Vector Space & Linear Transformation - 1 | Mathematics for Competitive Exams PDF Download

Vector Spaces 

1. one set and one binary composition in a Group and
2. one set and two binary compositions in a Group.
   Now here we shall study the structure consisting of two sets, (a field and an abelian group) and a composition combining an element of the field to an element of the abelian group. In fact, this is the most fundamental and basic structure in which the concept of distance and hence the concept of limit,s etc can be introduced, leading to the study of analysis in a much wider prospect. Thus this serves as a link between algebraic and topological structure

Important Definitions

1. Ring

Definition: A non-empty set R is called a ring if 

1. R is Abelian group under the addition 
2. R is semi-group under multiplication 
3. Distribution law holds
   a(b + c) = a • b + a • c
   (a + b)c = a • c + b • c
4. Example 
   (i) (Z, +, •) is a ring where Z = {0, ±1, ±2, ...}
   (ii) (Q, +, •), where Q is the set of rational numbers.
   (iii) (R, +, •), where R is set of real numbers.
   (iv) (Zn, +, •), Z_n = residue classes of module n.

2. Field

Definition: A non-empty set F is called a field if 

1. F is Abelian group under the addition
2. F - {0} is an abelian group under multiplication
3. The right distribution law holds in F.
   i.e. a, b, c, e F
   (a + b)c = ac + be
4. Example 
   (i) (R, +, •) is a field
   (ii)(C, +, •) is a field
   (iii) (Q, +, •) is a field
   (iv)(Z, +, •) is not field as (Z - {0}, •) is not group under multiplication. 
5. Vectors: Let V be a non-empty set whose element α,β, γ,.... or v_1, v₂, v₃, ........ etc. will be called vectors. 
6. Scalars: Let (F, +, .) be a field whose element α,β, γ..... will be called Scalars. 
7. Internal Binary Composition (Vector addition): A map * : V * V →V is called an internal composition in V and denoted by '+’ . i.e., for every α ∈ V , β ∈ V , α * γ∈ V
   Example: In the set Z of integers, the addition and multiplication are internal binary compositions. 
8. External Binary Composition (Scalar Multiplication): A map o : F * V → V is called an external composition in V and denoted by 'o’ or '•'.
   i.e., ∀ a ∈ F and ∀ a ∈ V,  a o α ∈ V,
   Example:In the set of matrices, scalar multiplication of matrices is an external binary operation on the R of real number

3. Polynomial Space

1. P(t) Let P(t) denote the set of all polynomials of the form
   
   where the coefficient a, belong to a field K. 
2. Then P(t) is a vector space over K using the following operations:
   (i) Vector Addition: Here p(t) + q(t) in P(t) is the usual operation of addition of polynomials.
   (ii) Scalar Multiplication: Here kp(t) in P(t) is the usual operation of the product of a scalar k and a polynomial p(t).
3. The zero polynomial 0 is the zero vector in P(t).

Vector and Scalar Addition and Multiplication

1. The following defines the notion of a Vector Space V where K is the field of scalars.
2. Definition: Let V be a nonempty set with two operations:
   (i) Vector Addition: This assigns to any u, v, ∈ V a sum u + v in V.
   (ii) Scalar Multiplication: This assigns to any u ∈ V, k ∈ K a product ku ∈ V.
3. Then V is called a vector space (over the field K) if the following axioms hold for any vectors u, v, w ∈ V:
   [A₁] (u + v) + w = u + (v + w)
   [A₂] There is a vector in V, denoted by 0 and called the zero vector, such that, for any u ∈ V, u + 0 = 0 + u = u
   [A₃] For each u ∈ V, there is a vector in V, denoted by - u and called the negative of u, such that u + ( - u) = ( - u) + u = 0
   [A₄] u + v = v + u
   [M₁] k(u + v) = ku + kv, for any scalar k ∈ K.
   [M₂] (a + b)u = au + bu, for any scalar a, b ∈ K.
   [M₃] (ab)u = a(bu), for any scalar a, b ∈ K.
   [M₄] 1 u = u, for the unit scalar 1 ∈ K.
4. Notation: 
   (i) Generally, the vector space V over the field F is expressed by V(F). 
   (ii) When the field F itself is clear, then it is expressed simply by V only.
   (iii) Throughout the text, V(F) will mean a vector space V of vectors over a field F of scalars 
   (iv)If F = R (the field of real numbers), then V(R) is called a Real vector over a field F of scalars.
   (v) Similarly, V(ℂ) is called the Complex Vector Space and V(ꙩ) is called the Rational Vector Space. 
   (vi)The set of three-dimensional vectors of geometry is called Three Dimensional Vector Space and is written as V₃(F) or V³(F) or simply V₃.

