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In $\mathbb{R}^n$, consider the closed cone $C^+ = \{ (x_1, \ldots, x_n) : x_i \geq 0,~~i= 1, \ldots, n\}.$ Let $S \subseteq \mathbb{R}^n$ be a subspace (of any dimension) such that $S \cap C^+ = \{0\}$. Prove that $S^{\perp}$ has non-empty intersection with the interior of $C^+$.

The orthogonal complement is taken with respect to the canonical inner product.

It's not hard to see why this must be true, but a real proof has eluded me for some time now.

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    @Davide My statement, that you quote, is not precise and is not related to the problem. It's just intuition based on a number of particular cases.2012-01-12

2 Answers 2

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I think that this question is answered by Theorem 4 in A. Ben-Israel, Notes on linear inequalities, 1: The intersection of the nonnegative orthant with complementary orthogonal subspaces, J. Math. Anal. Appl. 9:303-314 (1964).

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Hint: WLOG assume $S$ is maximal subspace with $S \cap C^+ = \{0\}$ then $dim (S) = n-1$ So there exist a nonzero vector say $a \in \Bbb{R^n} $ such that $ S = \{ x \in \Bbb R^n ~ : ~~ a^T x = 0\} $

Since $S \cap C^+ = \{0\}$ then $C^+$ has to live on side of $S$, so let assume,

$a^T y \geq 0 \quad \quad \forall y \in C^+ $ So this shows all component of $a$ are nonnegative. They actually strictly positive too! otherwise WLOG assume $a_1 = 0$ then $(1,0,0,...,0) \in S \cap C^+ $ which is contradiction.

Therefore $a >0$ i.e., $a \in \text{int}C^+$ and also $a^T x = 0$ for all $x \in S$, thus $a \in S^{\bot}$.

Hence we proved $ a \in S^{\bot} \cap \text{int} C^+ $