413 chapter 8 ism linear algebra v 2 e 5 e 1 1 v 1 5

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Chapter 8 ISM: Linear Algebra v 2 E 5 E 0 1 1 v 1 5 1 v 1 5 v 1 Figure 8.7: for Problem 8.2.19. v 2 E –4 E 6 1 v 1 6 1 v 1 6 v 1 Figure 8.8: for Problem 8.2.20. 414
ISM: Linear Algebra Section 8.2 21. a. In each case, it is informative to think about the intersections with the three coordinate planes: x 1 - x 2 , x 1 - x 3 , and x 2 - x 3 . For the surface x 2 1 + 4 x 2 2 + 9 x 2 3 = 1, all these intersections are ellipses , and the surface itself is an ellipsoid . This surface is connected and bounded; the points closest to the origin are ± 0 0 1 3 , and those farthest ± 1 0 0 . (See Figure 8.9.) x 3 x 1 x 2 Figure 8.9: for Problem 8.2.21a: x 2 1 + 4 x 2 2 + 9 x 2 3 = 1 , an ellipsoid (not to scale). In the case of x 2 1 + 4 x 2 2 - 9 x 2 3 = 1, the intersection with the x 1 - x 2 plane is an ellipse, and the two other intersections are hyperbolas. The surface is connected and not bounded; the points closest to the origin are ± 0 1 2 0 . (See Figure 8.10.) In the case - x 2 1 - 4 x 2 2 + 9 x 2 3 = 1, the intersection with the x 1 - x 2 plane is empty, and the two other intersections are hyperbolas. The surface consists of two pieces and is 415
Chapter 8 ISM: Linear Algebra x 3 x 1 x 2 Figure 8.10: for Problem 8.2.21a: x 2 1 +4 x 2 2 - 9 x 2 3 = 1 , a hyperboloid of one sheet (not to scale). unbounded. The points closest to the origin are ± 0 0 1 3 . (See Figure 8.11.) b. A = 1 1 2 1 1 2 2 3 2 1 3 2 3 is positive definite, with three positive eigenvalues λ 1 , λ 2 , λ 3 . The surface is given by λ 1 c 2 1 + λ 2 c 2 2 + λ 3 c 2 3 = 1 with respect to the principal axes, an ellipsoid . To find the points closest to and farthest from the origin, use technology to find the eigenvalues and eigenvectors: eigenvalues: λ 1 0 . 56 , λ 2 4 . 44 , λ 3 = 1 unit eigenvectors: v 1 0 . 86 0 . 19 - 0 . 47 , v 2 0 . 31 0 . 54 0 . 78 , v 3 = 1 6 1 - 2 1 Equation: 0 . 56 c 2 1 + 4 . 44 c 2 2 + c 2 3 = 1 Farthest points when c 1 = ± 1 0 . 56 and c 2 = c 3 = 0 Closest points when c 2 = ± 1 4 . 44 and c 1 = c 3 = 0 416
ISM: Linear Algebra Section 8.2 x 3 x 2 x 1 Figure 8.11: for Problem 8.2.21a: - x 2 1 - 4 x 2 2 + 9 x 2 3 = 1 , a hyperboloid of two sheets (not to scale). Farthest points ± 1 0 . 56 0 . 86 0 . 19 - 0 . 47 ± 1 . 15 0 . 26 - 0 . 63 Closest points ± 1 4 . 44 0 . 31 0 . 54 0 . 78 ± 0 . 15 0 . 26 0 . 37 22. A = - 1 0 5 0 1 0 5 0 - 1 ; eigenvalues λ 1 = 4, λ 2 = - 6, λ 3 = 1 Equation with respect to principal axes: 4 c 2 1 - 6 c 2 2 + c 2 3 = 1, a hyperboloid of one sheet (see Figure 8.10). Closest to origin when c 1 = ± 1 2 , c 2 = c 3 = 0. A unit eigenvector for eigenvalue 4 is v = 1 2 1 0 1 , so that the desired points are 417
Chapter 8 ISM: Linear Algebra ± 1 2 1 2 1 0 1 ± 0 . 35 0 0 . 35 . 23. Yes; M = 1 2 ( A + A T ) is symmetric, and x T Mx = 1 2 x T Ax + 1 2 x T A T x = 1 2 x T Ax + 1 2 x T Ax = x T Ax Note that x T Ax is a 1 × 1 matrix, so that x T Ax = ( x T Ax ) T = x T A T x .

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