# Semi-Iterative Method for Computing the Generalized InverseA

Authors: Xiaoji Liu; Caijing Jiang
Journal: Abstract and Applied Analysis
 (2015)

---

## Abstract

The main aim of this paper is to compute the generalized inverseA
T
,
S
(
2
) over Banach spaces by using semi-iterative method and to present the error bounds of the semi-iterative method for approximating A
T
,
S
(
2
).


---

## Body

## 1. Introduction

The semi-iterative method(
SIM
) was originally inspired by the classical theory of summation and had been discussed by numerous authors [1, 2], which played an important role in many applications (see [3]). In 1962, Varga [3] defined the SIM:
(1)
y
k
=
∑
j
=
0
k
‍
π
k
,
j
x
j
,
∑
j
=
0
k
‍
π
k
,
j
=
1
;
k
≥
0
for the vectors x
0
,
x
1
,
x
2
,
…
,
x
k. The weights π
k
,
j
∈
C form an infinite lower triangular matrix
(2)
ψ
=
π
k
,
j
k
≥
j
≥
0
.
This semi-iterative formula is used to solve the Drazin inverse solution of linear equations in Banach spaces in [1] and the solution of singular linear systems of algebraic equations in [4]. In 1996, Chen [5] defined the iterative formula:
(3)
X
k
+
1
=
X
k
+
β
Y
I
-
A
X
k
,
k
=
0,1
,
2
,
…
,
β
∈
C
∖
0
.
In 2010, Liu et al. [2] extended the iterative method to compute the generalized inverse A
T
,
S
(
2
) over Banach spaces.As we know, the iterative method in Liu et al. [2] can be used to compute the generalized inverse A
T
,
S
(
2
) when ρ
(
I
-
β
Y
A
)
<
1. However, we can not apply it to compute the generalized inverse A
T
,
S
(
2
) when ρ
(
I
-
β
Y
A
)
=
1. Therefor, it is necessary to study the semi-iterative method for computing the generalized inverse A
T
,
S
(
2
) when ρ
(
I
-
β
Y
A
)
=
1. In this paper, we use the semi-iterative method to compute the generalized inverse A
T
,
S
(
2
) over Banach spaces and present the error bounds of the semi-iterative method for approximating A
T
,
S
(
2
).Now we list some notations used in this paper.LetX and Y be arbitrary Banach spaces. The symbol B
(
X
,
Y
) denotes the set of all bounded linear operators from X to Y. Let B
(
X
)
∶
=
B
(
X
,
X
). For any A
∈
B
(
X
,
Y
), we denote its range, null space, and norm by R
(
A
), N
(
A
), and A, respectively. If A
∈
B
(
X
), then we denote its spectrum and spectral radius by σ
(
A
) and ρ
(
A
). If A
∈
B
(
X
,
Y
) and L
⊂
X, then the restriction A
L of A on L is defined by x
↦
A
x
,
x
∈
L.LetL
,
M
⊂
X with L
⊕
M
=
X. Denote P
L
,
M by the projection from M onto L.The paper is organized as follows. Some lemmas will be presented in the remainder of this section. In Section2, we reconsider the method to compute the generalized inverse A
T
,
S
(
2
) on a Banach space, and we also give some conditions for the existence of semi-iterative convergence to the generalized inverse A
T
,
S
(
2
) and its existence and estimate the error bounds of the semi-iterative method for approximating A
T
,
S
(
2
). In Section 3, we give an example for computing the generalized inverse A
T
,
S
(
2
) when ρ
(
I
-
β
Y
A
)
=
1 in our semi-iterative method.The following lemmas are needed in what follows.Lemma 1 (see [6, Section 4]).
LetX and Y be Banach spaces, A
∈
B
(
X
,
Y
), and T and S closed subspaces of X and Y, respectively. Then the following statements are equivalent:(i)
Ahas a {
2
}-inverse B
∈
B
(
Y
,
X
) such that R
(
B
)
=
T and N
(
B
)
=
S.
