PUBLICATIONS DE L’INSTITUT MATHÉMATIQUE Nouvelle série, tome 79(93) (2006), 109–114 DOI: 10.2298/PIM0693109S

COMMON SPECTRAL PROPERTIES OF LINEAR OPERATORS A AND B SUCH THAT ABA = A^{2} AND BAB = B^{2}

Christoph Schmoeger

Communicated by Stevan Pilipović

Abstract. Let A and B be bounded linear operators on a Banach space such that ABA = A^{2} and BAB = B^{2}. Then A and B have some spectral properties in common. This situation is studied in the present paper.

1. Terminology and motivation

Throughout this paper X denotes a complex Banach space and L(X) the Banach algebra of all bounded linear operators on X. For A ∈ L(X), let N(A) denote the null space of A, and let A(X) denote the range of A. We use

σ(A), σ_{p}(A), σ_{ap}(A), σ_{r}(A), σ_{c}(A) and ρ(A)

to denote spectrum, the point spectrum, the approximate point spectrum, the residual spectrum, the continuous spectrum and the resolvent set of A, respectively. An operator A ∈ L(X) is semi-Fredholm if A(X) is closed and either α(A) := dim N(A) or β(A) := codim A(X) is finite. A ∈ L(X) is Fredolm if A is semi-Fredholm, α(A) < ∞ and β(A) < ∞. The Fredholm spectrum σ_{F} (A) of A is given by

σ_{F} (A) = {λ ∈ ℂ : λI − A is not Fredholm}.

The dual space of X is denoted by X^{∗} and the adjoint of A ∈ L(X) by A^{∗}. The following theorem motivates our investigation:

Theorem 1.1. Let P, Q ∈ L(X) such that P^{2} = P and Q^{2} = Q. If A = P Q and B = QP, then

(1) ABA = A^{2} and BAB = B^{2};

(2) σ(A) ∖ {0} = σ(B) ∖ {0};

(3) σ_{p}(A) ∖ {0} = σ_{p}(B) ∖ {0};

(4) σ_{ap}(A) ∖ {0} = σ_{ap}(B) ∖ {0};

2000 Mathematics Subject Classification: Primary 47A10.

Key words and phrases: operator equation, spectrum.

(5) σ_{r}(A) ∖ {0} = σ_{r}(B) ∖ {0};
(6) σ_{c}(A) ∖ {0} = σ_{c}(B) ∖ {0};
(7) σ_{F} (A) ∖ {0} = σ_{F} (B) ∖ {0}.

Proof. (1) ABA = P QQP P Q = P QP Q = A^{2}, BAB = QP P QQP = QP QP = B^{2}. (2) follows from [2, Proposition 5.3], (3), (4), (5) and (6) are shown in [1, Theorem 3] and (7) follows from [1, Theorem 6]. □

The main result of this paper reads as follows:

Theorem 1.2. Let A, B ∈ L(X) such that ABA = A^{2} and BAB = B^{2}. Then (1) σ_{p}(A) ∖ {0} = σ_{p}(AB) ∖ {0} = σ_{p}(BA) ∖ {0} = σ_{p}(B) ∖ {0};

(2) σ_{ap}(A) ∖ {0} = σ_{ap}(AB) ∖ {0} = σ_{ap}(BA) ∖ {0} = σ_{ap}(B) ∖ {0};

(3) σ_{r}(A) ∖ {0} = σ_{r}(AB) ∖ {0} = σ_{r}(BA) ∖ {0} = σ_{r}(B) ∖ {0};

(4) σ_{c}(A) ∖ {0} = σ_{c}(AB) ∖ {0} = σ_{c}(BA) ∖ {0} = σ_{c}(B) ∖ {0};

(5) σ(A) = σ(B) = σ(AB) = σ(BA);

(6) σ_{F} (A) = σ_{F} (B) = σ_{F} (AB) = σ_{F} (BA).

A proof of Theorem 1.2 will be given in Section 2 of this paper.

For results concerning the operator equations ABA = A^{2} and BAB = B^{2} see [4], [6] and [7].

2. Proofs

Throughout we assume that A, B ∈ L(X) and that ABA = A^{2} and BAB = B^{2}. It it easy to see that if 0 ∈ ρ(A) or 0 ∈ ρ(B), then A = B = I. So we always assume that 0 ∈ σ(A) and 0 ∈ σ(B).

Proposition 2.1. σ_{p}(A)∖{0} = σ_{p}(AB)∖{0} = σ_{p}(BA)∖{0} = σ_{p}(B)∖{0}.

