Vector Spaces: Theorems Part 6 - Example Solved Problems

Vector Spaces: Theorems Part 6

Example Problems

Example 11

Let V=R³ and S₁ = {(1, 0, 0), (2, 2, 0), (5, 7, 2)}. Show that S₁ is a minimal generating set.

Solution:

 x(1, 0, 0) + y(2, 2, 0) + z (5, 7, 2) = (a, b, c)

(x, 0, 0) + (2y, 2y, 0) + (5z, 7z, 2z) = (a, b, c)

⇒  x+2y+5z=a, 0+2y+7z=b & 0+0+2z = c

 x+2y+5z=a             ...(1)

2y+7z b                   ...(2)

2z = c                      ...(3)

From equation (3), we have z = c/2.

From equation (2); 2y = b − (7c/2)

y = (2b−7c) / 4

From equation (1); x=a−2y−5z

= a − 2(2b−7c / 4) − 5(c/2)

= (4a−4b+ 14c − 10c) / 4

= (4a−4b+ 4c) / 4

x = a−b+ c

L(S₁) = V.

S₂ = {(1, 0, 0) } is a proper subset of S₁.

then

L (S₂) = { α (1, 0, 0)}  → x - axis

S₃ = {(2,2, 0)}

L(S₂) = {α (2, 2, 0)) → xy plane

Hence S₁ is a minimal generating set.

Example 12

Determine whether the vectors v₁ = (1, − 2, 3), v₂ = (5, 6, − 1), v₃ = (3, 2, 1) form a linearly dependent or linearly independent set in R³.

Solution:

We know that

 av₁ + bv₂ + cv₃ = 0

a(1, −2,3) + b(5, 6−1) + c(3, 2, 1) = 0

(a,− 2a, 3a)+(5b, 6b,− b) + (3c, 2c, c) = 0

a+5b+3c=0

 −2a + 6b+2c=0

 3a−b+c=0

= 1 [6+2]−5 [−2−6]+3 [2−18]

= 1 [8] − 5[8] + 3[16]

= 8+40−48

= 48−48=0

The given vectors are linearly dependent.

Example 13

In a matrix M_m×n (F), let E^ij denote the matrix whose only non−zero entry is 1 in the ith row and jth column. Show that {E^ij/1≤i≤m,1≤j≤n} is linearly independent.

Solution:

Let S = {E^ij/1≤i≤m,1≤j≤n}

Given that E^ij is a matrix whose only non−zero entry is 1 in the i^th row and j^th column. All the remaining entries in that matrix are zero.

Let us consider the matrices

The show that S is linearly independent, we must find scalars

 a₁₁, a₁₂, a₁₃, a₁₄ …..a_mn

Let the linear combination of the elements of S as

  a₁₁E¹¹ + a₁₂E¹² + a₁₃E¹³ + ….. + a_mnE^mn = 0

By adding all the above matrices we get

Equating the corresponding terms on both sides, we get the only solution is

a₁₁=0, a₁₂=0, a₁₃=0... a_mn = 0

  S is linearly independent,

Example 14

Prove that the set {1,x,x²...xⁿ } linearly independent in P_n(F).

Solution:

Let us consider P_n(F) = { a₀ + a₁x + a₂x² + a₃x³ + … + a_nx^x }

  where a₁, a₂, α³... a_n ∈ F.

Let S = {1, x, x², x³... xⁿ }

suppose that v₁ = 1 = (1, 0, 0, 0 ... 0)

v₂=x=(0, x, 0, 0... 0)

v₃ = x²= (0, 0, x², 0, 0, ... 0)

v₄ = x³ = (0, 0, 0, x³,0, ... 0)

v_i = xⁱ = (0, 0, 0, ..... 0, xⁱ, 0 ...... 0)

:

:

v_n=xⁿ = (0, 0, 0, 0, ... 0, xⁿ)

Suppose that a₁, a₂, a₃ ... a_n are scalars such that

a₁v₁ + a₂v₂ +  a₃v₃ +…. + a_nv_n=0.

a₁ (1, 0, 0 ... 0) + a₂ (0, x, 0 ... 0) + ... + a_n (0, 0, 0 ... xⁿ) = 0

⇒ (a₁, 0, 0 ... 0) + (0, a₂x, 0 ... 0) + (0, 0, a₃x², 0... 0) + (0, 0, 0, a₄x³, 0 ... 0) + ... + (0, 0, ... a_nx^n-1)=0

⇒  a₁=0, a₂ = 0, a₃ =0,…. a_n = 0

Therefore v₁, v₂, v₃ …. v_n are linearly independent

Hence S={1, x, x², x³, ...xⁿ} is linearly independent.

