2 an ellipsoid or 2 2 2 1 y1 2 y2 3 y3 1 where 1

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Unformatted text preview: unit-length vector, and can therefore be written in the form b1 = cos θ , sin θ for some choice of θ. The second column b2 is a unit-vector perpendicular to b1 and hence − sin θ b2 = ± . cos θ We must take the plus sign in this last expression, because det B = 1. Thus B 1 0 = cos θ , sin θ B 0 1 = − sin θ , cos θ or equivalently, B takes the standard basis vectors for R2 to vectors which have been rotated counterclockwise through an angle θ. By linearity, B= cos θ sin θ − sin θ cos θ must rotate every element of R 2 counterclockwise through an angle θ. Thus once we have sketched the conic in (y1 , y2 )-coordinates, we can obtain the sketch in (x1 , x2 )-coordinates by simply rotating counterclockwise through the angle θ. 37 Example. Let’s consider the conic 5x2 − 6x1 x2 + 5x2 = 1, 1 2 or equivalently, x1 −3 5 5 −3 x2 x1 x2 (2.6) = 1. The characteristic equation of the matrix −3 5 5 −3 A= is (5 − λ)2 − 9 = 0, λ2 − 10λ + 16 = 0, (λ − 2)(λ − 8) = 0, and the eigenvalues are λ1 = 2 and λ2 = 8. Unit-length eigenvectors corresponding to these eigenvalues are √...
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This document was uploaded on 01/12/2014.

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