Proof. It suffices to show that σ_{p}(A) ∖ {0} ⊆ σ_{p}(AB) ∖ {0} ⊆ σ_{p}(B) ∖ {0}. To this end let λ ∈ σ_{p}(A) ∖ {0}. Hence there is x ∈ X ∖ {0} such that Ax = λx. Then BAx = λBx and A^{2}x = λ^{2}x, thus λABx = ABAx = A^{2}x = λ^{2}x; this gives

(2.1)  ABx = λx,

hence λ ∈ σ_{p}(AB) ∖ {0} and B(Bx) = B^{2}x = BABx = λBx. Because of (2.1), Bx ≠ 0, therefore 0 ∈ σ_{p}(B) ∖ {0}. □

Corollary 2.2. If λ ≠ 0, then

N(A − λI) = N(AB − λI) = A(N(B − λI)),
N(B − λI) = N(BA − λI) = B(N(A − λI))

and

α(A − λI) = α(AB − λI) = α(BA − λI) = α(B − λI).

Proof. The proof of Proposition 2.1 shows that N(A − λI) ⊆ N(AB − λI). Let x ∈ N(AB −λI), thus ABx = λx, hence λAx = A^{2}Bx = ABABx = (AB)^{2}x = λ^{2}x. This gives x ∈ N(A − λI). Hence we have N(A − λI) = N(AB − λI). Similar arguments show that N(B − λI) = N(BA − λI). From [1, Proposition 2] we see that N(AB − λI) = A(N(BA − λI)), thus N(AB − λI) = A(N(B − λI)).

It is easy to see that N(A) ∩ N(B − λI) = {0}, hence the restriction of A to N(B − λI) is injective, thus

α(A − λI) = α(AB − λI) = α(A(N(B − λI))) = α(B − λI).  □

Proposition 2.3. We have

σ_{ap}(A) ∖ {0} = σ_{ap}(AB) ∖ {0} = σ_{ap}(BA) ∖ {0} = σ_{ap}(B) ∖ {0}.

Proof. It suffices to show that σ_{ap}(A)∖{0} ⊆ σ_{ap}(AB)∖{0} ⊆ σ_{ap}(B)∖{0}. To this end let λ ∈ σ_{ap}(A)∖{0}. Then there is a sequence (x_{n}) in X with ∥x_{n}∥ = 1 for all n ∈ ℕ and (λI − A)x_{n} → 0 (n → ∞). Let z_{n} = (λI − A)x_{n}; hence Ax_{n} = λx_{n} − z_{n} and z_{n} → 0. Then

A^{2}x_{n} = λAx_{n} − Az_{n} = λ(λx_{n} − z_{n}) − Az_{n} = λ^{2}x_{n} − λz_{n} − Az_{n}

and

BAx_{n} = λBx_{n} − Bz_{n},

thus

A^{2}x_{n} = ABAx_{n} = λABx_{n} − ABz_{n},

therefore

(2.2)  λ^{2}x_{n} − λABx_{n} = (λI + A − AB)z_{n},

this gives (AB − λI)x_{n} → 0, hence λ ∈ σ_{ap}(AB) ∖ {0}. From(2.2) we get

λ^{2}Bx_{n} − λB^{2}x_{n} = w_{n},

where w_{n} = (λB + AB − B^{2})z_{n} → 0 (n → ∞). Hence

(2.3)  (λI − B)Bx_{n} = λ^{−1}w_{n}.

Because of (2.2) there is m ∈ ℕ such that

Bx_{n} ≠ 0 for n ⩾ m and (∥Bx_{n}∥^{−1})_{n⩾m} is bounded.

For n ⩾ m let y_{n} = ∥Bx_{n}∥^{−1}Bx_{n}. Then ∥y_{n}∥ = 1 and, by (2.3)

(λI − B)y_{n} = (λ∥Bx_{n}∥)^{−1}w_{n} (n ⩾ m).

Therefore (λI − B)y_{n} → 0 (n → ∞), and so λ ∈ σ_{ap}(B) ∖ {0}. □

Remark. The proof of Proposition 2.3 also follows from Proposition 2.1 if we apply Berberian–Quigley functor (see e.g. [5, Theorem 1-5.11]).

Proposition 2.4. σ_{r}(A) ∖ {0} = σ_{r}(AB) ∖ {0}.

