BASES AND DIMENSIONS

 BASES AND DIMENSIONS

(a) Bases.

A non empty subset of S a vector space V(F) is said to be a (linear) basis or a base ( or coordinate system) of V iff

(i) S is L.I

(ii) V is generated or spanned by S

i.e. V = [S] i.e. each element of V is a linear combination of elements of S.

(b). Dimension of a vector space.

If a basis of vector space V(F) is a finite set containing n elements, then V is called an n-dimensional vector space or a vector space of dimension n and we write dim V = n

or

A vector space V(F) is said to be finite dimensional or finitely generated if there exists a finite subset S of V such that V = L (S).

A vector space V(F) which is not finitely generated is known as a finitely dimensional vector space.

Example 1:

The set S = {(1, 2, 1), (2, 1, 0), (1, -1, 2)} forms a basis for R³(R).

Solution:

For L.I let a₁, a₂, a₃ ∈ R

then a₁(1, 2, 1) + a₂(2, 1, 0) + a₃(1, -1, 2) = (0,0,0)

⟹ (a₁ + 2a₂ + a₃, 2a₁ + a₂ + -a₃, a₁ + 2a₃) = (0,0,0)

⟹ a₁ + 2a₂ + a₃ = 0,

a₁ + 2a₃ = 0

⟹ a₁ = a₂ = a₃

therefore S in L.I

for any vector (a₁, a₂, a₃) ∈ R³(R)

(a₁, a₂, a₃) = a₁(1, 0, 0) + a₂(0, 1, 0) + a₃(0, 0, 1)

where 

(1,0,0) = x(1, 2, 1) + y(2, 1, 0) + z(1, -1, 2)

= ( x + 2y + z, 2x + y -z, x + 2z)

i.e. x + 2y + z = 1,  2x + y -z = 0,  x + 2z = 0

Solving these we get 

x = -2/9, 

y = 5/9, 

z = 1/9

therefore (1,0,0) = -2/9(1, 2, 1) + 5/9(2, 1, 0) + 1/9(1, -1, 2)

Similarly (0,1,0) = 4/9(1, 2, 1) - 1/9(2, 1, 0) - 2/9(1, -1, 2)

(0,0,1) = 1/3(1, 2, 1) -1/3(2, 1, 0) + 1/3(1, -1, 2)

Using these unit vectors in (1), we get that (a₁, a₂, a₃) can be expressed as linear combination of elements of S.

Hence S is a basis of R³.

Example 2:

Show that the set of vectors

e₁= (1, 0, 0, ....0), e₂= (0, 1, 0, ....0), ....., eₙ =(0, 0, 0, ....1)

is a basis of Vₙ (F). This is known as standard basis of Vₙ (F).

Solution:

Let S = { e ₁, e ₂, ...... e ₙ}

Let ( a ₁, a ₂, ......,a ₙ) ∈ V ₙ, then for L.I

a ₁e ₁, a ₂e ₂, ......,a ₙe ₙ = 0 = (0, 0, 0, ....0)

or a ₁(1, 0, 0, ....0) + a ₂(0, 1, 0, ....0) + .....+ a ₙ(0, 0, 0, ....1) = (0, 0, 0, ....0)

⟹ ( a ₁, a ₂, ......,a ₙ) = (0, 0, ....0)

⟹ a ₁ = 0, a ₂ = 0, ......,a ₙ = 0

Hence the set S is L. I

Also we have

( a ₁, a ₂, ......,a ₙ) = a ₁e ₁ + a ₂e ₂ + ......+ a ₙe ₙ

i.e. x = ( a ₁, a ₂, ......,a ₙ) ∈ V ₙ is expressed as a linear combination of elements of S.

Hence S is a basis of Vₙ (F).

Note: (i) Also dim Vₙ (F) = n

(ii) We can prove that the set of unit vectors

{ (1, 0, 0),  (0, 1, 0), ( 0, 0, 1)} is basis of Vₙ (F).

and { (1, 0),  (0, 1)} is a basis of Vₙ (F).

Example 3:

If { x, y, z} is basis of R³, where R is the set of real numbers, then { x + y, y + z, z + x} is also a basis of R³.

