# On the Covariance of Moore-Penrose Inverses in Rings with Involution

Authors: Hesam Mahzoon
Journal: Abstract and Applied Analysis
 (2014)

---

## Abstract

We consider the so-called covariance set of Moore-Penrose inverses in rings with an involution. We deduce some new results concerning covariance set. We will show that ifa is a regular element in a C
∗-algebra, then the covariance set of a is closed in the set of invertible elements (with relative topology) of C
∗-algebra and is a cone in the C
∗-algebra.


---

## Body

## 1. Introduction

Suppose thatℜ is a ring with unity 1
≠
0. A mapping *
:
x
↦
x
* of ℜ into itself is called aninvolution if
(1)
(
x
*
)
*
=
x
,
(
x
+
y
)
*
=
x
*
+
y
*
,
(
x
y
)
*
=
y
*
x
*
,
for allxandyin ℜ. A ring ℜ with an involution * is called *-ring. Throughout this paper ℜ is a *-ring.An elementa
∈
ℜ is calledregular if it has a generalized inverse (in the sense of von Neumann) in ℜ; that is, there exists b
∈
ℜ such that
(2)
a
b
a
=
a
.
Note that such b is not unique [1, 2].Definition 1.
Letℜ be a *-ring and a
∈
ℜ.(i)
ais called Moore-Penrose invertible if there exists b
∈
ℜ such that
(3)
a
b
a
=
a
,
b
a
b
=
b
,
(
a
b
)
*
=
a
b
,
(
b
a
)
*
=
b
a
.
(ii)
a is called Drazin invertible if there exists b
∈
ℜ such that
(4)
b
a
b
=
b
,
a
b
=
b
a
,
a
k
+
1
b
=
a
kfor some nonnegative integer k. The least such k is the Drazin index of a, denoted by ind
(
a
).Obviously,ind
(
a
)
=
0 if and only ifais invertible and in this case the Drazin inverses of a and a
-
1 coincide. If ind
(
a
)
≤
1, then the Drazin inverse is known as thegroup inverse.It is well known that the Moore-Penrose inverse (briefly, MP-inverse) and the Drazin inverse are unique if they exist. We reserve the notationsa
† and a
D for the MP-inverse and Drazin inverse of a, respectively. According to the uniqueness of the notion under consideration, if a has a MP-inverse, then a
* and a
† also have MP-inverses. Moreover
(5)
(
a
†
)
†
=
a
,
(
a
†
)
*
=
(
a
*
)
†
,
a
*
=
a
†
a
a
*
=
a
*
a
a
†
.In what follows, we will denote byℜ
-
1 the subset of invertible elements of  ℜ and by ℜ
† the set of all MP-invertible elements of ℜ. An element x in ℜ is calledidempotent if x
2
=
x. Aprojection
p
∈
ℜ satisfies p
=
p
*
=
p
2. Note that if x
∈
ℜ
†, then x
x
† and x
†
x are projections. In addition,
(6)
(
x
x
†
)
†
=
x
x
†
,
(
x
†
x
)
†
=
x
†
x
.
Thecommutator of a pair of elements x and y in ℜ is given by
(7)
[
x
,
y
]
=
x
y
-
y
x
.
Note that [
x
,
y
]
=
0 if and only if x andycommute. Also, it is well known that if x
,
y, and z are in ℜ, then
(8)
[
x
,
y
z
]
=
[
x
,
y
]
z
+
y
[
x
,
z
]
,
[
x
y
,
z
]
=
x
[
y
,
z
]
+
[
x
,
z
]
y
.Letabe an element in ℜ
-
1; its inverse a
-
1 iscovariantwith respect to ℜ
-
1; that is, for all b
∈
ℜ
-
1, we have
(9)
(
b
a
b
-
1
)
-
1
=
b
a
-
1
b
-
1
.In general, the elements ofℜ
† are not covariant under ℜ
-
1 (see [2–4]). For a given element a
∈
ℜ
† with MP-inverse a
† we define itscovariance set
(10)
ℭ
(
a
)
=
{
b
∈
ℜ
-
1
:
(
b
a
b
-
1
)
†
=
b
a
†
b
-
1
}
.Schwerdtfeger [4] described the class ℭ
(
a
) for the matrices of rank 1 or 2. The characterization of the covariance set ℭ
(
a
) for an algebra of matrices was studied by Robinson [2] and some interesting results of ℭ
(
a
) were presented by Meenakshi and Chinnadurai [3].The paper is organized as follows. The endeavour in Section2 is to show how the results of [3] can be extended to MP-inverses in *-rings. Moreover, we show that Drazin inverses are covariant under the group of invertible elements of *-rings. In Section 3 we prove that the covariance set is aclosed set in 𝔄
-
1 and is aconein 𝔄. Furthermore, we show that if {
a
n
} is a sequence of MP-invertible elements of a C
*-algebra such that their MP-inverses norm is bounded and a
n converges to a, then there is some kind of convergence of ℭ
(
a
n
) to ℭ
(
a
).


