is to approximate theleast square solution of minimal norm of

(1) Tx = y

To do this we find least square solution of minimal norm xn of thefinite system of equations

(2) Tnx = Pny

and expect that xn = T †nPny → x = T †y for every y ∈ H2. This is

the idea of the projection methods. The operators Tn are known assections. If these sections are finite dimensional, these are known as


finite sections and the method of approximating the solution with thehelp of these finite sections is called a finite section method.

We now give a formal definition of the convergence of this generalizedprojection method.

Definition 3.2. Suppose T ∈ C(H1, H2) is densely defined and has aclosed range. Let Pn and Qn be bounded projections on H2 and H1

respectively with dim R(Pn) = n = dim R(Qn) such that

(1) Pny → PR(T )y for all y ∈ H2

(2) Qnx → PN(T )⊥x for all x ∈ H1

(3) Qnu ∈ N(Tn)⊥ for all u ∈ C(T )(4) Qnx ∈ D(T ) for all x ∈ D(T ).(5) TQnx → Tx for all x ∈ D(T ).

Then the sequence of pairs {Pn, Qn} is said to be admissible for T .The generalized projection method for T is said to convergewith respect to {Pn, Qn} if for each y ∈ H2, T †

nPny → T †y.

Remark 3.3. The condition that Qnx ∈ D(T ) for all x ∈ D(T ) isequivalent to the condition R(Qn) ⊆ D(T ) for all n [15, Proposition2.7].

The following theorem provides a criterion for the convergence of thegeneralized projection method. This is analogous to [15, Theorem 3.1],which was proved for closed operators with bounded inverse.

Theorem 3.4. Let T ∈ C(H1, H2) be densely defined with closed andseparable range. Let {Pn, Qn} be an admissible sequence for T . Thenthe generalized projection method for T is convergent with respect to the

pair {Pn, Qn} if and only if the operators T †n are uniformly bounded.

Proof. Assume that the generalized projection method for T is con-

vergent. That is T †nPny → T †y for all y ∈ H2 . Hence by Theorem

2.12, m := supn||T †

nPn|| < ∞. Next, let z ∈ R(Pn). Then Pnz = z.

Hence

||T †nz|| = ||T †

nPnPnz|| ≤ m||Pnz|| = m||z||.

Hence ||T †n|| ≤ m.


For the converse part, assume that supn ||T †n|| := M < ∞. Consider

||xn −QnT†y|| = ||xn − PN( bTn)⊥QnT

†y|| (by condition 3 Definition 3.2)

= ||T †nPny − T †

nTnQnT†y||

≤ ||T †n||||Pny − TnQnT

†y||

≤ M ||Pny − TnQnT†y||

= M ||Pny − TnT†y|| (since Q2

n = Qn)

→ M ||PR(T )y − PR(T )y|| = 0

Now ||xn−T †y|| ≤ ||xn−QnT†y||+||QnT

†y−T †y|| → 0 as n →∞. �

Example 3.5. Let H1 = H2 = `2,

D(T ) = {(x1, x2, . . . ) ∈ `2 : (x1, 0, 3x3, 0, . . . , (2n− 1)x2n−1 . . . ) ∈ `2}.

Define T : D(T ) → `2 be given by

Tx =

{jxj if j is odd

0 if j is even.

As D(T ) contains C00, the space of all complex sequences having at mostfinitely many non-zero terms, T is densely defined. We can easily showthat T = T ∗, It can be shown that R(T ) is closed. Let E := {e1, e2, . . .}be the standard orthonormal basis for `2. Then {e2n−1 : n ∈ N} isan orthonormal basis for R(T ) and for y = (y1, y2, . . . ) ∈ `2, T †y =

(y1, 0,1

3y3, 0,

1

5, 0 . . . ).

Let Yn := span({e1, e3, . . . , e2n−1)}. Define Pnx =2n−1∑j=1

〈x, ej〉ej for

all x ∈ H1. Let us take Xn = Yn and Qn = Pn for each n. Also

Tn(x1, x2, . . . ) = PnTPn(x1, x2, . . . ) =

{jxj j is odd and 1 ≤ j ≤ n

0 else

and

Tn(x1, x2, . . . , xn) =

{jxj j is odd, 1 ≤ j ≤ n

0 j is even, 1 ≤ j ≤ n.

