Extremal inscribed and circumscribed complex ellipsoids
aa r X i v : . [ m a t h . M G ] D ec Extremal inscribed and circumscribedcomplex ellipsoids
Jorge L. Arocha, Javier Bracho and Luis MontejanoJanuary 1, 2021
Abstract
We prove that if a convex set in C n contains two inscribed complexellipsoid of maximal volume then one is a translate of the other. Onthe other hand, the circumscribed complex elipsoid of minimal volume isunique. As application we prove the complex analoge of Brunn’s charac-terization of ellipsods. Contents
Let A a non-flat compact subset of the euclidean space R n . Denote by b A theconvex closure of A . A solid ellipsoid E is called circumscribed if A ⊂ E . On theother hand, it is called inscribed if E ⊂ b A . If among all circumscribed ellipsoids E has the minimal volume we say that E is a minimal circumscribed ellipsoid1MiCE). On the other hand, if among all inscribed ellipsoids E has the maximalvolume we say that E is a maximal inscribed ellipsoid (MaIE).The existence of these ellipsoids is proved using standard arguments. Due tothe properties of A there is a finite sphere big enough that contains A and a noncero sphere which is contained in b A . All ellipsoids are easily parametrized by amatrix and a vector and among them we can consider only those that containthe small sphere and are contained in the big sphere. This is a compact in theparameter space. Moreover the volume function is continuous and the existencefollows.The real interesting thing about MiCE and MaIE is that they are unique.The first proofs of this fact in its full generality seems to appear independentlyin [3] and [7]. MiCE and MaIE are known as L¨owner–John Ellipsoids and havemany applications in several areas of mathematics (see [6] and the referencesthere).In this paper we deal with similar questions in the space C n . In this spacethe ellipsoids are the unit balls for the norms defined by inner products. Theonly result previously available seems to be in the paper by Gromov [4] wherehe proved the uniqueness of the circumscribed ellipsoid minimal among thosecentered at the origin, when A is an unit ball of a Banach space over C ([4]Lemma 1).We prove in section 3 that If a convex set (convexity inherited from R n )contains two complex MaIE then one is a translate of the other. In section 4 weprove that the complex MiCE is unique. In section 5 we give some applications.The most robust result there, is the analog of the Brunn’s Theorem for complexellipsoids.Whenever possible, we use a unifying approach for the real and complexcases. We denote by K a field. Moreover, here K is R or C . The elements of K n willbe denoted in boldface. Often, the geometric terminology is used. Points arethe vectors of K n , affine subspaces of dimension 1 of K n are called lines; affinesubspaces of codimension 1 of K n are called hyperplanes. We emphasize that,unless the contrary is explicitly stated, all objects are general i.e. for example,a line is not specifically a real or complex line; it is a line in K n . For x = ( x , ..., x n ) ∈ K n denote by x = (¯ x , ..., ¯ x n ), where the bar aboveis the complex conjugate. Also, we denote by x ⊙ y the Hadamar product of x and y . That is, the coordinatewise product (see for example [5] Chapter 5).An scalar product in K n is a sesquilinear, Hermitian, definite positive func-tional K n × K n → K denoted by h x · y i for any x and y in K n . For eachscalar product there is a matrix A (Hermitian, definite positive) such that h x · y i = x T A y . In the case that A is the identity matrix, the inner product isthe usual Hermite’s product in K n . It is very well known that the eigenvaluesof A must be real positive numbers and also its determinant. It is also known2hat A is diagonalizable by a unitary transformation.An ellipsoid (centered at the origin) is a set (cid:8) x ∈ K n | x T A x ≤ (cid:9) . (1)For the case that A is the identity matrix this ellipsoid is the unit ball B ( K n )and its boundary is the unit sphere S ( K n ). The set of unitary transformationsis the subgroup of GL ( K n ) that preserves the unit sphere. The modulus of anscalar λ ∈ K will be denoted by | λ | . The set S (cid:0) K (cid:1) is the multiplicative groupof scalars with modulus 1 in K . We will denote by k·k the usual norm in K n ,i.e. the norm defined by the Hermite’s product.Denoting B def = √ A T , we have A T = BB and therefore A = B T B T = B T B .From this we obtain x T A x = x T B T B x = ( B x ) T ( B x ) = k B x k and therefore, the ellipsoids in K n can be written in the form (cid:8) B − u | u ∈ B ( K n ) (cid:9) If we use a unitary transformation to bring A to the diagonal form then thiscan be rewritten as El ( λ ) def = { λ ⊙ u | u ∈ B ( K n ) } (2)where λ is a vector in R n + . Since A = B , the two forms 1 and 2 are relatedby the fact that λ is the diagonal of A − . The map x λ ⊙ x = A − x is aninvertible linear map in GL ( K n ) which maps the unit sphere B ( K n ) into theellipsoid El ( λ ) . ThereforeVol El ( λ ) = det A − Vol B ( K n ) . (3)The results that follow do not depend on the measure chosen to define thevolume. We just need the validity of the equation 3.We will denote det λ the product of coordinates of λ . Of course we havedet λ = det A − and Vol El ( λ ) = det λ Vol B ( K n ) . The translates of ellipsoids centered at the origin are also ellipsoids. Trans-lations do not change volume.
