Beiträge zur Algebra und Geometrie Contributions to Algebra and Geometry Volume 37 (1996), No. 1, 41-49.

Redundance of Vertices of the Cube relatively to its Minimal Ellipsoid

Olaf Sarges

Fakultät für Mathematik, Otto-von-Guericke-Universität Magdeburg Postfach 4120, D-39016 Magdeburg e-mail: Olaf.Sarges@Mathematik.Uni-Magdeburg.de

Abstract. Each convex body K has a unique ellipsoid of minimal volume containing K (Minimal ellipsoid, Loewner ellipsoid). In this paper we look for subsets of the vertex set of the cube, which have the same minimal ellipsoid as the cube itself. All subsets with this property are determined in dimensions 2,3 and 4. For some higher dimensions (for example 4 to 423) we present the minimal cardinal number of such subsets. MSC 1991: 52B12, 52A20, 90C34

1. Introduction

Each convex body in ℝ^{n} has a uniquely determined minimal ellipsoid, cf. [3],[6]. Let K be a polytope given as the convex hull of its vertices. According to [9],[10],[11] there is a number r, n+1 ≤ r ≤ \frac{n(n+3)}{2}, of vertices of K whose convex hull has the same minimal ellipsoid as K. Therefore, one can remove some vertices of ”most” polytopes without changing their minimal ellipsoid. In this paper we investigate the cube in ℝ^{n} intending to find sets of vertices that have the same minimal ellipsoid as the cube. All such sets are presented in dimensions 2,3 and 4. Finally, we ask for the minimal number of vertices necessary to obtain this property. This question can be answered for a large class of dimensions because it is closely connected with the existence of Hadamard matrices. In theorem 1 we prove the following: If an Hadamard matrix exists for n = 4(m + 1), then one can figure out the minimal number for dimensions n = 4m to 4m + 3. This includes the solution of the problem up to dimension 423.

0138-4821/93 $ 2.50 \begin{matrix}c \\ ◯ \end{matrix} 1996 Heldermann Verlag, Berlin

2. General results

In the following let E(X^{0}^{−1}, x^{0}) := {x ∈ ℝ^{n}|(x − x^{0})^{T} X^{0}^{−1}(x − x^{0}) ≤ 1} be an ellipsoid with centre x^{0} where X^{0} is a positive definite symmetric matrix. B_{n} = {x ∈ ℝ^{n}| ∥x∥ ≤ 1} denotes the unit ball, ∂B_{n} stands for the boundary of B_{n} and I_{n} is the n-dimensional unit matrix. Let Q_{n} = conv{q_{1}, ..., q_{2^{n}}} be the n-dimensional cube with minimal ellipsoid B_{n}. In this connection q_{i}^{T} = (q_{i1}, ..., q_{in}), i = 1, ..., 2^{n}, are vectors with q_{ik} ∈ \{1/\sqrt{n}, −1/\sqrt{n}\} for k = 1, ..., n. The following result, see [10],[11], will prove to be useful: An ellipsoid E(X^{0}^{−1}, x^{0}) is the minimal ellipsoid of a convex body K if and only if E(X^{0}^{−1}, x^{0}) ⊇ K and the conditions (1) to (5) are fulfilled.

There are vectors

u_{1}, ..., u_{r} ∈ ∂B_{n} with n + 1 ≤ r ≤ \frac{n(n + 3)}{2}  (1)

and scalars

μ_{0} > 0, μ_{1} > 0, ..., μ_{r} > 0  (2)

with

u_{j}^{T} x^{0} + \sqrt{u_{j}^{T} X^{0}u_{j}} = k_{K}(u_{j}) for j = 1, ..., r  (3)
∑_{j=1}^{r} μ_{j}u_{j} = 0  (4)

(2μ_{0}detX^{0})I_{n} = ∑_{j=1}^{r} \frac{μ_{j}X^{0}u_{j}u_{j}^{T}}{\sqrt{u_{j}^{T} X^{0}u_{j}}} ,  (5)

where k_{K} is the Minkowski support function of K , cf. [4]. If K is the cube Q_{n}, then E(X^{0}^{−1}, x^{0}) = B_{n} holds. This implies X^{0} = I_{n} and x^{0} = 0. Then (3) and (5) change into