(ii)
T is a complemented subspace of X, A
(
T
) is closed, A
T
:
T
→
A
(
T
) is invertible, and A
(
T
)
⊕
S
=
Y.In the case when (i) or (ii) holds, B is unique and one denotes it by A
T
,
S
(
2
).Lemma 2 (see [7, Section 3]).
Suppose that the conditions of Lemma2 are satisfied. If one takes T
1
=
N
(
A
T
,
S
(
2
)
A
), then X
=
T
⊕
T
1 holds and A has the following matrix form:
(4)
A
=
A
1
0
0
A
2
:
T
T
1
⟶
A
T
S
,
where A
1 is invertible. Moreover, A
T
,
S
(
2
) has the following matrix form:
(5)
A
T
,
S
(
2
)
=
A
1
-
1
0
0
0
:
A
T
S
⟶
T
T
1
.
Consequently,
(6)
P
A
(
T
)
,
S
=
A
A
T
,
S
(
2
)
=
I
0
0
0
:
A
T
S
⟶
A
T
S
,
P
T
,
T
1
=
A
T
,
S
(
2
)
A
=
I
0
0
0
:
T
T
1
⟶
T
T
1
.Lemma 3 (see [2, Section 2]).
LetA
∈
B
(
X
,
Y
) and Y
∈
B
(
Y
,
X
). Define the sequence {
X
k
} in B
(
Y
,
X
) in the following way:
(7)
X
k
=
X
k
-
1
+
β
Y
I
Y
-
A
X
k
-
1
,
k
=
1,2
,
3
,
…
,
where β
∈
C
∖
{
0
} and X
0
∈
B
(
Y
,
X
) with Y
≠
Y
A
X
0. Then the iteration (7) converges if and only if ρ
(
I
X
-
β
Y
A
)
<
1; equivalently, ρ
(
I
Y
-
β
A
Y
)
<
1.
In this case, assume thatT and S are closed subspaces of X and Y, respectively. If R
(
Y
)
=
T and N
(
Y
)
=
S and R
(
X
0
)
⊂
T, then A
T
,
S
(
2
) exists and {
X
k
} converges to A
T
,
S
(
2
), and when q
=
min
⁡
{
I
X
-
β
Y
A
,
I
Y
-
β
A
Y
}
<
1,
(8)
A
T
,
S
(
2
)
-
X
k
≤
β
q
k
1
-
q
Y
I
Y
-
A
X
0
.


## 2. Semi-Iterative Method for Computing Generalized InversesA
T
,
S
(
2
)

In the section, we will discuss semi-iterative method for computing generalized inversesA
T
,
S
(
2
). First, we deduce convergent conditions and error bounds of our semi-iterative method for computing generalized inverse A
T
,
S
(
2
).Theorem 4.
LetA
∈
B
(
X
,
Y
) and Y
∈
B
(
Y
,
X
). Define the sequence {
X
k
} in B
(
Y
,
X
) in the following way:
(9)
X
k
=
X
k
-
1
+
β
Y
I
X
-
X
k
-
1
A
,
k
=
1,2
,
3
,
…
,
y
m
=
∑
j
=
0
m
π
m
,
j
x
j
,
m
≥
0
,
where β
∈
C
∖
{
0
} and X
0
∈
B
(
Y
,
X
) with Y
≠
Y
A
X
0. Then the semi-iteration (9) converges if and only if
(10)
ρ
I
X
-
β
Y
A
≤
1
,
e
q
u
i
v
a
l
e
n
t
l
y
,
ρ
I
Y
-
β
A
Y
≤
1
.