Theorem 1
Let A a non-flat compact in K n . Let E and E be two MaIEcontained in b A . Then, there is a vector c ∈ K n such that E = E + c . If K = R then E = E . roof. Using a suitable affine transformation we can suppose that E is theunit ball El ( ) . Suppose that the center of E is the vector c ∈ K n . We canuse a unitary transformation to diagonalize the matrix of E − c . And therefore E = El ( λ ) + c for some λ ∈ R n + .Let us prove first that the ellipsoid E = El (cid:0) ( λ + ) (cid:1) + c is contained inthe convex closure of E ∪ E , i.e. El (cid:18) λ + (cid:19) + c ⊂ \ El ( ) ∪ ( El ( λ ) + c ) . Indeed if x ∈ E , then for some u ∈ B ( K n ) = El ( ) = E we have x = λ + ⊙ u + c y = λ ⊙ u + c is in E and y + u λ ⊙ u + c + u x This means that x is the middle point of the segment joining the points y and u .This proves that E ⊂ \ E ∪ E . Since E ∪ E ⊂ b A we have E ⊂ \ E ∪ E ⊂ b A .i.e. E is also contained in b A .We know that Vol ( E ) = Vol ( E ) = det λ Vol ( E ) therefore det λ = 1 . Moreover Vol ( E ) = det (cid:0) ( λ + ) (cid:1) Vol ( E ) . If λ = then by lemma 12 onpage 9 det (cid:0) ( λ + ) (cid:1) > E ) > Vol ( E ) . This contradictsthat E is a MaIE. So, we conclude that λ = and therefore E = E + c . Thisconcludes the first part of the theorem.For the second, K = R and we can suppose that E and E are two unitballs whose centers are at distance 2 α. Let e = (1 , , ....,
0) be the first basisvector. After a suitable unitary affine transformation we can suppose that E has its center in α e and E has its center in − α e . Now, we shall prove that E = El ( α e + ) is contained in the convex closure of E ∪ E i.e. El ( α e + ) ⊂ \ ( El ( ) + α e ) ∪ ( El ( ) − α e ) . Indeed if x ∈ El ( α e + ) , then there exists u = ( u , ..., u n ) ∈ B ( R n ) suchthat x = ( α e + ) ⊙ u .Let y = u + α e ∈ E and z = u − α e . We shall see that x is in the segmentjoining the points y and z , i.e. we have to find t ∈ [0 , ⊂ R such that x = t y +(1 − t ) z . In any coordinate but the first any value of t is appropriate. In thefirst coordinate we have the equation ( α + 1) u = ( u + α ) t + (1 − t ) ( u − α ).If E = E then α = 0 and the solution of the equation is t = ( u + 1). Thecoordinates of vectors in the unit ball can be any real number in the interval[ − ,
1] hence t is in the interval [0 , E ⊂ \ E ∪ E andtherefore E ⊂ b A . We haveVol ( E )Vol ( E ) = det ( α e + ) = α + 1 . E = E , then α > E ) > Vol ( E ) , which contradicts themaximality of E . This proves that E = E . Remark 2
The proof of the second part of the theorem does not work for com-plex ellipsoids because we are using essentially that u is a real number. More-over, the second part of this theorem is not true for complex ellipsoids. In C the complex ellipsoids are just the disks. A rectangle of sides 2 and 4 containsmany unit disks which are of the maximal volume. It is easy to generalize thiscounterexample to all dimensions. A subset of K n is symmetric (centrally) if it is preserved by the multiplicativegroup of all scalars of modulus 1 in K . It is clear that the origin is the centerof any symmetric set. A translate of a symmetric set is also called symmetric.It is easy to see that an affine image of a symmetric set is also symmetric. Allellipsoids are symmetric as they are affine images of the unit ball.In C n , additionally to the complex ellipsoids, there are real ellipsoids whichare inherited from R n . Observe that all complex ellipsoids are real but not theother way around. The difference between real and complex ellipsoids is clearlyexplained by the following result from [2]. Theorem 3
Any symmetric ellipsoid in C n is complex. Now we shall see that symmetry guarantees uniqueness of the MaIE.
Theorem 4 If A is a compact convex symmetric set in C n , then its complexMaIE is unique. Proof.
We can suppose that the center of A is the origin. Let E be theunique real MaIE in A (Theorem 1). Let ζ be a scalar of modulus 1. Since ζ E ⊂ ζA = A and Vol ( ζ E ) = Vol ( E ) hence the uniqueness of E implies that E is symmetric. Using Theorem 3 we get that E is complex. There is not anothercomplex MaIE because any complex ellipsoid is real. Theorem 5
Let A be a non-flat compact set in K n . Let E and E be twoMiCE containing A . Then, E = E . Proof.
Again, using an affine transformation we make E the unit sphere anddiagonalize the matrix of E using an unitary transformation. After that, wecan translate so that the center of E is the origin and the center of E is somevector c ∈ K n .Then, E = n x ∈ K n | P λ i | x i | ≤ o = El ( β ) E = n x ∈ K n | P | x i − c i | ≤ o = El ( ) + c β = λ − . Therefore A ⊂ E ∩ E ⊂ E = n x ∈ K n | X λ i | x i | + | x i − c i | ≤ o . Using Lemma 13 the inequality is transformed into X ( λ i + 1) (cid:12)(cid:12)(cid:12)(cid:12)(cid:18) x i − c i ( λ i + 1) (cid:19)(cid:12)(cid:12)(cid:12)(cid:12) ≤ − X (cid:18) λ i λ i + 1 (cid:19) | c i | ≤ . (4)Let E = ( x ∈ K n | X ( λ i + 1) (cid:12)(cid:12)(cid:12)(cid:12)(cid:18) x i − c i ( λ i + 1) (cid:19)(cid:12)(cid:12)(cid:12)(cid:12) ≤ ) which is an ellipsoid. From inequality 4 we obtain that A ⊂ E ∩ E ⊂ E ⊂ E . Let z ∈ K n be the vector with coordinates c i / ( λ i + 1). We have E = z + n x ∈ K n | X ( λ i + 1) | x i | ≤ o = El (cid:18) λ + 12 (cid:19) − ! + z . Since det β = 1, we obtain det λ = 1 . If λ = the hypothesis of Lemma 12 aresatisfied and then ∆ def = det (cid:18) λ + 12 (cid:19) > . Therefore, Vol E = ∆ − Vol E < Vol E which contradicts the minimality of E . So we conclude that λ = , E is the unit ball and that E = E + c . Inthis case the inequality 4 transforms to the following X (cid:12)(cid:12)(cid:12)(cid:16) x i − c i (cid:17)(cid:12)(cid:12)(cid:12) ≤ − X (cid:12)(cid:12)(cid:12) c i (cid:12)(cid:12)(cid:12) . Therefore E is a ball with center in c /
2, and radius r def = r − X (cid:12)(cid:12)(cid:12) c i (cid:12)(cid:12)(cid:12) . If c is not the origin then r < E . Therefore c = and E = E . There are other extremal ellipsoids when the extremum is searched for ellipsoidsthat have a fixed center (say the origin). In this setting all extremal ellipsoidsare unique, maximal inscribed or minimal circumscribed; real or complex.6o see this, one can modify the proofs of the previous theorems (which is notdifficult) or use the symmetrization of sets. If A is a set in K n , its symmetrizationaround the origin is S ( A ) = [ | ζ | =1 ζA It is not hard to prove that if A is non-flat and compact then S ( A ) also isnon-flat and compact.Applying previous theorems we obtain unique extremal ellipsoids circum-scribed or inscribed in S ( A ) . These extremal ellipsoids contain the origin andit is not difficult to see that they are the unique extremal ellipsoids centered atthe origin.
