Representing point in pointed polyhedral cone
Dependencies:
- Polyhedral set and polyhedral cone
- Extreme directions of a convex set
- Rank of a matrix
- Bounded section of pointed cone
- Point in polytope is convex combination of BFS
- Extreme direction of convex cone as extreme point of intersection with hyperplane
- Extreme point iff BFS
$\newcommand{\xhat}{\widehat{x}}$ $\newcommand{\defeq}{=}$ $\newcommand{\rank}{\operatorname{rank}}$ Let $C = \{x \in \mathbb{R}^n: (a_i^Tx \ge 0, \forall i \in I) \textrm{ and } (a_i^Tx = 0, \forall i \in E)\}$ be a pointed polyhedral cone.
Let $\xhat \in C$. Let $T \defeq \{i \in I \cap E: a_i^T\xhat = 0\}$ and $B \defeq \{a_i: i \in T\}$. Then $\xhat$ can be represented as a non-negative combination of at most $n-\rank(B)$ extreme directions of $C$.
Proof
This is trivially true when $\xhat = 0$, so assume $\xhat \neq 0$.
Since $C$ is pointed, $\exists c$ such that $c^Tx > 0$ $\forall x \in C - \{0\}$. Without loss of generality, assume $c^T\xhat = 1$ (because we can scale $c$). Let $P \defeq \{x \in C: c^Tx = 1\}$. Then $\xhat \in P$ and $P$ is bounded.
Lemma: $c$ is not a linear combination of $B$.
Proof: Suppose $c$ is a linear combination of $B$. Let $c = \sum_{i \in T}\alpha_i a_i$. Then \[ 1 = c^T\xhat = \sum_{i \in T}\alpha_i(a_i^T\xhat) = 0. \] This is a contradiction. Hence, $c$ is not a linear combination of $B$. □
Since $P$ is bounded and $\xhat$ is tight at $\rank(B) + 1$ linearly independent constraints of $P$, we can represent $\xhat$ as a convex combination of $n - \rank(B)$ BFSes of $P$. The BFSes of $P$ are extreme points of $P$, which are extreme directions of $C$.
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Info:
- Depth: 13
- Number of transitive dependencies: 59
Transitive dependencies:
- /linear-algebra/vector-spaces/condition-for-subspace
- /linear-algebra/matrices/gauss-jordan-algo
- /complex-numbers/conjugation-is-homomorphic
- /sets-and-relations/equivalence-relation
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