BFS is vertex
Dependencies:
- Vertex of a set
- Basic feasible solutions
- Rank of a matrix
- Rank of a homogenous system of linear equations
$\newcommand{\eps}{\varepsilon}$ $\newcommand{\xhat}{\widehat{x}}$ $\newcommand{\rank}{\operatorname{rank}}$ Let $m$ and $n$ be positive integers. Let $I$ and $E$ be sets such that $I \cap E = \emptyset$ and $I \cup E = \{1, 2, \ldots, m\}$. Let $a_1, a_2, \ldots, a_m$ be vectors in $\mathbb{R}^n$ and $b_1, b_2, \ldots, b_m$ be real numbers. Let $P = \{x: (a_i^Tx \ge b_i, \forall i \in I) \textrm{ and } (a_i^Tx = b_i, \forall i \in E)\}$. Let $\xhat$ be a BFS of $P$. Then $\xhat$ is also a vertex of $P$.
Proof
Let $T = \{i: a_i^T\xhat = b_i\}$. Let $B$ be a matrix whose rows are $\{a_i^T: i \in T\}$. Since $\xhat$ is a BFS, $\rank(B) = n$.
Let $c = \sum_{i \in T} a_i$. Let $y \in P - \{\xhat\}$. Suppose $a_i^Ty = b_i$ for all $i \in T$. Then $B(y - \xhat) = 0$. Since $\rank(B) = n$, $y = \xhat$, which is a contradiction. Hence, $\exists k \in T$ such that $a_i^Ty \neq b_i$. Since $y \in P$, $\exists k \in T \cap I$ such that $a_i^Ty > b_i$. \begin{align} c^Ty &= \sum_{i \in T} a_i^Ty \\ &> \sum_{i \in T} b_i \\ &= \sum_{i \in T} a_i^T\xhat \\ &= c^T\xhat. \end{align} Hence, $c^T\xhat < c^Ty$ for all $y \in P - \{\xhat\}$. Hence, $\xhat$ is a vertex of $P$.
Dependency for:
Info:
- Depth: 9
- Number of transitive dependencies: 44
Transitive dependencies:
- /linear-algebra/vector-spaces/condition-for-subspace
- /linear-algebra/matrices/gauss-jordan-algo
- /sets-and-relations/equivalence-relation
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