Consider a region S defined by a set of linear constraints
 S ={x∈Rn :Ax≤b},
 where A is an m×n matrix and b ∈ Rm. Prove that S is convex.

To prove that the region \( S \) defined by the set of linear constraints

\[
S = \{ x \in \mathbb{R}^n : A x \leq b \},
\]

is convex, we need to show that for any two points \( x_1 \in S \) and \( x_2 \in S \), the line segment joining \( x_1 \) and \( x_2 \) is entirely contained within \( S \). This means that for any \( \lambda \in [0, 1] \), the point \( \lambda x_1 + (1 - \lambda) x_2 \) should also belong to \( S \).

### Step-by-Step Proof:

#### 1. Definition of convexity:
A set \( S \subseteq \mathbb{R}^n \) is convex if, for any two points \( x_1, x_2 \in S \), the point
\[
x_\lambda = \lambda x_1 + (1 - \lambda) x_2
\]
also belongs to \( S \) for every \( \lambda \in [0, 1] \).

#### 2. Region \( S \) and its constraint form:
The set \( S \) is defined by the inequality \( A x \leq b \), where:
- \( A \in \mathbb{R}^{m \times n} \) is a matrix of coefficients,
- \( b \in \mathbb{R}^m \) is a vector, and
- \( x \in \mathbb{R}^n \).

The inequality \( A x \leq b \) means that each component of the vector \( A x \) is less than or equal to the corresponding component of \( b \), i.e., for each \( i = 1, \dots, m \),
\[
(Ax)_i \leq b_i.
\]

#### 3. Assume \( x_1, x_2 \in S \):
Since \( x_1 \in S \), we know that:
\[
A x_1 \leq b.
\]
Similarly, since \( x_2 \in S \), we also have:
\[
A x_2 \leq b.
\]

#### 4. Convex combination of \( x_1 \) and \( x_2 \):
We need to check whether any convex combination of \( x_1 \) and \( x_2 \), given by
\[
x_\lambda = \lambda x_1 + (1 - \lambda) x_2
\]
for \( \lambda \in [0, 1] \), also belongs to \( S \), i.e., whether \( A x_\lambda \leq b \).

#### 5. Apply \( A \) to \( x_\lambda \):
Consider the action of \( A \) on \( x_\lambda \):
\[
A x_\lambda = A (\lambda x_1 + (1 - \lambda) x_2) = \lambda A x_1 + (1 - \lambda) A x_2.
\]

#### 6. Use the fact that \( A x_1 \leq b \) and \( A x_2 \leq b \):
Since \( A x_1 \leq b \) and \( A x_2 \leq b \), each component of \( A x_1 \) and \( A x_2 \) satisfies:
\[
(A x_1)_i \leq b_i \quad \text{and} \quad (A x_2)_i \leq b_i \quad \text{for all } i = 1, \dots, m.
\]

Now, for each \( i \),
\[
\lambda (A x_1)_i + (1 - \lambda) (A x_2)_i \leq \lambda b_i + (1 - \lambda) b_i = b_i.
\]

Thus,
\[
A x_\lambda \leq b.
\]

#### 7. Conclusion:
Since \( A x_\lambda \leq b \), it follows that \( x_\lambda \in S \) for any \( \lambda \in [0, 1] \). Therefore, the set \( S \) is convex.

### Conclusion:
The region \( S = \{ x \in \mathbb{R}^n : A x \leq b \} \) is convex because any convex combination of two points in \( S \) also lies in \( S \).

---
Would you like more details or clarification on any part of the proof?

Here are five questions to expand on this topic:
1. What are some examples of convex sets besides polyhedra?
2. How does convexity relate to optimization problems?
3. Can a non-convex set be described using linear constraints?
4. What is the geometric interpretation of the set \( S = \{x \in \mathbb{R}^n : Ax \leq b \} \)?
5. How does the concept of convex hull relate to convex sets?

**Tip:** Convexity is a fundamental concept in optimization because any local minimum of a convex function on a convex set is also a global minimum.

Consider a region S defined by a set of linear constraints S ={x∈Rn :Ax≤b}, where A is an m×n matrix and b ∈ Rm. Prove that S is convex.

Explore how to prove that the region S = { x ∈ ℝ^n : A x ≤ b } is convex. This proof covers key concepts in linear algebra, convexity, and inequalities, making it suitable for students studying convex optimization or advanced linear algebra.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation