Ellipsoid: Recall that an ellipsoid is a set of the form
n > P = {x ∈ R :(x − a) A(x − a) ≤ 1} (1)
where A is a (positive) definite matrix and a ∈ Rn. Here the point a is called the center of n the ellipsoid. Note that the unit ball Bn = {x ∈ R : ||x|| ≤ 1} is an ellipsoid.
Proposition 5.1
n (i) A set P ⊆ R is an ellipsoid if and only if P is the image of Bn under an affine transformation.
− 1 (ii) If P is an ellipsoid given by A, a as in (1), then Vol(P ) = (det(A)) 2
Proof: Let P be given by A, a as in (1). Since A is positive definite, we may write A = U >U where U is invertible. Now, consider applying the affine transformation x → U(x − a). This maps P to the set
> > > U(P ) = {U(x − a):(x − a) U U(x − a) ≤ 1} = {y : y y ≤ 1} = Bn
(here Bn is the n-dimensional ball of radius 1). So, this mapping sends P to Bn. The proof of (i) follows immediately from this. For (ii), note that the volume of P is
−1 − 1 Vol(P ) = det(U )Vol(Bn) = (det(A)) 2 Vol(Bn).
Theorem 5.2 (L¨owner-John) If K ⊆ Rn is compact, then there is a unique ellipsoid LJ(K) of minimum volume containing K. Furthermore, if K is convex, then scaling LJ(K) 1 by a factor of n results in an ellipsoid which is contained in K.
Lemma 5.3 Let P ⊆ Rn be an ellipsoid centered at a and let H be a halfspace defined by a hyperplane which contains a. Then there is an ellipsoid P 0 with the following properties:
(i) P ∩ H ⊆ P 0
− 1 (ii) Vol(P 0) ≤ e 2(n+1) Vol(P ) 2
Proof: First consider the special case that P = Bn. In this case, we may assume (by possibly n performing a rotation) that H = {x ∈ R : x1 ≥ 0}. We claim that
( 2 n ) (n + 1)2 1 n2 − 1 X P 0 = x ∈ n : x − + x2 ≤ 1 R n2 1 n + 1 n2 i i=2
0 1 has the desired properties. First, note that P is an ellipsoid with center ( n+1 , 0,..., 0) and (n+1)2 n2−1 matrix A a diagonal matrix with n2 in the first position and n2 in the others. To prove (ii) we need to compute Vol(P 0). Let U be the diagonal matrix with U 2 = A. Then we have
0 Vol(P ) − 1 = (det(A)) 2 Vol(Bn) n−1 n n 2 = n + 1 n2 − 1 n−1 − 1 − 1 1 − 1 ≤ e n+1 e (n2−1)2 = e n+1 e 2(n+1) = e 2(n+1)
Here we have used the fact that 1 + x ≤ ex for al x. Now, to prove (i), let x ∈ H ∩ P and observe that
2 n (n + 1)2 1 n2 − 1 X x − + x2 n2 1 n + 1 n2 i i=2 n 2(n + 1) 2(n + 1) 1 n2 − 1 X = x2 − x + + x2 n2 1 n2 1 n2 n2 i i=1 2(n + 1) 1 n2 − 1 ≤ x (x − 1) + + n2 1 1 n2 n2 ≤1.
This completes the proof in the special case when P = Bn. For the general case, choose
an affine linear transformation f(x) = Bx + b so that f maps P to Bn. Then apply the special case to choose P 0 and then apply the transformation f −1. Since affine transformations preserve volume ratios, this yields a suitable ellipsoid for P .