PUBLICATIONS DE L’INSTITUT MATHÉMATIQUE Nouvelle série, tome 78(92) (2005), 127–133

ON THE OPERATOR EQUATIONS ABA = A^{2} AND BAB = B^{2}

Christoph Schmoeger

Communicated by Stevan Pilipović

Abstract. We generalize a result of I. Vidav concerning the operator equations ABA = A^{2} and BAB = B^{2}.

1. Introduction

In [7] I. Vidav proved the following result:

Theorem 1.1. Let H be a complex Hilbert space and let A and B be bounded linear operators on H. Then the following assertions are equivalent:

(a) There is a uniquely determined bounded linear operator P on H such that P^{2} = P and A = P P^{∗} and B = P^{∗}P.

(b) A and B are selfadjoint and satisfy the relations ABA = A^{2} and BAB = B^{2}.

Vidav gave two proofs of Theorem 1.1; the first proof is geometrical and the second one is algebraic. In [6] Rakočević gave another proof of Theorem 1.1.

The aim of this paper is to prove a result, which implies Theorem 1.1. Section 2 deals with Drazin invertible elements of rings. In Section 3 we consider bounded linear operators on Banach spaces. Operators on Hilbert spaces are considered in Section 4, where we will give a proof of Theorem 1.1. In the final section we investigate several special classes of operators.

2. Drazin inverses in rings

In this section R denotes an associative ring. An element A ∈ R is said to be Drazin invertible if there exists C ∈ R such that

(1) A^{m} = A^{m+1}C for some integer m ⩾ 0,

(2) C = AC^{2}

(3) AC = CA.

2000 Mathematics Subject Classification: Primary 47A05.

Key words and phrases: selfadjoint operator, Drazin inverse.

In this case C is called a Drazin inverse of A and the smallest integer m ⩾ 0 in (1) is called the index i(A) of A.

If R has a neutral element I and if we define A^{0} = I, then (1), (2) and (3) hold with m = 0 if and only if A is invertible.

Proposition 2.1. If A ∈ R is Drazin invertible, then A has a unique Drazin inverse.

Proof. [4]. □

Our main result in this section is:

Theorem 2.2. (a) If P, Q ∈ R, P^{2} = P, Q^{2} = Q, A = P Q and B = QP, then ABA = A^{2} and BAB = B^{2}.

(b) Suppose that A, B ∈ R are Drazin invertible, i(A) = i(B) = 1, ABA = A^{2} and BAB = B^{2}. Then there are P, Q ∈ R such that P^{2} = P, Q^{2} = Q, A = P Q and B = QP.

Proof. (a) We have ABA = P Q^{2}P^{2}Q = (P Q)^{2} = A^{2} and BAB = QP^{2}Q^{2}P = (QP)^{2} = B^{2}.

(b) Since i(A) = i(B) = 1, there are C, D ∈ R with

ACA = A, CAC = C, AC = CA
BDB = B, DBD = D, BD = DB.

Let P := CAB, Q := BAC and R := DBA. Then

P^{2} = CABCAB = C(ABA)CB = CA^{2}CB = ACACB = ACB = CAB = P,
R^{2} = DBADBA = D(BAB)DA = DB^{2}DA = BDBDA = BDA = DBA = R,
Q^{2} = BACBAC = BC(ABA)C = BCA^{2}C = BCACA = BCA = BAC = Q.

Furthermore we have

P Q = CABBAC = CAB^{2}AC = CA(BAB)AC = C(ABA)BAC
= CA^{2}BAC = ACABAC = ABAC = A^{2}C = ACA = A,
RP = DB(ACA)B = DBAB = DB^{2} = BDB = B.

It follows that

QP = BACCAB = B(ACA)CB = BACB = BCAB
= BP = (RP)P = RP^{2} = RP = B.  □

3. Bounded linear operators

ON THE OPERATOR EQUATIONS ABA = A^{2} AND BAB = B^{2} 129

In this section X denotes a complex Banach space and L(X) the Banach algebra of all bounded linear operators on X. If A ∈ L(X), then σ(A), ρ(A) and r(A) denote the spectrum, the resolvent set and the spectral radius of A, respectively. We write N(A) for the kernel of A and A(X) for the range of A. Define p(A) [resp. q(A)], the ascent [resp. the descent] of A, to be the smallest integer n ⩾ 0 such that N(A^{n+1}) = N(A^{n}) [resp. A^{n+1}(X) = A^{n}(X)] or ∞ if no such n exists. It follows from [5, Satz 72.3] that if p(A) and q(A) are both finite, then they are equal and, if p = p(A) = q(A) < ∞, then X = N(A^{p}) ⊕ A^{p}(X).

A Drazin invertible operator A ∈ L(X) with i(A) ⩽ 1 is called simply polar. The following proposition tells us exactly which operators are Drazin invertible.

