What is a Vector Space? How to determine if its a Vector Space or not?

VECTOR SPACES AND SUBSPACES

 

INTRODUCTION

         The definition of a vector space V, whose elements are called vectors, involves arbitrary filed K, whose elements are called scalars.

         The following notation will be used (unless otherwise stated or implied)

 V given vectors space

u,v,w vectors in V

K given number field

a,b,c, or k scalars in K

VECTOR SPACES

                  The following define the notion of a vector space V, where K is the field of scalars.

Definition: Let V be a nonempty set with two operations:

1. Rule of vector addition, which assigns to any u,v∈V  a sum u+v in V.

2. Rule of vector multiplication, which assigns to any u∈ V and 

   k∈K a product ku∈K.

Then V is called a vector space (over the field K) if the following axioms hold:

        [A1]    For any vectors u,v,w∈V, (u+v)+w = u+(v+w).

        [A2]  There is a vector in V, denoted by 𝟢 and called the zero vector, for which u+𝟢 = 𝟢+u = u for any vector u∈V.

        [A3]  For each vector u∈V there is a vector in V, denoted by −u, for which u+(−u) = (−u) + u = 𝟢

        [A4]  For any vectors u,v∈V, u+v = v+u .

        [M1]     For any scalar k∈K and any vectors u,v∈V, k(u+v) = ku+kv.

        [M2]    For any scalars a,b∈K and any vector u∈V, (a+b)u = au+bu.  

        [M3]    For any scalars a,b∈K and any vector u∈V, (ab)u = a(bu).

        [M4]    For the unit scalar 1∈ K, 1u=u  for any vector u∈V.

    The above axioms naturally split into two sets. The first four are only concerned with the additive structure of V and can be summarized by saying that V is a commutative group under addition. It follows that any sum of vectors of the form:

v₁+v₂+⋯+aₙ

requires no parentheses and does not depend on the order of the summands, the zero vector 𝟢 is unique, the negative −u of u is unique, and the cancellation law holds, that is, for any vectors u,v,w∈ V, we have  

u+w = v+w implies u=v

Also, subtraction is defined by

u−v = u+(−v)

where −v is the unique negative of the vector v.

    The remaining four axioms are concerned within the “action” of the field K of scalars on the vectors space V.

Theorem: Let V be a vector space over a field K.

(i).                   For any scalar k∈K and 0∈V, k𝟢=𝟢.

(ii).                For 𝟢∈K and any vector u∈V, 𝟢u=𝟢.

(iii).             If ku=𝟢, where k∈K and u∈V, then k=𝟢 or  u=𝟢.

(iv).              For any k∈K and any u∈V, (−k)u=k(−u)=−ku.

EXAMPLES OF VECTOR SPACES

 

Space Rⁿ

    Recall that Rⁿ denoted the n-tuples of real numbers. Here Rⁿ is viewed as a vector over R, where vectors addition and scalar multiplication are defined by

(a₁,a₂,⋯,aₙ )+(b₁,b₂,⋯,bₙ )=(a₁+b₁,a₂+b₂,⋯,aₙ+bₙ)

and

k(a₁,a₂,⋯,aₙ)=(ka₁,ka₂,⋯,kaₙ)

The zero vector in Rⁿ is the n-tuples of zeros,

 𝟢=(𝟢,𝟢,⋯,𝟢)

and the negative of vector is defined by

−(a₁,a₂,⋯,aₙ )=(−a₁,−a₂,⋯,−aₙ)

 

Polynomial Space P(t)

         Let P(t) denote the set of all real polynomials

p(t)=a₀+a₁t+a₂t²+⋯+aₙtⁿ

where the coefficients a₁ belong to the real field R. Then P(t) is a vetors space over R using the operations:

1)   Vector addition p(t)+q(t) in P(t) is the usual operation of addition of polynomials.

2)  Scalar multiplication kp(t) in P(t) is the usual operation of the product of a scalar k and a polynomial p(t).

Polynomial Space Pₙ(t)

         Let Pₙ(t) denote the set of all real polynomials

                                        p(t)=a₀+a₁t+a₂t²+⋯+aₛtˢ

where the degree of p(t) is less than or equal to n, that is, s≤n. Then Pₙ(t) is a vector space over R with respect to the usual operations of addition of polynomials and of multiplication of polynomial by a constant. We include the zero polynomial 𝟢 as an element of Pₙ(t) even though its degree is undefined.

 
Matrix Space Mₘ̦ₙ

The notation Mₘ̦ₙ or simply M, will be used to denote the set of all m×n matrices whose entries are real numbers. Then Mₘ̦ₙ is a vector space over R with respect to the usual operations of matrix addition and scalar multiplication of matrices.

Fields and Subfields 

         Suppose a field E is an extension of a field K, that is, suppose E is a field which contains a subfield K. Then E may be viewed as a vector space over K using the operations:

              1)   Vector addition u+v in E is the usual addition in E..

2) Scalar multiplication ku in E, where k∈K and u∈E, is the usual operation of k and u as elements of E.     

Then the eight axioms of a vector space are satisfied by E and K and the above two operations. That is, E is a vector space over its subfield K. 

Solve Problems on Vector Spaces

Example 1. Suppose u and v belong to a vector space V. Simply each expression:

Solution:

Example 2. Show that for any scalar k and any vectors u and v, we have k(u-v)=ku-kv.

Solution:

         Using the definition of subtraction, that u−v=u+(−v) and the theorem: For any k∈K and any u∈V, (-k)u=k(−u)=−ku, that is,

k(−v)=−kv, we have 

k(u−v)=k[u+(−v)]=ku+k(−v)=ku+(−kv)=ku−kv.

Example 3. Show that u+u=2u for any vector u.

Solution:

           Using axiom M4: For the unit scalar 1∈K,1u=u for any vector u∈V and then axiom M2: For any scalars a,b∈K and any vector u∈V, (a+b)u=au+bu, we have

u+u=1u+1u=(1+1)u=2u.

Example 4. Let V be the set of ordered pairs of real numbers. Show that V is NOT a vector space over R with respect to each of the following operations of additions in V and scalar multiplication on V.

Solution: