v2007.09.13 - Convex Optimization

5.8. EUCLIDEAN METRIC VERSUS MATRIX CRITERIA 3475.8.3.1.1 Example. Small completion problem, II.Applying the inequality for λ 1 in (876) to the small completion problem onpage 292 Figure 74, the lower bound on √ d 14 (1.236 in (698)) is tightenedto 1.289 . The correct value of √ d 14 to three significant figures is 1.414 .5.8.4 Affine dimension reduction in two dimensions(confer5.14.4) The leading principal 2×2 submatrix T of −VN TDV N haslargest eigenvalue λ 1 (864) which is a convex function of D . 5.39 λ 1 can neverbe 0 unless d 12 = d 13 = d 23 = 0. Eigenvalue λ 1 can never be negative whilethe d ij are nonnegative. The remaining eigenvalue λ 2 is a concave functionof D that becomes 0 only at the upper and lower bounds of inequality (865a)and its equivalent forms: (confer (867))| √ d 12 − √ d 23 | ≤ √ d 13 ≤ √ d 12 + √ d 23 (a)⇔| √ d 12 − √ d 13 | ≤ √ d 23 ≤ √ d 12 + √ d 13 (b)⇔| √ d 13 − √ d 23 | ≤ √ d 12 ≤ √ d 13 + √ d 23 (c)(877)In between those bounds, λ 2 is strictly positive; otherwise, it would benegative but prevented by the condition T ≽ 0.When λ 2 becomes 0, it means triangle △ 123 has collapsed to a linesegment; a potential reduction in affine dimension r . The same logic is validfor any particular principal 2×2 submatrix of −VN TDV N , hence applicableto other triangles.5.39 The largest eigenvalue of any symmetric matrix is always a convex function of itsentries, while the ⎡ smallest ⎤ eigenvalue is always concave. [46, exmp.3.10] In our particulard 12case, say d =∆ ⎣d 13⎦∈ R 3 . Then the Hessian (1532) ∇ 2 λ 1 (d)≽0 certifies convexityd 23whereas ∇ 2 λ 2 (d)≼0 certifies concavity. Each Hessian has rank equal to 1. The respectivegradients ∇λ 1 (d) and ∇λ 2 (d) are nowhere 0.