## 2. Covariance Set of Moore-Penrose Inverses in*-Rings

Many of the results of this section are essentially due to [3], withthe main difference being that in [3] one considers covariance set for matrices. In this section we generalized these results to any *-ring.The next proposition describes a relation between the covariance setℭ
(
a
) and commutators. It was also shown in [2–4] in the special case of matrices. Here, we include a shorter proof for the sake of completeness.Proposition 2.
Letℜ be *-ring and a
∈
ℜ
† with MP-inverse a
†. Then the following statements are equivalent:(i)
b
∈
ℭ
(
a
)
;
(ii)
[
b
*
b
,
a
a
†
]
=
0
and
[
b
*
b
,
a
†
a
]
=
0
.Proof.
(i)⇒(ii) Suppose that b
∈
ℭ
(
a
). Then (
b
a
b
-
1
)
†
=
b
a
†
b
-
1. Set p
=
(
b
a
b
-
1
)
(
b
a
b
-
1
)
†. Then p is projection, so p
=
p
* and p
=
b
a
a
†
b
-
1. From here we get b
a
a
†
b
-
1
=
(
b
-
1
)
*
a
a
†
b
*. This implies that [
b
*
b
,
a
a
†
]
=
0. Similarly by putting q
=
(
b
a
b
-
1
)
†
(
b
a
b
-
1
), we conclude that [
b
*
b
,
a
†
a
]
=
0.
(ii)⇒(i) From the assumptions it is not hard to see that b
a
†
b
-
1 is the MP-inverse of b
a
b
-
1. By the uniqueness of Moore-Penrose inverse we get (
b
a
b
-
1
)
†
=
b
a
†
b
-
1; that is, b
∈
ℭ
(
a
).From Proposition2 we deduce the following result.Corollary 3.
Letℜ be *-ring and a
∈
ℜ
† with MP-inverse a
†. Then
(11)
b
-
1
∈
ℭ
(
a
)
i
f
f
[
b
b
*
,
a
a
†
]
=
0
,
b
-
1
∈
ℭ
(
a
)
i
f
f
[
b
b
*
,
a
†
a
]
=
0
.Combining the above corollary and Proposition2, we get the following corollary.Corollary 4.
Ifb is normal, then
(12)
b
∈
ℭ
(
a
)
i
f
f
b
-
1
∈
ℭ
(
a
)
.We now have some equalities for the covariance sets. See also [3].Proposition 5.
Letℜ be *-ring and a
∈
ℜ
† with MP-inverse a
†. Then
(13)
ℭ
(
a
)
=
ℭ
(
a
†
)
=
ℭ
(
a
*
)
=
ℭ
(
a
a
†
)
∩
ℭ
(
a
†
a
)
.Proof.
By replacinga with a
†, part (ii) of Proposition 2 does not change so the first equality holds. Since (
a
*
)
†
a
*
=
a
a
† and a
*
(
a
*
)
†
=
a
†
a, Proposition 2 yields the second equality. Also a
=
a
a
†
a and a
†
a
a
†
=
a
†, again from Proposition 2 we get the last equality.Note that ifu is any unitary element in ℜ
-
1, the u
*
u
=
u
u
*
=
1; thus u
∈
ℭ
(
a
) for every a
∈
ℜ
†
.This implies that ℭ
(
a
)
≠
∅ for each a
∈
ℜ
†.In the next proposition, we will show that ifa
∈
ℜ is Drazin invertible with Drazin inverse a