\sqrt{u_{j}^{T} u_{j}} = 1 = k_{Q_{n}}(u_{j})  (3’)

and

2μ_{0}I_{n} = ∑_{j=1}^{r} μ_{j}u_{j}u_{j}^{T} .  (5’)

Due to (3′) only the vertices of Q_{n} are possible candidates for u_{1}, ..., u_{r}. Each subset {u_{1}, ..., u_{r}}, n + 1 ≤ r ≤ n(n + 3)/2, of the vertex set {q_{1}, ..., q_{2^{n}}}, which fulfils (4) and (5′) with suitable μ_{i} > 0, i = 0, ..., r, has the same minimal ellipsoid (B_{n}) as the cube Q_{n}, cf. [10], [11]. Which different (in particular minimal) subsets of this type exist in a given dimension? To answer this question we investigate (similar to [2]) the system

λ_{i} ≥ 0 for i = 1, ..., 2^{n}
∑_{i=1}^{2^{n}} λ_{i}p_{i} = 0  (6)

∑_{i=1}^{2^{n}} λ_{i}p_{i}p_{i}^{T} = ρI_{n},

where p_{i} := \sqrt{n}q_{i} = (p_{i1}, ..., p_{in})^{T}. Therefore p_{ik} is either 1 or -1 for k = 1, ..., n and i = 1, ..., 2^{n}. Let Λ := (λ_{1}, ..., λ_{2^{n}})^{T} be a solution of this system and

J(Λ) := {i | λ_{i} > 0} = {i_{1}, ..., i_{k}}  (7)

the index set of its positive components. Each solution Λ ≠ 0 with k = |J(Λ)| ≤ \frac{n(n+3)}{2} corresponds to a solution of (1),(2),(3′),(4),(5′). Just choose r := k, (u_{1}, .., u_{r}) := (\frac{1}{\sqrt{n}}p_{i_{1}}, .., \frac{1}{\sqrt{n}}p_{i_{k}}), (μ_{1}, .., μ_{r}) := (λ_{i_{1}}, .., λ_{i_{k}}) and μ_{0} = ρ/2n (k ≥ n + 1 holds automatically, cf. [10],[11]). This vertex set {q_{i_{1}}, ..., q_{i_{k}}} of Q_{n} has the same minimal ellipsoid B_{n} as the initial cube Q_{n}. Let L_{j} = p_{j}p_{j}^{T} be the matrix with components l_{ik}^{j} := p_{ji}p_{jk}. It will turn out to be helpful to define L^{j} := (l_{21}^{j}, l_{31}^{j}, l_{32}^{j}, l_{41}^{j}, ..., l_{n1}^{j}, l_{n2}^{j}, ..., l_{nn−1}^{j}) = (L_{1}^{j}, L_{2}^{j}, ..., L_{N}^{j}) ∈ ℝ^{N}, where N := (n^{2} − n)/2. The matrices p_{j}p_{j}^{T} are symmetric and the elements of the diagonal all equal 1, implying ρ = ∑_{i=1}^{2^{n}} λ_{i}. Thus it is sufficient to consider the following homogeneous linear system instead of (6):

( \begin{matrix}p_{1} & p_{2} & ... & p_{2^{n}} \\ L^{1} & L^{2} & ... & L^{2^{n}} \end{matrix} ) ( \begin{matrix}λ_{1} \\ : \\ λ_{2^{n}} \end{matrix} ) = 0, λ_{j} ≥ 0.  (8)

For a given solution Λ ≠ 0 the set M_{n}(Λ) := {p_{j} | j ∈ J(Λ)} = {p_{i_{1}}, ..., p_{i_{k}}} is called a corresponding solution set and λ := (λ_{i_{1}}, ..., λ_{i_{k}})^{T} is said to be the corresponding solution vector. The matrix whose columns correspond to the elements of the solution set is called a solution matrix.

For n = 2 this leads to the system

( \begin{matrix}1 & 1 & −1 & −1 \\ 1 & −1 & 1 & −1 \\ 1 & −1 & −1 & 1 \end{matrix} ) ( \begin{matrix}λ_{1} \\ λ_{2} \\ λ_{3} \\ λ_{4} \end{matrix} ) = ( \begin{matrix}0 \\ 0 \\ 0 \end{matrix} ) , λ_{j} ≥ 0, λ ≠ 0,

with the unique solution λ_{1} = λ_{2} = λ_{3} = λ_{4} > 0. Thus it is impossible to remove one vertex of the cube without getting a smaller minimal ellipsoid.