Important Remarks

1. Here by vector, we do not mean the vector quantity which we have defined in Vector algebra as a directed line segment.
   Here the word vectors and scalars will used ina  more general sense.
2. Four operations have been used in the vector space V(F), Two different internal operations (+), (•) in the field F,
   Third internal operations (+) in the set V and the fourth external operations (•) in V over F.
   Though in these four operations, two are expressed by the symbol (+) and the remaining two by the symbol (•), it should be clear that under the reference which symbol (+), or (•) is being used for which operation.
   For Example, if a, b ∈ F, then a + b ∈ F express the addition in the field F which is often called the scalar addition in the field F, and if α, β ∈ V then α + β ∈ V express the addition in the set V which is generally called the vector addition. Similarly, a • b G f expresses the multiplication in the field F and a • b ∈ V expresses the external operation scalar multiplication of the vector space.
3. Two types of zero elements have been used in every vector space. Zero element i.e. additive identity of V will be indicated by hold face type ‘0’ and is called zero vector and zero element of the field F is generally indicated by ‘0’ which is called zero scalar. In future text we shall indicated both of these by the same symbol ‘0’ and the context will make it clear, which zero we intend.
4. The vector space which contains only zero elements is called Null space or Trivial space and is written as {0}.

General Properties of Vector Space

If V be the vector space over the field F and 0 ∈ V be its zero vector, then the following properties are obvious because of V being an abelian group :

(i) For any α ∈ V, α + β = α β = 0      [Uniqueness of the zero element]
(ii) α ∈ V , β ∈ V , a + β = 0 = > β = - α     [Uniqueness of inverse]
(iii) For any α, β, γ ∈ V , α + β = α + γ = > β = γ     [cancellation law]
Now we shall establish few more properties of V : 

Vector Subspaces

Let V(F) be a vector space and W ⊂ V. We have also seen that the set W is closed for the binary operation +, if for any

and in that case restriction of + to W is a binary operation in W.
Similarly we say that a subset W of vector space V(F) is closed for the external composition if for any

In such a case the restriction of the mapping of F * V to F * W is a mapping from F * W to W and is therefore an external composition for W. We shall therefore say that W is closed for the external composition in V, then external composition in V induces an external composition in W. If W itself be a vector space for these induced compositions, then we say that W is a subspace of V.

Definition: Let V(F) be a vector space. A non void subset W of V is a subspace of V, if W itself is a vector space over F for the restrictions to W of the addition and multiplication by scalar defined for V.

Improper or Trivial subspaces
Every vector space V(F) has atleast two subspaces viz,
(i) V itself
(ii) {0} zero vector space
These space are called Improper subspaces.

Proper subspaces:The subspaces other than the improper subspaces are called proper subspaces.
Example: The set of real numbers is a subspace of the vector space
Example: The set is a subspace of the vector space .

Criteria for a Subspace 

The necessary and sufficient conditions for a non void subset W of a vector space V(F) to be a subspace of V(F) is that W is closed under vector addition and scalar multiplication:
i.e., α g W, β g W ⇒ α + β ∈ W ...(1)
a ∈ F, β ∈ W ⇒ aα ∈ W ...(2)

Proof. 
The condition is necessary (⇒): Let W be a non empty subspace of V(F). By definition of subspace W itself is also a vector space over F wrt vector addition and scalar multiplication in V. So it implies that W is closed wrt these two binary compositions.
The condition is sufficient (⇐): Let W be closed for vector addition and scalar multiplication i.e. (1) and (2) hold. Now we have to prove that W is a subspace of V(F).
∵ W ≠ φ, therefore let α ∈ W and 1 ∈ F be unity element of the field F.
∵ 1 ∈ F ⇒ - 1 ∈ F
∴ -1 ∈ F, α ∈ W

Therefore every element of W has its additive inverse in W.
Now α ∈ W, -α ∈ W

Therefore zero vector of V is also the zero vector of W.
Since the elements of W are also the elements of v, therefore they also obey the commutative as well as associative laws as are true in V.
Therefore (W, +) is an abelian group.
Moreover V being the vector space satisfies remaining axioms V₂, V₃, V₄, V₅ so is in W and W is a subset of V.
Therefore W is a subspace of the vector space V(F).

Intersection of Two Subspaces

Definition: The intersection of two subspaces U and V of a vector space W is the set of all vectors that are in both U and V. In other words, it consists of every vector that belongs to U and also belongs to V.

Theorem: The intersection of two subspaces W₁ and W₂ of vector space V(F) is also a subspace of V(F).
Proof:

 

Since W₁, and W₂ are subspaces, therefore for any a, b ∈ F

Therefore a, b ∈ F and α, β ∈ W₁ ∩ W₂ ⇒ aα + bβ ∈ W₁ ∩ W₂
⇒ W₁ ∩ W₂ is also a subspace of V(F).

Generalization:The intersection of an arbitrary family of subspaces of a vector space is also a subspace.

Proof: Let V(F) be a vector space and { W₁, W₂, ...} be a family of its subspace, then to prove that ∩ W_i is also a subspace of V(F).

Now since every W is a subspace of V(F), therefore for any a, b ∈ F

⇒ ∩ W_i is also a subspace of V(F).