D, then {
b
∈
ℜ
-
1
:
(
b
a
b
)
D
=
b
a
D
b
-
1
}
=
ℜ
-
1. For this reason, the notion of covariance sets is not studied to Drazin inverses.Proposition 6.
Suppose thatℜ is a *-ring and a is a Drazin invertible element in ℜ. Then a
D is covariant under ℜ
-
1; that is,
(14)
(
b
a
b
-
1
)
D
=
b
a
D
b
-
1
,
∀
b
i
n
ℜ
-
1
.Proof.
Suppose thata
D is the Drazin inverse of a and b is an arbitrary element in ℜ
-
1. For simplicity of calculations, set X
=
b
a
b
-
1 and Y
=
b
a
D
b
-
1
.By hypothesis, a
D
a
a
D
=
a
D, a
D
a
=
a
a
D, and a
k
+
1
a
D
=
a
k; thus
(15)
Y
X
Y
=
(
b
a
D
b
-
1
)
(
b
a
b
-
1
)
(
b
a
D
b
-
1
)
=
b
a
D
a
a
D
b
-
1
=
b
a
D
b
-
1
=
Y
;
Y
X
=
(
b
a
D
b
-
1
)
(
b
a
b
-
1
)
=
b
a
D
a
b
-
1
=
b
a
a
D
b
-
1
=
X
Y
;
X
k
+
1
Y
=
b
a
k
+
1
a
D
b
-
1
=
b
a
k
b
-
1
=
(
b
a
b
-
1
)
k
=
X
k
.
Now the uniqueness of the Drazin inverse implies that Y
=
X
D; that is, a
D is covariant under ℜ
-
1.In particular, by applying the above proposition, ifa is group invertible with the group inverse a
♯
∈
ℜ, then a
♯ is also covariant under ℜ
-
1.We reproduce the following definition from [5].Definition 7.
Letℜ be a ring; a
∈
ℜ is called simply polar if it has a commuting generalized inverse (in the sense of von Neumann); that is, if b is any generalized inverse of a, then [
a
,
b
]
=
0.Some authors used the expression EP instead of simply polar. Indeed, they calleda
∈
ℜ
† with MP-inverse a
† is EP if and only if a
a
†
=
a
†
a.The next remark provides a large class of simply polar elements and some related properties.Remark 8.
Leta
∈
ℜ
† with MP-inverse a
†.(i)
Ifa is self-adjoint, then it is simply polar, since
(16)
a
a
†
=
(
a
a
†
)
*
=
(
a
†
)
*
a
*
=
a
†
a
.
(ii)
Ifa is normal, then it is simply polar, since
(17)
a
=
a
(
a
†
a
)
*
=
a
a
*
(
a
†
)
*
=
a
*
a
(
a
†
)
*
=
(
a
†
a
)
*
a
*
a
(
a
†
)
*
=
(
a
†
a
)
(
a
*
a
)
*
(
a
†
)
*
=
(
a
†
a
)
(
a
†
a
a
*
)
*
=
(
a
†
a
)
(
a
*
)
*
=
a
†
a
2
;
thus a
=
a
†
a
2. In a similar manner we get a
=
a
2
a
†. Therefore
(18)
a
a
†
=
a
†
a
2
a
†
=
a
†
a
.
(iii)
It is easy to check that simply polar properties ofa,  a
* and a
† are equivalent; that is, if one of them is simply polar, then two others are also simply polar.
(iv)
Ifa is simply polar, then
(19)
(
a
a
†
)
2
=
a
2
(
a
†
)
2
=
(
a
†
)
2
a
2
.
(v)
Ifa is simply polar, then Proposition 5 implies that ℭ
(
a
)
=
ℭ
(
