Vector Space & Linear Transformation- 2 | Mathematics for Competitive Exams PDF Download

Linear Combination

Let V be a vector space over a field K. A vector v in V is a linear combination of vectors u₁, u₂, .... , u_m in V if the exist scalars a₁, a₂, ...... a_m in K such that

Alternatively, v is a linear combination of u₁, u₂, .... , u_m  if there is a solution to the vector equation

where x₁, x₂, ..... x_m are unknown scalars.

Example: (Linear Combinations in ℝⁿ) Suppose we want to express v = (3, 7, - 4) in ℝ³ as a linear combination of the vectors
u₁ = (1, 2, 3), u₂ = (2, 3, 7) u₃ = (3, 5, 6)
We seek scalar x, y, z such that v = xu₁ + yu₂ + zu₃; that is,

(For notational convenience, we have written the vectors in ℝ³ as columns, because it is then easier to find the equivalent system of linear equations.) Reducing the system to echelon form yields

Back-substitution yields the solution
x = 2, y = - 4 , z = 3
Thus v = 2u₁ — 4u₂ + 3u₃.

Spanning Set

Let V be a vector space over K. Vectors u₁, u₂, .... , u_m in V are said to span V or to form a spanning set of V if every v in V is a linear combination of the vectors u₁, u₂, .... , u_m that is, if there exist scalars a₁, a₂, ...... a_m in K such that

The following remarks follow directly from the definition.

Remark

1. Suppose u₁, u₂, .... , u_m span V. Then, for any vector w, the w,u₁, u₂, .... , u_m also spans V.
2. Suppose u₁, u₂, .... , u_m span V and suppose u_k is a linear combination of some of the other u’s. Then u’s without u_k also span V.
3. Suppose u₁, u₂, .... , u_m span V and suppose one of the u’s is the zero vector. Then the u’s without the zero vector also span V.

Example: Consider the vector space V = ℝ³.
(a) We claim that the following vectors form a spanning set of ℝ³ :
e₁ = (1,0,0),   e₂ = (0,1,0), e₃ = (0, 0, 1)
Specifically, if v = (a, b, c) is any vector in ℝ³,
then v = ae₁ + be₂ + ce₃
For example, v = (5, - 6, 2) = - 5e₁ - 6e₂ + 2e₃.
Note : L(ϕ) = {0}
Example: If S = {(1, 0, 0)} ⊂ V₃(ℝ),
then L(S) = {(a, 0, 0) I a ∈ ℝ}
which is the set of all points of x-axis.
Example: If S = {(1, 0, 0), (0, 1, 0)} ⊂ V₃(ℝ), then
L(S) = {(a, b, 0) I a, b ∈ ℝ}
which is the set of all points of xy plane.
Example: If S = {(1, 0, 0), (0, 1, 0), (0, 0, 1)} ⊂ V₃(ℝ),
then for every vector α = (a, b, c) of V₃(ℝ)

(a, b, c) = a(1, 0, 0) + b(0, 1, 0) + c(0, 0, 1)
⇒  (a, b, c) ∈ L(S) ⇒ V₃(ℝ) ⊂ L(S)
but    L(S)⊂  V₃(ℝ)
L(S) = V₃(ℝ)

Theorem: The linear span L(S) of any non empty subset S of a vector space V(F) is the smallest subspace of V(F) containing S
i.e.,    L(S) = {S}

Proof. Let α, β ∈ L(S), then


where

and

If a, b are any arbitrary elements of the field F, then

which shows that aα + bβ is a LC of the vectors of finite number of subsets

Hence aα + bβ ∈ L(S)

∴ α, β ∈ F and α, β ∈ L(S) ⇒ aα + bβ ∈ L(S)
Therefore L(S) is a subspace of V(F).

Again for every vector α of S

Now let W be any subspace of V containing S(S ⊂ W), then element of L(S) must be in W, for W being subspace, is closed for vector addition and scalar multiplication. Therefore L(S) ⊂ W.
Consequently, L(S) = {S}
i.e. L(S) is the smallest subspace of V containing S.

Linear Sum of two subspaces
Let W₁ and W₂ be the two subspaces of V(F). Then the linear sum of the subspaces W₁ and W₂ is the set of all possible sums (a, + a2), where a, ∈ W. and a2 ∈ W₂ The linear sum is denoted by W₁ + W₂.

Symbolically, W₁ + W₂ = {α₁ + α₂ I α₁ ∈ W₁, α₂ ∈ W₂}

Since 0 ∈ W₁, and 0 ∈ W₂, therefore

Remark:

Theorem: The linear sum of two subspaces of a vector space is also a subspace.
Proof: Let W₁ and W₂ be the two subspaces of V(F).
Let  and
where
Now since W₁ and W₂ are two subspaces, therefore
 
Again



Therefore a, b ∈ F and
∴ (W₁ + W₂) is also a subspace of V(F).

Theorem: If W₁ and W₂ are subspace of a vector space V(F), then their sum is generated by their union i.e.

Proof: Since W₁ and W₂ are the subspaces of V(F), therefore
 
Let

⇒ W₁ ⊂  (W₁ + W₂)    ...(1)
Similarly W₂ ⊂ (W₁ + W₂)     ...(2)
(1) and (2) ⇒ W₁ ∪ W₂ ⊂ (W₁ + W₂)     ...(3)
But by theorem (W₁ + W₂) is a subspace of V(F) containing W₁ ∪ W₂.

To prove W₁ + W₂ = L (W₁ ∪ W₂) :
Let  where α₁ ∈ W₁ and α₂ ∈ W₂ 
then clearly
Now since
⇒ α₁ + α₂ can be expressed as a LC of the vectors α₁, α₂ ∈ W₁ ∪ W₂
   .... (4)
But we know that L(W₁ ∪ W₂) is the smallest subset of V(F) containing W₁ ∪ W₂ and by (3)
 
therefore
   .... (5)
(4) and (5)

Theorem: If S and T are subsets of a vector space V(F) then:
(a) S ⊂ T u L(S) ⊂ L(T)
(b) S ⊂ L(T) L(S) ⊂ L(T)
(c) L(S ∪ T) ⇒ L(S) + L(T)
(d) S is a subspace of V ⇔ L(S) = S
(e) L{L(S)} = L(S)

Proof: 
(a) Let   where a_i ∈ F and {α₁, α₂, .... α_n} is a finite subset of S.
Now since S ⊂ T , therefore
S ⊂ T ⇒ a_i ⊂ T    (i = 1, 2, n)

Therefore

(b) Let   where a_i ∈ F and {α₁, α₂, .... α_n} is a finite subset of S.
Now since S ⊂ L(T) ,
therefore


Therefore

(c) Let  be any element of the set L(S∪ T) where  are the elements of the field F and  is a limits subset of S ∪ T such that




