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Tiêu đề Direct methods and iterative techniques for solving linear systems
Tác giả Dr. Lê Xuân Đại
Trường học Ho Chi Minh City University of Technology
Chuyên ngành Applied Mathematics
Thể loại Lecture
Năm xuất bản 2019
Thành phố Ho Chi Minh City
Định dạng
Số trang 90
Dung lượng 402,3 KB

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DIRECT METHODS AND ITERATIVETECHNIQUES FOR SOLVING LINEAR SYSTEMS Dr.. Lê Xuân Đại HoChiMinh City University of Technology Faculty of Applied Science, Department of Applied Mathematics E

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DIRECT METHODS AND ITERATIVE

TECHNIQUES FOR SOLVING LINEAR SYSTEMS

Dr Lê Xuân Đại

HoChiMinh City University of Technology Faculty of Applied Science, Department of Applied Mathematics

Email: ytkadai@hcmut.edu.vn

HCMC — 2019.

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mathematics to the social sciences and the

quantitative study of business and economic

problems.

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1 We consider a linear system of n

A = (a ij ) ∈ M n (K ) and detA 6= 0. So this

methods of approximating the solution

to linear systems using iterative

methods.

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E LEMENTARY ROW OPERATIONS

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We use 3 operations to simplify the linear system (1):

1 Equations can be transposed in order (r i ↔ r j ).

2 Equation can be multiplied by any non-zero

constant λ 6= 0(r i → λr i ).

3 Equation can be multiplied by any constant λ

and added to another equation (r i → r i + λr j ).

By a sequence of these operations, a linear system will be systematically transformed into a new linear system that is more easily solved and has the same solutions.

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3 Rewrite the new system corresponding to matrix

of row echelon form.

4 Backward substitution can be performed to

solve the n th equation for x n , the (n − 1)st

equation for x n−1 Continuing this process, we obtain the unique solution.

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The corresponding system is

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LU D ECOMPOSITIONS

3 Doolittle’s Method LU factorization

when the diagonal elements of lower

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L andU have the form

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The entries of matrices L and U can be

defined by the formula

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Multiplying 2 matrices L and U, we have

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M ULTIPLE CHOICE E XERCISES

Factor A into the LU

decomposition A = LU using Doolittle’s

Method Find the entry ` 32 of matrixL

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Factor A into the LU

decomposition A = LU using Doolittle’s

Method Find the sum u 11 + u 22 + u 33 of

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T HEOREM 4.2

The square matrix A is positive definite if and only if A can be factored in the form A = B.B T , where B is lower triangular with nonzero diagonal entries

( b ii > 0, i = 1 n ) and is defined by the formula:

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The system can be rewritten in matrix form

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E XAMPLE 4.2

Determine matrix B in Cholesky

factorization of the positive definite matrix

p 7 7

2 p 21 7

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5 The above answers

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2 p

2 .

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D EFINITION 5.1

A vector norm X ∈ R n is a function, denoted

by ||X||, from R n into R with the following

properties:

1 ∀X ∈ R n , ||X|| Ê 0,||X|| = 0 ⇔ X = 0

2 ∀X ∈ R n , ∀λ ∈ R,||λX|| = |λ|.||X||

3 ∀X, Y ∈ R n , ||X + Y || É ||X|| + ||Y ||.

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We will need only two specific norms on R n :

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D EFINITION

D EFINITION 6.1

A sequence (X (m) ) ∞ m=0 of vectors X (m) ∈ R n is

said to converge to X with respect to the

norm || · || as m → +∞ if and only if

||X (m) − X|| → 0 as m → +∞.

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T HEOREM 6.1

The sequence of vectors (X (m) ) ∞ m=0 converges

(x (m) k ) converge tox k , ∀k = 1,2, ,n.

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The system AX = B(det(A) 6= 0) has a unique

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T HEOREM 6.2

A matrix A is well-conditioned if k(A) is close

to 1 , and is ill-conditioned when k(A) is

significantly greater than 1.

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Now we consider the system A X = e e B where B = e

µ 3 3.1

This system has unique solution X = e µ −17

10

We can find k(A) = 1207.01 >> 1 Therefore, B ≈ e B, however X and X e are as much as different.

