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An iterative method is developed for solving the solution of the general restricted linear equation. The convergence, stability, and error estimate are given. Numerical experiments are presented to demonstrate the efficiency and accuracy.

Let ℂ r m × n be the set of all m × n complex matrices with rank r. For any A ∈ ℂ r m × n , let ‖ A ‖ 2 , R ( A ) , and N ( A ) be matrix spectral norm, range space and null space, respectively. Let ρ ( A ) be the spectral radius of the matrix A. For any A ∈ C r m × n , if there exists a matrix X such that X A X = X , then X is called a {2}-inverse (or an outer inverse) of A [

The restricted linear equation is widely applied in many practical problems [

A x = b , x ∈ T , (1)

where A ∈ ℂ r m × n and T is a subspace of ℂ n . As the conclusion given in [

b ∈ A T , T ∩ N ( A ) = { 0 } . (2)

In recent years, some numerical methods have been developed to solve such as problems (1). The Cramer rule method is given in [

The paper is organized as follows. In Section 2, an iterative method for the general restricted linear equation is developed. The convergence analysis of our method is considered, an error estimate is also given in Section 3. In Section 4, some numerical examples are presented to test the effectiveness of our method.

In this section, we develop an iterative method for computing the solution of the general restricted linear Equation (1).

Lemma 1 ( [

A T ⊕ S = ℂ m ,

in which case X is unique ( denoted by A T , S ( 2 ) ).

Proposition 2 ( [

x = A T , S ( 2 ) b . (3)

Let L and M be complementary subspaces of ℂ m , i.e., L ⊕ M = ℂ m , the projection P L be a linear transformation such that P L x = x , x ∈ L and P L y = 0 , y ∈ M .

Lemma 3 ( [

In this paper, we construct our iterative scheme as follows:

{ Z k = [ t I − C t 2 Z k − 1 A + ⋯ + ( − 1 ) t − 1 ( Z k − 1 A ) t − 1 ] Z k − 1 , x k = x k − 1 + Z k ( b − A x k − 1 ) , (4)

where k = 1 , 2 , 3 , ⋯ , t ∈ ℕ , and t ≥ 2 . Here, we take the initial value Z 0 = β Y in our scheme (4), where β is a relaxation factor. Thus, if t = 2 , then (4) degenerates to the non-stationary Richardson iterative method given in [

Lemma 4 Let A ∈ ℂ m × n , T and S be subspaces of ℂ n and ℂ m , respectively. Assume that Z 0 = β Y and R ( Y ) ⊆ T , where β is a nonzero constant and Y ∈ ℂ n × m . For any initial x 0 ∈ T , the iterative scheme (4) converges to some solution of (1) if and only if

ρ ( P T − Z 0 A ) < 1 ,

where a projection P T from ℂ m onto T.

Proof. The proof can be given as following the line of in [

Now, we consider the convergence analysis of our iterative method (4).

Theorem 5 Let A ∈ ℂ m × n , T and S be subspaces of ℂ n and ℂ m , respectively. Assume that A T ⊕ S = ℂ m and Y ∈ ℂ n m satisfies R ( Y ) ⊆ T and N ( Y ) ⊆ S , where dim T = dim S ⊥ . If b ∈ A T , for the given initial value Z 0 = β Y , β ≠ 0 and x 0 ∈ T , then the sequence { x k } generated by iteration (4) converges to the unique solution of (1) if and only if ρ ( P T − Z 0 A ) < 1 , where P T is a projection. In this case, we have

lim k → ∞ x k = ( I − P T + Z 0 A ) − 1 Z 0 b . (5)

Further, we have

‖ x k − x ∞ ‖ ≤ ‖ q ‖ t ( t k − 1 ) t − 1 ( ‖ x 0 − Z 0 b ‖ + ‖ q ‖ 1 − ‖ q ‖ ‖ Z 0 b ‖ ) . (6)

where q = P T − A Z 0 .

Proof. For any x ∈ T ∩ N ( A ) , we have A x = 0 . By A T ⊕ S = ℂ m and Lemma 1, there exists a matrix X such that R ( X ) = T and X A X = X . Now, assume that y ∈ ℂ m satisfies x = X y , we have

x = X y = X A X y = X A x = 0 , T ∩ N ( A ) = { 0 } .

