Extreme point iff BFS

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$\newcommand{\eps}{\varepsilon}$ $\newcommand{\xhat}{\widehat{x}}$ $\newcommand{\rank}{\operatorname{rank}}$ 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 $P = \{x: (a_i^Tx \le b_i, \forall i \in I) \textrm{ and } (a_i^Tx = b_i, \forall i \in E)\}$. Then $\xhat$ is an extreme point of $P$ iff $\xhat$ is a basic feasible solution of $P$.

Proof that extreme point is BFS

Let $\xhat$ be an extreme point of $P$. Let $T = \{i: a_i^T\xhat = b_i\}$. Let $B$ be a matrix whose rows are $\{a_i^T: i \in T\}$.

Assume that $\xhat$ is not a BFS. Then $B$ doesn't contain $n$ linearly independent rows. Hence, $\rank(B) < n$. Hence, $Bx = 0$ has a non-zero solution. Let $z$ be such a solution, i.e., $z \neq 0$ and $Bz = 0$. Thus, $\forall i \in T$, $a_i^Tz = 0$.

Let $\eps$ be an infinitesimally small positive value. Let $\delta_i = b_i - a_i^T\xhat$. Then $i \not\in T \implies \delta_i > 0$. Let $x^{(1)} = \xhat - \eps z$ and $x^{(2)} = \xhat + \eps z$. Since $z \neq 0$, we get that $x^{(1)} \neq x^{(2)}$. For $i \in T$, we get $a_i^Tx^{(1)} = a_i^T\xhat - \eps a_i^Tz = b_i$. For $i \not\in T$, we get $a_i^Tx^{(1)} = a_i^T\xhat - \eps a_i^Tz = b_i - (\delta_i + \eps a_i^Tz)$. Since $\eps$ is small enough and $\delta_i > 0$, we get that $a_i^Tx^{(1)} \le b_i$. Hence, $x^{(1)} \in P$. Similarly, $x^{(2)} \in P$.

$\xhat = (1/2)x^{(1)} + (1/2)x^{(2)}$, so $\xhat$ is not an extreme point solution, which is a contradiction. Hence, $\xhat$ is a BFS.

Proof that BFS is extreme point

Let $\xhat$ be a BFS of $P$. 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, $\rank(B) = n$.

Assume $\xhat$ is not an extreme point of $P$. Then $\exists y_1 \in P$, $\exists y_2 \in P$, $\exists \alpha \in (0, 1)$, such that $y_1 \neq y_2$ and $\xhat = (1-\alpha)y_1 + \alpha y_2$.

Since $y_1 \in P$ and $y_2 \in P$, we get $b_i \ge a_i^Ty_1$ and $b_i \ge a_i^Ty_2$ for all $i \in I$ and $a_i^Ty_1 = b_i$ and $a_i^Ty_2 = b_i$ for all $i \in E$. Let $i \in T$. Then $a_i^Tx = b_i \implies (1-\alpha)(b_i - a_i^Ty_1) + \alpha(b_i - a_i^Ty_2) = 0$. Hence, $a_i^Ty_1 = b_i$ and $a_i^Ty_2 = b_i$, so $a_i^T(y_1 - y_2) = 0$. Since this is true for all $i \in T$, we get $B(y_1 - y_2) = 0$. But $\rank(B) = n$, which implies that $y_1 = y_2$. This is a contradiction. Hence, $\xhat$ is an extreme point of $P$.

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Depth: 9
Number of transitive dependencies: 43

Transitive dependencies:

/linear-algebra/matrices/gauss-jordan-algo
/linear-algebra/vector-spaces/condition-for-subspace
/sets-and-relations/equivalence-relation
Group
Ring
Polynomial
Field
Vector Space
Cone
Convex combination and convex hull
Convex set
Extreme point of a convex set
Linear independence
Span
Integral Domain
Comparing coefficients of a polynomial with disjoint variables
Semiring
Matrix
Polyhedral set and polyhedral cone
Basic feasible solutions
Stacking
System of linear equations
Product of stacked matrices
Matrix multiplication is associative
Reduced Row Echelon Form (RREF)
Elementary row operation
Every elementary row operation has a unique inverse
Row equivalence of matrices
Matrices over a field form a vector space
Row space
Row equivalent matrices have the same row space
RREF is unique
Identity matrix
Inverse of a matrix
Inverse of product
Elementary row operation is matrix pre-multiplication
Row equivalence matrix
Equations with row equivalent matrices have the same solution set
Rank of a matrix
Basis of a vector space
Linearly independent set is not bigger than a span
Homogeneous linear equations with more variables than equations
Rank of a homogenous system of linear equations