LINEARLY DEPENDENT AND INDEPENDENT SETS

 LINEARLY DEPENDENT AND INDEPENDENT SETS

A finite se S = { α₁, α₂, α₃, .......,αₙ } of vectors in a vector space V is said to be linearly dependent ( denoted by K.D) if there exist α₁, α₂, α₃, ........αₙ, not all zero such that

a₁α₁ + a₂α₂ + a₃α₃ + ......+ aₙαₙ

If however

a₁α₁ + a₂α₂ + a₃α₃ + ......+ aₙαₙ = 0 ⟹ a₁ = 0, a₂ = 0,........, aₙ = 0 then the set S = { α₁, α₂, α₃, .......,αₙ }  is said to be linearly independent. ( L.I)

From above definition of L.I of a finite set, we conclude that an infinite set of vectors of V is L.I if every finite subset is L.I because otherwise it is L.D

Example 1:

The set of vectors {(1,0,0) ,( 0,1,0) , (1,1,1) , (-1,1,-1)} of R³ 

Solution: Let

a₁(1,0,0) + a₂( 0,1,0) + a₃(1,1,1) + a₄(-1,1,-1) = (0,0,0) 

where a₁, a₂, a₃, a₄ ∈ R.

⟹ a₁ + a₃ - a₄ = 0,

a₂ + a₃ + a₄ = 0 ⟹ a₃ = a₄, a₁ = 0

a₃ - a₄ = 0,  a₂ = -2a

Taking a₄ = 1, a₃ = 1, a₂ = -2 we find that

The set is L.D

Example 2:

If F is the field of real numbers, prove that the vectors (1, 1, 0, 0), ( 0, 1, -1, 0) and (0, 0, 0, 3) in F⁴ are linearly independent.

Solution:

Let a₁, a₂, a₃ are real numbers.

Then for L.I we have

a₁(1, 1, 0, 0) + a₂( 0, 1, -1, 0) + a₃(0, 0, 0, 3) = (0, 0, 0, 0)

then a₁ + 0 + 0 = 0

a₁ + a₂ + 0 = 0

0 - a₂ + 0 = 0

0 + 0 + 3a₃ = 0

The only solution of above equation is a₁ = a₂ = a₃ = 0

Hence the given three vectors in F⁴ are linearly independent.

Example 3:

The set of vectors

{(1, 1, 1), ( -1, 0, 1) and (0, -2, 1)} in R³ is L.I

Solution:

Let a(1, 1, 1) + b( -1, 0, 1) + c(0, -2, 1) = (0, 0, 0) where a, b, c ∈ F

⟹ a - b = 0, a - 2c = 0, a + b + c = 0

⟹ a = b = c = 0

Hence the set is L.I

Example 4: The set { e₁, e₂,......,eᵢ,.....eₙ} of Rⁿ, Where eᵢ = (0, 0,....0, 1, 0, ....0) in which the ith coordinate is 1 and others are zero is L.I

Solution: 

Let a₁e₁ + a₂e₂ + ......+ aₙeₙ = 0

⟹ a₁(1, 0, ...,0) + a₂(0, 1, 0,...,0) + ....+ aₙ(0,0,...1)

= (0,0,.....,0)

⟹ ( a₁, a₂, ....,aₙ) = (0, 0, ......,0)

⟹ aᵢ = 0 for i = 1,2,.....,n.

Hence the set is L.I

THEOREMS ON LINEARLY DEPENDENT AND INDEPENDENT SETS

Theorem 1:

A set consisting of a single non zero vector L.I

Proof:

Let S = {x} where x ≠ 0

and let ax = 0 for some scalar a.

Then ax = 0 ⟹ a = 0, since x ≠ 0

Therefore S is L.I

Theorem 2:

The set {0} is L.D

Proof:

For every non zero scalar a, we have a0 = 0

so, {0} is L.D

Theorem 3:

A set of vectors containing at least one zero vector is L.D

Proof:

Let S = { x₁, x₂, x₃, ...,xᵢ.....xₙ }, where at least one of them say xᵢ = 0. Then it is clear that

0.x₁ + 0.x₂ + 0.x₃ + ...+ 0.xᵢ +.....+ 0.xₙ

Thus ∑aᵢ xᵢ= 0, where aᵢ= 1 ≠ 0

Hence S is L.D

Theorem 4:

Every subset of a L.I set, is L.I

Proof:

Let S = { x₁, x₂, x₃, .....xₙ }, be L.I

and T = { x₁, x₂, x₃, .....xₖ }, ( 1 ≤ k ≤ n) be subset of S. Then ax₁ + ax₂ + .....a ₖ xₖ+ ....+ aₙ xₙ 

⟹ a₁ = 0, a₂ = 0, ......,a ₖ = 0,.....,aₙ = 0

Let a₁ x₁ + a₂x₂ + .....+ a ₖ xₖ = 0 for some scalar a₁ ,a₂ ......, a ₖ

Then a₁ x₁ + a₂x₂ + .....+ a ₖ xₖ + 0 xₖ₊₁ + 0xₙ= 0

So a₁ = 0, a₂ = 0, ......,a ₖ = 0 Since S is L.I

Hence T is L.I

 

Theorem 5:

Every super set of a L.D set, is L.D

Proof:

Let S = { x₁, x₂, x₃, .....xₙ }, be L.D set of vectors and T = { x₁, x₂, x₃, .....xₖ, xₖ ₊ ₁, ....x ₙ } be super set of S.