Solution:

For L.I , let

a(x + y) + b(y + z) + c(z + x) = 0 = (0, 0, 0)

⟹ (a + c)x + (a + b)y + (b + c)z = (0, 0, 0)

⟹ a + c = 0, a + b = 0 , b + c = 0, Since { x, y, z} is L.I

⟹ a = 0, b = 0, c = 0

Therefore the set of vectors {x + y, y + z, z + x} is L.I

Since { x, y, z} is a basis of R³

and so there exists a ₁, a ₂, a ₃ ∈ R s.t.

any vector p ∈ R³ can be expressed as

p = a₁x + a₂y + a₃z   ..........(1)

Again suppose that

p = b ₁ (x + y) + b ₂ (y + z) + b ₃ (z + x) ......(2)

where b ₁, b ₂, b ₃ ∈ R³

From (1) and (2)

b ₁ (x + y) + b ₂ (y + z) + b ₃ (z + x) = a₁x + a₂y + a₃z

or (b ₁+ b ₃ - a ₁)x + ( b ₁+ b ₂ - a ₂)y + (b ₂+ b ₃ - a₃)z = 0 = (0, 0, 0)

Therefore

b ₁ + b ₃ = a ₁

b ₁ + b ₂ = a ₂

b ₂ + b ₃ = a ₃

solving these equations, we get

b ₁ = (a ₁ + a ₂ - a ₃)/2

b ₂ =  (a ₂ + a ₃ - a ₁)/2

b ₃ =  (a ₃ + a ₁ - a ₂)/2

Thus any p ∈ R³ can be expressed as a linear combination of vectors of the set {x + y, y + z, z + x}

Hence the set {x + y, y + z, z + x} is a basis of R³.

Example 4:

Determine whether or not the following vectors from a basis of R³.

(1, 1, 2) , (1, 2, 5), (5, 3, 4)

Solution:

We know that dim R³ = 3

For L.I , we have

a ₁ (1, 1, 2) + a ₂(1, 2, 5) + a ₃ (5, 3, 4) = (0, 0, 0)

a ₁, a ₂, a ₃ ∈ R s.t.

⟹ ( a ₁ + a ₂ + 5a ₃,  a ₁ + 2a ₂ + 3a ₃,  2a ₁ + 5a₂  + 4a ₃) = (0, 0, 0)

therefore 

a ₁ + a ₂ + 5a ₃ = 0  ...........(1)

a ₁ + 2a ₂ + 3a ₃ = 0  .........(2)

2a ₁ + 5a₂  + 4a ₃ = 0 ........(3)

From (1) and (2) on subtracting, we get

- a ₂ + 2a ₃ = 0 .........(4)

Multiplying (1) and (2) and subtracting from (3) we get 

3a ₂ - 6a ₃ = 0 or  a ₂ - 2a ₃ = 0 .........(5)

which is same as (4)

putting a ₂ = 2a ₃ in (1) we get

a ₁ = -7a ₃

If we put a ₃ = 1, we get a ₂ = 2 and a ₁ = -7

Thus a ₁ = -7 , a ₂ = 2,  a ₃ = 1is not a zero solution of equations (1), (2) and (3).

Hence the given set is linearly dependent and so it does not form a basis of R³.

Example 5:

Show that the vectors (2, -1, 0) , (3, 5, 1), (1, 1, 2) from a basis of R³(R).

Solution:

We have dim R³ = 3

For L.I , let a ₁, a ₂, a ₃ ∈ R then 

a ₁ (2, -1, 0) + a ₂(3, 5, 1) + a ₃ (1, 1, 2) = (0, 0, 0)

⟹ ( 2a₁+ 3a₂+ a₃,  -a ₁ + 5a₂ + a₃,  0 + a₂ +2a₃) = (0, 0, 0)

therefore

2a ₁ + 3a ₂ + a ₃ = 0  ...........(1)

-a ₁ + 5a ₂ + a ₃ = 0  .........(2)

a₂  + a ₃ = 0 ........(3)

From (1) and (2) , we have

13a ₂ + 3a ₃ = 0 .........(4)

Multiplying (3) by 3 and subtracting from (4), we get a₂ = 0 and so a ₃ = 0

therefore from (1), a ₁ = 0

Thus the three given vectors are L.I and so they form a basis for R³(R).

