Problem 1 - Advanced Engineering Mathematics

Question

In Problems 1-10, determine whether the given set is a vector space. If not, give at least one axiom that is not satisfied. Unless stated to the contrary, assume that vector addition and scalar multiplication are the ordinary operations defined on that set. $$ \text { The set of vectors }\left\langle a_{1}, a_{2}\right\rangle, \text { where } a_{1} \geq 0, a_{2} \geq 0 $$

Step-by-Step Solution

Answer

The set is not a vector space because it does not satisfy the existence of additive inverses axiom.

Step 1: Identify the Operations

First, understand that the operations mentioned are ordinary vector addition and scalar multiplication. For vectors \( \langle a_1, a_2 \rangle \) and \( \langle b_1, b_2 \rangle \), vector addition is defined as \( \langle a_1 + b_1, a_2 + b_2 \rangle \), and scalar multiplication for a scalar \( c \) is \( \langle c a_1, c a_2 \rangle \).

Step 2: Verify the Zero Vector Existence

In a vector space, there must exist a zero vector (\( \vec{0} \)) such that for any vector \( \vec{v} \), the equation \( \vec{v} + \vec{0} = \vec{v} \) holds true. The vector \( \langle 0, 0 \rangle \) meets this requirement since adding it to any vector \( \langle a_1, a_2 \rangle \) yields the same vector. Therefore, the set satisfies this axiom.

Step 3: Check for Additive Inverses

Each vector \( \langle a_1, a_2 \rangle \) in the set must have an additive inverse \( \langle -a_1, -a_2 \rangle \) such that their sum equals the zero vector \( \langle 0, 0 \rangle \). If either \( a_1 \) or \( a_2 \) is positive, their negative counterparts would not satisfy the requirement \( a_1 \geq 0 \) and \( a_2 \geq 0 \). Hence, this axiom does not hold.

Key Concepts

Vector Addition

Vector addition is a fundamental operation for vector spaces. It involves combining two vectors to produce another vector. If you have two vectors, say \( \langle a_1, a_2 \rangle \) and \( \langle b_1, b_2 \rangle \), vector addition is performed by adding corresponding components: \( \langle a_1 + b_1, a_2 + b_2 \rangle \). This operation should satisfy specific properties like commutativity \((\vec{u} + \vec{v} = \vec{v} + \vec{u})\) and associativity \((\vec{u} + (\vec{v} + \vec{w}) = (\vec{u} + \vec{v}) + \vec{w})\), ensuring that the order of addition does not affect the result.

As specified in the exercise, ordinary vector addition is assumed. However, in the given set, since the vectors must have non-negative components \((a_1 \geq 0, a_2 \geq 0)\), vector addition is inherently restricted. This is because adding two vectors could lead you outside the set if any resultant component turns negative. Therefore, it must be checked that performing vector addition on any two vectors in the set still results in a vector within the set.

Scalar Multiplication

Scalar multiplication involves multiplying each component of a vector by a scalar (a real number). For a vector \( \langle a_1, a_2 \rangle \) and a scalar \( c \), the scalar multiplication is \( \langle c a_1, c a_2 \rangle \). This operation should also satisfy specific properties such as distributivity \((c(\vec{u} + \vec{v}) = c\vec{u} + c\vec{v})\), and associativity with respect to multiplication \(((ab)\vec{u} = a(b\vec{u}))\).

In ordinary scenarios, scalar multiplication by a positive number retains the sign of the vector components, but multiplying by a negative scalar can result in negative vector components. Given the condition \( a_1 \geq 0 \) and \( a_2 \geq 0 \), the operation of scalar multiplication is limited to non-negative scalars unless the vector itself is the zero vector. Multiplying a non-zero vector with a negative scalar can lead to a violation of the set condition, indicating that the set can't fulfill all properties required by scalar multiplication in a vector space.

Zero Vector

The zero vector plays a crucial role in the structure of a vector space. It is essentially the neutral element in vector addition. For a set to be considered a vector space, it must have a zero vector \( \vec{0} \) that satisfies \( \vec{v} + \vec{0} = \vec{v} \) for any vector \( \vec{v} \).

In this exercise, the zero vector for the given set is \( \langle 0, 0 \rangle \). Adding this vector to any other vector \( \langle a_1, a_2 \rangle \) in the set satisfies the vector addition identity \( \langle a_1 + 0, a_2 + 0 \rangle = \langle a_1, a_2 \rangle \). Thus, the set correctly includes a zero vector, ensuring it meets this particular vector space requirement.

Additive Inverse

The additive inverse of a vector is vital; it ensures the possibility of subtraction within the vector space by equating the sum of a vector and its inverse to the zero vector. For any vector \( \langle a_1, a_2 \rangle \), its additive inverse would be \( \langle -a_1, -a_2 \rangle \) such that \( \langle a_1, a_2 \rangle + \langle -a_1, -a_2 \rangle = \langle 0, 0 \rangle \).

In the given set condition \( a_1 \geq 0 \) and \( a_2 \geq 0 \), achieving a vector with negative components \( -a_1 \) and \( -a_2 \), required for the additive inverse, violates the non-negative restriction of the set. As such, not every vector in this set can have an additive inverse that also belongs to the set. This restriction shows a critical failure to satisfy one of the necessary axioms for defining a complete vector space. Hence, the provided set fails to be a vector space due to the absence of valid additive inverses.