For n = 3 one obtains

( \begin{matrix}1 & 1 & 1 & 1 & −1 & −1 & −1 & −1 \\ 1 & 1 & −1 & −1 & 1 & 1 & −1 & −1 \\ 1 & −1 & 1 & −1 & 1 & −1 & 1 & −1 \\ 1 & 1 & −1 & −1 & −1 & −1 & 1 & 1 \\ 1 & −1 & 1 & −1 & −1 & 1 & −1 & 1 \\ 1 & −1 & −1 & 1 & 1 & −1 & −1 & 1 \end{matrix} ) ( \begin{matrix}λ_{1} \\ : \\ λ_{8} \end{matrix} ) = 0, λ_{j} ≥ 0, λ ≠ 0.

The vectors (0, 1, 1, 0, 1, 0, 0, 1) and (1, 0, 0, 1, 0, 1, 1, 0) form a basis of the solution space of the previous system. Hence the ball turns out to be the minimal ellipsoid of the two vertex disjoint regular simplices conv((1, 1, −1)^{T} , (1, −1, 1)^{T}, (−1, 1, 1)^{T} , (−1, −1, −1)^{T}), cf. Fig. 1, and conv((1, 1, 1)^{T} , (1, −1, −1)^{T}, (−1, 1, −1)^{T} , (−1, −1, 1)^{T}). These inscribed simplices are known since the days of Euclid, see [7]. Furthermore, each solution set in ℝ^{3} contains the vertices of one of these simplices.

[figure: ^{( \begin{matrix}1 \\ 1 \\ 1 \end{matrix} )}; Fig. 1]

One easily gets

Lemma 1: Let A = {p_{i_{1}}, ..., p_{i_{k}}} be a solution set in ℝ^{n}. One obtains a solution set in ℝ^{n−1} by removing the l-th component, l ∈ {1, ..., n}, of the vectors p_{i_{j}}, j = 1, ..., k. This set consists of at most k elements.

Proof: Let (λ_{i_{1}}, ..., λ_{i_{k}})^{T} be the corresponding solution vector of A. Then we have ∑_{j=1}^{k} λ_{i_{j}}p_{i_{j}} = 0 and ∑_{j=1}^{k} λ_{i_{j}}p_{i_{j}}p_{i_{j}}^{T} = ρI_{n} holds. Removing the l-th component , l ∈ {1, ..., n}, of the vectors p_{i_{j}}, we obtain vectors \overline{p}_{i_{j}}, which satisfy ∑_{j=1}^{k} λ_{i_{j}} \overline{p}_{i_{j}} = 0 and ∑_{j=1}^{k} λ_{i_{j}} \overline{p}_{i_{j}} \overline{p}_{i_{j}}^{T} = ρI_{n−1}. Note that two vectors \overline{p}_{i_{j}} and \overline{p}_{i_{r}} with j ≠ r may be equal and recall that the matrix \overline{p}_{i_{j}} \overline{p}_{i_{j}}^{T} results from p_{i_{j}}p_{i_{j}}^{T} by removing the l-th row and the l-th column. In this way we obtain a solution set in ℝ^{n−1} with at most k elements.

Lemma 2 : A subset of the vertex set of the 4m-dimensional cube, m ≥ 1, which has the same minimal ellipsoid as the cube contains at least 4(m + 1) elements.