N(Tn) = {(0, x2, 0, . . . , xn) : (x1, x2, . . . , xn) ∈ Yn} = N(T ∗n) = R(Tn)⊥.

Here R(Tn) is closed and for x ∈ C(T ), Qnx ∈ Yn = N(Tn)⊥.Also for x ∈ D(T ), TQnx = TPnx = (x1, 0, 3x3, . . . , nxn, 0, 0, 0 . . . ) →


(x1, 0, x3, 0, . . . , 0, xn, 0, . . . , ) = Tx. Hence {Pn, Qn} is admissible for

T . The matrix of Tn with respect to the basis of Yn is given by

1 0 0 . . . 00 0 0 . . . 00 0 3 . . . 0...

......

0 0 0 . . . 00 0 0 . . . 2n− 1

and the matrix of T †

n is

1 0 0 . . . 00 0 0 . . . 00 0 1

3. . . 0

......

...0 0 0 . . . 00 0 0 . . . 1

2n−1

.

Here ||T †n|| = 1 for each n and

xn = T †nPny

= (y1, 0,1

3y3, 0, . . . ,

1

2n− 1yn, 0, . . . ) → T †y for all y ∈ H2.

4. Existence of admissible sequence of projections

We have observed in the previous section that the generalized pro-jection methods depend on the given operator and the admissible se-quences of projections {Pn, Qn}. In the case of a bounded operator, itis easy to find such an admissible sequence. We can choose any pair ofsequences {Pn, Qn} satisfying the conditions

1. Pny → PR(T ) y for every y ∈ H2 and2. Qnx → PN(T )⊥ for every x ∈ H1.

Then, in view of the continuity of T , the assumptions (3), (4) and 5of Definition 3.2 are satisfied. Hence a natural question one can askis whether it is possible to find an admissible sequence {Pn, Qn} for agiven densely defined operator with closed and separable range. In thissection we answer this question affirmatively.

Proposition 4.1. [10, 11, 19, 12] Let T ∈ C(H1, H2) be densely defined.Then

(1) (I + T ∗T )−1 ∈ B(H1), (I + TT ∗)−1 ∈ B(H2).


(2) (I + TT ∗)−1T ⊆ T (I + T ∗T )−1 and ||T (I + T ∗T )−1|| ≤ 1

2

(3) (I + T ∗T )−1T ∗ ⊆ T ∗(I + TT ∗)−1 and ||T ∗(I + TT ∗)−1|| ≤ 1

2.

Lemma 4.2. Let T ∈ C(H1, H2) be densely defined. Let C := T (I +T ∗T )−1. Then

(1) R(T ) = R(C) i.e., N(T ∗) = N(C∗)(2) R(T ) is closed if and only if R(C) is closed.

Proof. Proof of (1)

It is clear that R(C) ⊆ R(T ). Hence R(C) ⊆ R(T ). To prove the other

way, it suffices to prove R(T ) ⊆ R(C).Let y ∈ R(T ). Then there exists x ∈ D(T ) such that Tx = y.

Choose a sequence {xn} ⊆ D(T ∗T ) such that xn → x and Txn → Tx.

This is possible since, G(T |D(T ∗T )) = G(T ) [20, Theorem, Page 336]. Since I + T ∗T is bijective, xn = (I + T ∗T )−1un, where un ∈ H1.Hence Txn = T (I + T ∗T )−1un = Cun. As y = lim

n→∞Txn = lim

n→∞Cun,

y ∈ R(C). It follows that R(T ) ⊆ R(C). Hence R(T ) ⊆ R(C).Proof of (2):Note that C∗ = T ∗(I +T ∗T )−1, C∗C = (I +T ∗T )−1T ∗T (I +T ∗T )−1 =

(I+T ∗T )−1−((I+T ∗T )−1)2. Hence σ(C∗C) = { λ

(1 + λ)2: λ ∈ σ(T ∗T )}.