a
a
†
).
For finding more equivalent statements about the simply polar elements see [1, Theorem 2.3 and final remark].Proposition 9.
Leta
,
b
∈
ℜ
† with MP-inverses a
† and b
†, respectively. If a
†
b
=
0
=
a
b
† and b
a
†
=
0
=
b
†
a, then ℭ
(
a
)
∩
ℭ
(
b
)
⊂
ℭ
(
a
+
b
).Proof.
The assumptions, after some easy calculations, imply thata
†
+
b
† is the MP-inverse of a
+
b. Thus (
a
+
b
)
†
=
a
†
+
b
†. Suppose that x
∈
ℭ
(
a
)
∩
ℭ
(
b
). Then Proposition 2 implies that
(20)
[
x
*
x
,
a
a
†
]
=
0
,
[
x
*
x
,
a
†
a
]
=
0
,
[
x
*
x
,
b
b
†
]
=
0
,
[
x
*
x
,
b
†
b
]
=
0
.
Since a
†
b
=
0
=
a
b
† and b
a
†
=
0
=
b
†
a, we have (
a
+
b
)
(
a
†
+
b
†
)
=
a
a
†
+
b
b
† and (
a
†
+
b
†
)
(
a
+
b
)
=
a
†
a
+
b
†
b. From the linearity of commutator we obtain
(21)
[
x
*
x
,
(
a
+
b
)
(
a
†
+
b
†
)
]
=
0
,
[
x
*
x
,
(
a
†
+
b
†
)
(
a
+
b
)
]
=
0
.
Again by applying Proposition 2, we get x
∈
ℭ
(
a
+
b
).Corollary 10.
Leta
,
b
∈
ℜ
† with MP-inverses a
† and b
†, respectively. If a and b are self adjoint and b
a
†
=
0
=
b
†
a, then ℭ
(
a
)
∩
ℭ
(
b
)
⊂
ℭ
(
a
+
b
).Proof.
By assumptiona and b are self adjoint. Thus b
a
†
=
0
=
b
†
a implies that a
†
b
=
0
=
a
b
†. The result now follows from Proposition 9.The next example shows that in Proposition9 inclusion can be proper.Example 11.
Seta
=
[
0
0
1
0
] and b
=
[
0
1
0
0
]. Then a
†
=
[
0
1
0
0
]
=
b, b
†
=
a,  and  a
†
b
=
0
=
a
b
†, and a
+
b
=
[
0
1
1
0
] is invertible; thus ℭ
(
a
+
b
)
=
ℜ
-
1. Now if we set y
=
[
1
1
0
1
] thenyis invertible:
(22)
y
*
=
[
1
0
1
1
]
,
y
y
*
=
[
2
1
1
1
]
.
On the other hand a
a
†
=
[
0
0
0
1
]; therefore
(23)
a
a
†
y
y
*
=
[
0
0
1
1
]
,
but
y
y
*
a
a
†
=
[
0
1
0
1
]
.
From here we conclude that [
a
a
†
,
y
y
*
]
≠
0. Thus y
∉
ℭ
(
a
).LetX and Y be two subsets of ℜ. We recall that
(24)
X
+
Y
=
{
x
+
y
:
x
∈
X
,
y
∈
Y
}
,
X
Y
=
{
x
y
:
x
∈
X
,
y
∈
Y
}
.Note that the reverse order rule for the MP-inverse, that is,(
a
b
)
†
=
b
†
a
†, is valid under certain conditions on MP-invertible elements; see [6].Remark 12.
Leta
,
b
∈
ℜ
† with MP-inverses a
† and b
†, respectively. One can easily check the following.(i)
Ifa
†
b
=
0
=
a
b
† and b
a
†
=
0
=
b
†
a, then ℭ
(
a
)
∩
ℭ
(
b
)
∩
(
ℭ
(
a
)
+
ℭ
(
b
)
)
=
ℭ
(
a
)
∩
ℭ
(
b
)
∩
ℭ
(
a
+
b
).
(ii)
If(
a
b
)
†
=
b
†
a
†, then ℭ
(
a
)
∩
ℭ
(
b
)
∩
(
ℭ
(
b
)
ℭ
(
a
)
)
=
ℭ
(
a
)
∩
ℭ
(
b
)
∩
ℭ
