1 2 43 226 suppose that 0 a 2 0 2 3 0 0 0 9 a

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Unformatted text preview: eigenvalue also in this case. In either case, λ = 1 is an eigenvalue and W1 = {x ∈ R 3 : B x = x} is nonzero. It is easily verified that if dim W1 is larger than one, then B must be the identity. Thus if B = I , there is a one-dimensional subspace W1 of R 3 which is left fixed under multiplication by B . This is the axis of rotation. ⊥ Let W1 denote the orthogonal complement to W1 . If x is a nonzero element ⊥ of W1 and y ∈ W1 , it follows from (2.8) that (B y) · x = (B y) · (B x) = yT B T B x = yT I x = y · x = 0, ⊥ ⊥ so B y ∈ W1 . Let y, z be elements of W1 such that y · y = 1, y · z = 0, z · z = 1; ⊥ we could say that {y, z} form an orthonormal basis for W1 . By (2.8), (B y) · (B y) = 1, (B y) · (B z) = 0, (B z) · (B z) = 1. ⊥ Thus B y must be a unit-length vector in W1 and there must exist a real number θ such that B y = cos θy + sin θz. ⊥ Moreover, B z must be a unit-length vector in W1 which is perpendicular to B y and hence B z = ±(− sin θy + cos θz). However, we cannot have B z = −(...
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This document was uploaded on 01/12/2014.

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