By Proposition 2.8, γ(C)2 =γ(T )2

(1 + γ(T )2)2. Hence γ(T ) > 0 if and only

if γ(C) > 0. Now the conclusion follows from Proposition 2.7. �

Lemma 4.3. Let T ∈ C(H1, H2) be densely defined. Let D := T ∗(I +TT ∗)−1. Then

(1) R(T ∗) = R(D) i.e., N(T ) = N(D∗)(2) R(T ∗) is closed if and only if R(D) is closed.

Proof. The proof this lemma follows from the previous lemma by re-placing T ∗ in place of T and observing that T ∗∗ = T . �

Lemma 4.4. Let T ∈ C(H1, H2) be densely defined. Let Yn ⊆ R(T ) besuch that

(a) Yn ⊆ Yn+1 for each n ∈ N(b) dim Yn = n

(c) ∪∞n=1Yn = R(T ).

Let Zn := (I + TT ∗)−1Yn and Xn := T ∗Zn = T ∗(I + TT ∗)−1Yn. Then

(1) Xn ⊆ Xn+1 · · · ⊆ N(T )⊥, dim Xn = n and

(2) ∪∞n=1Zn = R(T )


(3) ∪∞n=1Xn = R(T ∗)

(4) ∪∞n=1TXn = R(T ).

Proof. By the definition of Xn, Xn ⊆ C(T ) ⊆ N(T )⊥ = R(T ∗) for alln and Xn ⊆ Xn+1. Since the operator T ∗(I + TT ∗)−1|R(T ) is injectivedim Xn = n = dim Yn.

For a proof of (2), we make use of the following observation:

(I + TT ∗)−1(R(T )) = R(T ).

It can be proved easily that (I+TT ∗)−1(N(TT ∗)) = N(TT ∗). By theProjection Theorem [16, 21.1, Page 420 ], H2 = N(TT ∗)⊕⊥ N(TT ∗)⊥.

That is H2 = N(TT ∗)⊕⊥ R(TT ∗). But

(I + TT ∗)−1(H2) = D(TT ∗) = N(TT ∗)⊕⊥ C(TT ∗).

Hence

(I + TT ∗)−1H2 = (I + TT ∗)−1(N(TT ∗)⊕⊥ R(TT ∗))

= N(TT ∗)⊕⊥ (I + TT ∗)−1(R(TT ∗)).

From this we can conclude that (I + TT ∗)−1(R(TT ∗)) = C(TT ∗) and

as C(TT ∗) = N(TT ∗)⊥, we have (I + TT ∗)−1(R(TT ∗)) = R(TT ∗).

Hence (I + TT ∗)−1(R(T )) = R(T ), by Proposition (2.9). Thus

R(T ) = (I + TT ∗)−1(R(T )) = (I + TT ∗)−1(∪∞n=1Yn)

= ∪∞n=1(I + TT ∗)−1Yn

= ∪∞n=1Zn.

This proves (2).

It is clear that ∪∞n=1Xn ⊆ R(T ∗) = N(T )⊥.Suppose ∪∞n=1Xn ( N(T )⊥. Then there exists a 0 6= z0 ∈ N(T )⊥

such that z0 ∈ (∪∞n=1Xn)⊥. That is

〈z0, T∗(I + TT ∗)−1y〉 = 0 for all y ∈ R(T )

By the continuity of T ∗(I + TT ∗)−1, this holds for all y ∈ R(T ).We claim that this holds for all y ∈ H2. Let y ∈ H2. Then y = u+ v

for some u ∈ R(T ) and v ∈ R(T )⊥ = N(T ∗) ⊆ D(T ∗). Hence byProposition 4.1, T ∗(I + TT ∗)−1v = (I + T ∗T )−1T ∗v = 0. Hence

〈z0, T∗(I + TT ∗)−1y〉 = 〈z0, T

∗(I + TT ∗)−1u〉 = 0.

This proves the claim.


Next, since C(T ) = N(T )⊥[14], there exists a sequence {zn} ⊆ C(T )such that zn → z0. Hence for all y ∈ H2,

0 = 〈z0, T∗(I + TT ∗)−1y〉 = lim

n→∞〈zn, T

∗(I + TT ∗)−1y〉

= limn→∞

〈Tzn, (I + TT ∗)−1y〉

= limn→∞

〈(I + TT ∗)−1Tzn, y〉

= limn→∞

〈T (I + T ∗T )−1zn, y〉.

