# Tut05

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Unformatted text preview: A−1 ∞ ∞ Q UESTION 8 With the same A as above, ﬁnd the condition number of A−1 in the inﬁnity norm. 8 0 . Find the condition number of A in the inﬁnity norm. 02 ∞ 1/8 0 ANSWER: Note that κ(A) = A A−1 = κ(A−1). −1 So κ(A ) = 4, and we don’t need to do any further calculations. 1/8 0 −1 gives rise to maximal stretching (minimal shrinking) of 1/2 and Notice also that A = 0 1/2 minimal stretching (maximal shrinking) of 1/8. . 0 1/2 = max{1/8, 1/2} = 1/2. = 8 · 1 = 4. 2 Note: The condition number of a matrix denotes the ratio of the maximal stretching over the minimal stretching (or maximal shrinking) that the matrix gives rise to, when applied to any non-zero vector. General note: For diagonal (and some other special) matrices, it is easy to calculate max x=0 Ax Ax and min , so it x=0 x x is easy to calculate the condition number through the ratio. We can use this interpretation of the condition number to ﬁnd the condition number of A. For arbitrary matrices, it is not straightforward (and it may not be possible). First, as an example for this A, take x = Then Ax = 8 0 1 02 Now take x = 0 1 0 = 8 0 1 , for which we have x 0 . Thus Ax ∞ , for which we also have x ∞ = 1. 1 c C. Christara, 2012-13 9 8 0 0 02 = 1. = 8, which means that A stretches x by a factor or 8. Tut5 – Norms, condition numbers Then Ax = ∞ = 0 2 . Thus Ax ∞ = 2, which means that A stretches x by a factor or 2. We can show that, for this A, 8 and 2 are the maximal and minimal stretching A produces, when applied to any x = 0: Consider x = x1 x2 , then Ax = 1 Case 4 |x2| ≤ |x1| < |x2| x ∞ = |x2|, Ax ∞ = |x1|, Ax ∞ = |x2|, Ax , Ax ∞ = max{8|x1|, 2|x2|} and x 8|x1 | |x2 | = 8|x1|. Then Ax ∞ x∞ = ∞ = 8|x1|. Then Ax ∞ x∞ ∞ = 2|x2|. Then Ax ∞ x∞ ∞ = max{|x1|, |x2|}. = 8. = 2. 1 Case |x1| ≤ 4 |x2| x 2x2 ∞ Case |x2| ≤ |x1| x 8x1 Thus, for any x = 0, we have 2 ≤ Ax ∞ x∞ is ≥ 2 and < 8. ≤ 8, and there exist vectors for which the maximal and minimal stretchings 8 and 2 are obtained (see above x = (1, 0)T and x = (0, 1)T ), so κ∞ (A) = Tut5 – Norms, condition numbers 10 8 2 = 4. c C. Christara, 2012-13 Tut5 – Norms, condition numbers 11 c C. Christara, 2012-13...
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## This document was uploaded on 01/27/2014.

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