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Choose the elements of y such that ρj yj = |ρj |. That is, if ρj = 0, then yj = 0, and otherwise yj = |ρj |/ρj . Then y ∞ = 1 and |rk∗ y| = nj=1 ρj yj = nj=1 |ρj | = rk 1 . Hence A ∞ ≥ |rk∗ y| = rk 1 = max ri 1≤i≤m 1. The p-norms satisfy the following submultiplicative inequality. 18 (Norm of a Product). If A ∈ Cm×n and B ∈ Cn×p , then AB Proof. 15 twice gives AB p = ABx p ≤ A p p p ≤ A B p p. = ABx p and x ≤ A p B Bx p p p x p = 1. Applying Remark = A p B p. Since the computation of the two norm is more involved, we postpone it until later.
30 (Relative Distance to Singularity). Let A ∈ Cn×n be nonsingular. Then E A min where κ2 (A) = A A−1 2 2 2 : A + E is singular = 1 , κ2 (A) 2. Therefore, matrices that are ill-conditioned with respect to inversion are close to singular, and vice versa. In other words, matrices that are close to being singular have sensitive inverses. The example below illustrates that absolute and relative distance to singularity are not the same. Example. Just because a matrix is close to singularity in the absolute sense does not imply that it is also close to singularity in the relative sense.
13. (b) Suppose we want to compute an approximation to the solution of Ax = en when α = 2 and n = 100. 1? 2. Componentwise Condition Numbers. Let A ∈ Cn×n be nonsingular, b = 0, and Ax = b. Prove: If xj = 0, then b − Az |zj − xj | ≤ κj |xj | b p p , where κj = x p ∗ −1 ej A |x|j p A p. We can interpret κj as the condition number for xj . Which components of x would you expect to be sensitive to perturbations? 3. Condition Estimation. Let A be nonsingular. Show how to determine a lower bound for κp (A) with one linear system solution involving A.
Calculus on the HP-48G by Grapevine Publications