Since S is L.D and so there exist scalar a₁ ,a₂ ......, a ₖ not all zero such that 

a₁ x₁ + a₂x₂ + .....+ a ₖ xₖ = 0

Therefore

a₁ x₁ + a₂x₂ + .....+ a ₖ xₖ + aₖ₊₁ xₖ₊₁ + aₙxₙ= 0

Where

 aₖ₊₁ = aₖ₊ ₂= ....... = a ₙ = 0

Thus there are scalar a₁, a₂,.....,a ₙ not all zero s.t

a₁ x₁ + a₂x₂ + .....+ a ₖ xₖ + a ₙ x ₙ = 0

Hence T is L.D

Theorem 6:

If a vector x is a linear combination of vectors x₁, x₂, x₃, .....xₙ , then the set { x₁, x₂, x₃, .....xₙ } is L.D

Proof:

Since x is linear combination of x₁, x₂, x₃, .....xₙ , so there exist scalar a₁ ,a₂ ......, a ₙ such that 

a₁ x₁ + a₂x₂ + ....+ a ₙ x ₙ 

or a₁ x₁ + a₂x₂ + ....+ a ₙ x ₙ + (-1)x = 0

Thus there exist scalar a₁ ,a₂ ......, a ₙ, a = 0 not all zero such that

a₁ x₁ + a₂x₂ + ....+ a ₙ x ₙ + ax = 0

Hence the set { x₁, x₂, x₃, .....xₙ } is L.D

Theorem 7:

If the set { x₁, x₂, x₃, .....xₘ } be L.I and the set { x₁, x₂, x₃, .....xₘ , x } be L.D , then x is a linear combination of vectors x₁, x₂, x₃, .....xₘ .

Proof:

Since the set { x₁, x₂, x₃, .....xₘ , x }is L.D, there exist scalar a₁ ,a₂ ......, a ₘ, a not all zero s.t

a₁ x₁ + a₂x₂ + ....+ aₘ xₘ + ax = 0 ......(1)

if a = 0, then

a₁ x₁ + a₂x₂ + ....+ aₘ xₘ + ax = 0

⟹ a₁ x₁ + a₂x₂ + ....+ aₘ xₘ = 0

⟹ a₁ = 0, a₂ = 0, ......, a ₘ= 0 since the set

{ x₁, x₂, x₃, .....xₘ } is L.I

Therefore a ≠ 0

so (1) can be written as

x = (- a₁/a) x₁ + (-a₂/a) x₂ + .....+ (-aₘ/a) xₘ

where a₁ /a stand for a₁a ⁻¹ etc

Hence x is linear combination of x₁, x₂, x₃, .....xₘ

Theorem 8:

A set {x₁, x₂, x₃, .....x ₙ} of non zero vectors is L.D iff there exist some vector xₖ (2 ≤ k ≤ n) of the which is linear combination of it's proceeding vectors.

Proof:

Let {x₁, x₂, x₃, .....x ₙ} be L.D. set of non zero vectors.

Consider the set {x₁}.

Since x₁ ≠ 0, the set {x₁} is L.I

Again consider { x₁, x₂ }. If it is L.I, then consider {x₁, x₂, x₃}. Proceeding in this way, let k be the first position integer (2 ≤ k ≤ n) for which the set {x₁, x₂, x₃, .....x ₖ} is L.D

Hence there exist scalar a₁, a₂, ....a ₖ not all zero such that a₁ x₁ + a₂x₂ + ....+ aₖ xₖ= 0

Here a ₖ ≠ 0 for otherwise, if a ₖ ≠ 0 , the set {x₁, x₂, x₃, .....xₖ₋₁ } will be L.D contradicting the definition of k .

Hence 

xₖ= (- a₁/aₖ) x₁ + (-a₂/aₖ) x₂ + ....+ (-aₖ₋₁/aₖ) xₖ₋₁

where 1/aₖ= aₖ₋₁ is the multiplicative inverse of aₖ in the field and a₁/aₖ = a₁.aₖ₋₁ etc

This shows that aₖ is linear combination of preceding vectors x₁, x₂, x₃, .....xₖ₋₁

Conversely suppose that xₖ (2 ≤ k ≤ n) is a linear combination of it's preceding vectors

x₁, x₂, x₃, .....xₖ₋₁ so that

x = a₁ x₁ + a₂x₂ + ....+ aₖ ₋₁ xₖ₋₁= 0

or a₁ x₁ + a₂x₂ + ....+ aₖ ₋₁ xₖ₋₁ + aₖ xₖ= 0

where aₖ = -1 ≠ 0

This shows that {x₁, x₂, x₃, .....xₖ } is L.D

Since every super set of L.D set is L.D, it follow that {x₁, x₂, x₃, .....xₙ } is L.D

Theorem 9:

The set {x₁, x₂, x₃, .....x ₙ} of non zero vectors is L.D iff at least one of these vectors is linear combination of remaining ( n- 1) vectors.

Proof:

By above theorem, the set of non zero vectors {x₁, x₂, x₃, .....x ₙ} is L.D iff there exists aₖ (2 ≤ k ≤ n) which is linear combination of it's preceding vectors i.e.

xₖ = a₁ x₁ + a₂x₂ + ....+ aₖ ₋₁ xₖ₋₁

i.e 

xₖ = a₁ x₁ + a₂x₂ + ....+ aₖ ₋₁ xₖ₋₁ + 0xₖ₊₁ + 0xₙ 

i.e xₖ is linear combination of the remaining (n - 1) vectors.