In this case, assume that T and S are closed subspaces of X and Y, respectively. If R
(
Y
)
=
T and N
(
Y
)
=
S and R
(
X
0
)
⊂
T, then A
T
,
S
(
2
) exists and {
y
m
} converges to A
T
,
S
(
2
), and when q
=
min
⁡
{
I
X
-
β
Y
A
,
I
Y
-
β
A
Y
}
<
1,
(11)
A
T
,
S
(
2
)
-
y
m
≤
1
2
m
q
m
I
X
+
β
Y
A
-
1
β
Y
+
1
2
m
+
1
I
X
-
q
m
β
Y
A
-
1
β
Y
+
1
2
m
+
1
q
m
+
1
I
X
+
β
Y
A
-
1
y
0
.Proof.
From (7),
(12)
X
k
=
I
X
-
β
Y
A
k
X
0
+
I
X
-
β
Y
A
k
-
1
+
I
X
-
β
Y
A
k
-
2
+
I
X
-
β
Y
A
k
-
3
+
⋯
+
I
X
-
β
Y
A
+
1
β
Y
=
I
X
-
β
Y
A
k
X
0
+
∑
j
=
0
k
-
1
‍
I
X
-
β
Y
A
j
β
Y
.
Let H
=
I
X
-
β
Y
A, C
=
β
Y, X
0
=
y
0. Then we have
(13)
X
k
=
H
k
X
0
+
∑
j
=
0
k
-
1
‍
H
k
C
;
hence,
(14)
X
j
=
H
j
X
0
+
∑
i
=
0
j
-
1
‍
H
i
C
,
j
=
1,2
,
3
…
,
y
m
=
∑
j
=
0
m
‍
π
m
,
j
X
j
=
π
m
,
0
X
0
+
∑
j
=
1
m
‍
π
m
,
j
X
j
=
π
m
,
0
X
0
+
∑
j
=
1
m
‍
π
m
,
j
H
j
X
0
+
∑
i
=
0
j
-
1
‍
H
i
C
=
π
m
,
0
X
0
+
∑
j
=
1
m
‍
π
m
,
j
∑
i
=
0
j
-
1
‍
H
i
C
+
∑
j
=
1
m
‍
π
m
,
j
H
j
X
0
=
∑
j
=
1
m
‍
π
m
,
j
∑
i
=
0
j
-
1
‍
H
i
C
+
∑
j
=
0
m
‍
π
m
,
j
H
j
X
0
=
∑
i
=
0
m
-
1
‍
∑
j
=
i
+
1
m
‍
π
m
,
j
H
i
C
+
∑
j
=
0
m
‍
π
m
,
j
H
j
y
0
.
Let π
m
,
n
=
(
1
/
2
)
n
+
1 (n
=
1,2
,
…). We have
(15)
∑
i
=
0
m
-
1
‍
∑
j
=
i
+
1
m
‍
π
m
,
j
H
i
=
∑
j
=
1
m
‍
π
m
,
1
H
0
+
∑
j
=
2
m
‍
π
m
,
j
H
1
+
∑
j
=
3
m
‍
π
m
,
j
H
2
+
⋯
+
∑
j
=
m
m
‍
π
m
,
j
H
m
-
1
=
π
m
,
1
+
π
m
,
2
+
π
m
,
3
+
⋯
+
π
m
,
m
H
0
+
π
m
,
2
+
π
m
,
3
+
π
m
,
4
+
⋯
+
π
m
,
m
H
1
+
π
m
,
3
+
π
m
,
4
+
π
m
,
5
+
⋯
+
π
m
,
m
H
2
+
⋯
+
π
m
,
m
H
m
-
1
=
1
2
2
+
1
2
3
+
⋯
+
1
2
m
+
1
H
0
+
[
1
2
3
+
1
2
4
+
⋯
+
1
2
m
+
1
]
H
1
+
1
2
4
+
1
2
5
+
⋯
+
1
2
m
+
1
H
2
+
⋯
+
1
2
m
+
1
H
m
-
1
=
1
2
