Denition a collection of n vectors b1 b2 bn in

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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.

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