(
a
b
).
(iii)
Generally, there is no subset relation betweenℭ
(
a
+
b
) and ℭ
(
a
)
+
ℭ
(
b
). For instance, if we put b
=
-
a, then 0
∈
ℭ
(
a
)
+
ℭ
(
-
a
) which is not a subset of ℜ
-
1 but ℭ
(
a
+
b
)
=
ℭ
(
0
)
=
ℜ
-
1.
(iv)
Generally, there is no subset relation betweenℭ
(
a
b
) and ℭ
(
a
)
ℭ
(
b
). Set a
=
[
0
0
1
0
] as Example 11. Then a
2
=
0, and so ℭ
(
a
2
)
=
ℜ
-
1
≠
ℭ
(
a
)
ℭ
(
a
).Proposition 13.
Leta
,
b
∈
ℜ
† with MP-inverses a
† and b
†, respectively. If a
ℜ
=
b
ℜ, then a
a
†
=
b
b
†, where a
ℜ
=
{
a
x
:
x
∈
ℜ
}.Proof.
By assumptiona
ℜ
=
b
ℜ, so there exists x in ℜ such that a
=
b
x
=
b
b
†
b
x. Therefore a
=
b
b
†
a, and so a
a
†
=
b
b
†
a
a
†. In a similar manner we get b
b
†
=
a
a
†
b
b
†. Since a
a
† is projection, a
a
†
=
b
b
†.Corollary 14.
Leta
,
b
∈
ℜ
† with MP-inverses a
† and b
†, respectively. If a
ℜ
=
b
ℜ and a
†
ℜ
=
b
†
ℜ, then ℭ
(
a
)
=
ℭ
(
b
).Proof.
The proof is an immediate consequence of Propositions5 and 13.The following corollary was also proved for matrices in [3].Corollary 15.
Leta
,
b
∈
ℜ
† be simply polar and a
ℜ
=
b
ℜ. Then ℭ
(
a
)
=
ℭ
(
b
).According to the above corollary and Remark8, we have the following.Corollary 16.
Ifa
∈
ℜ
† and a is simply polar, then ℭ
(
a
)
=
ℭ
(
a
2
)
=
ℭ
(
a
4
)
=
⋯
=
ℭ
(
a
2
n
) for each n
∈
ℕ.Corollary 17.
Ifa
∈
ℜ
† and a is normal, then ℭ
(
a
)
=
ℭ
(
a
2
) = ℭ
(
a
4
)
=
⋯
=
ℭ
(
a
2
n
) for each n
∈
ℕ.Note that Example11 shows that the converses of the two last corollaries do not hold. Indeed, if we set a
=
[
0
0
1
0
], then a is neither simply polar nor normal and y
∉
ℭ
(
a
) but y
∈
ℭ
(
a
2
)
=
ℭ
(
0
)
=
ℜ
-
1.We know that if eithera
=
0 or a
∈
ℜ
-
1, then ℭ
(
a
)
=
ℜ
-
1. One can easily check that if ℜ is a *-ring with no nonzero nilpotent element, then ℭ
(
p
)
=
ℜ
-
1 where p
∈
ℜ
† and it is an idempotent element of ring. In all cases, we consider that ℭ
(
a
) has a group structure. But in general ℭ
(
a
) is not a group; see for instance [3]. Our purpose is to find a subset of ℭ
(
a
) which has mathematical (group) structure. For this purpose, let a be an element in ℜ
†, with MP-inverse a
†. We define H
(
a
) (as it is defined in [3] for matrices) by
(25)
H
(
a
)
=
{
x
∈
ℜ
-
1
:
[
x
,
a
a
†
]
=
0
,
[
x
,
a
†
a
]
=
0
}
.In the next proposition we collect some interesting properties ofH
(
a
).Proposition 18.
