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Unformatted text preview:  +  sin θ,  − sin θ +  cos θ} =  cos θ +  sin θ, Then = [x2 cos2 θ + x2 sin2 θ − 2x1x2 cos θ sin θ
1
2 Thus A 1 −1 and = [x2 + x2 ]1/2 = x
1
2 terms of θ). 2 ≤ A 1, and κ2(A) ≤ κ1(A), but for a general matrix, we cannot tell which norm or condition number is larger. 2 Ax 2
= max 1 = 1.
x=0
x2
= 1. Then κ2(A) = 1.
A geometric way of viewing this:
First, note that the condition number of a matrix denotes the ratio of the maximal stretching over the
Tut5 – Norms, condition numbers c C. Christara, 201213 5 Tut5 – Norms, condition numbers 7 c C. Christara, 201213 Tut5 – Norms, condition numbers 8 c C. Christara, 201213 minimal stretching (or maximal shrinking) that the matrix gives rise to, when applied to any nonzero
vector: κa (A) = AaA−1a =
= max
x=0 Ax
maxx=0 xa
A−1xa
Axa
y a
Axa
a
max
= max
max
=
Ay a
x=0
x=0 xa y =0 Ay a
xa
xa
miny =0 y a Then, notice that A represents a counterclockwise rotation of x by θ radians. We can see that the
Euclidian norm (length) of x does not change if A is applied to it. Since A produces neither stretching
nor shrinking when applied to any vector, it has condition number 1 with respect to Euclidian norm.
Another way of viewing this:
(Ax)T Ax
Ax2
It is easy to see that A is orthogonal, i.e., AT A = I. Then A2 = max
= max √
=
x=0 x2
x=0
xT x
√
√
xT AT Ax
xT x
max √
= max √
= 1. Also, since A is square (and orthogonal), its inverse is its transx=0
x=0
xT x
xT x
pose, i.e., A−1 = AT . So we have AAT = I, i.e., AT is also orthogonal (and square). Thus AT 2 = 1,
and thus κ2(A) = 1. Note: For the above matrix A, it is easy to ﬁnd κ2(A). For arbitrary matrices, it is not straightforward.
Even if we have the inverse explicitly, it is not always easy to calculate A Tut5 – Norms, condition numbers 6 2 and A−1 2. c C. Christara, 201213 Q UESTION 7 Let A = ANSWER: It is easy to see that A−1 = A ∞ = max{8, 2} = 8, A−1 κ∞(A) = A...
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 Winter '14

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