1
-
1
2
m
H
0
+
1
2
2
1
-
1
2
m
-
1
H
1
+
1
2
3
1
-
1
2
m
-
2
H
2
+
⋯
+
1
2
m
+
1
H
m
-
1
=
1
2
T
0
-
1
2
m
+
1
H
0
+
1
2
2
H
1
-
1
2
m
+
1
H
1
+
1
2
3
H
2
-
1
2
m
+
1
H
2
+
⋯
+
1
2
m
1
-
1
2
m
-
m
-
1
H
m
-
1
=
1
2
H
0
+
1
2
2
H
1
+
1
2
3
H
2
+
⋯
+
1
2
m
H
m
-
1
-
1
2
m
+
1
H
0
-
1
2
m
+
1
H
1
-
⋯
-
1
2
m
+
1
H
m
-
1
=
1
2
H
0
+
1
2
2
H
1
+
1
2
3
H
2
+
⋯
+
1
2
m
H
m
-
1
-
1
2
m
+
1
H
0
+
H
1
+
T
2
+
⋯
+
H
m
-
1
,
H
0
+
H
1
+
H
2
+
⋯
+
H
m
-
1
=
H
0
+
H
1
+
H
2
+
⋯
+
H
m
-
1
I
-
H
I
-
H
-
1
=
H
0
+
H
1
+
H
2
+
⋯
+
H
m
-
1
-
H
1
-
H
2
-
⋯
-
H
m
-
1
-
H
m
I
-
H
-
1
=
H
0
-
H
m
I
-
H
-
1
,
∑
j
=
0
m
‍
π
m
,
j
H
j
=
π
m
,
0
H
0
+
π
m
,
1
H
1
+
π
m
,
2
H
2
+
⋯
+
π
m
,
m
H
m
=
1
2
H
0
+
1
2
2
H
1
+
1
2
3
H
2
+
⋯
+
1
2
m
+
1
H
m
=
1
2
H
0
+
1
2
2
H
1
+
1
2
3
H
2
+
⋯
+
1
2
m
+
1
H
m
·
I
-
1
2
H
I
-
1
2
H
-
1
=
1
2
H
0
+
1
2
2
H
1
+
1
2
3
H
2
+
⋯
+
1
2
m
+
1
H
m
-
1
2
2
H
1
-
1
2
3
H
2
-
⋯
-
1
2
m
+
2
H
m
+
1
·
I
-
1
2
H
-
1
=
1
2
H
0
-
1
2
m
+
2
H
m
+
1
I
-
1
2
H
-
1
.
From (15), we have
(16)
y
m
=
∑
i
=
0
m
-
1
‍
∑
j
=
i
+
1
m
‍
π
m
,
j
H
i
C
+
∑
j
=
0
m
‍
π
m
,
j
H
j
y
0
=
1
2
H
0
+
1
2
2
H
1
+
1
2
3
H
2
+
⋯
+
1
2
m
H
m
-
1
-
1
2
m
+
1
H
0
+
H
1
+
H
2
+
⋯
+
H
m
-
1
C
+
1
2
H
0
+
1
2
2
H
1
+
1
2
3
H
2
+
⋯
+
1
2
m
+
1
H
m
y
0
=
1
2
H
0
-
1
2
m
+
1
H
m
I
X
-
1
2
H
-
1
-
1
2
m
+
1
H
0
-
H
m
I
X
-
H
-
1
C
+
1
2
H
0
-
1
2
m
+
2
H
m
+
1
I
X
-
1
2
H
-
1
y
0
;
thus,
(17)
y
m
-
y
m
-
1
=
1
2
m
H
m
-
1
-
1
2
m
+
1
H
m
I
X
-
1
2
H
-
1
C
+
1
2
m
1
2
H
0
+
1
2
H
m
-
H
m
-
1
I
X
-
H
-
1
C
+
1
2
m
+
1
H
m
-
1
2
m
+
2
H
m
+
1
I
X
-
1
2
H
-
1
y
0
=
1
2
m
H
m
-
1
-
1
2
H
m
I
X
-
1
2
H
-
1
C
+
1
2
m
1
2
H
0
+
1
2
H
m
-
H
m
-
1
I
X
-
H
-
1
C
