Numerical Methods Contents - SAM

Solution of LŨx = b: x
( )
2ǫ
= 1 − 2ǫ
(meaningless result !)
9 A = [A ( 1 , : ) ; −fac , C ] ; %
10 end
LU-factorization after swapping rows:
( ) ( ) (
1 1
1 0 1 1
A = ⇒ L = , U =
ǫ 1 ǫ 1 0 1 − ǫ
Solution of LŨx = b: x
= (
1 + 2ǫ
1 − 2ǫ
)
)
( )
1 1
= Ũ := 0 1
(sufficiently accurate result !)
( )
0 0
no row swapping, → (2.3.1): LŨ = A + E with E = 0 1
( )
0 0
after row swapping, → (2.3.2): LŨ = Ã + E with E = 0 ǫ
in M . (2.3.2)
unstable !
stable !
Suitable pivoting essential for controlling impact of roundoff errors on Gaussian elimination (→
Sect. 2.5.2)
✸
Ôº ¾º¿
choice of pivot row index j:
j ∈ {i,...,n} such that
for k = j, k ∈ {i,...,n}. (relatively largest pivot element p)
Explanations to Code 2.3.3:
|a ki |
∑ nl=i
|a kl|
→ max (2.3.3)
Line 4: Find the relatively largest pivot element p and the index j of the corresponding row of the
matrix.
Line 5: If the pivot element is still very small relative to the norm of the matrix, then we have encountered
an entire column that is close to zero. The matrix is (close to) singular and LUfactorization
does not exist.
Line 6: Swap the first and the j-th row of the matrix.
Line 7: Initialize the vector of multiplier.
Line 8: Call the routine for the upper right (n − 1) × (n − 1)-block of the matrix after subtracting
suitable multiples of the first row from the other rows, cf. Rem. 2.1.3 and Rem. 2.2.4.
Ôº ¾º¿
Example 2.3.2 (Gaussian elimination with pivoting for 3 × 3-matrix).
⎛
A = ⎝ 1 2 2
⎞ ⎛
2 −7 2 ⎠ →
➊ ⎝ 2 −7 2
⎞ ⎛
1 2 2 ⎠ →
➋ ⎝ 2 −7 2
⎞ ⎛
0 5.5 1 ⎠ →
➌ ⎝ 2 −7 2
⎞ ⎛
0 27.5 −1 ⎠ →
➍ ⎝ 2 −7 2
⎞
0 27.5 −1 ⎠
1 24 0 1 24 0 0 27.5 −1 0 5.5 1 0 0 1.2
➊: swap rows 1 & 2.
➋: elimination with top row as pivot row
➌: swap rows 2 & 3
➍: elimination with 2nd row as pivot row
Algorithm 2.3.3.
MATLAB-code for recursive in place LU-factorization of A ∈ R n,n with partial pivoting (ger.: Spaltenpivotsuche):
Code 2.3.4: recursive Gaussian elimination with row pivoting
1 function A = gsrecpiv (A)
2 n = size (A, 1 ) ;
3 i f ( n > 1)
4 [ p , j ] = max( abs (A ( : , 1 ) ) . / sum( abs (A) ’ ) ’ ) ; %
Ôº ¾º¿
5 i f ( p < eps∗norm(A ( : , 1 : n ) ,1) ) , error ( ’ Nearly Singular m a trix ’ ) ; end %
6 A ( [ 1 , j ] , : ) = A ( [ j , 1 ] , : ) ; %
7 fac = A( 2 : end , 1 ) / A( 1 , 1 ) ; %
8 C = gsrecpiv (A ( 2 : end , 2 : end )−fac∗A( 1 , 2 : end ) ) ; %
✸
Line 9: Reassemble the parts of the LU-factors. The vector of multipliers yields a column of L, see
Ex. 2.2.1.
Algorithm 2.3.5.
C++-code for in-situ LU-factorization of
A ∈ R n,n with partial pivoting ➤
Row permutations recorded in vector p !
Usual choice of pivot:
j ∈ {i, ...,n} such that
|a ki |
∑ nl=i
|a kl|
→ max (2.3.4)
for k = j, k ∈ {i,...,n}.
(relatively largest pivot element)
Why relatively largest pivot element in (2.3.4)?
template
void LU(Matrix &A,std::vector &p) {
int n = A.dim();
for(int i=1;i