Proof. Let λ ∈ σ_{r}(A) ∖ \{0\}. Hence λ ∉ σ_{p}(A) and \overline{(λI − A)(X)} ≠ X. Thus N(λI^{∗} − A^{∗}) ≠ {0}. By Proposition 2.1, N(λI^{∗} − (AB)^{∗}) = N(λI^{∗} − B^{∗}A^{∗}) ≠ \{0\}, hence \overline{(λI − AB)(X)} ≠ X. Since λ ∉ σ_{p}(AB)(Proposition 2.1), we have λ ∈ σ_{r}(AB) ∖ {0}.

Now let λ ∈ σ_{r}(AB) ∖ \{0\}, hence λ ∉ σ_{p}(AB) and \overline{(λI − AB)(X)} ≠ X. It follows thatN(λI^{∗} − (AB)^{∗}) = N(λI^{∗} − B^{∗}A^{∗}) ≠ {0}. From Proposition 2.1 we get N(λI^{∗} −A^{∗}) ≠ \{0\}, thus \overline{(λI − A)(X)} ≠ X. Since λ ∉ σ_{p}(A) (Proposition 2.1), λ ∈ σ_{r}(A). □

Corollary 2.5. σ_{r}(A) ∖ {0} = σ_{r}(B) ∖ {0}.

Proof. By [1, Theorem 3], σ_{r}(AB) ∖ {0} = σ_{r}(BA) ∖ {0}. Now use Proposition 2.4. □

Let T ∈ L(X). The number

γ(T) = inf\{\frac{∥T X∥}{d (x, N(T))} : x ∈ X, x ∉ N(T)\}

is called the minimal modulus of T; d (x, T) denotes the distance of x from N(T). It is well known that T (X) is closed if and only if γ(T) > 0 (see [3, Satz 55.2]).

Proposition 2.6. σ(A) = σ(AB).

Proof. Let λ ∈ σ(A) ∖ {0} and assume to the contrary that λ ∈ ρ(AB). Then α(λI − AB) = 0 and λ ∉ σ_{ap}(AB). By Proposition 2.1 and Proposition 2.3, α(λI − A) = 0 and λ ∉ σ_{ap}(A). Therefore

γ(λI − A) = inf{∥(λI − A)x∥ : x ∈ X, ∥x∥ = 1} > 0

hence (λI − A)(X) is closed. Thus we have shown that λI − A is semi-Fredholm. Since λ ∈ ρ((AB)^{∗}) = ρ(B^{∗}A^{∗}), it follows from [2, Proposition 5.3] that λ ∈ ρ(A^{∗}B^{∗}). Since A^{∗}B^{∗}A^{∗} = (A^{∗})^{2} and B^{∗}A^{∗}B^{∗} = (B^{∗})^{2}, the same arguments as above show that α(λI^{∗} − A^{∗}) = 0 and that λI^{∗} − A^{∗} is semi-Fredholm. By [3, Satz 82.1] it follows now that β(λI − A) = α(λI^{∗} − A^{∗}) = 0, thus 0 ∈ ρ(A), a contradiction. Hence σ(A) ∖ {0} ⊆ σ(AB) ∖ {0}.

Now let λ ∈ σ(AB) ∖ {0} and assume that λ ∈ ρ(A). Then α(λI − A) = 0 and λ ∉ σ_{ap}(A). Proposition 2.1 and Proposition 2.3 show that α(λI − AB) = 0 and λ ∉ σ_{ap}(AB). As in the first part of the proof we conclude that γ(λI − AB) > 0. Thus λI − AB is semi-Fredholm. Since λ ∈ ρ(A^{∗}), the same arguments as above give α(λI^{∗} − (AB)^{∗}) = 0 and λI^{∗} − (AB)^{∗} is semi-Fredholm. From β(λI − AB) = α(λI^{∗} − (AB)^{∗}) = 0 we get the contradiction λ ∈ ρ(AB).

So far we have σ(A) ∖ {0} = σ(AB) ∖ {0}. It remains to show that 0 ∈ σ(AB). Assume to the contrary that there is C ∈ L(X) with ABC = I = CAB. Then N(B) = {0} and B^{2}C = B, therefore BC = I, hence A = I, a contradiction. □

Corollary 2.7. σ(A) = σ(B).

Proof. Since σ(AB) ∖ {0} = σ(BA) ∖ {0} [2, Proposition 5.3], Proposition 2.6 shows that σ(A) ∖ {0} = σ(B) ∖ {0}. Because of 0 ∈ σ(A) ∩ σ(B), we have σ(A) = σ(B). □

Proposition 2.8. σ_{c}(A) ∖ {0} = σ_{c}(AB) ∖ {0}.