This shows that T (I +T ∗T )−1znw−→ 0 (weakly), but since T (I +T ∗T )−1

is bounded, we have T (I + T ∗T )−1z0 = 0. That is (I + T ∗T )−1z0 ∈N(T ). Let y = (I + T ∗T )−1z0. Then Ty = 0. Hence z0 = (I +T ∗T )y = y ∈ N(T ). Thus z0 ∈ N(T ) ∩N(T )⊥ = {0}. Hence z0 = 0, acontradiction to our assumption. This proves (3).

Using a similar argument we can prove (4). �

Theorem 4.5. Let T ∈ C(H1, H2) be densely defined and with closedand separable range. Then there exists a sequence {Pn, Qn} of projec-tions with finite dimensional ranges which is admissible for T .

Proof. Choose a sequence {Xn} of subspaces of C(T ) such that

(a) Xn ⊆ Xn+1 for each n,(b) dim Xn = n(c) ∪∞n=1Xn = N(T )⊥.

Let Yn := TXn. Then

(1) Yn ⊆ Yn+1 for each n,(2) dim Yn = n(3) ∪∞n=1Yn = R(T ).

Since T |C(T ) : C(T ) → R(T ) is bijective, we have dim Yn = dim Xn =n. It is clear by the linearity of T that Yn ⊆ Yn+1 for each n. Also∪∞n=1Yn ⊆ R(T ). We prove that these two subspaces are equal. If∪∞n=1Yn ⊂ R(T ), then there exists 0 6= z0 ∈ R(T ) such that z0 ∈(∪∞n=1Yn)⊥. Hence for all x ∈ Xn, we have 〈z0, Tx〉 = 0. Since R(T ) =R(T (I+T ∗T )−1), there exists a x0 ∈ H1 such that z0 = T (I+T ∗T )−1x0.Next let x ∈ ∪∞n=1Xn. Then

0 = 〈z0, Tx〉= 〈T ∗z0, x〉= 〈T ∗T (I + T ∗T )−1x0, x〉.

Since T ∗T (I+T ∗T )−1 is continuous, we obtain 〈T ∗T (I+T ∗T )−1x0, x〉 =0 for all x ∈ ∪∞n=1Xn = N(T )⊥.


Let x ∈ H. Then x = u+v, where u ∈ N(T ) and v ∈ N(T )⊥. Hence

〈T ∗T (I + T ∗T )−1x0, x〉 = 〈T ∗T (I + T ∗T )−1x0, u + v〉= 〈T ∗T (I + T ∗T )−1x0, u〉+ 〈T ∗T (I + T ∗T )−1x0, v〉= 〈T ∗T (I + T ∗T )−1x0, u〉+ 0

= 〈T (I + T ∗T )−1x0, Tu〉 since u ∈ N(T )

= 0.

Therefore T ∗T (I + T ∗T )−1x0 = 0. That is z0 ∈ N(T ∗). But z0 ∈R(T ) = N(T ∗)⊥. Hence z0 = 0. This is a contradiction. Hence∪∞n=1Yn = R(T ).

Let Pn : H2 → H2 and Qn : H1 → H1 be sequence of orthogonalprojections such that R(Pn) = Yn and R(Qn) = Xn. Let Tn := PnTQn

and Tn := Tn|Xn . That is Tn ∈ B(Xn, Yn).Claim: {Pn, Qn} is admissible for T .

Since {Pn} and {Qn} are increasing projections by Theorem 2.11, itfollows that

(1) Pny → PR(T ) y for all y ∈ H2,(2) Qnx → PN(T )⊥ x for all x ∈ H1,

Next, we prove

(3) Qnx ∈ N(Tn)⊥ for every x ∈ C(T ).

First we observe that N(Tn) = N(TQn|Xn). Let x ∈ N(TQn|Xn). Then

TQnx = 0. Hence PnTQnx = 0. That is N(TQn) ⊆ N(Tn).

