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 vectors, with two operations. The first operation, called addition and denoted by +, assigns to every (u,v) ∈ V x V a vector (u+v) ∈ V. The second operation, called scalar multiplication and represented by a juxtaposition, assigns to every (k,u) ∈ FxV a vector ku ∈ 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) = ru + rv
(r + s)u = ru + su
(rs)u = r(su)
1u = u, where 1 is the unity in the field F, which satisfies 1a = a1 = 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 = ru + su. For example, if we set r = 2, s = 3, and u = (5,7) then:
(r+s)u = (2+3)u = 5u = 5(5.7) = (5⋅5,7) = (25,7)
ru = 2u = 2(5,7) = (2⋅5,7) = (10,7)
su = 3u = 3(5,7) = (3⋅5,7) = (15,7)
ru + su = (10,7) + (15,7) = (25,14)
By the above, it turns out that (r + s)u ≠ ru + su. 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 1u = u. For example, if u = (2,7) then 1u = 1(2,7) = (3⋅1⋅2,1⋅7) = (6,7). It turns out that 1u ≠ 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 .
