# Linear Algebra – Introduction to Vector Space

Linear Algebra / Tuesday, August 13th, 2019

# Introduction

Here we introduce a general mathematical concept, called vector space.

In the study of mathematics we encounter many examples of mathematical objects that can be added to each other and multiplied by real numbers. First of all, the real numbers themselves are such objects. Other examples are real valued functions, the complex numbers, infinite series, vectors in n-dimensional spaces, and vector valued functions. Vector space includes all these examples and many others as special cases.

We have already studied algebraic structures like groups and rings; the former has its motivation in the set of one-to-one mappings of a set onto itself, the latter, in the set of integers. The present algebraic structure which we are about to consider – vector space – can, in large part trace its origins to topics in geometry and physics.

A vector space differs from the structures of groups and rings due to the fact that one of the products defined on it uses elements outside of the set itself.

The importance of vector spaces lies on the fact that so many models arising in the solutions of specific problems turn out to be vector spaces. For this reason, the basic concepts introduced in them have certain universality.

The fundamental motions like linear independence, basis and dimension which will be discussed here are potent and effective tools in all branches of mathematics.

### Orderd n-tuples of real numbers

A set (x1, x2, …, xn) of n real numbers arranged in order is called an ordered n-tuple of reals or simply n-tuple.

Example:

In two-dimensional co-ordinate system, any poit (7, 15) is nothing but an ordered 2-tuple.

(-5, 7, 0, 8) is a 4-tuple.

Each row of a matrix Is a 5-tuple, each column of the matrix is 2-tuple.

## Equality of two n-tuples

For two n-tuples (x1, x2, …, xn) and (y1, y2, …, yn) we say (x1, x2, …, xn) = (y1, y2, …, yn) if x1= y1, x2= y2,…, xn= yn.

### Sum of two n-tuples

Let α = (x1, x2, …, xn), ξ = (y1, y2, …, yn) be two n-tuples. Then their sum α + ξ = (x1+ y1, x2+ y2, …, xn+ yn)

Example:

(1, 0, 5, ½) + (7, 8, -7, ½) = (8, 8, -2, 1)

### Multiplication of an ordered n-tuple by a real number

Let α = (x1, x2, …, xn ) be an n-tuple and c be a real number. Then the product of α and c is c.α = α = (cx1, cx2, …, cxn).

Example:

1. (7, 8, -7, ½) = (14, 16, -14, 1)

### Difference of two n-tuples

Difference of two n-tuples α and ξ is α – ξ is defined as α – ξ = α + (-1). ξ

## Definition of Vector Space

Let V be a non-empty set and R be the set of all real numbers. Let we have two composition, one is ‘+’ between two numbers of V and another is ‘.’ Between a member of R and a member of V. V is said to be a vector space (or, linear space) if the following axioms hold:

α + β Є V for all α and β in V [closure property under ‘+’]
α + (β + γ) = (α + β) + γ for all α, β, γ in V [Associative property under ‘+’]
V contains a member, say θ, such that α + θ = α for all α in V.
Corresponding to each α in V, there exists an element, say ‘-α’ in V such that α + (-α) = θ.
α + β = β + α for all α and β in V. [Commutative property under ‘+’]
Thus V is a abelian group under ‘+’
α Є V for all a in R and for all α in V.
1.α = α for all α in V.
(ab). α = a.(b. α) for all a, b in R and for all α in V.
a.( α+ β) = a. α + a.β for all a in R and for α, β in V.
(a + b). α = a. α + b. α for all a, b in R and for α in V.
The elements of vector space are called vectors and the elements of R are called scalars.

 Theorem

In a vector space V over a field F the following results hold

(i) 1.α = θ for all α in V
(ii) c.θ = θ for all c in F
(iii) -1. α = – α, αЄV
(iv) 1.α = θ ⇒ either c = 0 or α = θ.

Proof:

(i) In F, we have 0 + 0 = 0

Therefore, for α in V, we have

(0 + 0). α = 0. α

Or, 0. α + 0. α = 0.α

Or, 0. α + 0. α = 0.α + θ

⇒ 0. α = θ, by left cancellation rule in the abelian group.

(ii) We have in V, θ + θ = θ

Therefore for any c in F,

c.( θ + θ) = c.θ

⇒ c.θ + c.θ = c.θ

⇒ c.θ + c.θ = c.θ + θ

So by left cancellation rule in V, c.θ

(iii) We have in F, 0 = 1 + (-1)

Therefore, for α in V, we have

0.α = (1 + (-1)).α

⇒0.α = 1.α + (-1).α

Now 0.α = θ and 1.α = α, so

θ = α + (-1).α

⇒ -α + θ = -α + α + (-1).α

⇒ -α = (-1).α

(iv) Let c. α = θ and suppose that c ≠ 0. Then c-1 exists in F.

Hence, c-1 .(c. α ) = c-1 . θ

⇒ (c-1 c).α = θ

⇒1.α = θ

⇒ α = θ

Again, if c = 0, then c. α = θ for any α Є V

Therefore, c. α = θ implies either c = 0 or α = θ

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