For the reverse inclusion, let x ∈ N(Tn). Then

PnTQnx = PnTx = 0

⇒ Tx ∈ N(Pn) = Y ⊥n

⇒ Tx ∈ N(Pn) = Y ⊥n ∩ Yn since Yn = TXn

⇒ Tx = 0

⇒ Qnx ∈ N(T )

⇒ x ∈ N(TQn).

Now we prove (3). Let y ∈ N(Tn) and x ∈ C(T ). Since x ∈ R(T ∗) =R(T ∗(I +TT ∗)−1) there exists w ∈ H2 such that x = T ∗(I +TT ∗)−1w.


Thus

〈x, Qny〉 = 〈Qnx, y〉 = 〈T ∗(I + TT ∗)−1w,Qny〉= 〈(I + TT ∗)−1w, TQny〉

= 0 since N(TQn|Xn) = N(Tn).

Hence for all x ∈ C(T ), Qnx ∈ N(Tn)⊥.Also for for every x ∈ D(T ), Qnx ∈ Xn ⊆ D(T ). Hence the condition

(4) is satisfied.Next we prove that

(5) TQnx → Tx for all x ∈ D(T ).

If x ∈ D(T ), then Qnx ∈ Xn ⊆ N(T )⊥ = R(T ∗) = R(T ∗(I + TT ∗)−1).Therefore Qnx = T ∗(I + TT ∗)−1wn, for some wn ∈ N(T ∗)⊥. HenceTQnx = TT ∗(I+TT ∗)−1wn. As Qnx → PN(T )⊥x and R(T ∗(I+TT ∗)−1)is closed, there exists k > 0 such that

||T ∗(I + TT ∗)−1wn|| ≥ k ||wn||.Since the left hand side of the above equation is convergent, it followsthat wn is convergent. Assume that wn → w. As T ∗(I + TT ∗)−1 isbounded, Qnx = T ∗(I + TT ∗)−1wn → T ∗(I + TT ∗)−1w = PN(T )⊥ x.Now TQnx = TT ∗(I + TT ∗)−1wn → TT ∗(I + TT ∗)−1 = TPN(T )⊥ x.Since x ∈ D(T ), x = u + v, where u ∈ N(T ) and v ∈ C(T ). ThereforeTPN(T )⊥x = Tv = T (u+v) = Tx. Hence TQnx → Tx for all x ∈ D(T ).

Now for any x ∈ D(T ),

||PnTQn − Tx|| ≤ ||PnTQnx− PnTx||+ ||PnTx− Tx||≤ ||Pn|| ||TQnx− Tx||+ ||PnTx− Tx||≤ ||TQnx− Tx||+ ||PnTx− Tx|| → 0 as n →∞.

Hence {Pn, Qn} is admissible for T . �

5. Perturbation results for Moore-Penrose inverses

Our next goal is to investigate whether the generalized projectionmethods are stable under bounded perturbations. For this, first wehave to check whether the closed range operators are stable under per-turbations. Analogue results for square matrices were proved in [21].This sections contains a detailed information about these questions.First we prove some auxillary results. Analogous results for rectangu-lar matrices were proved in [21].

Proposition 5.1. Let T ∈ C(H1, H2) be densely defined and have aclosed and separable range . Let S ∈ B(H1, H2). Then


(1) ST †T = S|D(T ) ⇔ N(T ) ⊆ N(T + S) ⇔ N(T ) ⊆ N(S).(2) If ||ST †|| < 1 and ST †T = S|D(T ), then N(T + S) = N(T ).

Proof. It is easy to see that N(T ) ⊆ N(T + S) ⇔ N(T ) ⊆ N(S).Suppose ST †T = S|D(T ). If x ∈ N(T ), then Sx = ST †Tx = 0.

Hence N(T ) ⊆ N(S).Now, for the converse part, let us assume that N(T ) ⊆ N(S). Let

x ∈ D(T ). Then x = u + v, where u ∈ N(T ), v ∈ C(T ) and 〈u, v〉 = 0.Then

ST †Tx = ST †Tv = SPR(T †)

v = SPC(T )v

= Sv

= S(u + v), since N(T ) ⊆ N(S)

= Sx.