Example 6:

Prove that

S = {(1, 0, 0), (1, 1, 0), (1, 1, 1), (0, 1, 0)}

spans the vector space R³(R) but is not basis of R³.

Solution:

let (a ₁, a ₂, a ₃) ∈ R and ( a₁, a₂, a₃) = x₁ (1, 0, 0) + x₂ ( 1, 1, 0) + x₃ (1, 1, 1) + x₄ (0, 1, 0) where x₁ 

, x₂, x ₃, x₄ ∈ R.

Then  (a₁, a ₂, a ₃) = (x₁ + x₂ + x ₃,  x₂ + x ₃ + x₄, x ₃) 

Therefore 

x₁ + x₂ + x ₃ = a ₁

x₂ + x ₃ + x₄ = a ₂

x ₃ = a ₃

or 

x ₁ = a ₁ - a ₂ + x ₄

x ₂ = a ₂ - a ₃ + x ₄

x ₃ = a ₃

Taking x ₄ = 0, we get

x ₁ = a ₁ - a ₂,

x ₂ = a ₂ - a ₃

x ₃ = a ₃

and  (a₁, a ₂, a ₃) =  (a₁ - a ₂)(1,0, 0) + (a ₂ - a ₃)(1, 1, 0) + a ₃ (1, 1, 1) + 0(0, 1, 0)

Thus every vector (a₁, a ₂, a ₃) ∈ R³ is a linear combination of elements of S.

But (0, 0, 0) = 1(1,0, 0) + (-1)(1, 1, 0) + 0(1, 1, 1) + 1(0, 1, 0) which shows that S is L.D

Therefore S is not basis of R³.

Theorem 1:

Let V be a finitely generated vector space over F. Then the set S = { x ₁, x ₂, .....,x ₙ} is a basis of V iff every x ∈ V can be expressed as a unique linear combination of elements of S.

Proof:

Let S = { x ₁, x ₂, .....,x ₙ} be a basis of V. Then V = L(S). Therefore every x ∈ V can be expressed as a linear combination of elements of S.

If possible, let

x = a ₁ x ₁ + a ₂ x ₂+ ......+ a ₙ x ₙ, where a ₁, a ₂,... a ₙ ∈ F

x = b₁ x ₁ + b₂ x ₂+ ......+ bₙ x ₙ, where b ₁, b ₂,..... b ₙ ∈ F

Therefore 

a ₁ x ₁ + a ₂ x ₂+ ......+ a ₙ x ₙ = b₁ x ₁ + b₂ x₂+ ......+ bₙ x ₙ

or ( a₁ - b₁)x ₁ + ( a₂ - b₂)x₂ +....+ ( a ₙ - b ₙ)x ₙ  = 0

a₁ - b₁ = 0,  a₂ - b₂ = 0, ....... , a ₙ - b ₙ = 0

since S is L.I

i.e. a₁ = b₁ ,  a₂ = b₂, ...... , a ₙ = b ₙ 

Thus each element of V is uniquely expressible as a linear combination of elements of S.

Conversely suppose that x ∈ V is uniquely expressible as a linear combination of elements of S. Then V = L(S)

To prove that S is L.I

Let a ₁ x ₁ + a ₂ x ₂+ ......+ a ₙ x ₙ = 0

or a ₁ x ₁ + a ₂ x ₂+ ....+ a ₙ x ₙ = 0x ₁ + 0x ₂+ ......+ 0x ₙ

Therefore a₁ = 0 ,  a₂ = 0, ...... , a ₙ = 0

i.e. S = { x ₁, x ₂, .....,x ₙ} is L.I

Hence S is basis of V.

Theorem 2:

Every finitely generated vector space possesses a basis. 

Proof:

Let V be a vector space generated by a finite set S = { x ₁, x ₂, .....,x ₙ} of non zero vectors in V.

If S is L.I. then it is basis of V. On the other hand, if S is L.D. ∃ some xₖ, 2 ≤ k ≤ n in S s.t. xₖ is a linear combination of preceding vectors x ₁, x ₂, ......,xₖ ₋₁ 

Since S generates V so each element of V is uniquely expressible as a linear combination of x ₁, x ₂, ......,xₖ, ....x ₙ.

