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Unformatted text preview: sin θy + cos θz).
Indeed, this would imply (via a short calculation) that the vector
u = cos(θ/2)y + sin(θ/2)z
⊥
is ﬁxed by B , in other words B u = u, contradicting the fact that u ∈ W1 .
Thus we must have
B z = − sin θy + cos θz, 41 1
0
1 1
0.5
0
0.5
1
1
1
0
1 Figure 2.3: Hyperboloid of one sheet.
⊥
and multiplication by B must be a rotation in the plane W1 through an angle
θ. Moreover, it is easily checked that y + iz and y − iz are eigenvectors for B
with eigenvalues
e±iθ = cos θ ± i sin θ. We can therefore conclude that a 3 × 3 orthogonal matrix B of determinant
one represents a rotation about an axis (which is the eigenspace for eigenvalue
one) and through an angle θ (which can be determined from the eigenvalues of
B , which must be 1 and e±iθ ).
Exercises:
2.2.1. Suppose that
A= 2
3 3
2 . a. Find an orthogonal matrix B such that B T AB is diagonal.
b. Sketch the conic section 2x2 + 6x1 x2 + 2x2 = 1.
1
2
c. Sketch the conic section 2(x1 − 1)2 + 6(x1 −...
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This document was uploaded on 01/12/2014.
 Winter '14
 Equations

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