To prove (2), in view of (1), it is enough to prove that N(T + S) ⊆N(T ). If x ∈ N(T +S), by our assumption, we have Tx+ST †Tx = 0.That is (I + ST †)Tx = 0. Since (I + ST †)−1 ∈ B(H1), we get aconclusion that Tx = 0 and hence x ∈ N(T ). �

Theorem 5.2. Let T ∈ C(H1, H2) be densely defined and have a closedrange and S ∈ B(H1, H2) be such that

(a) ||S|| < 1

||T †||and

(b) N(T ) ⊆ N(T + S).

Then R(T + S) is closed.

Proof. Since R(T ) is closed, we have ||Tx|| ≥ γ(T )||x|| for each x ∈C(T ). Since C(T + S) ⊆ C(T ), for all x ∈ C(T + S)

||Tx + Sx|| ≥ |||Tx|| − ||Sx|||≥ |γ(T )||x|| − ||S||||x||||≥ (γ(T )− ||S||)||x||.

By Proposition 2.8 and our assumption γ(T )−||S|| > 0. Thus R(T +S)is closed by Proposition 2.7. �

Proposition 5.3. Let T ∈ C(H1, H2) be densely defined and have aclosed range. Let S ∈ B(H1, H2). Then

(1) TT †S = S if and only if R(S) ⊆ R(T )(2) If TT †S = S, then R(T + S) ⊆ R(T )(3) If ||T †S|| < 1 and TT †S = S, then R(T + S) = R(T ).

Proof. If TT †S = S, then it is obvious that R(S) ⊆ R(T ).On the other hand, if R(S) ⊆ R(T ), we have TT †S = PR(T )S = S.


To prove (2), let us assume that TT †S = S. Then T + S = T +TT †S ⊆ T (I + T †S). That is R(T + S) ⊆ R(T ).

One part of proof of (3), namely, R(T +S) ⊆ R(T ) follows from (2).The condition ||T †S|| < 1 implies that (I2 + T †S)−1 ∈ B(H2). Hence,if y = Tx for some x ∈ D(T ), then by the surjectivity of I + T †S,there exists a u ∈ D(T ) such that x = (I + T †S)u . This shows thaty = T (I + T †S)u = Tu + Su ∈ R(T + S). �

Theorem 5.4. Let T ∈ C(H1, H2) be densely defined and have a closedrange and S ∈ B(H1, H2) be such that ||T †S|| < 1 and TT †S = S. ThenR(T + S) is closed.

Proof. By Proposition 5.3, R(T + S) = R(T ). �

An analogue of the following Theorem for the case of rectangularmatrices was proved by Stewart [21]. The same result for boundedoperators on Banach spaces with a slight different assumption was ob-tained by Nashed [18].

Theorem 5.5. Let T ∈ C(H1, H2) be densely defined operator with aclosed range. Let S ∈ B(H1, H2) be such that ||T †S|| < 1 and satisfythe following conditions.

(1) TT †S = S and(2) ST †T = S|D(T ).

Then

(a) R(T + S) is closed and(b) (T + S)† = (I + T †S)−1T † = T †(I + ST †)−1 and hence

T † = (T + S)†(I + ST †).

Proof. First note that T + S is a closed operator with D(T + S) =D(T ). The proof of the first follows by showing R(T + S) = R(T ),which is proved in Proposition 5.3. It remains to deduce the formulafor T †.

We want to prove T † = (I + T †S)(T + S)†. It suffices to prove(T + S)† = (I + T †S)−1T †. Let U := (I + T †S)−1T †. We show that Usatisfies all the axioms of definition of the Moore-Penrose inverse.

Note that the condition ST †S = S|D(T ) together with ||T †S|| < 1,implies that N(T ) = N(T + S).

Let z ∈ R(U). That is there exists a y ∈ D(T †) such that z =(I +T †S)−1T †y. Hence T †y = z +T †Sz. So z = T †y−T †Sz ∈ C(T ) =C(T + S) ⊆ D(T + S).


Now for z ∈ D(T ),

U(T + S)z = (I + T †S)−1T †(T + S)z

= (I + T †S)−1T †(T + TT †S)z

= (I + T †S)−1T †T (I + T †S)z

= (I + T †S)−1(I + T †S)T †Tz

= T †Tz = PR(T †)z = PR(U)z.

In the fourth line of the above equations we have used the fact thatT †T (I + T †S) = (I + T †S)T †T on D(T ), which can be verified easily.

Now

(T + S)U = (T + S)(I + T †S)−1T †

= (T + TT †S)(I + T †S)−1T †

= T (I + T †S)(I + T †S)−1T †)

= TT †

= PR(T )

= PR(T+S).

The uniqueness of (T + S)† follows from [3, Theorem 1].Since R(S) ⊆ R(T ), by Lemma (2.10)

(I + ST †)−1T † =∞∑

n=0

(−T †S)nT † =∞∑

n=0

T †(−ST †)n

= T †(I + ST †).

�

Remark 5.6. Proposition (5.1) and Proposition (5.3) tell us how tochoose an operator S satisfying the hypotheses of Theorem (5.5). Let

α be such that 0 < α <2

||T †||. Let Sα := αT (I + T ∗T )−1 . Then by

[10, Theorem 3.1.], ||T (I + T ∗T )−1|| ≤ 1

2. Thus ||Sα|| ≤

α

2. It can be

verified that Sα satisfies the hypotheses of Theorem 5.5.

Remark 5.7. From Theorem 5.5, the following bounds for ||(T +S)†−T †|| can be deduced.

(T + S)† − T † = (I + ST †)−1T † − (I + ST †)(I + ST †)−1T †= [I − (I + ST †−)](I + ST †)−1T †

= (−ST †)(I + ST †)−1T †.


Hence

||(T + S)† − T †|| ≤ ||S||||T †||2

1− ||T †S||.

Remark 5.8. Assume that T and S satisfy the conditions of Theorem5.5. We can deduce the following formula.

|γ(S + T )− γ(T )| =∣∣∣∣ 1

||(S + T )†||− 1

||T †||

∣∣∣∣=

∣∣∣∣ ||T †|| − ||(T + S)†||||T †||||(T + S)†||

∣∣∣∣≤

∣∣||T †|| − ||(T + S)†||∣∣

||T †||||(T + S)†||

≤ ||S||||T †||2

||T †||||(T + S)†||[1− ||T †S]||= β||S||

where β =||T †||

||(T + S)†||[1− ||T †S||]> 0.

Corollary 5.9. Let T ∈ C(H1, H2) be densely defined and have a closedrange. Assume that Sn ∈ B(H1, H2) satisfy all the conditions of Theo-rem 5.5 and Sn → 0. Then

(1) (T + Sn)† → T † in the operator norm of B(H2, H1).(2) γ(T + Sn) → γ(T ) as n →∞.

Proof. The proofs follow from Theorem 5.5 and Remark 5.8. �

The following results which are proved in [5], can be obtained asparticular cases of the above proved results.

Corollary 5.10. [5, Lemma 3.3] Let T ∈ B(H1, H2) be injective withclosed range, and let S ∈ B(H1, H2) be such that

(a) R(S) ⊆ R(T )(b) ||T †S|| < 1.

Then

(1) T + S is injective(2) R(T + S) = R(T )(3) (T + S)† = (I + T †S)−1T † = T †(I + ST †)−1

(4) ||(T + S)†|| ≤ ||T †||1− ||T †S||

(5) ||(T + S)† − T †|| ≤ ||T †S||||T †||1− ||T †S||

.


Proof. By Proposition (5.1), (a) is equivalent to the condition TT †S =S. Since T is injective, T †T = I|D(T ). Hence ST †T = SI|D(T ) = S|D(T ).Now by Theorem (5.5), (T + S)† = (I + T †S)−1T † = T †(I + ST †)−1.The other relations can be proved by using Lemma 2.10. �

Corollary 5.11. [5, Lemma 3.4] Let T ∈ B(H1, H2) be surjective andS ∈ B(H1, H2) be such that

(a) N(T ) ⊆ N(S)(b) ||ST †|| < 1.

Then

(1) T + S is surjective(2) N(T + S) = N(T )(3) (T + S)† = T †(I + ST †)−1 = (I + T †S)−1T †

(4) ||(T + S)†|| ≤ ||T †||1− ||ST †||

(5) ||(T + S)† − T †|| ≤ ||ST †||||T †||1− ||ST †||

.

