Problem 3 - 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. The set of vectors \(\left\langle a_{1}, a_{2}\right\rangle\), scalar multiplication defined by \(k\left\langle a_{1}, a_{2}\right\rangle=\left\langle k a_{1}, 0\right\rangle\)

Step-by-Step Solution

Answer

The set is not a vector space as it doesn't satisfy closure under scalar multiplication.

Step 1: Define the Vector Space Axioms

A vector space needs to satisfy the following axioms: closure under addition and scalar multiplication, the existence of a zero vector, existence of additive inverses, commutative and associative properties for addition, distributive properties, and scalar multiplication identity.

Step 2: Evaluate Closure Under Addition

To test closure under addition, consider two arbitrary vectors \( \langle a_1, a_2 \rangle \) and \( \langle b_1, b_2 \rangle \). Their sum is \( \langle a_1 + b_1, a_2 + b_2 \rangle \). The sum still fits the form \( \langle c_1, c_2 \rangle \), verifying closure under addition.

Step 3: Evaluate Closure Under Scalar Multiplication

Consider an arbitrary scalar \( k \) and vector \( \langle a_1, a_2 \rangle \). According to the given definition, scalar multiplication yields \( k \langle a_1, a_2 \rangle = \langle k a_1, 0 \rangle \). This vector is always of the form \( \langle m, 0 \rangle \), not closing to \( \langle a_1, a_2 \rangle \) form for general vectors \( \langle a_1, a_2 \rangle \).

Step 4: Check Existence of a Zero Vector

A zero vector \( \langle 0, 0 \rangle \) should exist such that \( \langle a_1, a_2 \rangle + \langle 0, 0 \rangle = \langle a_1, a_2 \rangle \). The zero vector \( \langle 0, 0 \rangle \) exists for addition, satisfying this condition.

Step 5: Test Existence of Additive Inverses

For any vector \( \langle a_1, a_2 \rangle \), an additive inverse \( \langle -a_1, -a_2 \rangle \) should exist such that their sum is \( \langle 0, 0 \rangle \). Assuredly, \( \langle a_1, a_2 \rangle + \langle -a_1, -a_2 \rangle = \langle 0, 0 \rangle \).

Key Concepts

Vector Space Axioms

A vector space is essentially a set of vectors where you can perform operations like addition and scalar multiplication. For a set to qualify as a vector space, it must fulfill certain criteria known as axioms. These include:

• Closure under Addition and Scalar Multiplication: The outcome of adding two vectors or multiplying a vector by a scalar should also be a vector within the same set.
• Existence of a Zero Vector: A vector space must include a zero vector, the identity element for addition, so any vector added to the zero vector results in the original vector.
• Existence of Additive Inverses: For any vector, an opposite vector must exist such that their sum is the zero vector.
• Commutative and Associative Properties: Adding vectors should be commutative and associative, so the order doesn’t affect the sum.
• Distributive Properties: Multiplying a scalar across vector addition and distributing vector addition over scalar addition should hold.
• Scalar Multiplication Identity: Multiplying a vector by the scalar one should leave the vector unchanged.

These axioms assure that the vectors perform predictably and consistently under these operations. Understanding them helps articulate why some sets form vector spaces while others do not.

Closure Under Addition

Closure under addition means that if you take any two vectors from the set and add them together, the result should also be in the set. Imagine having two vectors: let's call them \( \langle a_1, a_2 \rangle \) and \( \langle b_1, b_2 \rangle \). If their addition—\( \langle a_1 + b_1, a_2 + b_2 \rangle \)—is a vector that belongs to the original set, then the set is closed under addition.

In our particular problem, we confirmed this because:\ \( \langle a_1, a_2 \rangle + \langle b_1, b_2 \rangle = \langle a_1 + b_1, a_2 + b_2 \rangle \) still maintains the form required by the set. This property ensures that you can freely add vectors within the space without stepping out of the boundaries of the vector space.

Closure Under Scalar Multiplication

Scalar multiplication involves taking a vector and multiplying it by a scalar (a number). Closure under this operation means that the result should still be a vector in the same set. However, in our specific exercise, we faced an issue. When we multiply a vector \( \langle a_1, a_2 \rangle \) by a scalar \( k \), the resulting vector is \( \langle k a_1, 0 \rangle \), differing from the original set form \( \langle a_1, a_2 \rangle \).

This mismatch indicated that the set does not maintain closure under scalar multiplication. Therefore, the set does not qualify as a vector space because it fails to satisfy one of the vital vector space axioms, creating inconsistency and narrowing its mathematical utility.

Existence of Zero Vector

The zero vector is the equivalent of zero for numbers, but in the world of vectors. It serves as the neutral element for vector addition. For the set to be a vector space, this vector must exist such that adding it to any vector results in the same vector. In mathematical terms, for any vector \( \langle a_1, a_2 \rangle \), there should be a zero vector \( \langle 0, 0 \rangle \) satisfying the equation:\

\( \langle a_1, a_2 \rangle + \langle 0, 0 \rangle = \langle a_1, a_2 \rangle \).

Fortunately, this condition holds true as per our exercise, confirming the existence of the zero vector in the set. Thus, our set meets this essential vector space axiom, helping it partially qualify as a vector space.

Existence of Additive Inverses

Additive inverses give each vector its counterpart so that when you add a vector to its inverse, you get the zero vector. This concept is similar to numbers, where adding 5 to -5 equals zero. For a vector \( \langle a_1, a_2 \rangle \), its additive inverse is \( \langle -a_1, -a_2 \rangle \).

In practice, you check this axiom by ensuring that any given vector can be negated such that their sum with the original vector yields the zero vector, like so: \( \langle a_1, a_2 \rangle + \langle -a_1, -a_2 \rangle = \langle 0, 0 \rangle \).

Our exercise confirms that additive inverses do exist for each vector in the set, meaning this axiom holds true. Together with the zero vector, additive inverses allow the vector space to demonstrate a complete arithmetic structure, reinforcing its mathematical versatility.