Proposition 3.1. For A ∈ L(X) and n ⩾ 1 the following assertions are equivalent:

(a) A is Drazin invertible and i(A) = n.

(b) p(A) = q(A) = n.

(c) The resolvent (λI − A)^{−1} has a pole of order n at λ = 0.

Proof. [2, Theorem 5.2], [5, Satz 101.2]. □

As an immediate consequence of Proposition 3.1 and Theorem 2.2 we get the main result of this section:

Theorem 3.2. Suppose that A, B ∈ L(X), p(A) = q(A) = 1 and p(B) = q(B) = 1. Then the following assertions are equivalent:

(a) There are P, Q ∈ L(X) such that P^{2} = P, Q^{2} = Q, A = P Q and B = QP.

(b) ABA = A^{2} and BAB = B^{2}.

We use σ_{p}(A), σ_{ap}(A), σ_{r}(A) and σ_{c}(A) to denote the point, approximate point, residual and continuous spectrum of A ∈ L(X), respectively.

Corollary 3.3. Suppose that A, B ∈ L(X), p(A) = q(A) = p(B) = q(B) = 1, ABA = A^{2} and that BAB = B^{2}. Then:

(a) σ(A) = σ(B);

(b) σ_{p}(A) = σ_{p}(B);

(c) σ_{ap}(A) = σ_{ap}(B);

(d) σ_{r}(A) = σ_{r}(B);

(e) σ_{c}(A) = σ_{c}(B).

Proof. Recall that σ_{p}(A), σ_{r}(A) and σ_{c}(A) are pairwise disjoint and that their union is σ(A). Thus (a) follows from (b), (d) and (e).

(b) Since p(A) = p(B) > 0, 0 ∈ σ_{p}(A) and 0 ∈ σ_{p}(B). From [1, Theorem 3] and Theorem 3.2 we get

σ_{p}(A) ∖ {0} = σ_{p}(P Q) ∖ {0} = σ_{p}(QP) ∖ {0} = σ_{p}(B) ∖ {0}

hence σ_{p}(A) = σ_{p}(B).

(c) Because of σ_{p}(A) ⊆ σ_{ap}(A) and σ_{p}(B) ⊆ σ_{ap}(B), it follows that 0 ∈ σ_{ap}(A) and 0 ∈ σ_{ap}(B). As in the proof of (b) we see with Theorem 3 in [1] that σ_{ap}(A) = σ_{ap}(B).

(d) Since 0 ∈ σ_{p}(A) and 0 ∈ σ_{p}(B), 0 ∉ σ_{r}(A) and 0 ∉ σ_{r}(A). Proceed as in the proof of (b), to obtain σ_{r}(A) = σ_{r}(B).

(e) Similar. □

An operator A ∈ L(X) is called a Fredholm operator if dim N(A) < ∞ and codim A(X) < ∞. In this case we set ind(A) = dim N(A) − codim A(X).

By F(X) we denote the ideal of all finite dimensional operators in L(X). Let L̂ denote the quotient algebra L(X)/F(X) and write Â for the coset A + F(X) of A ∈ L(X) in L̂. From [5, Satz 81.1] we have

A is a Fredholm operator ⟺ Â is invertible in L̂.

Corollary 3.4. Let A and B as in Corollary 3.3 and λ ∈ ℂ. Then:

λI − A is a Fredholm operator ⟺ λI − B is a Fredholm operator.

In this case ind(λI − A) = ind(λI − B).

Proof. We first consider the case λ = 0. Let A be a Fredholm operator, thus Â is invertible in L̂. From ÂB̂Â = Â^{2} we obtain B̂ = Î, hence B is a Fredholm operator. Since B̂ÂB̂ = B̂^{2}, it follows that Â = Î. Hence there are F_{1}, F_{2} ∈ F(X) such that A = I + F_{1} and B = I + F_{2}. By [5, Satz 81.3],

ind(A) = ind(I + F_{1}) = ind(I) = 0 = ind(I + F_{2}) = ind(B).

Now assume that λ ≠ 0. Our statements follow directly from [1, Theorem 6] and Theorem 3.2. □

4. Operators on Hilbert spaces

In this section we will give a proof of Theorem 1.1. H denotes a complex Hilbert space. If A ∈ L(H) we write iso σ(A) for the set of all isolated points of σ(A).

Proposition 4.1. Let A ∈ L(H) be normal and 0 ∈ iso σ(A).

(a) 0 is simple pole of the resolvent (λI − A)^{−1}.

(b) p(A) = q(A) = 1.

(c) A is Drazin invertible and i(A) = 1.

Proof. (a) follows from [5, Satz 112.2], (b) and (c) follow from Proposition 3.1. □

Theorem 4.2. Let A, B ∈ L(H) be selfadjoint, ABA = A^{2} and BAB = B^{2}.