Leta be an element in ℜ
† with MP-inverse a
†. Then(i)
if
b
∈
H
(
a
)
, then
b
*
∈
H
(
a
)
;
(ii)
H
(
a
)
⊂
ℭ
(
a
)
;
(iii)
H
(
a
)
is a group;
(iv)
a
†
is covariant under
H
(
a
)
;
(v)
if
b
,
c
∈
H
(
a
)
such that
b
+
c
∈
ℜ
-
1
, then
b
+
c
∈
H
(
a
)
;
(vi)
if
b
∈
H
(
a
)
, then
P
(
b
)
∈
H
(
a
)
, where
P
(
b
)
is a polynomial in
b
;
(vii)
if
b
∈
ℭ
(
a
)
and
c
∈
H
(
a
)
, then
b
c
∈
ℭ
(
a
)
.Proof.
(i) Assume thatb
∈
H
(
a
). Then [
b
,
a
a
†
]
=
0 and so b
a
a
†
=
a
a
†
b. By taking the adjoint it follows that a
a
†
b
*
=
b
*
a
a
†. Thus [
b
*
,
a
a
†
]
=
0
.In a similar manner, from [
b
,
a
†
a
]
=
0, we obtain [
b
*
,
a
†
a
]
=
0. Therefore b
*
∈
H
(
a
).
(ii) Letb
∈
H
(
a
) by part (i) and definition of H
(
a
); we have
(26)
[
b
,
a
a
†
]
=
0
,
[
b
*
,
a
a
†
]
=
0
,
[
b
,
a
†
a
]
=
0
,
[
b
*
,
a
†
a
]
=
0
.
From (8) and (26) we conclude that
(27)
[
b
*
b
,
a
†
a
]
=
0
,
[
b
*
b
,
a
a
†
]
=
0
.
Therefore b
∈
ℭ
(
a
).
(iii) Suppose thatb
,
c
∈
H
(
a
). Then
(28)
[
b
,
a
a
†
]
=
0
,
[
b
,
a
†
a
]
=
0
,
[
c
,
a
a
†
]
=
0
,
[
c
,
a
†
a
]
=
0
.
From (8) and (28) we get
(29)
[
b
c
,
a
a
†
]
=
0
,
[
b
c
,
a
†
a
]
=
0
.
This means that b
c
∈
H
(
a
). If b
∈
H
(
a
). Then [
b
,
a
a
†
]
=
0 and so b
a
a
†
=
a
a
†
b. Multiply this from left and right to b
-
1; we obtain [
b
-
1
,
a
a
†
]
=
0. Similarly we have [
b
-
1
,
a
†
a
]
=
0. This means that b
-
1
∈
H
(
a
). Therefore, H
(
a
) is subgroup of  ℜ
-
1.
(iv) It is easy to check that ifa
∈
ℜ
†, then for every b
∈
H
(
a
), we have
(30)
(
b
a
b
-
1
)
†
=
b
a
†
b
-
1
.
(v) Ifb
,
c
∈
H
(
a
), by linearity of the commutator we get [
b
+
c
,
a
a
†
]
=
0 and [
b
+
c
,
a
†
a
]
=
0. That is, b
+
c
∈
H
(
a
).
(vi) It follows from (ii) and (iv).
(vii) Using (8) and part (i), we see that [
(
b
c
)
*
b
c
,
a
a
†
]
=
0 and [
(
b
c
)
*
b
c
,
a
†
a
]
=
0; that is, b
c
∈
ℭ
(
a
).Letℜ be the set of all n
×
n matrices. It was shown that in [3] H
(
a
) is a nonabelian subgroup of ℜ
-
1 if and only if n
>
2.Proposition 19.
Assume thata is an element in ℜ
† with MP-inverse a
†. If b
∈
ℭ
(
a
) is normal, then 〈
b
〉
⊂
ℭ
(
a
) where 〈
b
〉 is the cyclic group generated by b.Proof.
Using Proposition2, Corollary 4, and induction, we can show that for all integer n, b
n
∈
ℭ
(
a
).Note that, in fact ifb
∈
ℭ
(
a
) is normal, then P
(
b
)
∈
ℭ
(
a
),where
P
(
b
)
is a polynomial in
b.