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M ULTIPLE CHOICE EXERCISES

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MatA x −1 ⇒ A −1 =

à 3 14

2 21

¯

¯

¯

¯ ,

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An iterative technique to solve the n × n

´ ∞

m=0

defined by the formula

X (m) = TX (m−1) + C, m = 1,2, (2)

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T HEOREM 6.3

If ||T|| < 1 then the sequence of vectors

³

X (m) ´ ∞

m=0 defined by the formula (2) will

converge to X , starting with an initial

approximation X (0) Then the error analysis is

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D EFINITION 6.3

diagonally dominant when

n

X

j=1,j6=i

|a ij | < |a ii |, i = 1, 2, , n

Note If A is a strictly diagonally dominant

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Consider the system (1) where Ais strictly diagonally dominant matrix Factorize the

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Since a ii 6= 0, ∀i = 1, 2, , n so detD 6= 0.

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J ACOBI ’ S M ETHOD

We have

AX = B ⇔ (DưLưU)X = B ⇔ (D)X = (L+U)X +B

⇔ X = D ư1 (L + U)X + D ư1 B.

Jacobi iterative method can give us

X (m) = T j X (mư1) + C j , m = 1,2,

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For each i Ê 1, generate the components x i (m)

method always converges starting with

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E XAMPLE 6.2

Use Jacobi’s iterative technique to find

approximation accurate to within 10 −4 ,

choosing norm infinity

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|0.09| + | − 0.15| < |3|; |0.04| + | − 0.08| < |4| so

matrix Therefore, Jacobi’s method always converges Rewrite the given system in the

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 and then evaluate X (1) , X (2) ,

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||T j || ∞ = max

i=1,2,3

3 P

j=1 |t ij | = max{|0| + | − 0.06| + |0.02|,| − 0.03| + |0| + |0.05|,

1 − 0.08 × 5.48 × 10

−4

≈0.4765 × 10 −4 < 10 −4

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4

à 0.390 0.170

!

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⇒ Answer 3

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G AUSS -S EIDEL ’ S METHOD

Rewrite the system (1) in matrix form

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Let T g = (D − L) −1 U, C g = (D − L) −1 B, we receive iterative formula Gauss-Seidel of the form

X (m) = T g X (m−1) + C g , m = 1,2,

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E XPLICIT ITERATIVE FORMULA G AUSS -S EIDEL

x (m) 1 = c 1 +

n

X

j=2 t1jx m−1 j ,

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Gauss-Seidel’s method is a possible

improvement of Jacobi’s method, but it

these most recently calculated values

x (m) 1 , x 2 (m) , , x i−1 (m)

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E XAMPLE 6.3

Use the Gauss-Seidel iterative technique to find approximate solutions to the following system accurate within 10 −4 Choose norm infinity

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SOLUTION We have |0.24| + | − 0.08| < |4|;

|0.09| + | − 0.15| < |3|; |0.04| + | − 0.08| < |4| so

matrix Rewrite the system in the form

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 and then evaluate X (1) , X (2) ,

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x (1) 1 = c 1 + t 12 x 2 (0) + t 13 x (0) 3 ,

x (1) 2 = c 2 + t 21 x 1 (1) + t 23 x (0) 3 ,

x 3 (1) = c 3 + t 31 x (1) 1 + t 32 x (1) 2

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A = (8 − 0.24B + 0.08C) ÷ 4 :

B = (9 − 0.09A + 0.15C) ÷ 3 :

C = (20 − 0.04A + 0.08B) ÷ 4

Press ”=” continuously until receiving the

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Error analysis ||X (3)

−X (2) || ∞ = max

i=1,2,3 |x i (3) −x i (2) | = max{| − 1.499.10 −4 |, |0.123.10 −4 |, |0.017.10 −4 |} = 1.499 × 10 −4

1 − 0.08 × 1.499 × 10

−4

≈0.1303 × 10 −4 < 10 −4

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4

à 0.759 1.093

!

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A = (5 + 6B) ÷ 15 : B = (5 + 5A) ÷ 8

Press ”=” continuously until receiving the

à 0.755 1.096875

! ⇒ Answer

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THANK YOU FOR YOUR ATTENTION

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