Proof: Let A = (p_{i_{1}}...p_{is}) be a solution matrix in ℝ^{4m}. We can assume ∑_{j=1}^{s} λ_{i_{j}} = 4(m + 1) due to the homogeneity of the solution vector corresponding to A (cf. (8)). Associate an index set S_{1}, S_{2}, S_{3}, S_{4}, to each constellation \binom{1}{1}, \binom{1}{−1}, \binom{−1}{1} and \binom{−1}{−1} so that i_{j} ∈ S_{1} iff p_{i_{j}} = (1, 1, ...)^{T} , i_{j} ∈ S_{2} iff p_{i_{j}} = (1, −1, ...)^{T}, etc. Actually all these constellations do occur in the first two components of the vectors p_{i_{j}}. Otherwise it would be possible to construct a solution set in ℝ^{2} which consists of less than four vertices applying the previous lemma n − 2 times. By the same argument and the uniqueness (except scalar multiples) of the solution vector in ℝ^{2} we get ∑_{j∈S_{1}} λ_{i_{j}} = ∑_{j∈S_{2}} λ_{i_{j}} = ∑_{j∈S_{3}} λ_{i_{j}} = ∑_{j∈S_{4}} λ_{i_{j}} = m + 1. Suppose s is smaller than 4(m + 1). Consequently, at least one constellation of \binom{1}{1}, \binom{1}{−1}, \binom{−1}{1} and \binom{−1}{−1} occurs at most m times in the first two components of the vectors p_{i_{j}}, or in other words at least one index set of S_{1}, S_{2}, S_{3}, S_{4} has a cardinal number less than m + 1. Let this be S_{1}. Thus there exists a λ_{i^{∗}}, i^{∗} ∈ S_{1}, with λ_{i^{∗}} ≥ (m + 1)/m; without loss of generality we can assume λ_{i^{∗}} = λ_{i_{1}} with p_{i_{1}} = (1, 1, ..., 1)^{T}. Now look at p_{i_{j}}p_{i_{j}}^{T}: Let l be the number of components of p_{i_{j}} which equal +1 and k the number of components of p_{i_{j}} which equal −1. The matrix p_{i_{j}}p_{i_{j}}^{T} consists of the elements +1 (exactly l^{2} + k^{2} times) and -1 (exactly 2lk times). Therefore, the sum of all n^{2} elements of p_{i_{j}}p_{i_{j}}^{T} turns out to be l^{2} + k^{2} − 2lk = (l − k)^{2}. Moreover we have

∑_{ν=1}^{N} L_{ν}^{i_{j}} = \frac{(l − k)^{2}}{2} − \frac{n}{2} ≥ −\frac{n}{2} = −2m,  (9)

where equality holds iff l = k.

In the case j = 1 this turns to ∑_{ν=1}^{N} L_{ν}^{i_{1}} = N = 2m(4m − 1) = 8m^{2} − 2m. Thus we obtain

∑_{j=2}^{s} λ_{i_{j}} = 4(m + 1) − λ_{i_{1}} ≤ 4m + 4 − \frac{m + 1}{m} = \frac{4m^{2} + 3m − 1}{m}.

Since ∑_{j=1}^{s} λ_{i_{j}}L^{i_{j}} equals 0, we also have ∑_{j=1}^{s} λ_{i_{j}} ∑_{ν=1}^{N} L_{ν}^{i_{j}} = 0. Using this one can show that

0 = ∑_{j=1}^{s} λ_{i_{j}} ∑_{ν=1}^{N} L_{ν}^{i_{j}} ≥ (\frac{m + 1}{m}) N + ∑_{j=2}^{s} λ_{i_{j}} ∑_{ν=1}^{N} L_{ν}^{i_{j}}
≥ 8m^{2} + 6m − 2 + ∑_{j=2}^{s} λ_{i_{j}}(−2m)
≥ 8m^{2} + 6m − 2 − 2m\frac{(4m^{2} + 3m − 1)}{m} = 0.

Hence the equality sign holds in all these inequalities, which requires ∑_{ν=1}^{N} L_{ν}^{i_{j}} = −2m for j = 2, ..., s. According to (9) this is only possible if the vectors p_{i_{j}}, j = 2, .., s, contain +1 and -1 both exactly 2m times. From this it follows that ∑_{k=1}^{n} λ_{i_{j}}p_{i_{j}k} = 0 holds for j = 2, ..., s. As a result we obtain ∑_{j=1}^{s} λ_{i_{j}} ∑_{k=1}^{n} p_{i_{j}k} = nλ_{i_{1}}, in contradiction to ∑_{j=1}^{s} λ_{i_{j}}p_{i_{j}} = 0. Thus s ≥ 4(m + 1) is proved, q.e.d. Our problem is closely connected with the existence of Hadamard matrices. (An Hadamard matrix H is an n × n-matrix (h_{ik}) with h_{ik} ∈ {1, −1} and H^{T} H = nI_{n}.) It is known that an Hadamard matrix can only exist for n = 1, 2 and n ≡ 0 mod 4. The problem whether such a matrix exists for all n ≡ 0 mod 4, first considered by Sylvester [14], remains unsolved to this day. Summaries about the results concerning this topic are given in [1],[12],[13]. 428 is the smallest number for which no Hadamard matrix has been found until now, cf. [12].