Now let α ± β + γ ∈ L(S) + L(T)
⇒ β ∈ L(S) and γ ∈ L(T)
⇒ β is LC of elements of the finite subset of S and  γ is a LC of elements of the finite subset of T
⇒ β + γ is a LC of finite elements of S ∪ T
⇒ β + γ ∈ L(S ∪ T)
⇒ L(S) + L(T) ⊂ L(S ∪ T)    ...(2)
Therefore (1) and (2) L(S ∪ T ) = L(S) + L(T)

(d) First suppose S is a subspace of V, then to show that L(S) = S
Let  where a_i ∈ F and {α₁, α₂, ..., αn} is a finite subset of S.
Now since S is a finite subset of V(F), therefore this is closed for vector addition and scalar and scalar multiplication. Consequently

Therefore α ∈ L(S) ⇒ a ∈ S    ...(1)
L(S) ⊂ S     ...(2)
But in each case,  S ⊂ L(S)
Therefore (1) and (2)  ⇒ L(S) = S

Conversely: Let L(S) = S, then to show that S is a subspace of V(F). By theorem L(S) is a subspace of V(F), therefore S(= L(S)) will also be subspace of V(F).

(e) Since L(S) is a subspace of V(F), therefore by (d) above,

L{L (S)} = L(S).

Example 1: Is the vector α = (2, -5, 3) <= V₃(R), a LC of the following vectors?
α₁ = (1, -3, 2), α₂ = (2, -4, -1), α₃ = (1, -5, 7)

Let
Now (2, - 5 , 3) = a₁(1, - 3 , 2) + a₂(2, - 4 , - 1 ) + a₃(1, - 5 , 7)


(1) and (2)
 and
(1) and (3)

which is not possible. Therefore equations (1), (2), (3) are inconsistent. Consequently it is not possible to express a is LC of α₁, α₂, α₃.

Example 2: If possible, express the vector  as a LC of the following vectors:
α₁ = (2, 1, - 1), α₂ = (- 1, 0, 3), α₃ = (0, 1, 5)

2a₁ - a₂ = 0    ...(1)
a₁ + a₃ = 4   ...(2)
 -a₁ + 3a₂ + 5a₃ = 20    ...(3)
Solving (1), (2), (3), we obtain a₁ = 1, a₂ = 2, a₃ = 3 all in the field R.
Therefore vector a can be expressible as a LC of α₁, α₂, α₃ 

Example 3: In the vector space V₃(ℝ), let α₁ = (1, 2, 1); α₂ = (3, 1,5); α₃ = (3, -4, 7). Then prove that the subspaces spanned by S = {α₁, α₂} and T = {α₁, α₂, α₃} are the same.

Since the linear span L(T) of T is the set of LC of the vectors α₁, α₂, α₃, therefore let



Substituting these values of b₁ and b₂ in (2), we get

(3, -4, 7) = -3(1, 2, 1) + 2(3, 1, 5)

⇒ a₃(3, -4, 7) = -3a₃(1, 2, 1) + 2a₃(3, 1, 5)   .... (3)
Now by (1) and (3),

Example 4: Show that the following vectors span the vector space ℝ³:

Let

then
 are scalars such that

Subspaces

Definition: Let V be a vector space over a field K and let W be a subset of V. Then W is a subspace of V if W is itself a vector space over K with respect to the operations of vector addition and scalar multiplication on V.
Theorem: Suppose W is a subset of a vector space V. Then 1 is a subspace of V if the following two conditions hold :
(a) The zero vector 0 belongs to W.
(b) For every u, v ∈ W, k ∈ K: 

1. The sum u + v ∈ W. 
2. The multiple ku ∈ W.

Example: Consider the vector space V = ℝ³.
(a) Let U consist of all vectors in ℝ³ whose entries are equal; that is
U = {(a, b, c) : a = b = c}
For example, (1, 1, 1), (- 3, - 3, - 3), (7, 7, 7), (- 2, - 2, - 2) are vectors in U. Geometrically, U is the line through the origin O and the point (1, 1, 1). Clearly O = (0, 0, 0) belongs to U, because all entries in O are equal. Further, suppose u and v are arbitrary vectors in U, say u = (a, a, a) and v = (b, b, b). Then, for any scalar k ∈ R, the following are also vectors in U : u + v = (a + b, a + b, a + b) and ku = (ka, ka, ka)

Thus, U is a subspace of ℝ³.

Linear Dependence and Independence

Let V be a vector space over a field K. The following defines the notion of linear dependence and independence of vector over K.

Definition: We say that the vectors v₁, v₂,......v_m in V are linearly dependent if there exist scalars a₁, a₂,......a_m in K, not all of them 0, such tha
   
Otherwise, we say that the vectors are linearly independent.

The above definition may be restated as follows. Consider the vector equatio
 .. (*)
where the x’s are unknown scalars. This equation always has the zero solution x₁ = 0, x₂ = 0, ..., x_m = 0. Suppose this is the only solution; that is, suppose we can show:
 implies
Then the vectors v₁, v₂,......v_m are linearly independent, On the other hand, suppose the equation (*) has a nonzero solution; then the vectors are linearly dependent.
A set S = {v₁, v₂,......v_m} of vectors in V is linearly dependent or independent according to whether the vectors v₁, v₂,......v_m are linearly dependent or independent.
An infinite set S of vectors is linearly dependent or independent according to whether there do or do not exist vectors  v₁, v₂,......v_k in S that are linearly dependent.
The following remarks follow directly from the above definition.

Remark: Suppose 0 is one of the vectors v₁, v₂,......v_m, say v₁ = 0. Then the vectors must be linearly dependent, because we have the following linear combination where the coefficient of v₁ ≠ 0:

Remark: Suppose v is a nonzero vector. Then v, by itself, is linearly independent, because kv = 0, v ≠ 0 implies k = 0

Remark: Suppose two of the vectors v₁, v₂,......v_m are equal or one is a scalar multiple of the other, say v₁ = kv₂. Then the vectors must be linearly dependent, because we have the following linear combination where the coefficient of v₁ ≠ 0:

Remark: Two vectors v₁ and v₂ are linearly dependent if and only if one of them is a multiple of the other.

Remark: If the set {v₁,......v_m} is linearly independent, then any rearrangement of the vectors  is also linearly independent.

Remark: If a set S of vectors is linearly independent, then any subset of S is linearly independent. Alternatively, if S contains a linearly dependent subset, then S is linearly dependent.

Theorems on Linear Dependence and Independence
Theorem: A set of vectors which contains atleast one zero vector is LD.

Proof: Let S = {α₁, α₂,......α_n} be the set of n vectors of any vector space V(F) in which α₁ = 0 and all the remaining are non zero vectors.
Now choose scalars 1, 0, 0, 0, ... (not all zero) so that
1α₁ + 0α₂ + 0α₃ +...+ α_n = 0     ..... (1)
Therefore from (1), it is clear that

where a, = 1 ∈ F unit element and a₁ = 0 ∈ F is zero scalar.
Hence S is LD.

Theorem: Every non-empty subset of a LI set of vectors is also LI.

Proof: Let S = {α₁, α₂,......α_n} be LI set of n vectors of any vector space V(F). Then there exist a₁, a₂, .... a_n ∈ F such that

Let S’ = {α₁, α₂,......α_m}, m ≤ n be the non empty subset of the vector set S. Now since α₁, α₂, ... α_n are LI, therefore


Therefore the subset S’ is also LI.

Theorem: Any superset of a LD set of vectors is also LD.
Proof: Let S = {a₁, a₂, .... a_n} be LD set of n vectors of vector space V(F). Then there exist a₁, a₂, .... a_n ∈ F (not all zero) such that
    ....(1)
Let S’  be a super of set S,
then by (1),


[∵  scalars a₁, a₂, .... a_n are not all zero]
⇒ S’ is also LD set.

Theorem: If V(F) is a vector space and S = {α₁, α₂,......α_n} is a subset of some non-zero vectors of V, then S is LD iff some element of S can be expressed as a LC of the others.

Proof: Let S = {α₁, α₂,......α} be LD. Then there exist a₁, a₂, .... a_n ∈ F (not all zero) such that a₁α₁₊ a₂α_2 + .... + a_nα_n = 0    ...(1)
Let a_k ≠ 0, then (1) can be written as

Therefore α_k ∈ S is expressible as a LC of the remaining vectors.

Conversely: Let α₁ ∈ S be expressible as a LC of the remaining vectors of S. Therefore let

⇒
⇒ α₁, α₂,......α_n  is LD    ( ∵ - 1 ≠ 0)
⇒ S is LD

Theorem: If V(F) is a vector space and S = (α₁, α₂,......α_n) is a subset of some non-zero vectors of V, then S is LD iff some of the vectors of S, say α_k where 2 ≤ k ≤ n, can be expressed as a LC of preceding one's.

Proof: Let α_k ∈ S be a vector expressible as a LC of the preceding vectors of S.

Therefore let there exist scalars a₁, a₂, .... a_k-1 ∈ F such that


Conversely: Let S be LD.

The set {a₁} is clearly LI because a₁ ≠ 0
Now consider the set { α₁, α₂}
If this is LD then there exist a₁, a₂ ∈ F such that
 where both a₁, a₂ are not zero.  ... (1)
But a₂ ≠ 0 because

Therefore by (1),

⇒ a₂ can be expressed as a LC of its preceding vector α₁
Now if the set  is LI, then we consider the set {α₁, α₂,α₃}
If this is LD then on the above lines, it can be easily shown that α₃ can be expressed as a LC of its preceding vectors α₁ and α₂.
If {α₁, α₂,α₃} is LI, then proceed as above,
Let k ∈ N such that 2 ≤ k ≤ n and set {α₁, α₂,......α_k-1} is LI
Now consider the set {α₁, α₂,......α_k-1,α_k }
If this is LD, then
    ..... (2)
where b_i ∈ F(i = 1, 2, ..., k) are scalars such that atleast one of them is non zero. But a_K ≠ 0, because if a_k = 0 then set {α₁, α₂,......α_k-1} will be LD which is contrary to our earlier assumption.
Therefore by (2)


⇒ α_k can be expressed as a LC of its preceding vectors.

Example 1: Show that the following vectors of V₃(ℝ) are LI :

Let a₁, a₂, a₃ ∈ ℝ be the real numbers such that

From (1) and (2), a₃ = 0, then by (3), a₂ = 0
Now using these in (2), we obtain a₂ = 0
Therefore

Therefore the given vectors α₁, α_2, α₃ are LI

Example 2: The following vectors of V₃(ℝ) are LD : 

a₁ = (1, 3, 2); a₂ = (1, -7, -8); a₃ = (2, 1, -1) 

Let a₁, a₂, a₃ ∈ ℝ be real numbers such that




Therefore a₁= 3, a₂= -2, a₃ = 1 is a non zero solution of the above equations. Consequently,

Example 3: Show that the following matrices in the vector space M(R) of all 2 * 2 matrices are LI:

Let  such that

Consequently the given matrices are LI.

Example 4: Show that the vectors  

⇒    aa₁ + bb₁ = 0    .... (1)

aa₂ + bb₂ = 0    ... (2)
The necessary and sufficient conditions for the existence of the non zero solution of (1) and (2) are:

⇒ The given vectors will be LD iff a₁b₂ - a₂b₁ = 0

Example 5: Show that in the vector space V(F) of all polynomials over a field F, the infinite set S = {1, x, x², x³, ...} is LI.

Let  be the finite subset of the given set S where m₁, m₂, .... m_n are non zero integers.
Let a₁, a₂, ..., a_n ∈ F such that
 

⇒ S_n is LI.
⇒ Every finite subset of S is LI
⇒ S is also LI.

Example 6: If α₁, α_2, α₃ are LI vectors in a vector space V(ℂ), then show that the following vectors are also LI:

(a) Let a₁, a₂, a₃ ∈ ℂ such that

  .... (1)
But α₁, α_2, α₃ are LI, therefore by (1)

Consequently the vectors α₁ + α₂, α₂ + α₃, α₃ + α₁ are also LI.
(b) Again let b₁, b₂, b₃ ∈ ℂ such that


But α₁, α_2, α₃ are LI, therefore by (2)

Consequently the vectors
 are also LI.

Example 7: If V be the vector space of all functions defined from ℝ to ℝ; then show that f, g, h ∈ V are LI where:
(a) f(x) = e^2x, g(x) = x², h(x) = x
(b) f(x) = sin x, g(x) = cos x, h(x) = x

(a) Let there exist a₁, a₂, a₃ ∈ ℝ exist such that
a₁f + a₂g + a₃h = 0, where 0 is zero function
 
but

From (1), (2), (3), a₁ = 0, a₂ = 0, a₃ = 0
⇒ f, g, h are LI.

(a) Let there exist b₁, b₂, b₃ ∈ ℝ exist such that
b₁f + b₂g + b₃h = 0, where 0 is zero function


From (4), (5), (6), b₁ = 0, b₂ = 0, b₃ = 0
⇒ f, g, h are LI.

Example 8: α₁, α_2, α₃ are vectors of V(F ) and a₁, a₂ ∈ F, then show that the set {α₁, α_2, α₃} is L D if the set {α₁ + a₁α₂ + a₂α₃, α₂, α₃} 

Let {α₁ + a₁α₂ + a₂α₃, α₂, α₃} be a LD set, then there exist a, b, c ∈ F (not all zero) such that

Now if in the above relation (1) if all the coefficients of the vectors α₁, α_2, α₃ are not zero, then the set {α₁, α_2, α₃} will also be LD.
If a ≠ 0, then for any value of b and c, the set {α₁, α_2, α_3} will be LD.
If a = 0, then at least one of b and c will not be zero,
(because if all the three are zero, then the other set will not be LD).

