Approximating a Convex Body by An Ellipsoid

Transcription

1 Chapter 1 Approximating a Convex Body by An Ellipsoid By Sariel Har-Peled, May 10, Is there anything in the Geneva Convention about the rules of war in peacetime? Stalnko wanted to know, crawling back toward the truck. Absolutely nothing, Caulec assured him. The rules of war apply only in wartime. In peacetime, anything goes. Gasp, Romain Gary. In this chapter, we show that any convex body can be approximated reasonably well by an ellipsoid. (A quick reminder of linear algebra and the notations we use is provided in Section 1.4 p67.) 1.1 Ellipsoids Definition Let b = { x IR n x 1 be the unit ball. Let c IR n be a vector and let T : IR n IR n be an invertible linear transformation. The set is called an ellipsoid and c is its center. E = T(b) + c Alternatively, we can write E = {x IR n T 1 (x c) 1. However, T 1 (x c) = T 1 (x c), T 1 (x c) = (T 1 (x c)) T T 1 (x c) = (x c) T( T 1) T T 1 (x c). 1 This work is licensed under the Creative Commons Attribution-Noncommercial 3.0 License. To view a copy of this license, visit or send a letter to Creative Commons, 171 Second Street, Suite 300, San Francisco, California, 94105, USA. 63

2 In particular, let Q = ( T 1) T T 1, and observe that Q is symmetric, and positive definite (see Claim p67 ). Thus, { E = x IR n (x c) T Q(x c) 1. If we change the basis of IR n to be the set of unit eigenvectors of Q, then Q becomes a diagonal matrix, and we have that { E = (y 1,..., y n ) IR n λ 1 (y 1 c 1 ) + + λ n (y n c n ) 1, where c = (c 1,..., c n ) and λ 1,..., λ n are the eigenvalues of Q (all positive since Q is positive definite). In particular, this implies that the points ( c 1,..., c i ± 1/ ) λ i,..., c n, for i = 1,..., n, lies on the boundary of E. Specifically, the eigenvectors are the main axes of the ellipsoid, and their corresponding eigenvalues specify how long is the ellipse along each one of these axes. That is, the ith axis of the ellipsoid E has length 1/ λ i. Visually, the ellipsoid is being created by taking the unit ball, and stretching it by 1/ λ i along its ith axis. In particular, Vol(E) = Vol(b) λ1 λ n = Vol(b) det(q). For a convex body K (i.e., a convex and bounded set), let E be a largest volume ellipsoid contained inside K. One can show that E is unique. Namely, there is a single maximum volume ellipsoid inside K. The following is obvious and is stated for the sake of clarity. Observation 1.1. Given two bodies X, Y IR n such that X Y and a non-singular affine transformation M, then Vol(X)/ Vol(Y) = Vol(M(X))/ Vol(M(Y)) and M(X) M(Y). Theorem Let K IR n be a convex body, and let E { K be its maximum volume ellipsoid. Suppose that E is centered at the origin, then E K n E = nx x E. Proof : By applying an affine transformation (and by Observation 1.1.), we can assume that E is the unit ball b. Assume, for the sake of contradiction, that there is a point p K, such that p > n. Consider the set C which is the convex-hull of {p b. Since K is convex, we have that C K. We will reach a contradiction, by finding an ellipsoid G which has volume larger than b and is enclosed inside C. By rotating space, we can assume that the apex p of C is the point (ρ, 0,..., 0), for ρ > n. We consider ellipsoids of the form G = (y 1,..., y n ) (y 1 τ) + 1 n y α β i 1. i= ( 1, 0) o β α { { (τ, 0) p = (ρ, 0) (Using the notations above we have λ 1 = 1/α, λ =... = λ n = 1/β, c 1 = τ and c = = c n = 0.) We need to pick the values of τ, α and β such that G C. Observe that by symmetry, it is enough to enforce that G C in the first two dimensions. 64

3 Thus, we can consider C and G to be in two dimensions. Now, the center of G is going to be on the x-axis, at the point (τ, 0). The set G is an ellipse with axes parallel to the x and y axes. In particular, we require that ( 1, 0) be on the boundary of G. This implies that ( 1 τ) = α. Now, the center point (τ, 0) can be anywhere on the interval between the origin and the midpoint between ( 1, 0) and p = (ρ, 0). As such, we conclude that 0 τ (ρ + ( 1))/. Namely, In particular, the equation of the curve forming (the boundary of) G is α = 1 + τ. (1.1) F(x, y) = (x τ) (1 + τ) + y 1 = 0. (1.) β We next compute the value of β. Consider the tangent l to the unit circle that passes through p = (ρ, 0). Let q be the point where l touches the unit circle. (See figure on the right.) We have poq oqs, since oq form a right angle with pq, where s = (0, s y ). Since q o = 1, we have s q = s q q o = q o o p = 1 ρ. ( 1, 0) G C s o q (τ, 0) l (u, v) p = (ρ, 0) Furthermore, since q is on the unit circle, we have As such, the equation of the line l is q = ( 1/ρ, 1 1/ρ ). q, (x, y) = 1 x ρ + 1 1/ρ y = 1 l y = 1 ρ 1 x + ρ ρ 1. Next, consider the tangent line l to G at some point (u, v). We will derive a formula for l as a function of (u, v) and then require that l = l. (Thus, (u, v) will be the intersection point of G and l, which exists as G is as large as possible.) The slope of the l is the slope of the tangent to G at (u, v). We now compute the derivative of the implicit function F (see Eq. (1.) above) at (u, v), which is dy dx = F ( ) ( ) x(u, v) (u τ) v F y (u, v) = / (1 + τ) β = β (u τ) v(1 + τ) by computing the derivatives of the implicit function F. Since α = 1 + τ (see Eq. (1.1)), the line l is (u τ) α (x τ) + v β y = 1, For the rusty reader (and the rustier author) here is a quicky proof of the formula for derivative of an implicit function in two dimensions: Locally near (u, v) the function F(x, y) behaves like the linear function G(x, y) = (x u)f x (u, v) + (y v)f y (u, v). The curve is defined by F(x, y) = 0, which locally is equivalent to G(x, y) = 0. But G(x, y) = 0 is a line (which is the tangent line to F(x, y) at (u, v)). Rearranging, we get that y = ( F x (u, v)/f y (u, v) ) x + whatever, establishing the claim. 65