(a) 0 ∈ ρ(A) or 0 is a simple pole of (λI − A)^{−1}.

(b) σ(A) ⊆ {0} ∪ [1, ∞) (hence A ⩾ 0).

(c) A is Drazin invertible and i(A) ⩽ 1.

(d) If C is the Drazin inverse of A, then C = C^{∗} and 0 ⩽ C ⩽ I

(e) If A ≠ 0, then ∥A∥ ⩾ 1.

(f) If ∥A∥ = 1, then A^{2} = A = B.

Proof. (a) and (b): From

A(B − I)^{2}A = A(B^{2} − 2B + I)A = AB^{2}A − 2ABA + A^{2}
= ABABA − 2A^{2} + A^{2} = A^{2}BA − A^{2}
= A^{3} − A^{2}

it follows that A^{3} −A^{2} = A(B −I) (A(B −I))^{∗} ⩾ 0, therefore σ(A^{3} −A^{2}) ⊆ [0, ∞). Now take λ ∈ σ(A)∖{0}. The spectral mapping theorem gives λ^{2}(λ−1)=λ^{3}−λ^{2} ⩾0, thus λ ⩾ 1. This shows (b) and 0 ∈ iso σ(A) or 0 ∈ ρ(A). Now use Proposition 4.1 to derive (a).

(c) follows from (a), (b) and Proposition 4.1.

(d) Because of ACA = A, CAC = C and AC = CA it follows that AC^{∗}A = A, C^{∗}AC^{∗} = C^{∗} and AC^{∗} = C^{∗}A, hence C^{∗} is a Drazin inverse of A. By Proposition 2.1, C = C^{∗}. If 0 ∈ ρ(A), then A = I, thus C = I, hence ∥C∥ = 1. Now let 0 ∈ σ(A). In [2, page 53] it is shown that r(C)^{−1} = dist(0, σ(A) ∖ {0}).

Now we see from (b) that r(C)^{−1} ⩾ 1, hence, since C = C^{∗}, ∥C∥ = r(C) ⩽ 1. We denote the inner product on H by (·|·). Take x ∈ H and let y = Cx. Then

(Cx|x) = (CACx|x) = (ACx|Cx) = (Ay|y) ⩾ 0,

since A ⩾ 0. Thus C ⩾ 0. From ∥C∥ ⩽ 1 we obtain 0 ⩽ C ⩽ I.

(e) If A ≠ 0, we have ∥A∥ = r(A) ⩾ 1, by (b).

(f) If ∥A∥ = 1, then r(A) = 1, thus we obtain from (b) that σ(A) ⊆ {0, 1}. By the spectral mapping theorem, σ(A^{2} − A) = {0}, hence ∥A^{2} − A∥ = r(A^{2} − A) = 0, this gives A^{2} = A. Since σ(A) = σ(B) (Corollary 3.3), we see that ∥B∥ = r(B) = r(A) = ∥A∥ = 1. Hence, by the same arguments as above, B^{2} = B. It follows that ABA = A and BAB = B, hence (AB)^{2} = AB, thus AB is a projection ≠ 0, therefore ∥AB∥ ⩾ 1. But ∥AB∥ ⩽ ∥A∥, ∥B∥ ⩽ 1. Consequently ∥AB∥ = 1. From [8, Satz V.5.9] we derive that AB = (AB)^{∗}. Hence AB = BA. We conclude that A = ABA = BA^{2} = BA = B^{2}A = BAB = B. □

Proof of Theorem 1.1. Theorem 2.2 (a) shows that (a) implies (b). Now suppose that (b) is valid. If 0 ∈ ρ(A), then A = B = I and we are done. Therefore we can assume that 0 ∈ σ(A) and 0 ∈ σ(B). By Theorem 4.2, A and B are Drazin invertible and i(A) = i(B) = 1. Let P and Q as in the proof of Theorem 2.2(b). Hence P = CAB, Q = BAC, P Q = A, QP = B and C is the Drazin inverse of A. From Theorem 4.2 we get C = C^{∗}, thus P^{∗} = BAC = Q.

It remains to show that P is uniquely determined. Suppose that R^{2} = R, P P^{∗} = RR^{∗} and P^{∗}P = R^{∗}R. Then P^{∗}P (I −R) = R^{∗}R(I −R) = R^{∗}R−R^{∗}R = 0, thus P (I − R)(X) ⊆ N(P^{∗}) = P (X)^{⊥}, hence P (I − R) = 0. Therefore we have P R = P. A similar argument gives R^{∗}P^{∗} = R^{∗}. Taking adjoints we obtain R = P R = P. □

5. Examples and remarks

In this section we give some examples of operators A which are Drazin invertible with i(A) = 1. X always denotes a complex Banach space.