Union of Two Subspaces

The union of two subspaces of a vector space V(F) is not necessarily a subspace of V(F). This result can be proved with the help of the following counter-example:
if
are subspaces of the vector space
but their union
W₁ ∪ W₂ = {α I α = (a, 0, 0) or (0, b, 0)}
is not subspace of  because if

Theorem: The union of two subspaces W₁ and W₂ of a vector space V(F) is a subspace iff either W₁ ⊂ W₂ or W₂ ⊂ W₁

Proof:Let W₁ and W2 be the subspaces of a vector space V(F) such that

W₁ ⊂ W₂ or W₂ ⊂ W₁, then
W₁ ⊂ W₂ or W₂ ⊂ W_{1 ⇒}  W₁ ∪ W₂ = W₁ or W₁
But W₁ and W₂ are subspaces, therefore W₁ ∪ W₂ will also be a subspace.

Conversely:
Let W₁ ∪ W₂ be the subspace of the vector space V(F).

Also suppose that then there exist α₁,α₂ ∈ V such that


Hence W₁ ⊆ W₂ or W₂ ⊂ W₁

Important Theorems

Theorem 1: Let V be a vector space over a field F. If 0 be the zero vector in V and 0 be the additive identity in F, then
(i) a 0 = 0, ∀ a ∈ F 

(ii) 0α = 0, ∀ α ∈ V
(iii) a(-α) = -(aα), ∀ a ∈ F, a ∈ V
(iv) (-a)α = -(aα), ∀ a ∈ F, a ∈ V
(v) a(α - β) = aα - αβ, ∀ a ∈ F, α, β ∈ V
(iv) For any a ∈ F, α ∈V
aα = 0 a = 0 or α = 0 

Proof,

(i) ∵ 0 zero element in V, therefore 0 + 0 = 0

⇒ a(0 + 0) = a0      V a ∈ F
⇒ a0 + a0 = a0 ...... [by V₂]
⇒ a0 + a0 = a0 + 0   ...... [∵ 0 is zero element in V]
⇒ a0 = 0
(ii) ∵ 0 ∈ F is additive identity in F, therefore
0 + 0 = 0

⇒ (0 + 0)α = 0α     ∀ a ∈ V
⇒ 0α+ 0α = 9α ...... [by V₃]
⇒ 0α + 0α = 0α + 0     ...... [∵ 0 is zero element in V]
⇒ 0α = 0   [cancellation law in V]
(iii) For any a ∈ F and a ∈ V
a[α + (-α) = aa + a(-α) ...... [by V₂]
⇒ a0 = aα + a(-α)
⇒ 0 = aα + a(-α)      ...... [by (1)]
⇒ a(-α) is additive inverse of aa
⇒ a (-α) = -(aα) 

(iv) For any a ∈ F and α ∈ V
[a + (-a)]α = aa + (-a)α ...... [by V₃]
⇒ 0a = aα + (-a)α
⇒ 0 = aα + (-a)α    ....... [by (ii)]
⇒ (-a)α is additive inverse of aa
⇒ (-a)α = -(aα)
(v) For any a ∈ F and α, β ∈ V
a(α - β) = a[α + (-p)]

= aα + a(-β) .......[by V2]

= aα + [-(aβ)] ...... [by (iii)]

= aα - aβ

(vi) Let a ∈ F and α ∈ V such that a ≠ 0 and aα = 0

Now since multiplicative inverse of every non zero element exist in F,
therefore a ∈ F, a ≠ 0 ⇒ a^-1 ∈ F

aα = 0 => a^-1 (aα) = a^-1 0

⇒ (a^-1a)α = 0 . .......[by V₄ and (i)]

⇒ 1α = 0

⇒ α = 0     ......[byV₅]

Again let aα = 0 and α ≠ 0, then to prove that a = 0.

If possible, let a ≠ 0, then a^-1 ∈ F
aα = 0 ⇒ a^-1(aα) = 0
⇒ (a^-1a)α = 0
⇒ 1α = 0
⇒ α = 0
which is contrary to the assumption α ≠ 0, therefore a = 0.

Hence aα = 0⇒ a = 0 or α = 0

Theorem 2: The necessary and sufficient conditions for a non void subset W of a vector space V(F) to be a subspace of V(F) are:

Proof.
The condition is necessary (⇒) :
Let W be a subspace of V(F). Then W itself is also a vector space wrt the operations defined for V. Therefore (W, +) is also an abelian group.
Consequently,

Therefore the condition (i) is necessary.
Again W is closed for the scalar multiplication, being subspace of V. Therefore

Therefore the condition (ii) is also necessary.
The condition is sufficient (⇐) : 
Now let W be a non empty subset of V(F) which satisfies the given conditions, then to prove that W is a subspace of V(F).
∵ W ≠ φ, therefore let α ∈ W, then by condition (i)

Therefore zero vector of V is also the zero vector of W.
Again by condition (i),

Therefore additive inverse of each element exist in W.
Now α ∈ W, β ∈ W ⇒ α ∈ W, -β ∈ W
⇒ α - (-β) ∈ W      [by condition (i)]
⇒ (α + β) ∈ W

Therefore W is closed for vector addition.

Since W ⊂ V and vector addition is associative and commutative in V, so also in W
Hence (W, +) is an abelian group.
Again by condition (ii), W is closed for the scalar multiplication and W ⊂ V , therefore W will also satisfy remaining axioms V₂, V₃, V₄, V₅. Consequently, W is a subspace of V(F).

Theorem 3: The necessary and sufficient condition for a non void subset W of a vector space V(F) to be a subspace of V(F) is:
a, b ∈ F and α, β ∈ W ⇒ (aα + bβ) ∈ W

Proof: The condition is necessary (⇒ ):
Let W be a subspace of V(F). Then W will be closed for the scalar multiplication. Therefore

Again, We being subspace, will be closed for the vector addition, therefore

Therefore the given condition is necessary.

The condition is sufficient (⇐): Now let W be a non empty subset of V(F) which satisfy the given conditions. Taking a = 1, b = 1, then by the given condition
[by V₅]
Therefore W is closed for the vector addition.
Now taking a = -1, b = 0, then again by the given condition

Therefore additive inverse of each element exist in W.
Finally, taking a = 0, b = 0, then by the given condition

Therefore zero vector of V also exists in W.
Since W ⊂ V and vector addition is associative and commutative in V, so is in W also.
Hence (W, +) is an abelian group.
Again taking b = 0, we see that if a ∈ F and α, β ∈ W, then by the given condition
a, 0 ∈ F and α, β ∈ W

Therefore W is closed for the scalar multiplication. But W ⊂ V, therefore W will satisfy the remaining axioms V₂, V₃, V₄, V₅.
Consequently W is a subspace of V(F).

Theorem 4: Let V be a vector space over a field K.

(i)Foranyscalark ∈ K and 0∈V, k0 = 0.

(ii)For0 ∈ K andany vector u∈V, 0u = 0.

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

(iv)Foranyk ∈ Kand any u ∈V,(- k)u = k(-u)=-ku.

Examples of Vector Spaces

Space Kⁿ

Let K be an arbitrary field. The notation Kⁿ is frequently used to denote the set of all n-tuples of elements in K. Here Kⁿ is a vector space over K using the following operations :
(i) Vector Addition:
(ii) Scalar Multiplication:
k(a₁, a₂, ..... a_n) = (ka₁, ka₂,........., ka_n)
The zero vector in Kⁿ is the n-tuple of zeros,
0 = (0, 0, .... , 0)

and the negative of a vector is defined by

Theorem 5: The necessary and sufficient conditions for a non void subset W of a vector space V(F) to be a subspace of V(F) are:

Proof.
The condition is necessary (⇒) :
Let W be a subspace of V(F). Then W itself is also a vector space wrt the operations defined for V. Therefore (W, +) is also an abelian group.
Consequently,

Therefore the condition (i) is necessary.
Again W is closed for the scalar multiplication, being subspace of V. Therefore

Therefore the condition (ii) is also necessary.
The condition is sufficient (⇐) : 
Now let W be a non empty subset of V(F) which satisfies the given conditions, then to prove that W is a subspace of V(F).
∵ W ≠ φ, therefore let α ∈ W, then by condition (i)

Therefore zero vector of V is also the zero vector of W.
Again by condition (i),

Therefore additive inverse of each element exist in W.
Now α ∈ W, β ∈ W ⇒ α ∈ W, -β ∈ W
⇒ α - (-β) ∈ W      [by condition (i)]
⇒ (α + β) ∈ W

Therefore W is closed for vector addition.

Since W ⊂ V and vector addition is associative and commutative in V, so also in W
Hence (W, +) is an abelian group.
Again by condition (ii), W is closed for the scalar multiplication and W ⊂ V , therefore W will also satisfy remaining axioms V₂, V₃, V₄, V₅. Consequently, W is a subspace of V(F).

Theorem 6: The necessary and sufficient condition for a non void subset W of a vector space V(F) to be a subspace of V(F) is:
a, b ∈ F and α, β ∈ W ⇒ (aα + bβ) ∈ W

Proof: The condition is necessary (⇒ ):
Let W be a subspace of V(F). Then W will be closed for the scalar multiplication. Therefore

Again, We being subspace, will be closed for the vector addition, therefore

Therefore the given condition is necessary.

The condition is sufficient (⇐): Now let W be a non empty subset of V(F) which satisfy the given conditions. Taking a = 1, b = 1, then by the given condition
[by V₅]
Therefore W is closed for the vector addition.
Now taking a = -1, b = 0, then again by the given condition

Therefore additive inverse of each element exist in W.
Finally, taking a = 0, b = 0, then by the given condition

Therefore zero vector of V also exists in W.
Since W ⊂ V and vector addition is associative and commutative in V, so is in W also.
Hence (W, +) is an abelian group.
Again taking b = 0, we see that if a ∈ F and α, β ∈ W, then by the given condition
a, 0 ∈ F and α, β ∈ W

Therefore W is closed for the scalar multiplication. But W ⊂ V, therefore W will satisfy the remaining axioms V₂, V₃, V₄, V₅.
Consequently W is a subspace of V(F).

Solved Examples

Example 1:Every field is a vector space over its subfield.

Sol: Let F be a field and H be its subfield. Then F is an abelian group for the usual addition.

Also a ∈ H, x ∈ F ⇒ a ∈ F, x ∈ F ⇒ ax ∈ F
Hence usual multiplication in F can be taken as external composition.
Now we verify the various postulates of a vector space in F(H):
V_1. (F, +) is an abelian group .

V₂. For every a ∈ Hand α,β ∈ F

a(α + p) = aα + ap    [By distributivity in F]

V₃.For every a, b ∈ Hand α,∈ F

(a + b) a = aα + bα   [By distributivity in F]

V₄.For every a, b∈ H and a, ∈ F

(ab)α = α(bα)   [By multiplicative associativity]

V₅. If 1 is the unity element of R then

1α = α,∀α∈ F [∵ 1 is multiplicative identity in F]
Therefore F(H) is a vector space.

Special Case: Since every field is also a subfield of itself, therefore F(F) is a vector space. Consequently, ℂ(ℂ), ℝ (ℝ ), ꙩ(ꙩ) are also vector space for the ordinary and multiplication.
Remark: C(R) and R(ꙩ) are vector space, but R(C) and ꙩ(R) are not vector spaces, because they are not closed for scalar multiplication.

Example 2: The set Fⁿ of all ordered n-tuples of a field F is a vector space over the field F.

Sol: Let Fn = {(a₁, a₂,......,a_n) I a₁, a₂,.......,a_n ∈ F}
and α = (a₁,a₂,.....,  a_n), β = (b₁, b₂,.....,b_n) and a∈F.
We define the addition of n-tuples as

α + β = (a₁ + b₁, a₂ + b₂,....a_n + b_n) ...(1)
and scalar multiplication of n-tuples as
aα = a(a₁, a₂,..., a_n) = (aa₁, aa₂,......,aa_n) ...(2)
Clearly, α, β ∈ F and a∈ Fⁿ.
α + β ∈ Fⁿ ∀ α, β ∈ Fⁿ and a ∈ F, α ∈ Fⁿ => aα ∈ Fⁿ

Therefore Fⁿ(F) is closed for the above defined addition and scalar multiplication operations.

Verification of Space axioms: 
V₁. It can also be easily verified that Fⁿ is an abelian group for the addition defined above and 0 = (0, 0 .....0) ∈ Fⁿ being the additive identity and -- α = ( -a₁, -a_2,. . . , - a_n) ∈ Fⁿ being the additive identity in verse of α = (a₁, a₂,...., a_n) ∈ Fⁿ.
V₂. Scalar Multiplication is distributive over vector addition:
Let a ∈ F and α.β ∈ Fⁿ; then

a = (α + β) = a {(a₁, a₂,...., a_n) + (b₁, b₂,...., b_n)}
= a (a₁ + b₁, a₂ + b₂,...a_n + b_n) [by (1) ]
= {a(a₁ + b₁), a(a₂ + b₂),.....a(a_n + b_n)} [by (2) ]
= (aa₁ + ab₁, aa₂ + ab₂,..., aa_n + ab_n) [by distributivity in F]
= (aa₁, aa₂,..., aa_n) + (ab₁,ab₂,.....,ab_n)
= a(a₁, a₂,...., a_n) + a(b₁, b₂,...., b_n)
= aα + aβ
Therefore scalar multiplication is distributive over vector addition in Fⁿ.
V₃. Scalar addition is distributive over scalar Multiplication :
Let a, b ∈ F and α ∈ Fⁿ; then
(a + b) α = (a + b)(a,, a2 an)
= {(a + b) a₁, (a + b) a₂,.....,(a + b) a_n [by (2)]
= (aa₁ + ba₁, aa₂ + ba₂,....aa_n + ba_n) [by distributivity in F]
= (aa₁, aa₂,..., aa_n) + (ba₁, ba_2,......,ba_n)
= a (a₁, a₂,...., a_n) + b(a₁, a₂,...., a_n) [by (2)]
= aα + bα
Therefore addition is distributive over scalar multiplication in Fⁿ
V₄. Associativity for Scalar Multiplication:

Let a,b ∈ F, then α ∈ Fⁿ ;
(ab) α = (ab) (a₁, a₂,...., a_n)

= (aba₁, aba₂,..., aba_n) [by (2)]

= a (ba₁, ba₂,......, ba_n) [by (2)]

= a {b(a₁, a₂,...., a_n)} [by (2)]

= a (bα)
Therefore scalar multiplication is associative in Fⁿ
V₅. Let I be the unity element of the field F, then for α ∈ Fⁿ

1α = 1 (a₁, a₂,...., a_n) = (1a₁, 1a₂,...., 2a_n) [by (2)]

= (a₁, a₂,...., a_n)

= α
From the above discussion, it is clear that Fn satisfies all the axioms for the vector space, therefore Fⁿ(F) is a vector space.

Remarks: 

1. The above vector space is also expressed by V_n(F). 
2. If F = ℝ (Field of real number), then the corresponding vector space ℝⁿ is called the Real Euclidean n space.
   Similarly, ℂⁿ is called the Complex Euclidean n space. 
3. The set of all ordered pairs on F i.e. F²(F) = {( a₁, a₂) I a₁, a₂ ∈ f } is the vector space. Similarly the set of all ordered triads on F i.e. F₃(F) = {(a₁, a₂, a₃) I a₁, a₂, a₃ ∈ F } is the vector space
   These vector spaces are denoted by V₂(F) and V₃(F) respectively.