As one is introduced in the subject of L¨owner–John Ellipsoids there is a strongfeeling of certain symmetry in the concepts and proofs. This subsection partiallyexplains that feeling.Let A be a non-flat compact convex subset of R n containing the origin inits interior . The polar of A is A ∗ = { x ∈ R n | x · a ≤ ∀ a ≤ } . Polarity reverses the inclusion relation. It is a fact that A ∗∗ = A . (see forexample [1] Chapter IV). If B is a symmetric definite positive matrix then { B u | u ∈ B ( R n ) } ∗ = (cid:8) B − u | u ∈ B ( R n ) (cid:9) i.e. the polars of ellipsoids centered at the origin are ellipsoids centered at theorigin. Moreover, if E is an ellipsoid centered at the origin thenVol ( E ) Vol ( E ∗ ) = 1 . (5)Denote by MI ( A ), the ellipsoid of minimal volume among all ellipsoidscontaining A and centered at the origin. Denote by MA ( A ) ,the ellipsoid ofmaximal volume among all ellipsoids contained in A and centered at the origin. Theorem 6 MA ( A ) ∗ = MI ( A ∗ ) . Proof.
Denote E = MA ( A ) and F = MI ( A ∗ ). We have A ∗ ⊂ E ∗ and bythe minimality of F we have Vol F ≤ Vol E ∗ . By equation 5 Vol F Vol E ≤ F ∗ ⊂ A and by the maximality of E we haveVol E ≥ Vol F ∗ . By equation 5, Vol F Vol E ≥
1. Therefore, Vol F Vol E = 1.The ellipsoid E ∗ contains A ∗ and has the same volume as F . By the unique-ness of MI ( A ∗ ) we have E ∗ = F . Corollary 7
The two theorems about the uniqueness of extremal real ellipsoidscentered at the origin are polar of each other.
Remark 8
There is no polarity in C n . .3 A characterization of ellipsoids Theorem 9
Let A be a non-flat compact subset of K n . Then A is an ellipsoidif given any two non-interior points of A , there is an affine isomorphism whichmaps one of them to the other and preserves A. Proof.
Let f be an affine isomorphism that preserves A then, it must preservethe convex closure of A . Since A is non-flat, its convex closure has non-cerovolume. This implies that f preserves volumes. Let E be the unique ellipsoidof minimum volume containing A . The ellipsoid f ( E ) contains f ( A ) = A andhas the same volume than E . Since E is unique f ( E ) = E . Since any affinity iscontinuous f ( ∂ E ) = ∂ E (the boundary of E ).Denote by ∂A the set of non interior points of A. Since E is minimal, thereexists p ∈ ∂ E ∩ ∂A. By hypothesis, for any point q ∈ ∂A there is an affinity f which preserves A and f ( p ) = q . Since f ( ∂ E ) = ∂ E , then q ∈ ∂ E . Thismeans that ∂A ⊆ ∂ E . Reciprocally, let r be any point in ∂ E . If r is an interiorpoint of A, then there is a small ball with center in r contained in A but not in E which is not possible because A ⊂ E . This means that ∂A = ∂ E and provesthat A is convex. Therefore A = c ∂A = c ∂ E = E . Let A be a non-flat compact subset of K n . We will call A a puck if any nonempty intersection with a line is convex and symmetric i.e. if K = R then it isa line segment; if K = C then it is a disk. When K = R pucks are just convexbodies. Theorem 10
A puck A in K n is an ellipsoid if and only if for any line ℓ thecenters of all intersections of A with lines parallel to ℓ , lie in a hyperplane. Proof.