Consequently, at least one of the coefficient (aa₁ + b) and (aa₂ + c) will not be zero.
Therefore the set {α₁, α_2, α₃} will be LD.

From the above discussion, it is clear that if the set {α₁ + a₁α₂ + a₂α₃, α₂, α₃} is LD, then the set {α₁, α_2, α₃} will always be LD.

Example : (a) Let u = (1, 1, 0), v = (1, 3, 2), w = (4, 9, 5). Then u, v, w are linearly dependent, because 3u + 5v - 2w = (1, 1, 0) + 5(1, 3, 2) - 2(4, 9, 5) = (0, 0, 0) = 0
(b) We show that the vectors u = (1, 2, 3), v = (2, 5, 7), w = (1, 3, 5) are linearly independent. We form the vector equation xu + yv + zw = 0, where x, y, z are unknown scalars. This yields

Back-substitution yields x = 0, y = 0, z = 0. We have shown that xu + yv + zw = 0 implies x = 0, y = 0, z = 0
Accordingly, u, v, w are linearly independent.
Lemma: Suppose two or more nonzero vectors v₁, v₂, ......, v_m are linearly dependent. Then one of the vectors is a linear combination of the preceding vectors; that is, there exists k > 1 such that
v_k = c₁v₁ + c₂v₂ + ... + c_k-1v_k-1,
Theorem: The nonzero rows of a matrix in echelon form are linearly independent.
Proof: Every non-zero row of a matrix in reduced row-echelon form contains a leading 1, and the other entries in that column are zeroes. Then any linear combination of those other non-zero rows must contain a zero in that position, so the original non-zero row cannot be a linear combination of those other rows. This is true no matter which non-zero row you start with, so the non-zero rows of the matrix must be linearly independent.

Basis and Dimension

Basis of a Vector Space: 

Remark: 

• Zero vector can not be an element of a basis because then the vectors of the basis will not be LI.
• There may exist a number of bases for a given vector space.

Examples of Bases
Example 1: (Finite Basis) Let V_n(F) be a vector space, then S = {e₁, e₂,....., e_n} is a basis of V_n(F), where e₁ = (1, 0, 0, .... 0); e₂ = (0, 1, 0    0);....., e_n = (0, 0, 0,    1).

Earlier we have already proved that S is LI.
Again for every vector a = (a₁, a₂,....., a_n) of V_n,
there exist a₁, a₂,....., a_n ∈ F
such that a = a₁e₁ + a₂e₂ +...+ a_ne_n 
⇒ each vector of V_n is a LC of elements of S.

Therefore S is the basis of the vectors space V_n(F).

Remarks: 

1. The above basis of V_n(F) is called the Standard basis. 
2. The standard basis of V₂(F) = {(1, 0), (0, 1)} 
3. The standard basis of V₃(F) = {(1, 0, 0); (0, 1, 0); (0, 0, 1)} etc.

Example 2: (Finite basis) Every vector space F(F) has a basis {1} where 1 ∈ F is unit element. 

Clearly, the set {1} has a non zero element only.
Therefore {1} is LI.
Again for every a ∈ F, a = a1 i.e.
every element of F is LC of {1}.
Consequently {1} is a basis of F(F).

Remark: Corresponding to every non zero element a ∈ F, {a} is a basis of F(F)

Example 3: (Infinite basis) : Let F[x] be the vector space of all polynomials over the field F. Then the set S = {1, x, x₂, ..., x_n, ...} is a basis of F[x].

Let  be the finite subset of the set S and a₁, a₂,....., a_k ∈ F Such that

⇒ S’ is LI.
⇒ every finite subset of S is LI
⇒ S is also LI.
Now we will show that S generate F(x) i.e. every element (Polynomial) of F(x) can be expressed as a LC of finite number of elements of F.
Let
Then there exist a₀, a₁, a₂,....., a_k ∈ F such that

Therefore S is a basis of F[x].

Remark: F[x] has no finite basis.

Dimension of a Vector Space

Definition: The number of elements in the finite basis of a vector space V(F) is called the dimension of the vector space V(F).
The dimension of the vector space V(F) is denoted by dim V or Dim V.
If dim V = n, then V is called n-dimensional vector space.

Remark: The dimension of the zero space {0} is taken as zero (0).
Example: dim F(F) = 1, because every basis of F is a singleton set containing a nonzero element of F.
Example: dim V_n(F) = n, because {e₁, e₂,....., e_n} is a basis of V_n(F) containing n elements.
Similarly, R₂(R) = 2, dim R₃(R) = 3 etc.

➤ Finite Dimensional Vector Space (FDVS):
Definition: A vector space V(F) is said to be Finite Dimensional Vector Space, if it has a finite basis S such that V = L(S)
i.e. every vector of V is generated by S
The vector space which is not finite dimensional is called infinite dimensional vector space.
Remark: Throughout the text, we shall be using finite dimensional vector spaces (FDVS) only.
Remark: The FDVS is also called finitely generated vector space.
Example: The vector space V₃(R) is finite dimensional because S = {(1, 0, 0); (0, 1, 0); (0, 0, 1)} is a finite subset of V₃ such that V₃ = L(S).

➤ Infinite dimensional vector space: 
Definition: The vector space which is not finite dimensional is called infinite dimensional vector space.
Example: The vector space F[x] of all the polynomials in x over any field F is infinite dimensional because there does not exist any finite subset of F[x] which can generate F[x].

Basis of a Finite Dimensional Vector Space:

Theorem: [Existence theorem]:
Every finite dimensional vector space has a basis.
Proof. Let V(F) be a FDVS.
Then by definition, there exist a finite subset
S = α₁, α₂,......α_n of V which spans V i.e. V = L(S).
Without loss of generality, we may suppose that 0 ∉ S because its contribution in the LC of elements of S is zero.
If S is LI, then S itself is a basis of V and the theorem is established.

If S is LD, then by an earlier theorem there exist α_k, 2 ≤ k ≤ n is S such that either α_k = 0 or α_k is a LC of its preceding vectors α₁, α₂,......α_k-1

Let us remove α_k from S, and denote the remaining set by S₁.