Example 3: The set M of all m * n real matrices (having their elements as real numbers) is a vector space over the field ͹ of real numbers with respect to the addition of matrices and scalar multiplication of matrices.

Sol: Let M = {[a_ij]_m*n |a_ij ∈ R} . If A, B ∈ M
where A = [a_ij]_m*n ;B = [b_ij]_m*n and a∈ℝ, then matrix addition and scalar multiplication in M are defined as follows :

A + B= [a_ij] + [b_ij] = [a_ij+b_ij|]_m*n ...(1)

and aA= a[a_ij] = [aa_ij]_m*n ...(2)

Verification of space axioms:
V₁. (M, +) is additive abelian group: 
In Group theory, we have already seen that (M,+) is a commutative group where null matrix O mKn is the zero element and - A = [—a_ij] _m*n ∈ M is the addictive inverse of any matrix A = [a_ij] ∈ M From the theory of matrices, following can be easily verified:
V₂. Scalar Multiplication is distributive over vector addition

V₃. Scalar addition is distributive over Scalar Multiplication.

V₄. Scalar Multiplication is associative.

V₅. 1 is the unity element in ℝ, therefore  every A ∈ M
1A = 1[a_ij] = [1a_ij] = [a_ij] = A
From the above discussion, it is clear that M(ℝ) is a vector space.

Example 4: The set V of all real valued continuous (differential or integral) functions defined in a closed interval [0,1] is a vector space over the field ℝ of real numbers with addition and scalar multiplication of functions defined by:
(f + g ) ( x ) = f (x) + g (x) ∀ f , g ∈ V ; x ∈ [0,1]    ..... (1)
(af)(x) = af (x) ∀ f ∈ V ,a ∈ ℝ ; x ∈ [0,1]   .... (2)

Sol: Let V = (f I f :[0,1] → R is a continuous function}
From Calculus, we know that the sum of two continuous function is also a continuous function i.e. f ∈ V, g∈ V ⇒ (f + g) ∈ V
⇒ V is closed for addition of function defined above. Also we know that a ∈ ℝ,f ∈ V af ∈ V ⇒ V is closed for scalar multiplication of function.
Verification of space axioms V (R): 
V₁. In group theory, we have already seen that (V, +) is an additive abelian group.

V₂.Scalar multiplication is distributive over vector addition:
Let f, g ∈ V and a ∈ ℝ. then
[a(f + g)](x) = a[(f + g)(x)]     [by (2)]
= [a(f(x) + g(x))]    [by (1)]
= af(x) + ag(x)    [by distributivity in R]
= (af)(x) + (ag)(x)    [by (2)]
= (af + ag)(x) ∀x ∈ [0,1]    [by (1)]
⇒ a(f + g) = af + ag ,∀f ∈V and ∀a, b ∈ ℝ

V₃. Scalar addition is distributive over Scalar Multiplication.
Let f ∈ V and a ,b ∈ ℝ, then
[(a + b)]f(x) = [(a + b)f(x)]      [by (2)]
= [a f(x) + b f(x)]    [by (1)]

V₄. Scalar multiplication is associative :
Let a ,b ∈ ℝand f ∈ V, then
[(ab)f](x) = (ab)f(x)      [by (2)]
= a[(bf)(x)]       [by (2)]
= [a(bf)](x) ∀ x ∈ [0,1]     [by (2)]

Therefore scalar multiplication is associative in V.
V₅. If unity 1 is real number and f ∈ V, then
(1f) (x) = 1 f(x)     [by (2)]
= f(x) [v 1 is multiplicative identity in ℝ]
1 f = f        [∀ f ∈ V]
From the above discussion, it is clear that V(ℝ) satisfies all the axioms for the vector space, therefore V(R) is a vector space

Example 5: Let V(F) be a vector space, then :

(i) For any a, b ∈ F and a ∈ V, a ≠ 0,

aα = bα ⇒ a = b

(ii) For any α, β ∈ V and a ∈ F, a ≠ 0

aα = aβ ⇒ α = β

Sol:
(i) aα = bα ⇒ aα - b α = 0
(a - b)α = 0 .... [by V₃]

⇒ [a + (-b)]α = 0

⇒ a + (-b) = 0 [∵  α ≠ 0]

[by theorem (vi)]

⇒ a = -(-b) [by additive inverse]

⇒ a = b
(ii) aα = aβ => aα - aβ = 0

⇒ a(α - β) = 0 [by V₂]

⇒ α - β = 0       [∵  a ≠ 0, therefore by theorem (vi)]

⇒ α = β

Example 8: If V(F) is a vector space, then prove that:

(i) By theorem (iv), we have


(ii)

(iii)

Example 6: Let V be the set of all ordered pairs of real numbers and F be the field of real numbers. Examine whether V(F) is a vector space or not for the following defined operations :
(a) (a, b) + (C, d) = (0, b + d);     P(a, b) = (pa, pb)
(b) (a, b) + (C, d) == (a + c, b + d);    P(a, b) = (0, pb)
(c) (a, b) + (C, d) == (a + c, b + d); P(a, b) = (p² a, p² b)
(d) (a, b) + (C, d) == (a, b); P(a, b) = (pa, pb)
(e) (a, b) + (C, d) == (a + c, b + d); P(a, b) = (I p I a, I p I b)    ∀ a, b, c, d, p ∈ ℝ

Sol: 
(a) It can be easily seen that the additive identity for the defined vector addition does not exist. If suppose any element (c, d) ∈ V is taken as additive identity for vector addition, then
(a, b) + (c, d) = (0, b + d) * (a, b)
Therefore (V, +) is not an abelian group.

