Problem 1 Let \(V\) denote the set of orde... [FREE SOLUTION]

Chapter 6: Problem 1

Let \(V\) denote the set of ordered triples \((x, y, z)\) and define addition in \(V\) as in \(\mathbb{R}^{3}\). For each of the following definitions of scalar multiplication, decide whether \(V\) is a vector space. a. \(a(x, y, z)=(a x, y, a z)\) b. \(a(x, y, z)=(a x, 0, a z)\) c. \(a(x, y, z)=(0,0,0)\) d. \(a(x, y, z)=(2 a x, 2 a y, 2 a z)\)

Short Answer

Expert verified

None of the cases define a vector space.

Step by step solution

Understand the Vector Space Definition

A vector space, over a field (usually the real numbers), is a set equipped with two operations: vector addition and scalar multiplication, satisfying certain properties such as closure, commutativity, associativity, the existence of an additive identity, additive inverses, distributive properties, etc.

Check Closure under Scalar Multiplication for Case a

The scalar multiplication is defined as \( a(x, y, z) = (a x, y, a z) \). For the set \( V \) to be a vector space, multiplying any vector by a scalar should result in another vector in \( V \). Multiplying all components except the second component by a scalar violates this property. Thus, \( V \) is not closed under scalar multiplication with this definition.

Check the Vector Space Axioms for Case b

Scalar multiplication is \( a(x, y, z) = (a x, 0, a z) \). The zero vector \( (0,0,0) \) must be present in any vector space, but multiplying any vector by 1 does not give back the original vector due to the second component always being 0. Thus, \( V \) doesn't satisfy the identity property for scalar multiplication and is not a vector space.

Evaluate Scalar Multiplication for Case c

Here, \( a(x, y, z) = (0, 0, 0) \). Scalar multiplication returns the zero vector for any vector and any scalar, which violates the existence of an identity element in scalar multiplication (1 times a vector should be the vector itself). Hence, this does not form a vector space.

Verify Scalar Multiplication Properties for Case d

Scalar multiplication is defined as \( a(x, y, z) = (2 a x, 2 a y, 2 a z) \). This maintains closure, as multiplying by scalars remains within \( V \). However, scalar multiplication must satisfy the identity property: \(1 \cdot (x, y, z) = (x, y, z)\), which fails here since \( (1 \cdot (x, y, z)) = (2x, 2y, 2z)\). Thus, this is not a vector space.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Scalar Multiplication

Scalar multiplication is an operation where each component of a vector is multiplied by the same scalar. It is one of the fundamental operations required for a set to be considered a vector space. Proper scalar multiplication should satisfy specific rules:

When a vector is multiplied by 1, the vector remains unchanged. This is known as the identity property.
Multiplying a vector by 0 results in the zero vector, \((0, 0, 0)\) for ordered triples.
For any scalars \(a\) and \(b\), and any vector \(\textbf{v}\), \(a(b\textbf{v}) = (ab)\textbf{v}\).

In the given exercise, the examples provided (especially using definitions like \(a(x, y, z) = (a x, y, a z)\) and \(a(x, y, z) = (2 a x, 2 a y, 2 a z)\)) fail to meet these criteria, highlighting why the set \(V\) is not forming a valid vector space.

Vector Addition

Vector addition involves taking two vectors from the same space and combining them component-wise. For a set to be a vector space, vector addition must be defined so that:

It is closed - the sum of any two vectors is also a vector in the same set.
The result follows the associative property: \( (\textbf{u} + \textbf{v}) + \textbf{w} = \textbf{u} + (\textbf{v} + \textbf{w}) \).
The operation is commutative: \( \textbf{u} + \textbf{v} = \textbf{v} + \textbf{u} \).
There exists a zero vector (additive identity) so that \( \textbf{u} + \mathbf{0} = \textbf{u} \).
Each vector \(\textbf{v}\) has an additive inverse \(\textbf{-v}\), such that \(\textbf{v} + \textbf{-v} = \mathbf{0}\).

In \(\mathbb{R}^{3}\), vector addition is simply \((x_1, y_1, z_1) + (x_2, y_2, z_2) = (x_1 + x_2, y_1 + y_2, z_1 + z_2)\), and this process inherently follows all these requirements, forming the baseline understanding for other operations.

Closure Property

The closure property is crucial for both vector addition and scalar multiplication. It asserts that when these operations are applied within a set, the outcome must also belong to the same set. This means:

For addition, adding two vectors from the set results in another vector from the same set.
For scalar multiplication, multiplying a vector by a scalar results in another vector from the set.

In the exercise, some definitions of scalar multiplication, like \(a(x, y, z) = (a x, 0, a z)\), do not have closure because applying scalar multiplication does not always produce another vector within \(V\). If closure is violated, the structure cannot be a vector space.

Vector Space Axioms

The vector space axioms are a series of rules that a set must satisfy to be considered a vector space. These include both the operations of vector addition and scalar multiplication. Some of the relevant axioms are:

Closure under addition and scalar multiplication, meaning these operations should always result in vectors within the space.
Existence of an additive identity (zero vector) and an additive inverse for every vector.
Distributive laws, stating that scalar multiplication distributes over both vector addition and scalar addition.
Associative properties, where the grouping of vectors, or scalars with vectors, does not affect the result.
Commutative property of vector addition.

The various definitions of scalar multiplication examined in the exercise highlight scenarios where these axioms are violated, thereby preventing \(V\) from forming a legitimate vector space. Cases like \(a(x, y, z) = (a x, y, a z)\) fail due to the lack of an identity element in scalar multiplication.

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