The proof of Theorem 1 makes use of the following lemma, cf. for example [8]:

Lemma 3: An (n−1)-dimensional regular simplex can be inscribed in an (n−1)-dimensional cube such that each vertex of the simplex coincides with a vertex of the cube if and only if an Hadamard matrix of order n exists.

Summarizing lemma 1 and lemma 3 we obtain the following statement. If an Hadamard matrix of order n exists, then we can construct solution sets of at most n elements for all dimensions smaller than n. On the other hand, such sets consist of at least n elements in dimension n − 4, n ≥ 8 (n ≡ 0 mod 4), as proved in lemma 2. This yields

Theorem 1: If an Hadamard matrix of order n, n ≥ 8, exists, then the minimal number of vertices of the cube which have the same minimal ellipsoid as the cube itself amounts to n in the dimensions n − 4, n − 3, n − 2 and n − 1.

Thus we get the following table:

dim 1 2 3 4 5 6 7 8 ... 11 12 ... 15 ..... 420 ... 423 min 2 4 4 8 8 8 8 12 ... 12 16 ... 16 ..... 424 ... 424 , where dim denotes the dimension of the cube and min stands for the minimal number of vertices whose minimal ellipsoid is the ball, too.

3. Dimension n = 4

The determination of all solution sets in dimension 4 becomes more difficult than in dimensions 2 and 3. In this case (n = 4), the following six vectors, v_{1}, ..., v_{6}, form a basis of the solution space of (8):

\begin{matrix}v_{1}^{T} = ( & −1, & 1, & 1, & 0, & 1, & 0, & 0, & 0, & 1, & 0, & 0, & 0, & 0, & 0, & 0, & 1) \\ v_{2}^{T} = ( & 1, & −1, & 0, & 1, & 0, & 1, & 0, & 0, & 0, & 1, & 0, & 0, & 0, & 0, & 1, & 0) \\ v_{3}^{T} = ( & 1, & 0, & −1, & 1, & 0, & 0, & 1, & 0, & 0, & 0, & 1, & 0, & 0, & 1, & 0, & 0) \\ v_{4}^{T} = ( & −1, & 1, & 1, & −1, & 0, & 0, & 0, & 0, & 1, & −1, & −1, & 1, & 0, & 0, & 0, & 0) \\ v_{5}^{T} = ( & −1, & 1, & 1, & −1, & 1, & −1, & −1, & 1, & 0, & 0, & 0, & 0, & 0, & 0, & 0, & 0) \\ v_{6}^{T} = ( & 2, & −1, & −1, & 1, & −1, & 1, & 1, & 0, & −1, & 1, & 1, & 0, & 1, & 0, & 0, & 0) \end{matrix}

The order of the vectors p_{1}, ..., p_{16} is chosen in a canonical way. The vector p_{j}^{T}, regarded as a binary number, corresponds to the natural number j−1. In this connection the numbers 1 and −1 replace the dual digits 0 and 1, respectively, i.e. p_{1}^{T} = (1, 1, 1, 1), p_{2}^{T} = (1, 1, 1, −1), p_{3}^{T} = (1, 1, −1, 1), ..., p_{16}^{T} = (−1, −1, −1, −1). We have to consider all linear combinations Λ := Λ(μ) = ∑_{i=1}^{6} μ_{i}v_{i}, μ^{T} := (μ_{1}, ..., μ_{6}), of these basis vectors to get all possible solution sets in ℝ^{4}. Additionally, no component of the linear combination Λ is allowed to become negative, cf. (8). This implies μ_{i} ≥ 0, i = 1, ..., 6, because each vector v_{i}, i = 1, ..., 6, contains one positive component (+1) at such a place where all other vectors v_{j} have components equal to 0. Since we actually look for the solution sets, only the index sets J(Λ) (cf. (7)) of positive components of Λ are of interest. For example, consider all linear combinations with μ_{1}, μ_{2} > 0 and μ_{3} = ... = μ_{6} = 0. Since v_{11} = −1 and v_{21} = 1, we have μ_{2} ≥ μ_{1}. On the other hand, μ_{1} ≥ μ_{2} results from v_{12} = 1 and v_{22} = −1, which leads to μ_{1} = μ_{2}. Hence Λ((1, 1, 0, 0, 0, 0)^{T}) = (0, 0, 1, 1, 1, 1, 0, 0, 1, 1, 0, 0, 0, 0, 1, 1)^{T} holds and {p_{3}, p_{4}, p_{5}, p_{6}, p_{9}, p_{10}, p_{15}, p_{16}} is a solution set, denoted by S_{1}.