Consequently V(F) is not a vector space.
(b) In this case, (V, +) is an abelian group but space axiom V₅ is not satisfied, because
1(a, b) = (0, 1b) = (0, b) ≠ (a, b)
Therefore V(F) is not a vector space.
(c) In this case, (V +) is an abelian group but space axiom V3 is not satisfied, because

Therefore V(F) is not a vector space.
(d) In this case the defined vector addition is not commutative, because for
(a, b); (c, d) ∈ V
(a, b) + (c, d) = (a, b)
and (c, d) + (a, b) = (c, d)
(a, b) + (c, d) ≠ (c, d) + (a, b)
Therefore (V, +) is not an abelian group.

Consequently, V(F) is not a vector space.
(e) In this case, (V, +) is an abelian group but space axiom V₃ is not satisfied, because
(p + q)(a, b) = (I p + q I a, I p + q I b)        p, q ∈ ℝ
≠ (I p I a, I p I b) + (I q I a, I q I b)      [ ∵ l p + q l ≠ lp l + l q l , p, q ∈ ℝ]
⇒ (p + q)(a, b) ∈ p(a, b) + q(a, b)
Therefore V(F) is not a vector space.

Example 7 : Show that the set W = {(a, b, c) I a - 3b + 4c = 0; a, b, c ∈ R} of 3-tuples is a subspace of the vector space  V₃(ℝ)

Sol: V₃(ℝ) = {(a, b, c) I a, b, c ∈ ℝ}
and W = {(a, b, c) I a - 3b + 4c = 0; a, b, c ∈ ℝ} therefore, clearly W ⊂ V₃(ℝ)
Let


.....(1)
Now if, p, q ∈ ℝ, then

Therefore p, q ∈ ℝ and α, β ∈ W ⇒ pα + qβ ∈ W
∴ W is a subspace of the vector space V₃(ℝ).

Example 8 : Examine which of the following sets is a subspace of the vector space V₃(ℝ).
(a) W₁ = {(x, 2y, 3z) I x, y, z ∈ ℝ}
(b) W 2 = {(x, x, x) I x ∈ ℝ )
(c) W 3 = {(x, y, z) I x, y, z ∈ ℚ  }

Sol: 
V₃(ℝ) = {(x, 2y, 3z) I x, y, z ∈ ℝ}
Therefore observing 3-tuples of W₁, W₂, W₃ and V₃(ℝ), it is clear that

(a) Let α = (x₁, 2y₁, 3z₁) ∈ W₁ and β = (x₂, 2y₂, 3z₂) ∈ W₁
where x₁, x₂, y₁, y₂, z₁, z₂ are all real numbers.
if a, b ∈ R, then




∴ a, b ∈ R and α, β ∈ W₁ ⇒ aα + bβ ∈ W
Therefore W₁ is a subspace of the vector space V₃(ℝ).
(b) let
where x₁, x₂ ∈ ℝ. If a, b ∈ ℝ, then



Therefore W₂ is a subspace of the vector space V₃(ℝ).
(c) Let a = (3, 5, 4) ∈ W₃ then for a = √2 ∈ ℝ

Consequently, W₃ is not closed for the scalar multiplication.
Therefore W₃ is not a subspace of the vector space V₃(ℝ).

Example 9: If V(F) is a vector space of all n * n matrices over the field F, then prove that the set W of all n x n symmetric matrices over F will be a subspace of V(F).

Sol: 
Clearly W is a non empty subset of V(F), W = (A : A^T = A, A ∈ V}.

Let A, B ∈ W ⇒ A^T = A and B^T = B [being symmetric] ...(1)
If a, b ∈ F, Then aA ∈ V, bB ∈ V ⇒ aA + bB ∈ V
and
⇒ aA + bB is a symmetric matrix⇒ (aA + bB) ∈ W
a, b ∈ F and A, B ∈ W ⇒ aA + bB ∈ W
Therefore W is a subspace of V(F).

Example 10: If M₂(F) is a vector space of all 2 x 2 matrices over the field F, then prove that the set W of all matrices of the form over F will be a subspace of M₂(F).

Sol:Let

and p ∈ F, then



Therefore W is closed for the vector addition and scalar multiplication in M₂(F) Hence M₂(F) is a subspace.

Example 11: If a₁, a₂, a₃ are fixed elements of a field F; then prove that the set W = {(x₁, x₂, x₃) I x₁, x₂, x₂ ∈ F; a₁x₁  + a₂x₂ + a₃x₃ = 0} is a subspace of the vector space V₃(F).