By an earlier theorem S1 also spans V i.e. V = L(S₁).
Now if S₁ is LI, then it is a basis of V.
But if S₁ is LD, then again there exist α₁ ∈ S₁(l > k) is S₁ such that either α_i = 0 or α_i is a LC of its preceding vectors α₁, α₂,......α_i-1  of S₁.
Again let us remove a. from S₁ and denote the remaining set by S₂ i.e. S₂ = S - {α_k, α_i} which again spans V i.e. V = L(S₂).
Proceeding this way after a finite number of steps, either we get a basis for V in the mean while or we are left with the set consisting of a single non zero element which generates V and is LI, thus becomes the basis of V.
Consequently, V(F) has a basis.
Another Form: If a finite set S generates a FDVS V(F), then there exists a subset of S which is the basis of V(F).

Theorem: (Replacement theorem):
Let V(F) be a vector space which is generated by a finite set S = {v₁, v₂,......v_n} of V, then any LI set of V contains not more than n elements.
Proof: Let S = {v₁, v₂,......v_n} generates the vector space V.

S’ = {u₁, u₂,......u_n} be any LI set in V, then we have to prove that m ≤ n.
In order to prove this result, it sufficient to show that for m > n, S’ is L.D.
Now since L(S) = V, therefore any v ∈ V

In particular, u₁ ∈ S ’⊂ V
⇒ u₁ is a LC of v₁, v₂,......v_n
⇒ {u₁,v₁, v₂,......v_n} = S₁ say) is LD and L(S,) = V (Since the set S, is a spanning set for V)

Let us remove this vector v_k from S₁ and denote the remaining set by S₂ i.e.

Since L(S₁) = V, therefore any v ∈ V is a LC of vectors of S₁



Thus any v ∈ V is a LC of vectors belonging to S₂.
∴ L(S₂) = V
Next u₂ ∈ S’ ⊂ V and S₂ generates V therefore u₂ is a LC of vectors of S₂.

Therefore remove this vector from the set S₃ and denote the remaining set by S₄ which generates V.
Proceeding in this manner we find that each step consist in the exclusion of a v and the inclusion of u and the resulting set after each step generates V.
If m > n, then after n steps we obtain a set
generates V and therefore v_n+1, is a LC of the proceeding vectors leading us to the conclusion that the set
which contradicts the hypothesis that S’ is LI.
Hence m ≥ n i.e. m ≤ n.

Corollary. [Invariance of dimension]:
Any two bases of a FDVS have the same number of elements.
Proof: Let V(F) be a FDVS which has two bases

To prove m = n:
Now S₁ is basis ⇒ S₁ is LI and L(S₁) = V    ...(i)
and S₂ is a basis S₂ is LI and L(S₂) = V    ...(ii)
(i) and (ii)  ⇒  L(S₁) = V and S₂ is LI
Therefore by the above result, m ≤n    ...(1)
Also when L(S₂) = V and S₁ is LI
n ≤ m      ...(2)
(1) and (2) ⇒    m = n.

Example: For the vector space V₃(F), the sets
B₁ ={(1, 0, 0), (0, 1, 0), (0, 0, 1)}
and B₂ = {(1, 0, 0), (1, 1, 0), (1, 1, 1)}
are bases as can easily be seen. Both these bases contain the same number of elements i.e. 3.

Some properties of FDVS
Theorem : [Extension Theorem]:
Every LI subset of a FDVS V(F) is either a basis of V or can be extended to form a basis of V.
Proof: Let V(F) be a FDVS whose basis B is
Therefore dim V = n.
Let S = (α₁, α₂,......α_n) be a LI subset of V. Since B is a basis of the space V, therefore B is LI and L(B) = V.
Consider
then clearly, L(B’) and B’ are LD because every a can be expressed as a LC of β₁, β₂,......β_n.
Therefore there will exist β_k is B' which will be expressible as LC of α₁, α₂,......α_m, β₁, β₂,......β_k-1.
Clearly, β_k can not be any of the α₁, α₂,......α_m, because S is LI.
Now remove β_k from B’ and denote the remaining set by B”

Obviously, L(B”) = V.
Now if B” is LI, then this will be a basis of V and is the required extension set.
If B” is LD, then we repeat the process till after a finite number of steps, we obtain a LI set containing α₁, α₂,......α_m and generating V i.e. basis of V.
Since dim V = n, therefore every basis of V will contain n elements.
Thus exactly (n - m) elements of B will be adjoined to S.
 so as to form a basis of V which is the extended form of S. Hence either S is already a basis (when n = m) or it can be extended (when m < n) by adjoining (n - m) elements of B to form the basis of V.
Another form: “Any LI subset of a FDVSV is a part of a basis"
Theorem: In an n-dimensional vector space V(F), any set of (n + 1) or more vectors of V is LD.
Proof: Let V(F) be a vector space and dim V = n.
Therefore every basis of V will contain exactly n elements.
Let S be a subset of V containing (n + 1) or more vectors.
Let, if possible, S is LI, then either it is already a basis or it can be extended to form the basis of V.
Thus in both the cases, the basis will contain (n + 1) or more than (n + 1) vectors which contradicts the hypothesis that V is n-dimensional.
Therefore S is LD and so is every superset of the same.

Theorem: In an n-dimensional vector space V(F), any set of n LI vectors is a basis of V.
Proof: We have dim V = n.
Let S = {α₁, α₂,......α_n} be a LI subset of n-dimensional V(F).
S, being LI subset of V is
either (i) a basis of V,
or (ii) can be extended to form the basis of V.
In case (ii), the basis of V (extended form of S) will contain more than n vectors which contradicts the hypothesis that dim V = n.
Hence the case (i) is true i.e. S is basis.

Theorem: In an n-dimensional finite vector space V(F), any list of n vectors spanning V is a basis.
Proof: We have dim V = n.
Let S = {α₁, α₂,......α_n} be a subset of V such that L(S) = V.
If S is LI, then it will form a basis of V.
IF S is LD, then there exist a proper subset of S’ spanning V and will form a basis of V. This basis S’ will contain less than n elements which contradicts the hypothesis that dim V = n.
Hence S is LI and form a basis of V.

Dimension of a Subspace

Theorem: If W is a subspace of a FDVS V(F), then :
(a) dim W ≤ dim V
(b) W = V ⇔ dim W = dim V.
(c) dim W < dim V, whenever W is a proper subset of V.
Proof: (a) Let V(F) be a finite n-dimensional vector space and W be the subspace of V.
Since dim V = n, therefore any subset of V containing (n + 1) or more vectors is LD,
consequently a LI set of vectors in W contains at the most n elements.
Let S = {α₁, α₂,......α_m }, m ≤ n, be a maximal LI set of vectors in W.
Now if α is any element of W, then the set
S₁ = {α₁, α₂,......α_m} is LD (for S being maximum LI vectors of W)
Therefore α ∈ W is a LC of the vectors α₁, α₂,......α_m 
⇒    L(S) = W
⇒ S is a basis of the subspace W
⇒ dim W = m, where m ≤ n
⇒ dim W ≤ dim V

(b) (⇒) : Let W = V, then
W = V ⇒ W is a subspace of V and V is a subspace of W.
⇒ dim W ≤ dim V and dim V ≤ dim W    [by (a)]

⇒ dim W = dim V.

