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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 onedimensional 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 unitlength 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 unitlength 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.
 Winter '14
 Equations

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