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Unformatted text preview: ⇒ S1 ⊆ S3 ∞ κ∞(DA) = D A In 2, we have already shown that S3 ⊆ S1. Below, we show that S1 ⊆ S3, thus S1 = S3 . 1. x ∈ S1 ⇒ x = 1 ⇒ x = κ∞(A) = A κ∞(A) = A Note: We can avoid showing 1 above, if we show equality in 2. Tut5 – Norms, condition numbers −1 1. Let S1 = {x : x = 1}, S2 = {x : x = 0}. Clearly, S1 ⊂ S2. Then, we have
Ax
Ax
≥ max
= max Ax .
max
x=0
x =1
x =1
x
x Ax
x . So we essentially proved that there is an equivalent deﬁnition of the induced matrix norm, namely, A ≡ max x =1 { condition number because it is nearly singular. δ 0 for some δ ∼ 0 (but δ = 0). Find κ∞ (A) and det(A).
Q UESTION 4 Let A = 0δ
ANSWER: We have
κ∞(A) = A ∞ A−1 det(A) = δ 2 (small) ∞ = δ · 1 = 1 (small),
δ Ax }.
Note: In general, there is no relation between κ(A) and det(A). A singular matrix A has κ(A) = ∞ and det(A) = 0. However, a matrix A with small determinat is not necessarily nearly singular, and a
matrix with large condition number is not necessarily nearly singular. If a matrix has large condition
number after scaling, then it is nearly singular. Tut5 – Norms, condition numbers 2 c C. Christara, 201213 Tut5 – Norms, condition numbers 4 c C. Christara, 201213 Q UESTION 5 Consider A =
Euclidian norm. cos θ − sin θ
sin θ
−1 ANSWER: It is easy to see that A Note that Ax = x1 cos θ − x2 sin θ x1 sin θ + x2 cos θ
Ax 2 cos θ = cos θ Q UESTION 6 With the same A as before, ﬁnd the condition number of A in the onenorm (possibly in
for some θ. Find the condition number of A in the
sin θ − sin θ cos θ ANSWER: We have A . A Similarly, = max
x=0
A−1 2 1 = max{ cos θ +  sin θ,  − sin θ +  cos θ} =  cos θ +  sin θ. κ1(A) = = (x1 cos θ − x2 sin θ)2 + (x1 sin θ + x2 cos θ) 2 1/2 A 1 A−1 1 = ( cos θ +  sin θ)2 = cos2 θ + sin2 θ + 2 sin θ cos θ = 1 + 2 sin θ cos θ
Note: For this A, we have A +x2 sin2 θ + x2 cos2 θ + 2x1x2 cos θ sin θ]1/2
1
2 2 = max{ cos θ...
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 Winter '14

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