Sol:
Let α = (x₁, x_2, x₃) ∈ W and b = (y₁, y₂, y₃) ∈ W
and a₁x₁ + a₂x₂ + a₃x₃ = 0 and a₁y₁ + a₂y₂ + a₃y₃ = 0 ...(1)
If a, b ∈ F, then
(by the properties of the field F)= a0 + b0
= 0 + 0 = 0    ...(3)
Therefore by (2) and (3),
aα + bβ = (ax₁ +  by₁,ax₂ + by₂, ax₃ + by₃) ∈ W
Therefore W is a subspace of V₃(F).

Example 12: Let V(F) be a vector space of all polynomials of degree n in an indeterminate x, then prove that the set S of all polynomials of degree ≤ n over F is a subspace of V(F) where n is an arbitrary positive integer.

Sol:
Let f(x), g(x) ∈ S, so that f(x) and g(x) are polynomials over F of degree ≤ n. Now if a and b are any scalars ∈ F, then af(x) + bg(x) will also be a polynomials of degree ≤ n.

Thus f(x), g(x) ∈ S and a, b ∈ F ⇒ af(x) + bg(x)
Hence S is a subspace of V.

Example 13: Let S be the set of all solutions (x, y, z) satisfying the simultaneous equations ax + by + cz = 0 and dx + ey + fz = 0 ; a, b . c, d, e, f ∈ ℝ. Thus S is a subspace of ℝ³ over ℝ.

Sol:
If (x₁, y₁, z₁) and (x₂, y₂, z₂) be solutions of the given equations,



Hence if (x₁, y₁, z₂) ∈ V, (x₂, y₂, z₂) ∈ V, then (x₁ + x₂, y₁ + y₂, z₁ + z₂) is also in V.
Similarly a(ax₁) + b(αy₁) + c(αz₁) = (aαx₁ + bαy₁ + cαz₁)

Hence S is a subspace of ℝ³ over ℝ.

Example 14: Examine which of the following sets of vectors α = (a₁, a₂,....., a_n) in ℝⁿ are subspaces of ℝⁿ ? (n ≥ 3)

Sol:
(a) Let
Again let and b = (b₁, b₂,...... b_n) ∈ W₁,
then by the given condition a₁ ≥ 0, b₁ ≥ 0.

Now for a, b ∈ ℝ

But aa₁ + bb₁ is not necessarily non-negative,
for example If we take a = -2, b = -1, then for a, = 4, b = 3
aa₁ + bb₁ = (-2 )4 + (-1 )3 = -11 < 0
Therefore aα + bβ ∉ W,. Hence W₁ is not a subspace.

(b) Let W₂ = {α I α ∈ ℝⁿ and a₁ + 3a₂ = a_3}
Again let α = (a₁, a₂, ..., a_n) ∈ W₂ and β= (b₁, b₂, ..., b_n) ∈ W₂
Then by the given condition
a₁ + 3a₂ = a₃ and b, + 3b₂ = b₃ ...(1)
Now for a, b ∈ ℝ

= aa3 + bb3 [by (1)]
a, b ∈ ℝ and α, β ∈ W₂ ⇒ aα + gβ ∈  W₂
Therefore W₂ is a subspace of .

(c) Let Again let α = (a₁, a₂, ..., a_n) ∈ W₃ and β = (b₁, b₂, ..., b_n) ∈ W₃
Then by the given condition,
Now for a, b ∈ ℝ
But the square of (aa₁ + bb₁) is not necessarily (aa₂ + bb₂).
for example When a₂ = 4, a₁ = 2, b₂ = 9, b₁ = 3, a = 2, b = -2,
then aa₂ + bb₂ = 8 - 18 = - 10 when (aa₁ + bb₁)² = (4 - 6)2 = 4.
therefore Hence W₃ is not a subspace of ℝⁿ.
(d) Let
Again let α = (a₁, a₂, ..., a_n) ∈ W₄ and β = (b₁, b₂, ..., b_n) ∈ W₄
Then by the given condition a₁a₂ = 0 and b₁b₂ = 0.

Now for a, b ∈ ℝ

But (aa₁ + bb₁) (aa₂ + bb₂) is not necessarily zero,
for example Taking a₁ = 0, a₂ = 2, b₁ = 1, b₂ = 0, a = 2, b = 3
(aa₁ + bb₁)(aa₂ + bb₂) = (0 + 3)(4 + 0) = 12 * 0
Therefore aα + bβ ∈ W₄. Hence W. is not a subspace of ℝⁿ.
(e)Let

Again let α = (a₁, a₂, ..., a_n) ∈ W₅ and β = (b₁, b₂, ..., b_n) ∈ W₄
Then by the given condition a₂ b₂ are rational numbers.

Now for a, b ∈ ℝ

But aa₂ + bb₂ is not necessarily a rational number.

for example 
If we take a = √2, b = √3, a₂ = 3, b₂ = 4
aa₂ + bb₂ = 3√2 + 4√3 which is not rational
Therefore aα + bβ ∉ W₅. Hence W₅ is not a subspace of ℝⁿ.