For the only if part, observe that the property holds for balls and it’spreserved by affine transformations. We shall prove the if part using Theorem9. For this, let x and y be two different non-interior points of A an let ℓ be theonly affine line which contains x and y . The line ℓ defines an hyperplane H asstated in the hypothesis. Let o be the point ℓ ∩ H . Let us make a translation t o which sends o to the origin. Denote A ′ = A − o , x ′ = x − o , y ′ = y − o , etc.Since x ′ and y ′ belong to the same linear subspace ℓ ′ which is of dimension1, hence there exists an scalar λ ∈ K such that y ′ = λ x ′ . The center of ℓ ∩ A is o , therefore the center of ℓ ′ ∩ A ′ is the origin. Moreover since x ′ and y ′ arenon-interior points of A ′ they must lie in the boundary of ℓ ′ ∩ A ′ and therefore k x ′ k = k y ′ k . Hence we obtain | λ | = k y ′ k k x ′ k − = 1.The linear subspaces H ′ and ℓ ′ are complementary. Therefore, for any vector z ∈ K n there exists unique vectors z ℓ ′ ∈ ℓ ′ and z H ′ ∈ H ′ such that z = z ℓ ′ + z H ′ .Define φ ( z ) = λ z ℓ ′ + z H ′ which is a linear isomorphism of the whole space.Observe that φ ( x ′ ) = λ x ′ = y ′ . The spaces ℓ ′ and H ′ are invariant subspacesof φ ; and φ is the direct sum of the identity in H ′ and the multiplication by λ in ℓ ′ . Any parallel line L to ℓ ′ is equal to ℓ ′ + p where p = L ∩ H ′ . If z ∈ L φ ( z ) = λ z ℓ ′ + p . Therefore φ leaves invariant L, and it acts inside L likemultiplication by λ. Since L ∩ A ′ is symmetric, | λ | = 1 and the center of L ∩ A ′ lies in H ′ then φ ( L ∩ A ′ ) = L ∩ A ′ . We have that A ′ is the disjoint union of allthe L ∩ A ′ with L parallel to ℓ ′ . Since φ preserves each “slice” then φ preserves A ′ . Now consider the affine isomorphism f : K n ∋ z t − o ( φ ( t o ( z ))) ∈ K n .We have that f ( A ) = t − o ( φ ( A ′ )) = A and f ( x ) = t − o ( φ ( x ′ )) = t − o ( y ′ ) = y .So the hypothesis of theorem 9 is fulfilled and therefore A is an ellipsoid. Remark 11 If K = R then the affine isomorphism f from the previous proof isusually called skew reflection. If K = C then f has no settled down name andthe authors preferred to describe it than to name it. This section contains two lemmas needed in the paper whose proofs are straight-forward computations. Recall that is the vector which has all coordinatesequal to 1. Lemma 12 If λ ∈ R n + , det λ = 1 and λ = then det (cid:18) λ + (cid:19) > . Proof.
We have ( λ i − ≥ λ i + 1) ≥ λ i .With equality onlywhen λ i = 1 . Taking the product, we get det ( λ + ) > n det λ which is thesame as det ( λ + ) > n √ det λ = 2 n . This proves the lemma.
Lemma 13
Let c, x be complex numbers and λ a real number. Then λ | x | + | x − c | = ( λ + 1) (cid:12)(cid:12)(cid:12)(cid:12) x − c ( λ + 1) (cid:12)(cid:12)(cid:12)(cid:12) + (cid:18) λλ + 1 (cid:19) | c | Proof.
We have λ | x | + | x − c | = λ | x | + ( x − c ) ( x − c )= ( λ + 1) | x | − ( xc + cx ) + | c | and completing the square we obtain= ( λ + 1) | x | − ( xc + cx )( λ + 1) + (cid:12)(cid:12)(cid:12)(cid:12) c ( λ + 1) (cid:12)(cid:12)(cid:12)(cid:12) − (cid:12)(cid:12)(cid:12)(cid:12) c ( λ + 1) (cid:12)(cid:12)(cid:12)(cid:12) ! + | c | = ( λ + 1) (cid:12)(cid:12)(cid:12)(cid:12) x − c ( λ + 1) (cid:12)(cid:12)(cid:12)(cid:12) + (cid:18) λλ + 1 (cid:19) | c | . eferences [1] Barvinok, A. A Course in Convexity . Graduate Studies in Mathematics, vol.54.
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