Problem 6 - 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 \(\left\langle a_{1}, a_{2}\right\rangle\), addition and scalar multiplication defined by $$ \begin{aligned} \left\langle a_{1}, a_{2}\right\rangle+\left\langle b_{1}, b_{2}\right\rangle &=\left\langle a_{1}+b_{1}+1, a_{2}+b_{2}+1\right\rangle \\ k\left(a_{1}, a_{2}\right\rangle &=\left\langle k a_{1}+k-1, k a_{2}+k-1\right\rangle \end{aligned} $$

Step-by-Step Solution

Answer

The set is not a vector space; it violates the identity element for addition axiom.

Step 1: Review Vector Space Axioms

To determine if a set is a vector space, we must verify it satisfies all 10 vector space axioms. These include axioms for closure under addition and scalar multiplication, identity elements, inverses, distributive properties, and more.

Step 2: Check Closure Under Addition

For two vectors \( \langle a_{1}, a_{2} \rangle + \langle b_{1}, b_{2} \rangle = \langle a_{1} + b_{1} + 1, a_{2} + b_{2} + 1 \rangle \), determine if the result always belongs to the set. For closure, the operation must yield another element in the same set. Given the extra '+1' in each component, adding two vectors results in shifting from traditional addition.

Step 3: Examine Identity Element for Addition

A vector space must have a zero vector \( \langle 0, 0 \rangle \) where \( \langle a_{1}, a_{2} \rangle + \langle 0, 0 \rangle = \langle a_{1}, a_{2} \rangle \). Here, with modified addition, this condition is not satisfied since adding \( \langle 0, 0 \rangle \) results in \( \langle a_{1} +1, a_{2} +1 \rangle \), not \( \langle a_{1}, a_{2} \rangle \).

Step 4: Examine Scalar Multiplication Identity

We check if \(1 \cdot \langle a_{1}, a_{2} \rangle = \langle a_{1}, a_{2} \rangle\). The given definition is \(1 \cdot \langle a_{1}, a_{2} \rangle = \langle 1 \cdot a_{1} + 1 - 1, 1 \cdot a_{2} + 1 - 1 \rangle = \langle a_{1}, a_{2} \rangle \), so this holds.

Step 5: Identify Violated Axiom

The vector space axioms are disrupted by failure of the identity element for addition. The structure \( \langle a_{1} +1, a_{2} +1 \rangle \) upon adding the zero vector, instead of obtaining \( \langle a_{1}, a_{2} \rangle \) itself, violates Axiom of the existence of additive identity.

Key Concepts

Vector Addition

Vector addition is a fundamental operation in vector spaces where two vectors are combined to produce a third vector. This operation must satisfy certain properties for a set to qualify as a vector space. Let's take a closer look at these properties:

• Commutativity: Vector addition should be commutative, meaning \(<\langle a_1, a_2 \rangle + \langle b_1, b_2 \rangle = \langle b_1, b_2 \rangle + \langle a_1, a_2 \rangle\).
• Associativity: The operation should be associative, so that \(\cloudf (\langle a_1, a_2 \rangle + \langle b_1, b_2 \rangle) + \langle c_1, c_2 \rangle = \langle a_1, a_2 \rangle + (\langle b_1, b_2 \rangle + \langle c_1, c_2 \rangle)\).
• Identity Element: There must be a zero vector \(\langle 0, 0 \rangle\) such that \(\langle a_1, a_2 \rangle + \langle 0, 0 \rangle = \langle a_1, a_2 \rangle\).
• Inverse Elements: Each vector must have an inverse, so \(\langle a_1, a_2 \rangle + \langle -a_1, -a_2 \rangle = \langle 0, 0 \rangle\).

In the original exercise provided, the vector addition is defined as \(\langle a_1, a_2 \rangle + \langle b_1, b_2 \rangle = \langle a_1+b_1+1, a_2+b_2+1 \rangle\), which modifies the typical operation by adding an extra \(+1\) to each component. This modification affects the identity element property since adding \(\langle 0, 0 \rangle\) does not return \(\langle a_1, a_2 \rangle\) due to the additional \(+1\). Thus, this set fails to be a vector space based on the identity property of addition.

Scalar Multiplication

Scalar multiplication involves multiplying a vector by a scalar (a real number) to produce another vector. This operation must also satisfy specific properties in a vector space:

• Closure Under Scalar Multiplication: The product of a scalar and a vector should still be within the vector space.
• Distributive Property: Scalars should distribute across vector addition: \(k(\langle a_1, a_2 \rangle + \langle b_1, b_2 \rangle) = k\langle a_1, a_2 \rangle + k\langle b_1, b_2 \rangle\).
• Scalar Product Distributive Property: Adding two scalars, then multiplying with a vector should be the same as multiplying each scalar first: \((m + n)\langle a_1, a_2 \rangle = m\langle a_1, a_2 \rangle + n\langle a_2, a_2 \rangle\).
• Associative Law: \((km)\langle a_1, a_2 \rangle = k(m\langle a_1, a_2 \rangle)\).
• Multiplicative Identity: Multiplying by 1, the outcome should be the vector itself: \(1\langle a_1, a_2 \rangle = \langle a_1, a_2 \rangle\).

In the original exercise, scalar multiplication is defined as \(k\langle a_1, a_2 \rangle = \langle ka_1 + k - 1, ka_2 + k - 1 \rangle\). The identity check with scalar \(1\) satisfies the condition because it leads back to the vector components \(\langle a_1, a_2 \rangle\) itself, without any disruption. While the scalar multiplication identity holds, other properties need to be checked as well to ensure the vector space formation.

Vector Space Axioms

A vector space must adhere to a set of axioms, also referred to as properties or rules, that define its structure. Let's break down the core axioms necessary for a vector space:

• Closure Under Addition and Scalar Multiplication: Any sum of vectors and scalar multiple of a vector must also be a vector within the same set.
• Contains the Zero Vector: Every vector space must have an additive identity, or zero vector.
• Inverse and Identity Elements for Addition: There must be a way to "undo" addition, called the inverse, and an addition operation that doesn’t change the vector, called the identity.
• Associative and Commutative Properties: These ensure consistent combining of vectors and scalars.
• Distributive Laws: Scalars should distribute over both vector addition and their own addition.

In the exercise, one of the important axioms that fails is the existence of an identity element for addition. Normally, a zero vector \(\langle 0, 0 \rangle\) should satisfy the identity condition. However, due to the altered addition operation \(+1\) in each component, \(\langle a_1, a_2 \rangle + \langle 0, 0 \rangle e \langle a_1, a_2 \rangle\), violating this axiom and thus disproving the set's qualification as a vector space.