Chapter 8ISM:Linear Algebrav2E5E011v1√51v1–√5v1Figure 8.7: for Problem 8.2.19.v2E–4E61v1√61v1–√6v1Figure 8.8: for Problem 126.96.36.1994
ISM:Linear AlgebraSection 8.221. a. In each case, it is informative to think about the intersections with the three coordinateplanes:x1-x2,x1-x3, andx2-x3.•For the surfacex21+ 4x22+ 9x23= 1, all these intersections areellipses, and the surfaceitself is anellipsoid.This surface is connected and bounded; the points closest to the origin are±0013,and those farthest±100. (See Figure 8.9.)x3x1x2Figure 8.9: for Problem 8.2.21a:x21+ 4x22+ 9x23= 1,anellipsoid(not to scale).•In the case ofx21+ 4x22-9x23= 1, the intersection with thex1-x2plane is an ellipse,and the two other intersections are hyperbolas.The surface is connected and notbounded; the points closest to the origin are±0120. (See Figure 8.10.)•In the case-x21-4x22+ 9x23= 1, the intersection with thex1-x2plane is empty, andthe two other intersections are hyperbolas. The surface consists of two pieces and is415
Chapter 8ISM:Linear Algebrax3x1x2Figure 8.10: for Problem 8.2.21a:x21+4x22-9x23= 1,ahyperboloid of one sheet(not to scale).unbounded. The points closest to the origin are±0013. (See Figure 8.11.)b.A=1121122321323is positive definite, with three positive eigenvaluesλ1,λ2,λ3.The surface is given byλ1c21+λ2c22+λ3c23= 1 with respect to the principal axes, anellipsoid. To find the points closest to and farthest from the origin, use technology tofind the eigenvalues and eigenvectors:eigenvalues:λ1≈0.56,λ2≈4.44,λ3= 1unit eigenvectors:v1≈0.860.19-0.47,v2≈0.310.540.78,v3=1√61-21Equation: 0.56c21+ 4.44c22+c23= 1Farthest points whenc1=±1√0.56andc2=c3= 0Closest points whenc2=±1√4.44andc1=c3= 0416
ISM:Linear AlgebraSection 8.2x3x2x1Figure 8.11: for Problem 8.2.21a:-x21-4x22+ 9x23= 1,ahyperboloid of two sheets(not toscale).Farthest points≈±1√0.560.860.19-0.47≈±1.150.26-0.63Closest points≈±1√4.440.310.540.78≈±0.150.260.3722.A=-10501050-1; eigenvaluesλ1= 4,λ2=-6,λ3= 1Equation with respect to principal axes: 4c21-6c22+c23= 1, a hyperboloid of one sheet(see Figure 8.10).Closest to origin whenc1=±12,c2=c3= 0.A unit eigenvector for eigenvalue 4 isv=1√2101, so that the desired points are417
Chapter 8ISM:Linear Algebra±121√2101≈±0.3500.35.23. Yes;M=12(A+AT) is symmetric, andxTMx=12xTAx+12xTATx=12xTAx+12xTAx=xTAxNote thatxTAxis a 1×1 matrix, so thatxTAx= (xTAx)T=xTATx.