Conversely (⇐) : Let dim W = dim V = n (say)
Let B = {α₁, α₂,......α_n} be the basis of the space W, then
L(B) = W    ...(1)
But B being a LI subset of V containing n vectors, will also generate V i.e.
L(B) = V    ...(2)

(1) and (2) ⇒    W = V = L(B)

Therefore    W = V ⇔ dim W = dim V.

(c)    When W is a proper subspace of V, then there exist a vector α ∈ V but not in W and as such a cannot be expressed as a LC of vectors of W, the basis of W.
Consequently the set {α₁, α₂,......α_m} forms a LI subset of V, therefore the basis of V will contain more than m vectors
∴ dim W < dim V.

Dimension of a Linear Sum
Theorem: If W₁ and W₂ are two subspaces of a FDVS V(F), then:
dim(W₁ + W₂) = dim W₁ + dim W₂ - dim(W₁ ∩ W₂)

Proof: Let dim(W₁ ∩ W₂) = k and set S = {γ₁, γ₂,......, γ_k} be a basis of W₁ ∩ W₂, then S ⊂ W₁ and S ⊂ W₂.
Now since S is LI and S ⊂ W₁, therefore by the extension theorem, S can be extended to form the basis of W₁.
Let  be a basis of W₁.
Then dim W, = k + m
Similarly, let  be the basis of W₂.
The dim W₂ = k + n
∴ dim W₁ + dim W₂  - dim(W₁ ∩ W₂) = (k + m) + (k + n ) - k = k + m + n

To prove dim(W₁ + W₂) = k + m + n :
For this, we shall show that

is a basis of (W₁ + W₂).
Let there exist  such that

Now since
 

Therefore this can be expressible as a LC of the elements of the basis S of W₁ ∩ W₂.
Hence let d₁, d₂, d_k ∈ F such that
Substituting these values in (1).Therefore by (4) and (5),(2)
⇒ c₁ = 0, c₂ = 0, c_k = 0, a₁ = 0, a₂ = 0...... a_m = 0, b₁ = 0, b₂ = 0, b_n = 0
⇒ the set S, is LI.

To show L(S₁) = (W₁ + W₂):
(W₁ + W₂) is a subspace of V(F) and each element of S₁ belongs to (W₁ + W₂).
Hence  L(S,) c (W₁ + W₂)    ...(6)
Again let α ∈ (W₁ + W₂),
then    α = any element of W₁ + any element of W₂
= LC of elements of a basis of W₁ + LC of elements of a basis of W₂ = LC of elements of S₁.
⇒    α ∈ L(S₁)
⇒    (W₁ + W₂) ⊂ L(S₁)    ...(7)
Therefore (6) and (7)
⇒ L(S₁) = (W₁ + W₂)
Consequently, S. is a basis of (W₁ + W₂)

Direct sum of two spaces.

Let V(F) and V'(F) be two vector spaces, then the product set

is a vector space over F for the compositions defined as
This vector space V * V ’ over F is a called external direct sum o f two vector spaces V(F) and V'(F) and is written as V * V’.
(b) Internal Direct Sum or Direct Sum.
If W₁ and W₂ be the two subspaces of a vector space V(F), then V(F) is said to be the direct sum of its subspaces W₁ and W₂, if every α ∈ V can be uniquely expressed a

We denote the direct sum of W₁ and W₂ by W₁ ⊕ W₂
Then V = W₁ ⊕ W₂
⇒ v = W₁ + W₂
Complementary subspaces
If V = W₁ ⊕ W₂ then W₁ and W₂ are said to be complementary subspaces i.e. any α ∈ V is uniquely expressible as

Disjoint subspaces
Two subspaces W₁ and W₂ of the vector space V(F) are said to be disjoint if W₁ ∩ W₂ = {0} = zero space.
Theorem: The necessary and sufficient conditions for a vector space V(F) to be a direct sum of its two subspaces W₁ and W₂ are
Proof: The Conditions are necessary (⇒):
Let V = W₁ ⊕ W₂
By definition of direct sum, each element a ∈ V is uniquely expressed

Let, If possible

Evidently

Since the sum for a is unique and hence α = 0But α ∈ W₁ ∩ W₂ is arbitrary
∴ W₁ ∩ W₂ = {o}.    ...(2)
The conditions are sufficient (⇐):
Let V = W₁ + W₂ and W₁ ∩ W₂ = {0}
Let α ∈ V be arbitrary.

such thatα = α₁ + α₂
Now we have to show that this representation is unique.
If possible, let
 
being subspaces of V
     (given)
Hence V = W₁ ⊕ W₂

Dimension of direct sum.

Theorem: If a FDVS V(F) is a direct sum of its two subspaces W₁ and W₂, then
dim V = dim W₁ + dim W₂
i.e  Proof: We have
V = W₁ ⊕ W₂

By theorem we have
    ...(3)
Here we haveW₁ + W₂ =

Example 1: Prove that the set S = {(1,2, 1), (2, 1, 0), (1,-1, 2)} forms a basis of the vector space V₃(ℝ).

All the 3 elements of S are in V₃(ℝ), therefore S ⊂ V₃(ℝ).
To show that S is LI:
Let a₁, a₂, a₃ ∈ R such that

By (1) and (2),

By (1) and (3),

By (3) and (4) and (5),a₁ = 0, a₂ = 0, a₃ = 0

⇒ Vectors of S are LI.

Hence S is a set of 3 LI vectors of V₃(ℝ).

Example 2: If  V(ℝ) be the vector space of all ordered pairs of complex numbers over the real field R, then prove that the set S = {(1, 0); (i, 0); (0, 1), (0, i)} is a basis of  V(ℝ). To show that S is LI:

Let a₁, a₂, a₃, a₄ ∈ R Such that

⇒ S is LI.

To show that V = L(S):
Let α = (a + ib, c + id) ∈ V, where a, b, c, d ∈ R, then
α = (a + ib, c + id)

= a(1, 0) + b(i, 0) + c(0, 1) + d(0, i)
⇒ α ∈ V is LC of elements of S
⇒ V = L(S).
Hence S is a basis of V(R).

Example 3: Prove that the set S = {a + ib, c + id} is a basis set of the vector space ℂ(ℝ) iff (ad - be) ≠ 0.

To show that S is LI :

Let a₁, a₂ ∈ ℝ such that
  ...(1)Again when a₂ = 0 then by (1), a₁ = 0
Therefore S will be LI.
Therefore S is LI if (be - ad) ≠ 0.
To show that C(R) = L(S).