+
1
2
m
+
1
H
m
-
1
2
m
H
m
+
1
I
X
-
1
2
H
-
1
y
0
=
1
2
m
H
m
-
1
I
X
-
1
2
H
I
X
-
1
2
H
-
1
C
+
1
2
m
+
1
I
X
-
H
-
1
C
+
1
2
m
1
2
H
m
-
H
m
-
1
I
X
-
H
-
1
C
+
1
2
m
+
1
H
m
I
X
-
1
2
H
I
X
-
1
2
H
-
1
y
0
=
1
2
m
H
m
-
1
C
+
1
2
m
+
1
I
X
-
H
-
1
C
+
1
2
m
H
m
-
1
1
2
H
-
I
X
I
X
-
H
-
1
C
+
1
2
m
+
1
H
m
y
0
=
1
2
m
I
X
-
β
Y
A
m
-
1
β
Y
+
1
2
m
+
1
I
X
-
I
X
-
β
Y
A
-
1
β
Y
+
1
2
m
I
X
-
β
Y
A
m
-
1
1
2
I
X
-
β
Y
A
-
I
X
·
I
X
-
I
X
-
β
Y
A
-
1
β
Y
+
1
2
m
+
1
I
X
-
β
Y
A
m
y
0
=
1
2
m
I
X
-
β
Y
A
m
-
1
β
Y
+
1
2
m
+
1
A
-
1
+
1
2
m
I
X
-
β
Y
A
m
-
1
-
1
2
I
X
+
β
Y
A
A
-
1
+
1
2
m
+
1
I
X
-
β
Y
A
m
-
1
y
0
=
1
2
m
I
X
-
β
Y
A
m
-
1
β
Y
+
1
2
m
+
1
A
-
1
+
1
2
m
I
X
-
β
Y
A
m
-
1
-
1
2
A
-
1
+
1
2
m
I
X
-
β
Y
A
m
-
1
-
1
2
β
Y
A
A
-
1
+
1
2
m
+
1
I
X
-
β
Y
A
m
y
0
=
1
2
m
+
1
I
X
-
β
Y
A
m
-
1
β
Y
+
1
2
m
+
1
I
X
-
I
X
-
β
Y
A
m
-
1
A
-
1
+
1
2
m
+
1
I
X
-
β
Y
A
m
y
0
;
hence,
(18)
y
m
-
y
m
-
1
≤
1
2
m
+
1
I
X
-
β
Y
A
m
-
1
β
Y
+
1
2
m
+
1
I
X
-
I
X
-
β
Y
A
m
-
1
A
-
1
+
1
2
m
+
1
I
X
-
β
Y
A
m
y
0
.
Therefore, we have
(19)
lim
⁡
m
→
∞
y
m
=
lim
⁡
m
→
∞
1
2
H
0
-
1
2
m
+
1
H
m
I
X
-
1
2
H
-
1
-
1
2
m
+
1
H
0
-
T
m
I
X
-
H
-
1
C
+
1
2
H
0
-
1
2
m
+
2
H
m
+
1
I
X
-
1
2
H
-
1
y
0
=
1
2
H
0
I
X
-
1
2
H
-
1
C
+
1
2
H
0
I
X
-
1
2
H
-
1
y
0
=
1
2
H
0
I
X
-
1
2
H
-
1
C
+
y
0
.
Denotelim
⁡
m
→
∞
y
m
=
y
∞, and H
=
I
X
-
β
Y
A, C
=
β
Y, X
0
=
y
0. Then from (19), we have
(20)
y
∞
=
1
2
I
X
-
1
2
H
-
1
C
+
y
0
=
1
2
I
X
-
1
2
I
X
-
β
Y
A
-
1
β
Y
+
y
0
=
1
2
1
2
I
X
+
1
2
β
Y
A
-
1
β
Y
+
y
0
=
I
X
+
β
Y
A
-
1
β
Y
+
y
0
.
So
(21)
I
X
+
β
Y
A
y
∞
=
β
Y
+
y
0
,
y
∞
+
β
Y
A
y
∞
=
β
Y
+
y
0
,
y
∞
=
β
Y
+
y
0
-
β
Y
A
y
∞
.