If b ∈ A T , then (2) is satisfied. Therefore, by ( [

Since R ( Y ) ⊆ T , P T Z 0 = Z 0 , and then by (4), we obtain P T Z k = Z k . Since x 0 ∈ T , P T x k = x k by (4b), and therefore

x k = x k − 1 + Z k ( b − A x k − 1 ) = Z k b + ( P T − Z k A ) x k − 1 . (7)

If P T W = W , then

( I − Z k − 1 A ) W = ( P T − Z k − 1 A ) W . (8)

By (4) and (8), we obtain

Z k A = ∑ i = 1 t ( − 1 ) i − 1 C t i ( Z k − 1 A ) i = ∑ i = 0 t − 1 ( P T − Z k − 1 A ) i Z k − 1 A ,

( P T − Z k A ) W = ( P T − Z k − 1 A ) t W = ( P T − Z 0 A ) t k W . (9)

By induction on k, it leads to

Z k A = ∑ i = 0 t − 1 ( P T − Z 0 A ) i t k − 1 Z k − 1 A = ∑ i 1 = 0 t − 1 ( P T − Z 0 A ) i 1 t k − 1 ∑ i 2 = 0 t − 1 ( P T − Z 0 A ) i 2 t k − 2 Z k − 2 A = S k Z 0 A , (10)

where S k : = ∑ i = 0 t k − 1 ( P T − Z 0 A ) i . From b ∈ A T , w ∈ T , we have b = A w and it implies that Z k b = Z k A w = S k Z 0 A w = S k Z 0 b . By (9), we have

x k = S k Z 0 b + ( P T − Z 0 A ) t k x k − 1 = S k Z 0 b + ∑ i = 0 k − 1 ∏ j = 0 i ( P T − Z 0 A ) t k − j S k − 1 − i Z 0 b + ∏ i = 0 k − 1 ( P T − Z 0 A ) t k − i x 0 = ∑ i = 0 k ( P T − Z 0 A ) t k + 1 − t k + 1 − i t − 1 S k − i Z 0 b + ( P T − Z 0 A ) t k + 1 − t t − 1 x 0 . (11)

Note that ( P T − Z 0 A ) S k = S k ( P T − Z 0 A ) , [ I − ( P T − Z 0 A ) ] S k = I − ( P T − Z 0 A ) t k . From (11), we obtain

[ I − ( P T − Z 0 A ) ] x k = [ I − ( P T − Z 0 A ) t k + 1 − t t − 1 ] Z 0 b + [ I − ( P T − Z 0 A ) ] ( P T − Z 0 A ) t k + 1 − t t − 1 x 0 . (12)

If ρ ( P T − Z 0 A ) < 1 , then I − ( P T − Z 0 A ) is invertible and it implies that x k converges as k → ∞ . For convenience, let its limit denote by x ∞ . Thus, we have [ I − ( P T − Z 0 A ) ] x ∞ = Z 0 b . Since T is closed and x k ∈ T , we have x ∞ ∈ T and P T x ∞ = x ∞ . Thus, Z 0 ( A x ∞ − b ) = 0 and A x ∞ − b ∈ N ( Z 0 ) ∩ A T = N ( Y ) ∩ A T ⊆ S ∩ A T = { 0 } . Note that x ∞ is the unique solution of (1) and x ∞ = [ I − ( P T − Z 0 A ) ] − 1 Z 0 b . From (4), it follows that

x k + 1 − x ∞ = ( P T − Z k + 1 A ) ( x 0 − x ∞ ) = ( P T − Z 0 A ) t ( t k + 1 − 1 ) t − 1 ( x 0 − x ∞ ) . (13)

From Lemma 4, we have ρ ( P T − Z 0 A ) < 1 and

x k − x ∞ = ( P T − Z 0 A ) t ( t k − 1 ) t − 1 ( ( x 0 − Z 0 b ) − ( I − ( I − P T + Z 0 A ) − 1 ) Z 0 b ) .

Therefore,

‖ x k − x ∞ ‖ ≤ ‖ q ‖ t ( t k − 1 ) t − 1 ( ‖ x 0 − Z 0 b ‖ + ‖ I − ( I − q ) − 1 ‖ ‖ Z 0 b ‖ ) ≤ ‖ q ‖ t ( t k − 1 ) t − 1 ( x 0 − Z 0 b + ‖ q ‖ 1 − ‖ q ‖ Z 0 b ) ,

where q = P T − A Z 0 . □

Remark If N ( Y ) ⊆ S in Theorem 5 is removed and t = 2 , then the result degenerates into that given in ( [

Let A and b of the general restricted linear Equation (1) be

A = [ 2 2.5 0.2 0.3 0 0 1.5 0 0 0 0 0 0.2 0.2 0 0 0 0 0.25 0 0 0 0 0 0 0 0 0 0 0 ] ∈ ℂ 4 6 × 5 , b = [ 7 3 0.2 0.2 0 0 ] . (14)

The matrix Y is

Y = [ 1.2 2 0.2 − 2 1 0 0 2 5 − 2 0 0 0 0 0.25 0.1 0 0 0 − 0.1 0 1.3 0 0 1 0 0 0 0 0 ]

