The term * vector* is much found when we study physics or mathematics. What is the meaning of a vector, actually? In this post, the general definition of vectors will be discussed. Below is the definition of a vector space.

**Definition**

Let F be a field whose elements are called * scalars*. A vector space over F is a non-empty set V, whose members are called

*, with two operations. The first operation, called*

**vectors***and denoted by +, assigns to every (*

**addition****u**,

**v**) ∈ V x V a vector (

**u**+

**v**) ∈ V. The second operation, called

*and represented by a juxtaposition, assigns to every (k,*

**scalar multiplication****u**) ∈ FxV a vector k

**u**∈ V. In addition, the following conditions must be met.

Associative property of addition:

**u** + (**v** + **w**) = (**u** + **v**) + **w** for every **u**, **v**, **w** ∈ V

Commutative property of addition:

**u** + **v** = **v** + **u** for every **u**, **v** ∈ V

The existence of zero:

There is a vector **0** ∈ V such that **0** + **v** = **v** + **0** = **v** for every **v** ∈ V.

The existence of additive inverses:

For every **u** ∈ V there is a vector, denoted by **-u**, which satisfies **u** + (**-u**) = (**-u**) + **u** = **0**

Properties of scalar multiplication:

For every r, s ∈ F and every **u**, **v** ∈ V the following properties hold.

r(**u** + **v**) = r**u** + r**v**

(r + s)**u** = r**u** + s**u**

(rs)**u** = r(s**u**)

1**u** = **u**, where 1 is the unity in the field F, which satisfies 1*a* = *a*1 = *a* for every *a* ∈ F.

Note:

In this post, to distinguish vectors from scalars, vectors are expressed in bold letters. So, vector v is written as **v**. Letters denoting scalars are not in bold.

**Example 1**

Let A be the set of all ordered pairs (x,y) with x, y ∈ . The addition in A is defined by (x,y) + (x’,y’) = (x+x’,y+y’) and the scalar multiplication is defined by k(x,y) = (kx,y). Is A a vector space over the field ?

**Answer**

A is not a vector space over the field because it does not satisfy the property of scalar multiplication (r + s)**u** = r**u** + s**u**. For example, if we set r = 2, s = 3, and **u** = (5,7) then:

(r+s)**u** = (2+3)**u** = 5**u** = 5(5.7) = (5⋅5,7) = (25,7)

r**u** = 2**u** = 2(5,7) = (2⋅5,7) = (10,7)

s**u** = 3**u** = 3(5,7) = (3⋅5,7) = (15,7)

r**u** + s**u** = (10,7) + (15,7) = (25,14)

By the above, it turns out that (r + s)**u** ≠ r**u** + s**u**. Consequently, A is not a vector space over the field .

**Example 2**

Let B be the set of all ordered pairs (x,y) with x, y ∈ . The addition in B is defined by (x,y) + (x’,y’) = (x+x’,y+y’) and the scalar multiplication is defined by k(x,y) = (3kx,ky). Is B a vector space over the field ?

**Answer**

B is not a vector space over the field because it does not satisfy the condition 1**u** = **u**. For example, if u = (2,7) then 1**u** = 1(2,7) = (3⋅1⋅2,1⋅7) = (6,7). It turns out that 1**u **≠ **u**. Hence, B is not a vector space over .

**Example 3**

Let C be the set of all ordered pairs (x,y) with x, y ∈ . The addition in C is defined by (x,y) + (x’,y’) = (x+x’,y+y’) and the scalar multiplication is defined by k(x,y) = (kx,ky). Is C a vector space over the field ?

**Answer**

C, with the addition and scalar multiplication operations defined above, satisfies all conditions that must be satisfied by a vector space. (**Check it!**) – (An example of how to prove it can be obtained in Example 2 of the other post entitled Vector Subspaces). So, C is a vector space over .