4 since it has the required slope and it passes through (u, v), as (u, v) G and by Eq. (1.). Namely l y = β (u τ) (x τ) + β vα v. Setting l = l, we have that the line l passes through (ρ, 0). As such, (ρ τ)(u τ) (u τ) = 1 = α (u τ) α = α α ρ τ α (ρ τ). (1.3) Since l and l are the same line, we can equate their slopes; namely, However, by Eq. (1.3), we have Thus, β (u τ) vα β (u τ) vα = = β (u τ) vα α β v = 1 ρ 1. = β vα α ρ τ = β v(ρ τ). ρ τ ρ 1. Squaring and inverting both sides, we have v β = ρ 1 v and thus 4 (ρ τ) β = ρ 1 β. The point (ρ τ) (u τ) (u, v) G, and as such + v = 1. Using Eq. (1.3) and the above, we get α β and thus α (ρ τ) + ρ 1 β (ρ τ) = 1, β = (ρ τ) α = (ρ τ) (τ + 1) ρ 1 ρ 1 (ρ τ 1)(ρ + 1) = = ρ τ 1 ρ 1 ρ 1 = ρ ρτ τ 1 ρ 1 = 1 τ ρ 1. Namely, for n, and 0 τ < (ρ 1)/, we have that the ellipsoid G is defined by the parameters α = 1 + τ and β, and it is contained inside the cone C. It holds that Vol(G) = β n 1 α Vol(b), and thus µ = ln Vol(G) Vol(b) = (n 1) ln β + ln α = n 1 ln β + ln α. For τ > 0 sufficiently small, we have ln α = ln(1 + τ) = τ + O(τ ), because of the Taylor expansion ln(1 + x) = x x / + x 3 /3, for 1 < x 1. Similarly, ln β = ln ( ) 1 τ ρ 1 = τ + ρ 1 O(τ ). Thus, µ = n 1 ( τ ) ( ρ 1 + O(τ ) + τ + O(τ ) = 1 n 1 ) τ + O(τ ) > 0, ρ 1 for τ sufficiently small and if ρ > n. Thus, Vol(G)/ Vol(b) = exp(µ) > 1 implying that Vol(G) > Vol(b). A contradiction. 66

5 A convex body K centered at the origin is symmetric if p K, implies that p K. Interestingly, the constant in Theorem can be improved to n in this case. We omit the proof, since it is similar to the proof of Theorem Theorem Let K IR n be a symmetric convex body, and let E K be its maximum volume ellipsoid. Suppose that E is centered at the origin, then K n E. 1. Bibliographical notes We closely follow the exposition of Barvinok [Bar0]. The maximum volume ellipsoid is sometimes referred to as John s ellipsoid. In particular, Theorem is known as John s theorem, and was originally proved by Fritz John [Joh48]. One can approximate the John ellipsoid using interior point techniques [GLS88]. However, this is not very efficient in low dimensions. In the next chapter, we will show how to do this by other (simpler) techniques, which are faster (and simpler) for point sets in low dimensions. There are numerous interesting results in convexity theory about what convex shapes look like. One of the surprising results is Dvoretsky s theorem, which states that any symmetric convex body K around the origin (in high enough dimension) can be cut by a k-dimensional subspace, such that the intersection of K with this k-dimensional space, contains an ellipsoid E and is contained inside an ellipsoid (1 + ε)e (k here is an arbitrary parameter). Since one can define a metric in a Banach space by providing a symmetric convex body which defines the unit ball, this implies that any high enough dimensional Banach space contains (up to translation that maps the ellipsoid to a ball) a subspace which is almost Euclidean. A survey of those results is provided by Ball [Bal97]. 1.3 From previous lectures 1.4 Linear Algebra Claim If M is a positive definite matrix then there exists a matrix B such that M = N T B and B is not singular. Bibliography [Bal97] [Bar0] K. Ball. An elementary introduction to modern convex geometry. In Flavors of geometry, volume MSRI Publ. 31. Cambridge Univ. Press, publications/books/book31/files/ball.pdf. A. Barvinok. A course in convexity, volume 54 of Graduate Studies in Mathematics. American Mathematical Society, Providence, RI,

6 [GLS88] M. Grötschel, L. Lovász, and A. Schrijver. Geometric Algorithms and Combinatorial Optimization, volume of Algorithms and Combinatorics. Springer-Verlag, Berlin Heidelberg, nd edition, nd edition [Joh48] F. John. Extremum problems with inequalities as subsidary conditions. Courant Anniversary, pages , [Leo98] S. J. Leon. Linear Algebra with Applications. Prentice Hall, 5th edition,