Proving That ( mathbb{R}^n ) is a Subspace: A Comprehensive Analysis

Introduction

Linear algebra is a fundamental branch of mathematics with vast applications in various fields, including computer science, physics, and engineering. One of the core concepts in linear algebra is that of a subspace. A subspace is a subset of a vector space that is itself a vector space under the same operations. In this article, we will delve into the concept of ( mathbb{R}^n ), a n-dimensional space over the field of real numbers, and prove that it is indeed a subspace. We will explore the requirements and theorems that are necessary to establish this proof, emphasizing the importance of linearity in this context.

Understanding ( mathbb{R}^n )

Before we dive into the technicalities, let's understand what ( mathbb{R}^n ) represents. ( mathbb{R}^n ) is a n-dimensional vector space consisting of all n-tuples of real numbers, i.e., vectors with n components. Each vector in ( mathbb{R}^n ) can be represented as:

vec{v} begin{pmatrix} v_1 v_2 vdots v_n end{pmatrix} end{pmatrix}

where each ( v_i ) is a real number, and ( i 1, 2, ldots, n ).

The Requirement of Subspace: Linearity

To prove that ( mathbb{R}^n ) is a subspace of some vector space, we need to verify that it satisfies the requirements of a subspace. Specifically, we need to confirm that ( mathbb{R}^n ) meets the following criteria:

Zero Vector: The zero vector must be in ( mathbb{R}^n ). Closure Under Addition: If ( vec{u} ) and ( vec{v} ) are vectors in ( mathbb{R}^n ), then ( vec{u} vec{v} ) must also be in ( mathbb{R}^n ). Closure Under Scalar Multiplication: If ( c ) is a scalar (a real number) and ( vec{u} ) is a vector in ( mathbb{R}^n ), then ( c cdot vec{u} ) must also be in ( mathbb{R}^n ).

Proof That ( mathbb{R}^n ) is a Subspace

Zero Vector

The zero vector ( vec{0} ) in ( mathbb{R}^n ) is given by:

vec{0} begin{pmatrix} 0 0 vdots 0 end{pmatrix} end{pmatrix}

Clearly, ( vec{0} ) is an n-tuple of real numbers, so it belongs to ( mathbb{R}^n ).

Closure Under Addition

Let ( vec{u} (u_1, u_2, ldots, u_n) ) and ( vec{v} (v_1, v_2, ldots, v_n) ) be any two vectors in ( mathbb{R}^n ). Then, the sum of ( vec{u} ) and ( vec{v} ) is defined as:

vec{u} vec{v} begin{pmatrix} u_1 v_1 u_2 v_2 vdots u_n v_n end{pmatrix} end{pmatrix}

Since the sum of any two real numbers is also a real number, ( vec{u} vec{v} ) is an n-tuple of real numbers, and thus it belongs to ( mathbb{R}^n ).

Closure Under Scalar Multiplication

Let ( c ) be any scalar (a real number) and ( vec{u} (u_1, u_2, ldots, u_n) ) be any vector in ( mathbb{R}^n ). Then, the scalar multiple of ( vec{u} ) with ( c ) is defined as:

c cdot vec{u} begin{pmatrix} c u_1 c u_2 vdots c u_n end{pmatrix} end{pmatrix}

Since the product of any scalar and a real number is also a real number, ( c cdot vec{u} ) is an n-tuple of real numbers, and thus it belongs to ( mathbb{R}^n ).

Since ( mathbb{R}^n ) satisfies all the conditions of a subspace, we can conclude that ( mathbb{R}^n ) is indeed a subspace.

Conclusion

The proof that ( mathbb{R}^n ) is a subspace is a fundamental concept in linear algebra. It demonstrates the importance of linearity in vector spaces and helps us understand the structure and properties of higher-dimensional spaces. By ensuring that the zero vector, addition, and scalar multiplication operations are preserved, we validate the subspace requirement, leading to a deeper understanding of vector spaces and their applications.