Linear Dependence and Linear Independence: Example Solved Problems

LINEAR DEPENDENCE AND LINEAR INDEPENDENCE

WORKED EXAMPLE PROBLEMS

Example 1

Verify whether the following given sets are linearly dependent or linearly independent.

Solution:

(i) To show that it is linearly independent, we must find scalars a₁ and a₂ such that a₁u₁ + a₂u₂ = 0.

 a₁− a₂=0        ……. (1)

 −2a₁ + a₂ = 0        ……. (2)

−a₁+2a₂=0        ……. (3)

 4a₁−4a₂=0        ……. (4)

From equations (1) and (4), we have a₁ = a₂     ⇒ a₁ = 0 and a₂ = 0.

  a₁ = 0 and a₂ =0 satisfy all the above 4 equations.

Hence u₁ and u₂ are linearly independent.

(ii)

 u₂ = −2u₁

Here u₁ and u₂ are linearly dependent.

(iii) Let a₁u₁+ а₂u₂ + a₃u₃ + a₄u₄ = 0 where

 a₁−a₃+2a₄=0       ……(1)

−α₂+2a₃ +α₄=0       ……(2)

−2a₁ + a₂+ a₃ +2a₄ = 0       ……(3)

 a₁+a₂−2a₄=0       ……(4)

By solving these above four equations, we get

 a₁ = 0, a₂ = 0, a₃ = 0 and a₄ = 0.

. The given sets are linearly independent.

Example 2

The set of diagonal matrices of M_2×2(F) is a subspace. Find a linearly independent set that generates this subspace.

Solution:

Let S = {∈ M_2×2: a,b ∈ F }

To prove that it is linearly independent set that generates S.

Let us first find the set that generates S.

Let u ∈ S then u = 

The vector u can be written as 

 S = a₁u₁ + a₂u₂

The set generates S is {,}

Now we have to verify that the sets  and  are linearly independent or not.

Hence a₁ =0 and a₂=0 ⇒ a₁u₁ + а₂u₂ = 0.

  The given sets are linearly independent.

Example 3

Prove that if {A₁, A₂, A₃ ...  A_k} is a linearly independent subset of M_n×n (F), then {A₁^t, A₂^t, A₃^t, A₄ ^t ... A_k^t } is also linearly independent.

Solution:

Let A₁, A₂, A₃ ... A_k ∈ M_n×n (F).

Then a₁A₁+ a₂ A₂ + a₃A₃ + ... + a_kA_k = 0

which means a₁ = 0, a₂ = 0... a_k = 0.

Let a₁, a₂, a₃... a_k be the scalars such that

 a₁A₁^t + a₂A₂^t + a₃A₃^t + ... + a_kA₁^t = 0

By taking transpose on both sides we get

[ a₁A₁^t + a₂A₂^t + a₃A₃^t + ... + a_kA₁^t ]^t = 0

⇒ a₁(A₁^t)^t + a₂(A₂^t)^t + a₃(A₃^t)^t + …. + a_k(A_k^t)^t  = 0

⇒ a₁А₁ + a₂ A₂ + a₃А₃ + ... a_kA_k = 0

That means a₁ = a₂ = a₃ = a₄ …… = a_k = 0, since { A₁, A₂ ... A_k} is a linearly independent.

 { A₁^t, A₂^t, A₃^t ... A^t_k} is linearly independent.

Example 4

Let us consider the set S={(1, 3, −4,2), (2, 2, − 4, 0), (1, 3, 2, 4), (−1, 0, 1, 0)} in R⁴. Show that S is linearly dependent and express one of the vectors in S as a linear combination of the other vectors in S.

Solution:

To show that S is linearly dependent, we must find scalars a₁, a₂, a₃ and a₄ not all zero such that

 a₁(1,3,−4, 2) + a₂(2, 2, −4, 0) + a₃(1, −3, 2,−4) + a₄(−1, 0, 1, 0) = 0.

  (a₁, 3a₁, − 4a₁, 2a₁) + (2a₂, 2a₂, − 4a₂, 0) + (α₃, − 3a₃, 2a₃, − 4a₃) + (−α₄, 0, α₄, 0)=0

Finding such scalars amounts to find a non−zero solution to the system of linear equations

a₁+2a₂+ a₃−a₄=0

3a₁+2a₂−3a₃+0α₁ =0

−4a₁−4a₂+2a₃+a₄ = 0

 2a₁+0a₂ − 4a₃+0a₄=0

Solving these 4 equations for a₁, a₂, a₃ and a₄ we get

 a₁=4, a₂=−3, a₃ = 2 and a₄ = 0.

Thus S is a linearly dependent sub set of R⁴ and

 4 (1, 3, ‒4, 2) ‒3 (2, 2, ‒4, 0) + 2(1,−3, 2,−4) + 0(−1, 0, 1, 0) = 0

 (4, 12, ‒16, 8) + (−6, −6, 12, 0) + (2, ‒6,4,−8) = 0

 ⇒ 4−6+2=0, 12−6−6=0,− 16+ 12+4=0 and 8+0−8=0

Example 5

Verify whether the given set in M_2×3(R) is linearly dependent or not.

Solution:

To show that S is linearly dependent, we must find scalars a₁, a₂, a₃ such that

 а₁ − 3а₂ − 2а₃ = 0

−3a₁+7a₂ + 3a₃ = 0

2a₁ +4a₂+11a₃ = 0

−4a₁+6a₂−a₃ = 0

0a₁−2a₂−3a₃=0

 5a₁−7a₂+2a₃ = 0

By solving these 6 equations we get the values as

 a₁=5, a₂ = 3, a₃ = −2.

It is linearly dependent.

Example 6

Prove that the set

S={(1, 0, 0, −1), (0, 1, 0, −1), (0, 0, 1, −1), (0, 0, 0, 1)} is linearly independent.

Solution:

Now we must show that the only linear combinations of vectors in S that equals the zero vector is the one in which all the coefficients are zero.

Suppose that a₁, a₂, a₃ and a₄ are scalars such that

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

 (a₁, 0, 0, −a₁) + (0, a₂, 0, − a₂) + (0, 0, a₃, −a₃) + (0, 0, 0, a₄) = (0, 0, 0, 0)

⇒ a₁=0; a₂=0; a₃=0

 −a₁−a₂−a₃ + a₄ =0

⇒ а₄=0.

Clearly the only solution to this system is

 a₁ =0, a₂ = 0, a₃=0 and a₄ = 0 and so S is linearly independent.

Example 7

For k = 0, 1, 2, 3, ... n.

Let P_k(x) = x^k + x^k+1  + x ^k+2  + ...+ xⁿ. Then prove that the set {P₀(x), P₁(x), P₂(x), ... P_n(x)} is linearly independent in P_n(F).

Solution:

Let us consider

 a₀P₀(x) + a₁P₁(x) + a₂P₂(x) +...+ a_nP_n(x) = 0

for some scalars a₀, a₁, a₂, ... a_n.

Then

 a₀+(a₀+a₁)x + (a₀+a₁+a₂)x² + (a₀+a₁+a₂+a₃)x³ + … + (a₀ + a₁ + a₂ ... + a_n) xⁿ = 0.

By equating the coefficient of x^k on both sides of this equation for k = 1, 2, 3 ... n, we obtain    

 a₀ = 0, a₀ + a₁ = 0, a₀ + a₁ + a₂ = 0... and so on.

a₀ + a₁  + a₂ + …. + a_n =0

Clearly the only solution to this system of linear equation is a = 0, a₁ = 0, a₂ = 0 ... a_n = 0 and so it is linearly independent.

Example 8

If V=P (x) is a vector space over F. Let S = {1, 1+x, 1 + x + x² }. Check whether S is linearly independent or not.

Solution:

a (1) + b (1 + x) + c (1+x+x²)=0

c + a + b + bx + cx + cx² = 0x²+0x +0

(a+b+c) + (b + c) x + cx² = 0x²+0x+0.

Comparing the coefficients on both sides, we have

 a+b+c=0; b+c=0; c=0.

⇒  a=0;b=0; c=0

  S is linearly independent

Example 9

Let V=R⁴. Prove that V₁ = (1, 0, 1, 0); v₂ = (0, 1, 0, 1), v₃ = (0, 0, 0, 2) are linearly independent.

Solution:

We know that av₁+bv₂+cv₃=0

a(1, 0, 1, 0) + b(0, 1, 0, 1) + c(0, 0, 0, 2) = 0

(a, 0, a, 0) + (0, b, 0, b) + (0, 0, 0, 2c) = (0, 0, 0, 0)

(a, b, a, b + 2c) = (0, 0, 0, 0)

a=0; b=0; b+2c=0.

⇒ c = 0

 a=0=b=c

The vectors v₁, v₂, v₃ are linearly independent.

Example 10

Verify whether the following vectors are linearly dependent or linearly independent v₁ = (2, 4, 14), v₂ = (7, −3, 15), v₃ = (− 1, 4, 7)

Solution:

We know that av₁ + bv₂+cv₃ = 0

 a(2, 4, 14) + b(7,−3, 15) + c (−1, 4, 7) = (0, 0, 0)

 (2a, 4a, 14a) + (7b, − 3b, 15b) + ( ‒c, 4c, 7c) = (0, 0, 0)

2a+7b−c=0

4a−3b+4c=0

14a+15b+7c=0

 = 68 ≠ 0

The vectors v₁, v₂, v₃ are linearly independent.