Note that R ( Y ) ⊂ T , but N ( Y ) ⊆ S . If take β = 0.16 , then ρ ( P T − Z 0 A ) < 1 . Here, we choose t = 2 in (4). Thus, it can be seen as the method given in [

Theorem 6 Under the same conditions as in Theorem 5. The iterative scheme (4) is stable for solving (1), where

Proof. Let

By (4), we derive

From (9) and (4), we have

k = 11 | 3.3871e−10 | 8.3702e−09 | 21.8430 |

k = 12 | 6.8861e−16 | 3.5821e−15 | 21.8430 |

Therefore, we obtain

By (4), we have

By (17) and (16), we derive

Thus, by

If

where

In the section, we give an example to test the accuracy of our scheme (4), which is implemented by our main code given in Appendix, and make a comparison with the method given in [

Example 1 Consider the restricted linear system (1) with a coefficient matrix being random

Numerical results given in

method of [ | ||||
---|---|---|---|---|

t | MCT [ | |||

2.221434 | 1.4613e−12 | 3.4439e−12 | ||

3.412709 | 1.5764e−13 | 2.2203e−12 | ||

4.249560 | 9.7132e−13 | 7.3896e−12 | ||

14.14631 | 4.5823e−13 | 3.3269e−12 | ||

Our Scheme | ||||

t | MCT | |||

6 | 0.438568 | 9.2913e−13 | 8.3698e−13 | |

7 | 0.742810 | 1.3444e−12 | 1.2432e−12 | |

8 | 1.336598 | 1.4786e−12 | 6.8270e−13 | |

10 | 13.361577 | 4.3165e−12 | 3.0732e−12 |

Example 2 Consider the general restricted linear Equation (1), where A and b is given as in (14). Here, we use the scheme (4) to solve the example. Let

Obviously,

t | b | k | |||
---|---|---|---|---|---|

0.13 | 11 | 2.0035e−08 | 3.1905e−08 | 3.1905e−08 | |

12 | 9.9495e−16 | 6.1815e−16 | 3.6906e−15 | ||

2 | 0.14 | 11 | 5.1443e−09 | 8.1923e−09 | 8.1923e−09 |

12 | 5.9511e−17 | 2.2204e−16 | 1.8243e−15 | ||

0.15 | 11 | 1.3203e−09 | 2.1026e−09 | 2.1026e−09 | |

12 | 8.8861e−16 | 4.5776e−16 | 3.9315e−15 | ||

t | b | k | |||

0.13 | 11 | 0 | 0 | 4.1466e−15 | |

12 | 0 | 0 | 4.1466e−15 | ||

3 | 0.14 | 11 | 0 | 0 | 2.0562e−15 |

12 | 0 | 0 | 2.0562e−15 | ||

0.15 | 11 | 0 | 0 | 3.9440e−15 | |

12 | 0 | 0 | 3.9440e−15 |

To ensure

t | b | k | |||
---|---|---|---|---|---|

0.13 | k = 5 | 4.5219e−08 | 7.2011e−08 | 7.2011e−08 | |

k = 6 | 8.8861e−16 | 4.5776e−16 | 3.8357e−15 | ||

5 | 0.15 | k = 5 | 3.3786e−09 | 5.3804e−09 | 5.3804e−09 |

k = 6 | 0 | 0 | 3.8794e−15 | ||

0.16 | k = 5 | 9.2293e−10 | 1.4698e−09 | 1.4698e−09 | |

k = 6 | 2.7756e−17 | 1.1102e−16 | 2.6990e−15 | ||

t | b | k | |||

0.13 | k = 5 | 8.8818e−16 | 4.4409e−16 | 4.4270e−15 | |

k = 6 | 0 | 0 | 4.0550e−15 | ||

8 | 0.15 | k = 5 | 0 | 0 | 3.0562e−15 |

k = 6 | 0 | 0 | 3.7659e−15 | ||

0.16 | k = 5 | 8.8816e−16 | 4.5776e−16 | 2.2834e−15 | |

k = 6 | 0 | 0 | 2.7240e−15 |

From the numerical results given in

The high order iterative method has been derived for solving the general restricted linear equation. The convergence and stability of our method also have derived. Numerical experiments have presented to demonstrate the efficiency and accuracy.

This work was supported by the National Natural Science Foundation of China (No. 11061005, 11701119, 11761024), the Natural Science Foundation of Guangxi (No. 2017GXNSFBA198053), the Ministry of Education Science and Technology Key Project (210164), and the open fund of Guangxi Key laboratory of hybrid computation and IC design analysis (HCIC201607).

Liu, X.J., Du, W.R., Yu, Y.M. and Qin, Y.H. (2018) Efficient Iterative Method for Solving the General Restricted Linear Equation. Journal of Applied Mathematics and Physics, 6, 418-428. https://doi.org/10.4236/jamp.2018.62039

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