Proof. By Proposition 2.1, Proposition 2.4 and Proposition 2.6

σ_{c}(A) ∖ {0} = σ(A) ∖ [σ_{p}(A) ∪ σ_{r}(A) ∪ {0}]
= σ(AB) ∖ [σ_{p}(AB) ∪ σ_{r}(AB) ∪ {0}] = σ_{c}(AB) ∖ {0}.  □

Corollary 2.9. σ_{c}(A) ∖ {0} = σ_{c}(B) ∖ {0}.

Proof. Use Proposition 2.8 and [1, Theorem 3]. □

In what follows A denotes a complex unital Banach algebra. For a ∈ A we write σ(a) for the spectrum of a and λ_{a} for the bounded linear operator on A given by λ_{a}(x) = ax (x ∈ A).

Proposition 2.10. Let a, b ∈ A.

(1) σ(a) = σ(λ_{a});

(2) λ_{ab} = λ_{a}λ_{b};

(3) if aba = a^{2} and bab = b^{2}, then λ_{a}λ_{b}λ_{a} = λ_{a}^{2}, λ_{b}λ_{a}λ_{b} = λ_{b}^{2}, and

σ(a) = σ(b) = σ(ab) = σ(ba).

Proof. (1) [2, Proposition 3.19].

(2) Clear.

(3) follows from (1), (2), Proposition 2.6 and Corollary 2.7. □

Let K(X) denote the ideal of all compact operators in L(X) and let L̂ denote the quotient algebra L(X)/¸(X). By T̂ we denote the coset T + K(X) ∈ L̂ (T ∈ L(X)). Observe that L̂ is a Banach algebra with unit Î. Satz 81.2 in [3] shows that for T ∈ L(X) we have

T is Fredholm ⟺ 0 ∉ σ(T̂).

Hence

σ_{F} (T) = σ(T̂). (2.4)

Since ÂB̂Â = Â^{2} and B̂ÂB̂ = B̂^{2}, an immediate consequence of Proposition 2.10 and (2.4) is

Corollary 2.11. σ_{F} (A) = σ_{F} (B) = σ_{F} (AB) = σ_{F} (BA).

The proof of Theorem 1.2 is now complete. □

If T ∈ L(X) is Fredholm then the index ind(T) of T is defined by ind(T) = α(T) − β(T).

Corollary 2.12. Let λ ∉ σ_{F} (A).

(1) If λ ≠ 0, then ind(λI − A) = ind(λI − AB) = ind(λI − BA) = ind(λI − B).

(2) If λ = 0, then ind(A) = ind(B) = ind(AB) = ind(BA) = 0.

Proof. (1) Because of [1, Theorem 6] it suffices to show that ind(λI − A) = ind(λI − BA). By Corollary 2.2 and [3, Satz 82.1] we have

ind(λI − A) = α(λI − A) − α(λI^{∗} − A^{∗})
= α(λI − AB) − α(λI^{∗} − A^{∗}B^{∗})
= α(λI − AB) − α(λI^{∗} − (BA)^{∗})
= α(λI − AB) − β(λI − BA)
= α(λI − BA) − β(λI − BA) = ind I − BA).

(2) We have Â = B̂ = ÂB = B̂A = Î, thus, by [3, Satz 82.5] we get

ind(A) = ind(B) = ind(AB) = ind(BA) = ind(I) = 0.  □

An operator T ∈ L(X) is called a Riesz operator if σ_{F} (T) = {0}. From Corollary 2.11 we have:

Corollary 2.13. The following assertions are equivalent:

(1) A is a Riesz operator;

(2) B is a Riesz operator;

(3) AB is a Riesz operator;

(4) BA is a Riesz operator.

References

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[2] F. F. Bonsall, J. Duncan, Complete Normed Algebras, Springer-Verlag, 1973.

[3] H. Heuser, Funktionalanalysis, 2nd ed., Teubner, 1986.

[4] V. Rakočević, A note on a theorem of I. Vidav, Publ. Inst. Math. (Beograd) 68(82) (2000), 105–107.

[5] C. E. Rickart, General Theory of Banach Algebras, Van Nostrand, 1960.

[6] Ch. Schmoeger, On the operator equations ABA = A^{2} and BAB = B^{2}, Publ. Inst. Math. (Beograd) 78(92) (2005), 127–133.

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

Institut für Analysis (Received 15 03 2006) Universität Karlsruhe (TH) (Revised 06 09 2006) Englerstraße 2 76128 Karlsruhe Germany

    christoph.schmoeger@math.uni-karlsruhe.de