On weighted reverse order laws for the Moore-Penrose inverse and ...

E. Boasso, D. S. Cvetković-Ilić, R. Harte 8
Clearly b{1, 3}·a{1, 3}·c ⊆ (ab){1, 3} is equivalent to the fact that for any a (1,3) ∈ a{1, 3}
and any b (1,3) ∈ b{1, 3}:
b (1,3) a (1,3) c ∈ (ab){1, 3}. (2)
to
Using the matrix representations considered in Remark 3.1, we have that (2) is equivalent
(i) a 1 z 1 c 1 a 1 = a 1 , (ii) a 1 z 2 c 2 = 0, (iii) (a 1 z 1 c 1 ) ∗ = a 1 z 1 c 1 , (3)
where z 1 and z 2 are defined by (1). Now, using (1), the identity (3)(i) is equivalent to
a 1 a ∗ 1d † c 1 a 1 + (a 1 − a 1 a ∗ 1d † a 1 )x 1 c 1 a 1 − a 1 a ∗ 1d † a 2 x 3 c 1 a 1 = a 1 . (4)
Since, x 1 and x 3 are arbitrary elements from appropriate subalgebras, (3)(i) is equivalent to
(i) a 1 a ∗ 1d † c 1 a 1 = a 1 , (ii) (a 1 − a 1 a ∗ 1d † a 1 )x 1 c 1 a 1 = 0, (iii) a 1 a ∗ 1d † a 2 x 3 c 1 a 1 = 0. (5)
What is more, since a 1 = rap, x 1 = pxr and c 1 = rcr, (5)(ii) is equivalent to
(a 1 − a 1 a ∗ 1d † a 1 )xc 1 a 1 = 0,
where x ∈ R is arbitrary. However, since R is a prime ring, a 1 − a 1 a ∗ 1 d† a 1 = 0 or c 1 a 1 = 0.
Similarly, from (5)(iii), we get that a 1 a ∗ 1 d† a 2 = 0 or c 1 a 1 = 0.
Note that the case c 1 a 1 = 0 imply that ab = 0, which is not possible.
Therefore, c 1 a 1 ≠ 0, and equation (3)(i) is equivalent to
i.e.,
(i) a 1 a ∗ 1d † c 1 a 1 = a 1 , (ii) a 1 − a 1 a ∗ 1d † a 1 = 0, (iii) a 1 a ∗ 1d † a 2 = 0, (6)
(i) b † a † c ∈ ab{1}, (ii) b † a † ∈ ab{1}, (iii) a † ∈ a(e − bb † ){1}. (7)
Now, (6) imply that a 1 z 2 = 0 and the fact that a 1 z 1 c 1 = a 1 a ∗ 1 d† c 1 is hermitian is equivalent
to the fact that abb † a † c is hermitian, i.e. b † a † c ∈ ab{3}.
Remark that from the proof of Theorem 3.2, we have that under the assumption a 2 =
a(e − bb † ) ∈ R † , we get that condition (6)(ii) implies the condition (6)(iii):
a 1 − a 1 a ∗ 1d † a 1 = 0 ⇒ a 1 − dd † a 1 + a 2 a ∗ 2d † a 1 = 0 ⇒ a ∗ 2d † a 1 = 0,
so we get the following corollary.
Corollary 3.3. Let R be a prime ring with involution. Consider a, b ∈ R † such that a(e −
bb † ) ∈ R † and let c ∈ R such that c commutes with a and a ∗ . Then, following statements are
equivalent:
(i) b{1, 3} · a{1, 3} · c ⊆ (ab){1, 3};
(ii) b † a † c ∈ ab{1, 3}, b † a † ∈ ab{1}.