## 3. Covariance Set inC
*-Algebras

Given unitalC
*-algebras 𝔄 with the nonzero element 1
𝔄. We will denote by 𝔄
-
1 and 𝔄
† the subset of invertible elements and MP-invertible elements of 𝔄, respectively.In this section, we find some topological properties forℭ
(
a
); for instance, we will show that ℭ
(
a
) is a closed set in 𝔄
-
1 with respect to the relative topology.Theorem 20.
Suppose that𝔄 is a C
*-algebra and a
∈
𝔄
†. Then ℭ
(
a
) is closed in 𝔄
-
1 with respect to the relative topology.Proof.
Suppose thatb belongs to the closure of ℭ
(
a
) in 𝔄
-
1. Then there exists a sequence b
n
∈
ℭ
(
a
) such that b
n
→
b, from which it follows that b
n
*
→
b
*. Thus
(31)
[
b
n
*
b
n
,
a
a
†
]
=
0
,
[
b
n
*
b
n
,
a
†
a
]
=
0
∀
n
∈
ℕ
by Proposition 2. Therefore
(32)
b
n
*
b
n
a
a
†
=
a
a
†
b
n
*
b
n
,
b
n
*
b
n
a
†
a
=
a
†
a
b
n
*
b
n
∀
n
∈
ℕ
.
By taking limits in (32) as n
→
∞, we get
(33)
b
*
b
a
a
†
=
a
a
†
b
*
b
,
b
*
b
a
†
a
=
a
†
a
b
*
b
.
Since b and b
* are in 𝔄
-
1, again Proposition 2 implies that b
∈
ℭ
(
a
). This means that ℭ
(
a
) is closed in 𝔄
-
1 with respect to the relative topology.Note that generallyℭ
(
a
) is not a closed set in 𝔄. For example, if we set a
=
[
1
0
0
1
] and b
n
=
[
1
/
n
0
0
1
/
n
], then b
n
∈
ℭ
(
a
) for all n
∈
ℕ, but lim
⁡
n
→
∞
b
n
=
0
∉
ℭ
(
a
).We will now reproduce an important theorem of [7] that will be crucial to prove the next result.Theorem 21 ([see [7]).
Leta
n
,
abe nonzero elements of 𝔄 such that a
n
→
a in 𝔄. Then the following conditions are equivalent:(i)
a
n
†
→
a
†;
(ii)
a
n
†
a
n
→
a
†
a;
(iii)
a
n
a
n
†
→
a
a
†;
(iv)
sup
⁡
n
∥
a
n
†
∥
<
∞.The next theorem shows that the covariance set, seen as a multivalued map, has some kind of continuity.Theorem 22.
Let{
a
n
} be a sequence of MP-invertible elements in the C
*-algebra 𝔄 such that a
n
→
a and the norms ∥
a
n
†
∥ are bounded. If b
n
∈
ℭ
(
a
n
) and b
n
→
b
∈
ℜ
-
1 as n
→
∞, then b
∈
ℭ
(
a
).Proof.
By hypothesis,a
n’s are MP-invertible, a
n
→
a, and ∥
a
n
†
∥
<
∞. By Theorem 21, a is MP-invertible and a
n
†
→
a
†. Thus
(34)
a
n
†
a
n
⟶
a
†
a
,
a
n
a
n
†
⟶
a
a
†
.
Therefore by Proposition 2
(35)
b
n
∈
ℭ
(
a
n
)
⟺
b
n
b
n
*
a
n
†
a
n
=
a
n
†
a
n
b
n
b
n
*
,
b
n
b
n
*
a
n
a
n
†
=
a
n
a
n
†
b
n
b
n
*
.
Now, letting n
→
∞ in (35) we get
(36)
b
b
*
a
†
a
=
a
†
a
b
b
*
,
b
b
*
a
a
†
=
a
a
†
b
b
*
.
Again by applying Proposition 2 we conclude that b
∈
ℭ
(
a
).We recall that a setK
⊂
𝔄 is called acone
λ
x
∈
K whenever x
∈
K and λ
>
0.Proposition 23.
Suppose thata is a regular element in 𝔄 and λ is any nonzero scalar. Then b
∈
ℭ
(
a
) if and only if λ
b
∈
ℭ
(
a
).Proof.
Assume thatb
∈
ℭ
(
a
). Then by Proposition 2,
(37)
[
b
*
b
,
a
a
†
]
=
0
,
[
b
*
b
,
a
†
a
]
=
0
.
This is true if and only if
(38)
|
λ
|
2
[
b
*
b
,
a
a
†
]
=
0
,
|
λ
|
2
[
b
*
b
,
a
†
a
]
=
0
,
which is equivalent to
(39)
[
(
λ
b
)
*
(
λ
b
)
,
a
a
†
]
=
0
,
[
(
λ
b
)
*
(
λ
b
)
,
a
†
a
]
=
0
.
Again by Proposition 2, these hold if and only if λ
b
∈
ℭ
(
a
).Corollary 24.
Ifa is regular in 𝔄, then ℭ
(
a
) is a cone.Proof.
The proof is an immediate consequence of the above proposition.Proposition 25.
Suppose thata is a regular element in 𝔄 and λ is any nonzero scalar. Then ℭ
(
a
)
=
ℭ
(
λ
a
).Proof.
By assumptionλ
≠
0, thus (
λ
a
)
†
=
(
1
/
λ
)
a
† and so
(40)
(
λ
a
)
†
(
λ
a
)
=
a
†
a
,
(
λ
a
)
(
λ
a
)
†
=
a
a
†
.
By applying Proposition 5 we get
(41)
ℭ
(
a
)
=
ℭ
(
a
a
†
)
∩
ℭ
(
a
†
a
)
=
ℭ
(
(
λ
a
)
(
λ
a
)
†
)
∩
ℭ
(
(
λ
a
)
†
(
λ
a
)
)
=
ℭ
(
λ
a
)
.



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*Source: 309708-2014-05-05.xml*