 Also, x ₖ is a linear combination of x ₁, x ₂, ......,xₖ ₋₁. It follows that each element of V is expressible as a linear combination of x ₁, x ₂, ......,xₖ ₋₁ , x ₖ ₊₁, .....,x ₙ.

Thus the subset S = {x ₁, x ₂, ......,xₖ ₋₁, x ₖ ₊₁, .....,x ₙ} of n- 1 elements of S also generated If S ₁ is L.I, then it is a basis of V. 

On the other hand if S ₁ is L.D., then proceeding as above we can get subset S₂ of n - 2 elements of S which generates V. If S₂ is L.I., then S₂ is a basis of V. If not, then continuing this process, after a finite number of steps, we obtain a L.I. subset of S, which generates V. It may happen that we may be left with a single element generating V, which is L.I. and therefore it will become a basis of V.

 Hence V possesses a basis. 

Theorem 3:

Any two bases of a finitely generated vector space V have the same number of elements.

Proof. 

Let V have the following two beset: 

B ₁ = { x ₁, x ₂, .....,x ₘ} 

B ₂ = {y ₁, y ₂, .......,y ₙ}

 Since V is generated by B ₁ and so y ₁ ∈ V is a linear combination of elements of B ₁.

Therefore the set  {y ₁,  x ₁, x ₂, .....,x ₘ}  is L.D.

Hence ∃ a vector x ₖ (≠ y ₁) which is a linear combination of the proceding vectors i.e. linear combination of y ₁ ,x ₁, ....,x ₖ₋₁

Let S ₁ = {y ₁,  x ₁, x ₂, .....,x ₖ₋₁ , x ₖ ₊₁, .....,x ₘ}

obtained by deleting x ₖ from B ₁. 

It is clear that each clement of V is linear combination of elements of S ₁.

 Hence V is generated by S ₁.

 Consequently, the element y ₂ ∈ B ₂ is a linear combination of elements of S ₁. 

Therefore the set

{y ₂, y ₁,  x ₁, x ₂, .....,x ₖ₋₁ , x ₖ ₊₁, .....,x ₘ} is L.D.

Therefore ∃ a vector xⱼ different from y₁ or y₂ which is linear combination of the proceding vectors i.e. of y ₂, y ₁,  x ₁, x ₂, .....,x ₖ₋₁ , x ₖ ₊₁, .....,xⱼ ₋₁

Let S ₂ = {y ₂, y ₁,  x ₁, x ₂, .....,x ₖ₋₁ , x ₖ ₊₁, ...

., xⱼ ₋₁, x ⱼ ₊₁, .....,x ₘ} clearly this set also generates V. 

We continue to proceed in the above manner. Now if m ≺ n, then B₁ gets exhausted before B₂ and we will get. 

S = { y ₘ, y ₘ ₋₁, ....., y ₁}

which is obtained by excluding α x from B₁ and including a y from B₂. 

Thus Sₘ is a proper subset of B₂ generating V which is L.D. 

Therefore the set B₂ becomes L.D. contradicting the L.I. of B₂ 

Hence m ≥ n Similarly, by changing the roles of the bases, we get n ≥ m  

Hence m = n.

Theorem 4:

If a vector space V be generated by a set of n vectors, then every L.I. subset of V can have utmost n vectors. 

Proof. 

Suppose that S ₁ = { x ₁, x ₂, .....,x ₙ} is a set of generators of V and S₂ = {y ₁, y ₂, .....,y ₘ} is L.I. subset of V containing m vectors. 

Then we have proved in previous theorem that m ≤ n, which shows that S₂ have utmost n vectors. 

Theorem 5:

Every L.I subset of a finite dimensional vector space V(F) is either a basis of V or it can be extended to form a basis of V. 

Proof. 

