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2.13: How to compute matrix norms Matrix norms are computed by applying the following formulas: 1-norm (Th. 2.8): kAk1 = maxj=1:n Pn ∞-norm (Th. 2.7): kAk1 = maxi=1:n i=1 |aij | Pn j=1 |aij | maximal column sum maximal row sum p 2-norm (Th. 2.9): kAk2 = maxi=1:n λi(ATA) where λi(ATA) is the ith eigenvalue of ATA. C. Führer: FMN081-2005 45 2.14: Condition of a Problem A mathematical problem can be viewed as a function mapping in-data to out-data (solution): f :D⊂V→W Condition number is a measure for the sensitivity of the out-data with respect to perturbations in in-data. Scalar case: y = f (x) and ŷ = f (x + δx): 1 00 ŷ − y = f (x + δx) − f (x) = f (x)δx + f (x + θδx)(δx)2 2! 0 C. Führer: FMN081-2005 46 2.15: Condition of a Problem (Cont.) Scalar case, relative: ŷ − y = y absolut (local) condition number Condx = |f 0(x)| 0 xf (x) f (x) δx + O(δx2) x relative (local) condition number 0 xf (x) condx(f ) = f (x) C. Führer: FMN081-2005 47 2.16: Condition of a Problem (Cont.) In general absolut (local) condition number relative (local) condition number Condx = kf 0(x)k kxk kf 0(x)k condx(f ) = kf (x)k and we get krel.output errork ≤ condx(f )krel.input errork C. Führer: FMN081-2005 48 2.17: Examples Example 1: (Summation) x1 Problem: f : R2 −→ R1 with 7→ x1 + x2 x2 0 T Jacobian: f (x1, x2) = 1, 1 In 1-norm: Condx1,x2 (f ) = 1 |x1| + |x2| |x1 + x2| Problem if two nearly identical numbers are subtracted. condx1,x2 (f ) = C. Führer: FMN081-2005 49 2.18: Examples (Cont.) Example 2: (Linear Systems) Problem: f : R −→ R with b 7→ x = A−1b 0 Jacobian: f b = A−1 n In 1-norm: n Condb(f ) = kA−1k kbkkA−1k kAkkxkkA−1k −1 condb(f ) = ≤ = kAkkA k =: κ(A) −1 kA bk kxk C. Führer: FMN081-2005 50 2.19: Examples (Cont.) The estimate is sharp, i.e. there is a worst case perturbation Example: A= 3 10 0 0 10−3 1 0 b= δb = 0 10−5 (see the exercises of the current week and Exercise 2.14 in the book.) C. Führer: FMN081-2005 51 Unit 3: Systems of nonlinear functions Example (p. 107) F (x) = 1 2 2 f1(x1, x2) x1 + x2 − 1 = =0 f2(x1, x2) 5x21 + 21x22 − 9 f1 0.5 f2 p p T ξ1 = (− 3/2, 1/2) ξ2 = ( 3/2, 1/2)T p p T ξ3 = (− 3/2, −1/2) ξ4 = ( 3/2, −1/2)T 0 -0.5 -1 -1 -0.5 0 0.5 1 C. Führer: FMN081-2005 52 3.1 Fixed Point Iteration in Rn Definition. [4.1] Let g : D ⊂ Rn → Rn, D closed (cf. p. 105), x0 ∈ D and g(D) ⊂ D. The iteration xk+1 = g(xk ) is called fixed point iteration or simple iteration. and similar to the scalar case we define Definition. [4.2] Let g : D ⊂ Rn → Rn, D closed (cf. p. 105). If there is a constant L < 1 such that kg(x) − g(y)k ≤ Lkx − yk then g is called contractive or contractive. C. Führer: FMN081-2005 53 3.2 Contractivity and norms Contractivity depends on the choice of a norm. Here a function, which is contractive in one norm, but not in another g(x) = 3/4 1/3 x 0 3/4 It follows kg(x) − g(y)k = kA(x − y)k ≤ kAkkx − yk Thus L = kAk. But kAk1 = kAk∞ = 13 12 and kAk2 = 0.9350. g is contractive in the 2-norm and dissipative and the others. C. Führer: FMN081-2005 54 3.3 Contractivity and norms (Cont.) We start a fixed point iteration in the last example with x = (1, 1)T and get the following norms: 2 1 1.5 different norms of fixed point iterates 2 1 inf 0.5 0 0 1 2 3 4 5 6 7 8 9 10 (see also ”equivalence of norms”) C. Führer: FMN081-2005 55 3.4 Fixed Point Theorem in Rn Theorem. [4.1] Let g : D ⊂ Rn → Rn, D closed, g(D) ⊂ D. If there is a norm such that g is contractive, then g has a unique fixed point ξ ∈ D and the fixed point iteration converges. Let J(x) be the Jacobian (functionalmatrix → flerdim) of g. If kJ(ξ)k < 1 then fixed point iterations converges in a neighborhood of ξ. (Th. 4.2) C. Führer: FMN081-2005 56