By 28 b y b y 1 b y b z 0 b z b z

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Unformatted text preview: ral Theorem, there is a 3 × 3 orthogonal matrix B of determinant one such that B T AB is diagonal. We introduce new coordinates y1 by setting x = B y, y = y2 y3 and equation (2.7) becomes yT (B T AB )y = 1. Thus after a suitable linear change of coordinates, the equation (2.7) can be put in the form λ1 0 y1 0 y1 y2 y3 0 λ2 0 y2 = 1, y3 0 0 λ3 39 4 1 0.5 0 -0.5 -1 2 0 -2 -2 -2 0 2 -4 Figure 2.2: An ellipsoid. or 2 2 2 λ1 y1 + λ2 y2 + λ3 y3 = 1, where λ1 , λ2 , and λ3 are the eigenvalues of A. It is relatively easy to sketch the quadric surface in the coordinates (y1 , y2 , y3 ). If the eigenvalues are all nonzero, we ﬁnd that there are four cases: • If λ1 , λ2 , and λ3 are all positive, the quadric surface is an ellipsoid . • If two of the λ’s are positive and one is negative, the quadric surface is an hyperboloid of one sheet . • If two of the λ’s are negative and one is positive, the quadric surface is an hyperboloid of two sheets . • If λ1 , λ2 , and λ3 are all negative, the equation represents the empty set . Just as in the cas...
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