Let S = { x ₁, x ₂, .....,xₘ} be L.I. subset of a finite dimensional vector space V(F). If S is not basis of V, suppose dim V= n, then V has a finite basis say 

  { y₁, y₂, .....,yₙ}. It is clear that x ₘ is a linear combination of vectors y₁, y₂, .....,yₙ

Therefore the set { x ₘ, y₁, y₂, .....,yₙ} is LD. Since every super set of a L.D. set is L.D., and so the set. S = { x ₁, x ₂, .....,xₘ , y₁, y₂, .....,yₙ} is L.D. 

Hence there exists a vector in S ₁, which is linear combination of its preceding vectors. This vector cannot be any of the x's since S is L.I. So it must be some y ₖ say. 

Now each vector in V is linear combination of x ₁, x ₂, .....,xₘ , y₁, y₂, .....,yₙ and yₖ is a linear combination of vectors

x ₁, x ₂, .....,xₘ , y₁, y₂, .....,y ₖ ₋₁ 

Thesefore each element of V is expressible as a linear combination of elements of 

S₂= {x ₁, x ₂, .....,xₘ , y₁, y₂, .....,y ₖ₋₁, y ₖ₊₁, ...yₙ}

Now, if S₂ is L.I. than it is a basis of V containing S. 

Otherwise, the set S₃ can be obtained by excluding yⱼ (j > k) from S₂, generates V. 

Now if S₃ is L.I. then it is a basis of V containing S. Continuing this process, till after a finite number of steps, either we get L.I. set which generates V and contains S or we are left with S itself generating V and S being L.I. will be a basis of V.

Hence, either S is already a basis or it can be extended to form a basis of V 

Theorem 6: 

Every set of (n + 1) vectors or more vectors in an n dimensional vector space V is L.D. 

Proof:

Since dim V = n and so every basis of V will contain exactly n vectors. 

Suppose S is any L.I. subset of V containing (n + 1) or more vector, then by previous theorem either S forms a basis or it can be extended to form a basis of V. 

Therefore in each case the basis of V contains (n + 1) or more vectors, which is contrary to dim V = n 

Hence S is L.D. 

Theorem 7: 

If V is an n-dimensional vector space then a set of n vectors in V forms a basis of V iff it is L.I. or iff V is generated by the set. 

Proof:

Suppose a set of n vectors is L.I., then it must be a basis of V since dim V = n. For otherwise it can be extended to form a basis containing more than n elements which is not possible, since dim V = n. 

Conversely suppose that the set 

S = { x ₁, x ₂, .....,x ₙ} of n vectors of V forms a basis of V. 

If S is L.I., then it will form a basis of V, othewise there will be a proper subset of S which will form a basis. This is not possible, since dim V = n. 

Hence S is L.I. 

Theorem 8:

Every subspace W of a finite dimensional vector space V(F) of dimenstion n is a finite dimensional space with dim m ≤ n. 

Proof. 

Let W be a subspace of a finite dimensional vector space V(F) with dim = n. 

Let S = { x ₁, x ₂, .....,x ₙ} be a basis of V. Then

L(S)= V so that each element of V is a linear combination of elements of S. 

Also W ⊆ V. Hence in particular each element of W can also be a linear combination of element of S. Also S is L.I. Therefore S is a basis of W or any subset of S is a basis of W (As every subset of L.I. is L.I). 

In either case, the basis of W cannot have more than n vectors. 

Hence m ≤ n.  

Theorem 9:

Let W be a subspace of a finite dimensional vector space V(F). 

Then dim V= dim W iff V= W. 

Proof:

Let W be a subspace of a finite dimensional vector space V(F), then W ⊆ V. 

Let W = V, then W is a subspace of V and V is a subspace of W. 

Therefore, 

W ⊆ dim V and dim V ≤ dim W 

Hence dim V= dim W. 

Conversely suppose that dim V = dim W

Let dim V = din W = n and let

S = { x ₁, x ₂, .....,x ₙ} be a basis of W then

S ⊆ W and L(S) = W and S is L.I

W ⊆ V

Therefore, x ᵢ ∈ W ⟹ x ᵢ ∈ V

i.e. S = { x ₁, x ₂, .....,x ₙ} is a L. I subset of V.

dim V = n ⟹ every L.I. subset of V containing n is a basis of V.

Therefore S is basis of V i.e. L(S) = V

Hence W = L(S) = V. i.e. W = V