Let α = e + if ∈ ℂ(ℝ) such that

α = e + if = x(a + ib) + y(c + id), where x, y ∈ E
⇒ e = if = (xa + yc) + i(xb + yd)
⇒ c = ax + cy, f = bx + dy

Therefore each element of ℂ(ℝ) can be expressed as LC of elements of S if (ad - be) ≠ 0⇒ ℂ(ℝ) = L(S) if (ad - be) ≠ 0
Hence S is a basis of ℂ(ℝ) iff (ad - be) ≠ 0

Example 4: If the set S = {α, β, γ} be a basis of the vector space V₃(ℝ), then prove that the set  will also be a basis of V₃(ℝ).

Since there are 3 elements in the basis S of V₃(ℝ).
Therefore the set of any 3 LI vectors of V₃(ℝ) will be its basis.

To show that S’ is LI :

Let a₁, a_2, a₃ ∈ ℝ such that
  ...(1)

But α, β, γ are LI, therefore

Therefore S’ is also LI. Now since the basis S of V₃(ℝ) contains 3 vectors, therefore S’ is a subset of 3 LI vectors. As such S’ will also be basis of V₃(ℝ). 

Example 5: Show that the dimension of the vector space V(ℝ) of all 2 * 2 real matrices is 4. 

Let us consider the following 4 matrices of the order 2*2.

We will show that their set  is LI and generates V(ℝ)

Let

Again for any element  of V, there exist a, b, c, d ∈ ℝ such that

⇒ each element of V is a LC of elements of S⇒ V  = L(S)
⇒ S is a basis of V(ℝ).

But the number of elements in S is 4, therefore dim V = 4.

Example 6: Show that the set S = {(1, 0, 0); (1, 1, 0); (1, 1, 1); (0, 1, 0)} spans the vector spaceV₃(ℝ) but is not a basis set.

Here we will show that S is not LI, yet V₃(ℝ) = L(S).
Let a₁, a₂, a₃, a₄ ∈ ℝ such that
   ...(A)

The above system (A) has 4 unknown quantities and 3 equations.

Therefore there exist a non zero solution. Consequently, S is not LI.
Hence S’ is not a basis set of V₃(ℝ).
Again suppose that for any vector (a, b, c) of V₃(ℝ), there exist a₁, a₂, a₃, a₄ ∈ ℝ such that

Therefore for the relation (B), non zero values of a₁, a₂, a₃, a₄ exist.

Therefore the set S spans V₃(ℝ), but it is not a basis set. Hence Proved.

Example 7: Show that the set S = {1, x, x², ..., xⁿ} is a basis of the vector space F[x] of all polynomials of degree at most n, over the field ℝ of real numbers.

To show that S is LI:
Let a₁, a₂, a₃, a₄ ∈ ℝ such that
 (zero polynomial)

Again F[x] = L(S),

since

⇒ p(x) is a LC of elements 1, x, x², ..., xⁿ of S
Therefore S is a basis of F[x].
First we state two equivalent ways to define a basis of a vector space V.

Definition A: A set S = {u_1, u₂, ..., u_n} of vectors is a basis of V if it has the following two properties: (1) S is linearly independent. (2) S spans V.
Definition B: A set S = {u_1, u₂, ..., u_n} of vectors is a basis of V if every v e V can be written uniquely as a linear combination of the basis vectors.
Theorem: Let V be a vector space such that one basis has m elements and another basis has n elements. Then m = n.
A vector space V is said to be of finite dimension n or n-dimensional, written
dim V = n
If V has a basis with n elements. Theorem tells us that all bases of V have the same number of elements, so this definition is well defined.
The vector space {0} is defined to have dimension 0.
Suppose a vector space V does not have a finite basis. Then V is said to be of infinite dimension or to be infinite-dimensional.

Lemma: Suppose {v₁, v₂ ..... v_n} spans V, and suppose {w₁, w₂, ..., w_m} is linearly independent.

Then m ≤ n, and V is spanned by a set of the form

➤ Examples of Bases 

Vector space Kⁿ: Consider the following n vectors in Kⁿ:
e₁ = (1, 0, 0, 0, .... 0, 0), e₂ = (0, 1, 0, 0, .... 0, 0 ) ...... e_n = (0, 0 , 0 , 0 ...... 0 , 1)
These vectors are linearly independent. (For example, they form a matrix in echelon form.)
Furthermore, any vector u = (a₁, a₂......a_n) in Kⁿ can be written as a linear combination of the above vectors. Specifically,
 
Accordingly, the vectors form a basis of Kⁿ called the usual or standard basis of Kⁿ. Thus (as one might expect), Kⁿ has dimension n. In particular, any other basis of Kⁿ has n elements.

➤ Theorem on Bases
The following three theorems will be used frequently.

Theorem : Let V be a vector space of finite dimension n. Then:
(i) Any n + 1 or more vectors in V are linearly dependent.

The statement dim V = n means there exists a basis {v₁, .... v_n} for V. Let v_n+1 ∈ v\{v₁, .... v_n}. Then by the definition of a basis, there exist a1, ..., an ∈ F(F being the field over which V is defined) such that
 
So, {v₁, .... v_n+1} is linearly dependent.

(ii) Any linearly independent set S = {u₁, u₂, ...... ,u_n} with n elements is a basis of V.
We just need to show that S spans V. That is, each elements of V can be expressed as a linear combination of the elements of S. Let w be an element in V. Since S is a maximal linearly independent set, the elements w, u₁, u₂, ...... ,u_n are linearly dependent. Hence there exists λ₀, λ₁, ...... ,λ_n in K, not all 0_K, such that

It is clear that :
otherwise the elements u₁, u₂, ...... ,u_n would be linearly dependent. Thus, we obtain from (1)

This show that w is a linear combination of the elements of S. Thus, S is a basis of V.
(iii) Any spanning set  of V with n elements is a basis of V.
Suppose  and . Subtracting these equations side from side, we obtain  
Since the set {v₁, v₂, ...... ,v_n} is linearly independent we have a₁ - b₁ = 0, which means a_i = b_i for each i = 1, 2, ..., n.
Hence T is a basis of V.
Theorem: Suppose S spans a vector space V. Then :
(i) Any maximum number of linearly independent vectors in S form a basis of V.
(ii) Suppose one deletes from S every vector that is a linear combination of preceding vectors in S. Then the remaining vectors form a basis of V.
Theorem: Let V be a vector space of finite dimension and let S = {u₁, u₂, ...... ,u_n} be a set of linearly independent vectors in V. Then S is part of a basis of V; that is, S may be extended to a basis of V.

Example: (a) The following four vectors in ℝ⁴ form a matrix in echelon form:
(1, 1, 1, 1), (0, 1, 1, 1), (0, 0, 1, 1), (0, 0, 0, 1)
Thus, the vectors are linearly independent, and, because dim ℝ⁴ = 4, the four vectors form a basis of ℝ⁴.
(b) The following n + 1 polynomials in P_n(t) are of increasing degree:

Therefore, no polynomial is a linear combination of preceding polynomials; hence, the polynomials are linear independent. Furthermore, they form a basis of Pn(t), because dim Pn(t) = n + 1.