Proof. By Proposition (5.1), N(T ) ⊆ N(S) ⇔ ST †T = S|D(T ). SinceT is surjective, C(T ) = D(T ) and hence TT † = I, that is STT † = S.Now applying Theorem (5.5) we get the expression for (S + T )†. Thebounds for ||(S + T )†|| and ||(S + T )† − T †|| follows from Lemma 2.10and Theorem 5.5. �

Corollary 5.12. [5, Lemma 4.3] Let T ∈ B(H1, H2) have a closedrange and S ∈ B(H1, H2) be such that

(1) N(T ) ⊆ N(S)(2) ||S||||T †|| < 1.

Then

(i) N(T + S) = N(T )

(ii) ||(T + S)†|| ≤ ||T †||1− ||S||||T †||

.

Proof. The proof of (i), follows by Proposition (5.1) and Theorem(5.2)

Proof of (ii):By Theorem 5.2, we have

||Tx + Sx|| ≥ (γ(T )− ||S||)||x|| for all x ∈ C(T + S).

Hence γ(T + S) ≥ γ(T )− ||S||. That is

1

||(T + S)†||≥ 1

||T †||− ||S|| ≥ 1− ||S||||T †||

||T †||.


Hence ||(T + S)†|| ≤ ||T †||1− ||S||||T †||

. �

6. Stability of generalized projection methods

In this section we prove that the generalized projection methodsare stable under particular bounded perturbations with a restrictedadmissible sequence.

Let Xn ⊆ D(T ) be such that dim Xn = n, Xn ⊆ Xn+1 for eachn and ∪∞n=1Xn = N(T )⊥. Let Yn = TXn. Define Pn : H2 → H2

and Qn : H1 → H1 orthogonal projections such that R(Pn) = Yn

and R(Qn) = Xn. By the construction in Theorem 4.5, {Pn, Qn} isadmissible for T .

Theorem 6.1. Let T ∈ C(H1, H2) be densely defined and have a closedand separable range. Assume that the generalized projection method forT is convergent with respect to the admissible sequence {Pn, Qn} definedas above. Then there exists a γ > 0 such that for all S ∈ B(H1, H2)with ||S|| < γ and

(a) TT †S = S(b) ST †T = S|D(T ).

Then the generalized projection method for T + S is convergent.

Proof. It is clear that by Proposition 5.3, R(T + S) is closed. Sincethe generalized projection method for T is convergent with respect

to {Pn, Qn}, supn||T †

n|| < ∞. Let M = supn(||T †n|| + ||Pn||) < ∞.

Write γ := min { 12M2 ,

1||T †||}. We will complete the proof by showing

|| (T + S)n

†|| = L < ∞ for some L > 0. Hence we have to find the

expression for (Tn + Sn)† as in Theorem 5.5. Thus we have to prove

that Tn and Sn satisfy hypotheses of Theorem 5.5. Equivalently we

have to show that N(Tn) ⊆ N(Sn), R(Sn) ⊆ R(Tn) and ||T †n Sn|| < 1.

Let x ∈ Xn be such that Tnx = 0. That is PnTx = 0. Thus Tx ∈N(Pn)⊥ = Y ⊥

n . But by Hypothesis Tx ∈ Yn, hence x ∈ N(T ). As

x ∈ Xn ⊆ N(T )⊥, we have x = 0. This concludes that Tn is one-to-

one. Hence Tn is bijective. This also proves that R(Sn) ⊆ Yn = R(Tn).

Now ||T †n Sn|| ≤ ||T †

n||||Sn|| ≤ M ||S|| < 1. Hence by Proposition 2.10,

||(In + Sn T−1n )−1|| ≤ 2.

Now consider (T + S)n = Tn + Sn = Tn(In + T−1n Sn)

By Theorem 5.5, (Tn + Sn)† = T−1n (In + SnT

−1n )−1. Hence ||(Tn +

Sn)†|| ≤ 2M < ∞. The conclusion follows from Theorem 3.4 �


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Department of Mathematics, I. I. T. Madras, Chennai, India-600 036.E-mail address: [email protected]

Department of Mathematics, I. I. T. Madras, Chennai, India-600 036.E-mail address: [email protected]

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