Since y
0
∈
R
(
Y
) and R
(
Y
A
y
∞
)
⊂
R
(
Y
), then R
(
y
∞
)
⊂
R
(
Y
). On the other hand,
(22)
β
Y
-
β
Y
A
y
∞
+
y
0
=
y
∞
,
β
Y
I
X
-
A
y
∞
+
1
β
=
y
∞
,
R
(
Y
)
⊂
R
(
y
∞
)
,
so R
(
y
∞
)
=
R
(
Y
)
=
T. Since
(23)
y
∞
+
β
Y
A
y
∞
=
β
Y
+
y
0
,
N
y
∞
+
β
Y
A
y
∞
=
N
β
Y
+
y
0
,
N
Y
=
N
y
∞
+
β
Y
A
y
∞
,
then N
(
Y
)
⊂
N
(
y
∞
). As
(24)
y
∞
=
I
X
+
β
Y
A
-
1
C
+
y
0
=
I
X
+
β
Y
A
-
1
β
Y
+
y
0
,
and N
(
y
∞
)
⊂
N
(
Y
), then N
(
y
∞
)
=
N
(
Y
)
=
S; hence y
∞
=
A
T
,
S
(
2
). Furthermore,
(25)
lim
⁡
m
→
∞
y
∞
-
y
m
=
1
2
H
0
I
X
-
1
2
H
-
1
C
+
y
0
1
2
H
0
-
1
2
m
+
2
H
m
+
1
-
1
2
H
0
-
1
2
m
+
1
H
m
I
X
-
1
2
H
-
1
-
1
2
m
+
1
H
0
-
H
m
I
Y
-
H
m
-
1
C
-
1
2
H
0
-
1
2
m
+
2
H
m
+
1
=
1
2
I
X
-
1
2
H
-
1
C
+
1
2
I
X
-
1
2
H
-
1
y
0
-
1
2
I
X
-
1
2
H
-
1
C
+
1
2
m
+
1
H
m
I
X
-
1
2
H
-
1
C
+
1
2
m
+
1
I
X
-
H
-
1
C
-
1
2
m
+
1
H
m
I
X
-
H
-
1
C
-
1
2
I
X
-
1
2
H
-
1
y
0
+
1
2
m
+
2
H
m
+
1
I
X
-
1
2
H
-
1
y
0
=
1
2
m
+
1
H
m
I
X
-
1
2
H
-
1
C
+
1
2
m
+
1
I
X
-
H
m
I
X
-
H
-
1
C
+
1
2
m
+
2
H
m
+
1
I
X
-
1
2
H
-
1
y
0
=
1
2
m
+
1
I
X
-
β
Y
A
m
1
2
I
X
+
β
Y
A
-
1
β
Y
+
1
2
m
+
1
I
X
-
I
X
-
β
Y
A
m
I
X
-
I
X
-
β
Y
A
-
1
β
Y
+
1
2
m
+
2
I
X
-
β
Y
A
m
+
1
1
2
I
X
+
β
Y
A
-
1
y
0
=
1
2
m
I
X
-
β
Y
A
m
I
X
+
β
Y
A
-
1
β
Y
+
1
2
m
+
1
I
X
-
I
X
-
β
Y
A
m
β
Y
A
-
1
β
Y
+
1
2
m
+
1
I
X
-
β
Y
A
m
+
1
I
X
+
β
Y
A
-
1
y
0
=
1
2
m
q
m
I
X
+
β
Y
A
-
1
β
Y
1
2
m
+
1
+
1
2
m
+
1
I
X
-
q
m
β
Y
A
-
1
β
Y
+
1
2
m
+
1
q
m
+
1
I
X
+
β
Y
A
-
1
y
0
=
1
2
m
q
m
I
X
+
β
Y
A
-
1
β
Y
+
1
2
m
+
1
β
Y
A
-
1
β
Y
-
1
2
m
+
1
q
m
β
Y
A
-
1
β
Y
+
1
2
m
q
m
+
1
I
X
+
β
Y
A
-
1
y
0
=
1
2
m
q
m
I
X
+
β
Y
A
-
1
β
Y
+
1
2
β
Y
A
-
1
β
Y
-
1
2
q
m
β
Y
A
-
1
β
Y
+
1
2
q
m
+
1
I
X
+
β
Y
A
-
1
y
0
=
1
2
m
q
m
I
X
+
β
Y
A
-
1
β
Y
1
2
m
+
1
+
1
2
m
+
1
I
X
-
q
m
β
Y
A
-
1
β
Y
+
1
2
m
+
1
q
m
+
1
I
X
+
β
Y
A
-
1
y
0
≤
1
2
m
q
m
I
X
+
β
Y
A
-
1
β
Y
+
1
2
m
+
1
I
X
-
q
m
β
Y
A
-
1
β
Y
+
1
2
m
+
1
q
m
+
1
I
X
+
β
Y
A
-
1
y
0
.