An operator A ∈ L(X) is called hermitian if ∥ exp(itA)∥ = 1 for all t ∈ ℝ.

Example 5.1. If A ∈ L(X) is hermitian and if 0 ∈ iso σ(A), then A is Drazin invertible and i(A) = 1.

Proof. Let P_{0} be the spectral projection associated with {0}. Let M_{0} = P_{0}(X) and A_{0} = A_{|_{M_{0}}} . By [5, Satz 100.1] we have A(M_{0}) ⊆ M_{0} and σ(A_{0}) = {0}. Since A_{0} is hermitian operator on M_{0} [3, Proposition 4.12], we have ∥A_{0}∥ = r(A_{0}) = 0 [3, Theorem 4.10]. It follows that AP_{0} = 0. Now [5, (101.9)] shows that 0 is a simple pole of (λI − A)^{−1}. Proposition 3.1 completes the proof. □

An operator A ∈ L(X) is said to be paranormal if ∥Ax∥^{2} ⩽ ∥A^{2}x∥ ∥x∥ for all x ∈ X.

Example 5.2. If A ∈ L(X) is paranormal and if 0 ∈ iso σ(A), then A is Drazin invertible and i(A) = 1.

Proof. Let P_{0}, M_{0} and A_{0} as in the proof of 5.1. From [5, page 500] we get ∥A_{0}∥ = r(A_{0}) = 0. Now proceed as in the proof of 5.1. □

A bounded linear operator A on a Hilbert space H is called hyponormal if ∥A^{∗}x∥ ⩽ ∥Ax∥ for all x ∈ H. Since hyponormal operators are paranormal, we have by Example 5.2:

Example 5.3. If A ∈ L(H) is hyponormal and if 0 ∈ iso σ(A), then A is Drazin invertible and i(A) = 1.

Remark 5.4. If A, B ∈ L(X), ABA = A^{2}, BAB = B^{2}, AB = BA, p(A) ⩽ 1 and p(B) ⩽ 1, then A^{2} = A = B.

Proof. From A^{2} = A^{2}B = ABAB = AB^{2} = B^{2} it follows that A^{3} = AB^{2} = A^{2}, thus A^{2}(A − I) = 0. Since p(A) ⩽ 1, we get A(A − I) = 0, hence A^{2} = A. In the same way we derive B^{2} = B. Consequently

B = B^{2} = B(AB) = B(AB^{2}) = BA^{2} = BA = AB = A^{2}B = A^{2} = A.  □

Remark 5.5. Suppose that A, B ∈ L(X) are paranormal, ABA = A^{2}, BAB = B^{2} and AB = BA; then A^{2} = A = B.

Proof. Since ∥Ax∥^{2} ⩽ ∥A^{2}x∥ ∥x∥ for x ∈ X, it follows that p(A) ⩽ 1. Similarly p(B) ⩽ 1. Now use 5.4. □

Remark 5.6. Suppose that H is a complex Hilbert space, A, B ∈ L(H) are normal, ABA = A^{2}, BAB = B^{2} and AB = BA. Then A is selfadjoint and A^{2} = A = B.

Proof. Since normal operators are paranormal, it follows from 5.5 that A and B are normal projections, hence they are selfadjoint. □

Remark 5.7. If A ∈ L(X) is hermitian, then p(A) ⩽ 1.

Proof. Let x ∈ N(A^{2}) and ∥x∥ = 1. Then for t ∈ ℝ,

1 = ∥x∥ = ∥ exp(−itA) exp(itA)x∥ ⩽ ∥ exp(−itA)∥ ∥ exp(itA)x∥
= ∥ exp(itA)x∥ ⩽ ∥ exp(itA)∥ ∥x∥ = ∥x∥ = 1,

thus, since A^{n}x = 0 for n ⩾ 2,

1 = ∥ exp(itA)x∥ = ∥x + itAx∥.

Therefore |t| ∥Ax∥ − 1 ⩽ 1 for all t ∈ ℝ. This gives x ∈ N(A). □

Remark 5.8. Suppose that A, B ∈ L(X) are hermitian, ABA = A^{2}, BAB = B^{2} and that AB = BA; then A^{2} = A = B.

Proof. 5.7 and 5.4. □

References

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[3] H. R. Dowson, Spectral Theory of Linear Operators, Academic Press, 1978.

[4] M. P. Drazin, Pseudoinverse in associative rings and semigroups, Amer. Math. Monthly 65 (1958), 506–514.

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[7] I. Vidav, On idempotent operators in a Hilbert space, Publ. Inst. Math. (Beograd) 4(18) (1964), 157–163.

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Mathematisches Institut I (Received 16 08 2005) Universität Karlsruhe (TH) Englerstraße 2 76128 Karlsruhe Germany

    christoph.schmoeger@math.uni-karlsruhe.de