Point in polytope is convex combination of BFS

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$\newcommand{\eps}{\varepsilon}$ $\newcommand{\xhat}{\widehat{x}}$ $\newcommand{\xstar}{x^*}$ $\newcommand{\rank}{\operatorname{rank}}$ $\newcommand{\argmax}{\operatorname{argmax}}$ $\newcommand{\argmin}{\operatorname{argmin}}$ Let $P = \{x: (a_i^Tx \ge b_i, \forall i \in I) \textrm{ and } (a_i^Tx = b_i, \forall i \in E)\}$ be a non-empty polyhedron in $\mathbb{R}^n$ that does not contain a ray. A ray is a set of the form $\{z + \lambda d: \lambda \ge 0\}$, where $z \in \mathbb{R}^n$ and $d \in \mathbb{R}^n - \{0\}$. Let $S$ be the set of basic feasible solutions of $P$.

Let $\xhat \in P$, and $B$ be a matrix whose rows are $\{a_i^T: a_i^T\xhat = b_i\}$. Then $\xhat \in P$ can be represented as a convex combination of at most $n + 1 - \rank(B)$ points in $S$. (We can prove that $S$ is non-empty.)

Proof

Since $P$ does not contain a ray, it cannot contain a line. Hence, $S$ is non-empty.

We will prove this using induction on $\rank(B)$. If $\rank(B) = n$, then $\xhat$ is a BFS, so we are done. Now assume that this result is true for $\rank(B) \ge r+1$, where $r \le n-1$. We will prove this when $\rank(B) = r$. Since $\rank(B) \le n-1$, $\xhat$ is not a BFS.

Let $T = \{i: a_i^T\xhat = b_i\}$. Let $P' = \{x: (a_i^Tx = b_i, \forall i \in T) \textrm{ and } (a_i^Tx \ge b_i, \forall i \in (I \cup E) - T)\}$. Then $P' \subseteq P$ and $\xhat \in P'$. Since $P'$ is non-empty and cannot contain a line, it contains a BFS $y$. Since $y$ is tight at $n$ linearly independent constraints, it is also a BFS of $P$.

Let $d = \xhat - y$. Since $\xhat$ is not a BFS of $P$, $d \neq 0$. For each $i \in T$, we get $a_i^Ty = b_i$ and $a_i^T\xhat = b_i$. Hence, $Bd = 0$ and $d \neq 0$.

When $i \in T$, $a_i^Td = 0$, so $a_i^T(\xhat + \lambda d) = a_i^T\xhat = b_i$. If $a_i^Td \ge 0$ for all $i$, then for all $\lambda \ge 0$, we get $a_i^T(\xhat + \lambda d) = a_i^T\xhat + \lambda a_i^Td \ge a_i^T\xhat \ge b_i$. Thus, $\xhat + \lambda d \in P'$ for all $\lambda \ge 0$. But this isn't possible since $P$ does not contain a ray. Hence, $\exists k$ such that $a_k^Td < 0$. In fact, choose $k$ as $\argmin_{i: a_i^Td < 0} (b_i - a_i^T\xhat)/(a_i^Td)$. Let $\lambda^* = (b_k - a_k^T\xhat)/(a_k^Td)$. Note that $\lambda^* \ge 0$, since $\xhat \in P$. Let $\xstar = \xhat + \lambda^* d$. We will show that $\xstar \in P'.$

When $i \in T$, we have $a_i^Td = 0$. When $a_i^Td = 0$, we get $a_i^T\xstar = a_i^T\xhat$. When $a_i^Td > 0$, we get $a_i^T\xstar = a_i^T\xhat + \lambda^* a_i^Td \ge b_i$. Now let $a_i^Td < 0$. $\lambda^* \le (b_i - a_i^T\xhat)/(a_i^Td) \implies a_i^T\xhat + \lambda^* a_i^Td \ge b_i$. Hence, $a_i^T\xstar \ge b_i$. Threfore, $\xstar \in P'$. Furthermore, $a_k^T\xstar = a_k^T\xhat + \lambda^* a_k^Td = b_k$.

Since $a_i^Td = 0$ for $i \in T$, but $a_k^Td < 0$, we get that $k \not\in T$. Furthermore, $a_k$ doesn't lie in the row space of $B$, since $a_i^Td = 0$ for all $i \in T$ but $a_k^Td < 0$. Let $C$ be the matrix with rows $\{a_i^T: a_i^T\xstar = b_i\}$. Then $C$ contains the rows of $B$ and contains $a_k^T$. Hence, $\rank(C) > \rank(B) = r$.

By the inductive hypothesis, $\xstar$ can be represented as a convex combination of at most $n - \rank(C) + 1 \le n - r$ basic feasible solutions of $P$. \[ \xstar = \xhat + \lambda^* (\xhat - y) \implies \xhat = \frac{1}{1 + \lambda^*}\xstar + \frac{\lambda^*}{1+\lambda^*}y. \] Since $\xhat$ is a convex combination of $\xstar$ and $y$, by the transitivity of convexity, we get that $\xhat$ is a convex combination of at most $n - r + 1$ basic feasible solutions of $P$.

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Depth: 10
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
Group
Ring
Polynomial
Integral Domain
Comparing coefficients of a polynomial with disjoint variables
Field
Vector Space
Linear independence
Span
Semiring
Matrix
Stacking
System of linear equations
Product of stacked matrices
Matrix multiplication is associative
Reduced Row Echelon Form (RREF)
Matrices over a field form a vector space
Row space
Elementary row operation
Every elementary row operation has a unique inverse
Row equivalence of matrices
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
Cone
Convex combination and convex hull
Transitivity of convexity
Convex set
Polyhedral set and polyhedral cone
Basic feasible solutions
Condition for existence of BFS in a polyhedron