Numerical Methods Contents - SAM

Terminology:
Def. 2.5.5 introduces stability in the sense of
backward error analysis
x
̂x
X
F
y
˜F(x)
Y
Theorem 2.5.7 (Stability of Gaussian elimination with partial pivoting).
Let A ∈ R n,n be regular and A (k) ∈ R n,n , k = 1,...,n − 1, denote the intermediate matrix
arising in the k-th step of Algorithm 2.3.5 (Gaussian elimination with partial pivoting).
For the approximate solution ˜x ∈ R n,n of the LSE Ax = b, b ∈ R n , computed by Algorithm
2.3.5 (based on machine arithmetic with machine precision eps, → Ass. 2.4.2) there is
∆A ∈ R n,n with
2.5.3 Roundoff analysis of Gaussian elimination
max
‖∆A‖ ∞ ≤ n 3 3eps
1 − 3neps ρ ‖A‖ ∞ , ρ := i,j,k |(A(k) ) ij |
max |(A) ,
ij|
i,j
such that (A + ∆A)˜x = b .
F
˜F
✬
✩
Simplification:
equivalence of Gaussian elimination and LU-factorization extends to machine arithmetic,
cf. Sect. 2.2
✫
✪
ρ “small” ➥ Gaussian elimination with partial pivoting is stable (→ Def. 2.5.5)
Ôº½¾½ ¾º
If ρ is “small”, the computed solution of a LSE can be regarded as the exact solution of a LSE with
“slightly perturbed” system matrix (perturbations of size O(n 3 eps)).
Ôº½¾¿ ¾º
✬
Lemma 2.5.6 (Equivalence of Gaussian elimination and LU-factorization).
The following algorithms for solving the LSE Ax = b (A ∈ K n,n , b ∈ K n ) are
numerically equivalent:
✫
❶ Gaussian elimination (forward elimination and back substitution) without pivoting, see Algorithm
2.1.2.
❷ LU-factorization of A (→ Code 2.2.1) followed by forward and backward substitution, see
Algorithm 2.2.5.
Rem. 2.3.7 ➣ sufficient to consider LU-factorization without pivoting
A profound roundoff analysis of Gaussian eliminatin/LU-factorization can be found in [18, Sect. 3.3 &
3.5] and [24, Sect. 9.3]. A less rigorous, but more lucid discussion is given in [42, Lecture 22].
Here we only quote a result due to Wilkinson, [24, Thm. 9.5]:
✩
✪
Ôº½¾¾ ¾º
Bad news: exponential growth ρ ∼ 2 n is possible !
Example 2.5.2 (Wilkinson’s counterexample).
⎛
⎞
1 0 0 0 0 0 0 0 0 1
−1 1 0 0 0 0 0 0 0 1
n=10:
−1 −1 1 0 0 0 0 0 0 1
⎧
⎪⎨ 1 , if i = j ∨j = n ,
−1 −1 −1 1 0 0 0 0 0 1
−1 −1 −1 −1 1 0 0 0 0 1
a ij = −1 , if i > j , , A =
⎪⎩
−1 −1 −1 −1 −1 1 0 0 0 1
0 else.
−1 −1 −1 −1 −1 −1 1 0 0 1
⎜−1 −1 −1 −1 −1 −1 −1 1 0 1
⎟
⎝−1 −1 −1 −1 −1 −1 −1 −1 1 1⎠
−1 −1 −1 −1 −1 −1 −1 −1 −1 1
Partial pivoting does not trigger row permutations !
⎧
⎪⎨ 1 , if i = j ,
A = LU , l ij = −1 , if i > j ,
⎪⎩
0 else
Exponential blow-up of entries of U !
⎧
⎪⎨ 1 , if i = j ,
u ij = 2
⎪⎩
i−1 , if j = n ,
0 else.
Ôº½¾ ¾º
✸