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Gruber P. Convex and Discrete Geometry

Gruber P. Convex and Discrete Geometry

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78 <strong>Convex</strong> Bodies<br />

N =<br />

� �<br />

X, O<br />

Y, Z<br />

where X is an r × r matrix <strong>and</strong> Z a (d − r) × (d − r) matrix. By assumption on<br />

M, every diagonal of N contains 0. Thus, if there is a diagonal of X which does not<br />

contain 0, then every diagonal of Z contains 0. This shows that all diagonals of X<br />

contain 0 or all diagonals of Z contain 0. We may suppose that the former is the case.<br />

(If the latter is the case, the proof is almost identical.) By the induction hypothesis,<br />

X has a t × u zero submatrix with t + u = r + 1. Then N, <strong>and</strong> therefore also M, has<br />

a t × (u + d − r) zero submatrix. Since t + u + d − r = r + 1 + d − r = d + 1, the<br />

induction is complete. ⊓⊔<br />

Proof of the Theorem. Since a convex combination of doubly stochastic d × d<br />

matrices is also doubly stochastic, we see that<br />

(1) Ωd is convex.<br />

The main step is to prove that<br />

(2) Ωd is the convex hull of the d × d permutation matrices.<br />

Each doubly stochastic d × d matrix has n positive entries, where d ≤ n ≤ d 2 .<br />

For the proof of (2) it is thus sufficient to show the following proposition by induction<br />

on n:<br />

(3) Let d ≤ n ≤ d 2 . Then each doubly stochastic d × d matrix with n positive<br />

entries is a convex combination of d × d permutation matrices.<br />

Any doubly stochastic d × d matrix M with precisely d positive entries is a<br />

permutation matrix. Thus (3) holds for n = d. Assume now that n > d <strong>and</strong> (3)<br />

holds for n − 1. Let M be a doubly stochastic d × d matrix with n positive entries.<br />

We show that M has a diagonal with positive entries. For assume not, that is, each<br />

diagonal of M contains 0. Lemma 5.3 then shows that M has a p × q zero submatrix<br />

with p + q = d + 1, say Mστ. Since the non-zero entries of M in the rows with<br />

indices in σ are in Mστc, the sum of the entries in each row of Mστc is 1. Therefore<br />

the sum of all entries of Mστc is p. Similarly, the sum of all entries of Mσ cτ is q.<br />

Since Mστc <strong>and</strong> Mσ cτ are disjoint submatrices of M, the sum of all entries of M is<br />

at least p + q = d + 1. This contradiction shows that M has a diagonal with positive<br />

entries. Since, by assumption, M has n > d positive entries <strong>and</strong> the sum of all these<br />

n entries is d, among the d entries of the diagonal, there must be one entry less than<br />

1. Let 0

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