The proof is complete.Similarly, we can obtain the following theorem.Theorem 5.
LetA
∈
B
(
X
,
Y
) and Y
∈
B
(
Y
,
X
). Define the sequence {
X
k
} in B
(
Y
,
X
) in the following way:
(26)
X
k
=
X
k
-
1
+
β
Y
I
Y
-
A
X
k
-
1
,
k
=
1,2
,
3
,
…
,
y
m
=
∑
i
=
0
m
π
m
,
j
x
j
,
m
≥
0
,
where β
∈
C
∖
{
0
} and X
0
∈
B
(
Y
,
X
) with Y
≠
X
0
A
Y. Then the semi-iteration (26) converges if and only if
(27)
ρ
I
Y
-
β
A
Y
≤
1
,
e
q
u
i
v
a
l
e
n
t
l
y
,
ρ
I
X
-
β
Y
A
≤
1
.
In this case, assume that T and S are closed subspaces of X and Y, respectively. If R
(
Y
)
=
T and N
(
Y
)
=
S and N
(
X
0
)
⊃
S, then A
T
,
S
(
2
) exists and {
y
m
} converges to A
T
,
S
(
2
), and when q
=
min
⁡
{
I
X
-
β
Y
A
,
I
Y
-
β
A
Y
}
<
1,
(28)
A
T
,
S
(
2
)
-
y
m
≤
1
2
m
q
m
I
Y
+
β
A
Y
-
1
β
Y
+
1
2
m
+
1
I
Y
-
q
m
β
A
Y
-
1
β
Y
+
1
2
m
+
1
q
m
+
1
I
Y
+
β
A
Y
-
1
y
0
.


## 3. Examples

We give an example for computingA
T
,
S
(
2
) by the semi-iterative (9). Let the symbol · denote the Frobenius norm.Example 1.
Consider the matrix(29)
A
=
0.5
0
0
0
0
0.5
0
0
0
0
0.25
0
0
0
0
1
,
Y
=
X
0
=
y
0
=
0.5
0
0
0
0
2
0
0
0
0
2
0
0
0
0
0
,
T
=
C
4, e
=
(
0,0
,
0,0
,
1
)
T
∈
C
4, and S
=
span
⁡
{
e
}. Obviously, R
(
Y
)
=
T and N
(
Y
)
=
S. Moreover, we choose
(30)
I
=
1
0
0
0
0
1
0
0
0
0
1
0
0
0
0
1
.
Therefore, for any β
∈
C
∖
{
0
}, we have ρ
(
I
-
β
Y
A
)
=
1. If the scalar β
=
0.9, it is easily verified that
(31)
A
T
,
S
(
2
)
=
0.7755
0
0
0
0
2.0000
0
0
0
0
2.6207
0
0
0
0
0
.
The error estimateA
T
,
S
(
2
)
-
y
m for diverse scalar β and parameter m of (25) and y
m
-
y
m
-
1 of (18) is given in Table 1.Table 1
Results for computingA
T
,
S
(
2
) by using iteration (20).