➤ Quotient Space

Definition :

Let V(F) be a vector space and W be its subspace. Then for any α ∈ V, the set W + α  = {w + αlw ∈ W} is called the Right coset of W wrt a in V.
Also the set  is called the Left coset of W wrt α in V.

Since V is an abelian group wrt addition, therefore

i.e. each right coset of W is equal to its corresponding left cost.
Therefore this is simply called coset.
The set of all the cosets of W in V is denoted by VA/V i.e.

Some properties of cosets:
Let W be a subspace of the vector space V(F), then :

(i) W + 0 = 0 + W = 0, i.e. W is the right as well as left coset of itself.
Addition and Scalar multiplication for cosets :If W is a subspace of the vector space V(F), then the addition and scalar multiplication of cosets in V/W are defined as follows:

Theorem: If W is a subspace of vector space V(F), then the set V/W of all cosets of W in V is a Vector Space over F for the vector addition and scalar multiplication defined as follows:

Proof: Vector addition is well defined :
For any α , α', β, β' ∈ V

Therefore vector addition is well defined in V/W.

Scalar multiplication is well defined:
For any α ∈ V, α ∈ F

Therefore scalar multiplication is also well defined in V/W.

Again for each α, β ∈ V, a ∈ F

Therefore V/W is closed for both the above operations.

Verification of Space axioms: 
V₁. (V/W, +) is an abelian group.
In Group theory it has already been shown that (V/W, +) is an abelian group.
V₂. Scalar multiplication is distributive over addition of coset:


Therefore scalar multiplication is distributive over addition of cosets in V/W.
V₃. Scalar addition is distributive over scalar multiplication:


Therefore scalar addition is distributive over scalar multiplication in V/W.
V₄. Associativity for scalar multiplication:


Therefore scalar multiplication is associative in V/W.
V₅. Let 1 ∈ F, is the unit element of F, then for W + α ∈ V/W
From the above discussion, it is clear that V/W satisfy all the space axioms. Hence V/W is a vector space over the field F.V/W is called the Quotient space or Factor space of V relative to W Dimension of a Quotient Space 
Theorem: If W be a subspace of a FDVS V(F), then:

dim(V/W) = dimV - dimW.

Proof. Let dim V = n and dim W = m. Clearly, m ≤ n.
Let  be the basis of W.Since S is a LI subset of V, therefore this can be extended to form the basis of V. Therefore let the extended set  be the basis for V.
Now we will show that the set of (n - m) cosetsis a basis of V/W.C is LI: Let a_1, a₂...... a_n-m ∈ F such that

[since any element of W can be expressed as a LC of elements of its basis S]
V/W = L(C): Let W + α be any element of V/W where α ∈ V but S’ is a basis of V, therefore α = LC of elements of S’

where

since S is the basis of W.⇒ V/W = L(C)∴ C is a basis of V/W.
⇒  dim(V/W) = n - m = dimV - dimW.

Co-ordinate Representation of a vector 
With the help of the basis, we can determine the co-ordinates of any vector α ∈ V which is just similar to natural co-ordinates of α.
For example, if  {e₁, e₂,..... e_n} is a standard (natural) basis of V_n(F) where e₁ = (1, 0, 0......0), e₂ = (0, 1, 0, .... 0), e_n = (0, 0, 0, ..... 1)
then any  can be expressed as LC of basis vectors i.e.
 If  is an arbitrary basis of the FDVS V_n(F), then any vector α of V_n can be uniquely expressed as a LC of basis vectors α₁, α₂,..... α_n.

The n-tuple of scalars (a₁, a₂,..... a_n) used in this LC is called the co-ordinates of a relative to the basis B.
But in the basis B, the basis vectors can be written in any order. By changing the order of the vectors, the c-ordinates of α are changed.
Ordered Basis: If the vectors of B are kept in a particular form by a well defined way, then B is called the Ordered Basis for V. The co-ordinates of any vector of the space of unique wrt an ordered basis.
➤ Co-ordinates of a Vector. 
Definition:
Let V (F ) be a FDVS. Let B = ( α₁, α₂,..... α_n) be an ordered basis for V. Again in let α ∈ V, then ∃ a unique n-tuple (x₁, x₂, ..... ,x_n) of scalars such that

The n-tuple (x₁, x₂, ..... ,x_n) is called n-tuple of co-ordinates of relative to the ordered basis B. The scalar x_i is called i^th co-ordinate of relative to the ordered basis B.

Example 1: Show that S = {(1, 1, 1), (0, 1, 1), (0, 0, 1)} is a basis for the space V₃(ℝ). Also find the co-ordinates of α = (3, 1, -4) ∈ V₃(ℝ) relative to this basis.

Clearly S ⊂ V₃(ℝ).

S is LI: Let there exist a₁, a₂, a₃ ∈ ℝ such that

Therefore the required co - ordinates are (3, - 2 , - 5)

Example 2: Find the co-ordinates of the vector (a, b, c) of V₃(ℝ) relative to its basis S = {(1, 1, 1), (1, 1, 0), (1, 0, 0)}.

Let a₁, a₂, a₃ ∈ ℝ

Therefore the required co-ordinates are (c, b - c, a - b).

Example 3: Find the co-ordinates of the vector p(x) = 2x² - 5x + 6 of P₂[x] of all polynomials over R with degree < 2, relative to its basis S = {1, x - 1, x² - x + 1}

Let a₁, a₂, a₃ ∈ R such that
 

Therefore the required co -ordinates are (3, - 1 , 2).

Example 4: If V (R ) be the vector space over the real field of all 2*2 matrices and  be a basis of it, then find the co-ordinates of   relative to the basis.

Let a₁, a₂, a₃ ∈ R such that

Solving the equations,a, = -7, a2 = 11, a3 = -21, a, = 30
Therefore the required co-ordinates are (-7, 11, -21, 30).

Example 5: Let V₃(ℝ) be a FDVS. Find the co-ordinate vector of = (3, 1, -4) relative to the following basis: v = (0, 0, 1); v₂ = (0, 1, 1); v₃ = (1, 1, 1)

Let a₁, a₂, a₃ ∈ R such that
i.e. (3, 1, -4) = a₁(0, 0, 1) + a₂(0, 1, 1) + a₃(1, 1, 1)
⇒ a3 = 3, a₂ + a₃ = 1, a₁ + a₂ + a₃ = -4
⇒ a₁ = -5, a₂ = -2, a₃ = 3
 ∴ The required co-ordinates of v wrt the given basis are [v] = (-5, -2, 3)