If we consider all possible linear combinations this way, then we get the minimal solution sets S_{1}, ..., S_{26}. In this context minimal solution set means that no proper subset of S_{i}, i = 1, ..., 26, is a solution set itself.

p_{1} p_{2} p_{3} p_{4} p_{5} p_{6} p_{7} p_{8} p_{9} p_{10} p_{11} p_{12} p_{13} p_{14} p_{15} p_{16}
S_{1}  + + + +  + +  + +
S_{2} + + + + +  +  +  +
S_{3} +  + + +  + +  +  +
S_{4} + +  + + +  + +  +
S_{5} + + +  +  + +  +  +
S_{6} +  + + + +  +  + +
S_{7} +  + +  +  + +  + +
S_{8} + + +  + +  +  + +
S_{9} + +  + +  +  + +  +
S_{10} + +  + +  + + + +
S_{11} + + + + + + + + +  +  +
S_{12} +  + + + + + + +  + + +
S_{13} + + + + + + +  +  + + +
S_{14} + + + +  + + + +  + + +
S_{15} + + + + + + +  + +  + +
S_{16} + + + + + +  + +  +  + +
S_{17} + + + + + +  + + + +  +
S_{18} + + + + + +  + +  + +  +
S_{19} + + +  + + +  + + + +  +
S_{20} + + + + + + + +  +  + +
S_{21} + + + + + + +  + +  + +
S_{22} + + + + + + +  + +  + +
S_{23} + + + + + +  + + + +  +
S_{24} +  + + + +  + + + + + +
S_{25} + + +  + + + +  + + + +
S_{26} + + + +  +  + + + + + +

The sets S_{1} to S_{10} consist of eight elements whereas the cardinal number of S_{11} to S_{26} equals eleven. These (minimal) sets can be divided into three classes. (We call two vertices adjacent or neighbours if they are connected by an edge of the cube.)

1. The set contains no two adjacent vertices. This leads to the two vertex disjoint regular cross-polytopes, see [5], corresponding to S_{3} and S_{8}, cf. Fig. 2.

2. Each vertex p_{i} is related to exactly one other vertex p_{j} in the sense that p_{j} and p_{i} are neighbours (S_{1}, S_{2}, S_{4}, S_{5}, S_{6}, S_{7}, S_{9} and S_{10}), cf. Fig. 3.

3. Let p be any vertex of the cube. Remove −p and the vertices adjacent to p (S_{11} to S_{26}), cf. Fig. 4.

All other solution sets contain one of these minimal sets S_{1}, ..., S_{26}.

[figure: Fig. 2; Fig. 3; p; -p; Fig. 4]

References

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[7] Euclid, Euclidis Elementa XV.

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[11] Juhnke, F., Polarity of embedded and circumscribed ellipsoids, Beiträge zur Algebra und Geometrie 36 (1995), No. 1, 17-24.

[12] Seberry, J., Yamada, M., Hadamard matrices, sequences and block designs, in: Contemporary Design Theory, ed. by Dintz J., Stinson D., John Wiley & Sons, Inc. 1992, 431-560.

[13] Seberry-Wallis, J., Hadamard matrices, Springer Lecture Notes in Mathematics 292, Part 4 (1972), 273-489.

[14] Sylvester, J.J., Thoughts on inverse orthogonal matrices, Phil. Mag., (4) 24 (1867), 461-475.

Received: March 3, 1995; revised version: October 10, 1995