β
Step
E
(
m
)
=
A
T
,
S
(
2
)
-
y
m
y
m
-
y
m
-
1
β
=
0.9
m
=
6
3.820778657434200E − 02
3.753942246016400E − 02
m
=
10
2.398173793546000E − 03
2.387899534974000E − 03
m
=
11
1.198734582542000E − 03
1.195013235056000E − 03
m
=
17
1.869961136884640E − 05
1.868898369984000E − 05
m
=
21
1.168271567800560E − 06
1.168036362390750E − 06
m
=
26
3.650255486528950E − 08
3.650050835579940E − 08
m
=
31
1.140652761558560E − 09
1.140634888089180E − 09
m
=
32
5.703241064454260E − 10
5.703171804724460E − 10
m
=
64
1.327871055533770E − 19
1.327871050907670E − 19
β
=
1.0
m
=
3
3.012230808118550E − 01
2.881862247962140E − 01
m
=
5
7.658271287825600E − 02
7.500354183760500E − 02
m
=
7
1.921616264226400E − 02
1.904402615464200E − 02
m
=
10
2.398878340030000E − 03
2.392152983091000E − 03
m
=
13
2.994086021157650E − 04
2.990957245386160E − 04
m
=
14
1.496563849083790E − 04
1.495411757049050E − 04
m
=
30
2.281288673173540E − 09
2.281271586655230E − 09
m
=
32
5.703195890929840E − 10
5.703171861702270E − 10
m
=
64
1.327871051078640E − 19
1.327871050516590E − 19
β
=
1.2
m
=
3
3.074437258505530E − 01
2.943566993012280E − 01
m
=
5
7.720793593447200E − 02
7.622976300737400E − 02
m
=
8
9.608716640119000E − 03
9.584156541330000E − 03
m
=
10
2.397229438700000E − 03
2.394566834436000E − 03
m
=
14
1.495829839649130E − 04
1.495447708030870E − 04
m
=
25
7.300123408684290E − 08
7.300086429541690E − 08
m
=
31
1.140633927430880E − 09
1.140633247516680E − 09
m
=
32
5.703167566295210E − 10
5.703165186574720E − 10
m
=
64
1.327871050315660E − 19
1.327871050309540E − 19
β
=
1.25
m
=
4
1.545274774285740E − 01
1.521939388599310E − 01
m
=
6
3.856094399263400E − 02
3.834943769150000E − 02
m
=
8
9.606257104476000E − 03
9.586990214228000E − 03
m
=
10
2.396666491289000E − 03
2.394595536570000E − 03
m
=
13
2.991967359449660E − 04
2.991146333829520E − 04
m
=
14
1.495691457980860E − 04
1.495409285906900E − 04
m
=
25
7.300098966535540E − 08
7.300076512906710E − 08
m
=
32
5.703165607729410E − 10
5.703164334061720E − 10
m
=
64
1.327871050307390E − 19
1.327871050305550E − 19
β
=
1.3
m
=
3
3.110631215879030E − 01
2.986869244201630E − 01
m
=
4
1.547020913728830E − 01
1.533981948482430E − 01
m
=
6
3.855659973720000E − 02
3.840382345353100E − 02
m
=
12
5.984798515077750E − 04
5.983002588957640E − 04
m
=
14
1.495572846785640E − 04
1.495367682194500E − 04
m
=
15
7.477013266862850E − 05
7.476319565060200E − 05
m
=
30
2.281266572051920E − 09
2.281265986830480E − 09
m
=
32
5.703164418238200E − 10
5.703163751627730E − 10
m
=
64
1.327871050304590E − 19
1.327871050304050E − 19Table1 shows that within limits the larger the parameters m and β, the smaller the error bounds. But m can not be infinitely large, because when m is large enough, the error bounds are large as well. So we must choose the best β and m.



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*Source: 849487-2015-01-29.xml*