Example 15

In Fⁿ, let e_j denote the vector whose j^th coordinate is 1 and whose other coordinates are 0. Prove that {e₁, e₂, e₃... e_n } is linearly independent.

Solution:

In Fⁿ, let us consider

 e₁ = (1, 0, 0 ... 0), e₂ = (0, 1, 0... 0), e₃ = (0, 0, 1, 0 ... 0)

 e₄ = (0, 0, 0, 1, 0... 0) and in general, e_n=(0, 0, 0... 0, 1)

If a₁, a₂, a₃... a_n are scalars then

 a₁е₁ + a₂е₂ + a₃e₃ + ... + a_ne_n = 0

 a₁(1, 0, 0... 0) + a₂ (0, 1, 0... 0) + a₃(0, 0, 1, 0... 0) + a₄(0, 0, 0, 1, 0... 0) + ... + a_n(0, 0, ... 1) = 0

⇒ (a₁, 0... 0) + (0, a₂, 0, ... 0) + (0, 0, a₃ ... 0) + (0, 0, 0, a₄, 0... 0) + ... + (0, 0... a_n)=0

⇒ (a₁, a₂, a₃... a_n) = (0, 0, 0... 0) = 0.

  а₁ = 0, a₂ = 0, a₃ = 0 ... a_n = 0

Hence e₁, e₂, e₃ ... e_n are linearly independent.

Example 16

Let S = {(1, 1, 0), (1, 0, 1), (0, 1, 1)} be the subset of F³.

(i) Prove that if F=R, then S is linearly independent.

(ii) Prove that if F has characteristic 2, then S is linearly dependent.

Solution:

Given that S={(1, 1, 0), (1, 0, 1), (0, 1, 1)).

(i) To prove S is linearly independent, let us consider a₁, a₂ and a₃ are scalars such that

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

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

a₁+ a₂ = 0        ………..(1)

a₁+a₃= 0        ………..(2)

a₂+a₃=0        ………..(3)

By solving these above three equations.

We get a₁=0, a₂ =0 and a₃ = 0.

Hence (1, 1, 0),(1, 0, 1), (0, 1, 1) are linearly independent in R³.

(ii) Suppose that F has characteristic 2, then let us consider a₁ = 1, a₂ = 1 and a₃ = 1.

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

= (1, 1, 0) + (1, 0, 1) + (0, 1, 1)

= (2, 2, 2)

≠ (0, 0, 0)

The linear combinations

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

has non−trivial solution a₁ = 1, a₂ = 1 and a₃ = 1.

  The given set S={(1, 1, 0) (1, 0, 1) (0, 1, 1) } are linearly dependent.

Example 17

Verify the following sets P₃(R) are linearly dependent or linearly independent.

 (i) { x³ + 2x², − x² + 3x + 1, x³− x²+2x−1}

 (ii). {x³ −x, 2x²+4, −2x³+3x²+2x+6 }

Solution:

 (i) Let us consider v₁ = x³ +2x²

 v₂ = − x² + 3x + 1

 v₃ = x³− x²+2x−1

 Suppose that a₁, a₂, a₃ are scalars such that a₁v₁ + a₂ v₂ + a₃v₃ = 0

 a₁ (x³ + 2x²) + a₂ (− x² + 3x + 1) + a₃ (x³ − x²+2x−1)=0

 a₁x³ + 2α₁x² − a₂ x² + 3a₂x + a₂ + a₃x³ – a₃x² + 2a₃x – a₃ = 0

 x³[a₁ + a₃] + x²[2a₁−a₂−a₃] +x [3a₂+2a₃] + (a₂ − a₃) = 0

 By Comparing the coefficients of x on both sides, we have

 a₁+a₃=0          ... (1)

2a₁−a₂−a₃=0          ... (2)

3α₂+2a₃ = 0          ... (3)

a₂−a₃ = 0          ... (4)

Solving these equations we get a₁ =0, a₂ = 0 and a₃=0

Hence the given set is linearly independent.

(ii) Let us consider v₁ =x³−x,

 v₂=2x²+4

 v₃ = −2x³ +3x²+2x+6

Suppose that a₁, a₂, a₃ are scalars such that a₁v₁ + a₂v₂ + a₃v₃ = 0

 a₁ (x³ − x) + a₂ (2x² + 4) + a₃(− 2x³ + 3x² + 2x + 6) = 0

a₁x³ − a₁x + 2a₂x² + 4a₂ − 2a₃x³ + 3a₃x²+2a₃x+6a₃ = 0

 x³[a₁ − 2a₃] + x² [2a₂ + 3a₃] + x[− a₁ +2a₃] + [4a₂ +6a₃]=0

By comparing the coefficients of x on both sides, we have

   a₁−2a₃=0             ……..(1)

   2a₂+3a₃=0             ……..(2)

   −a₁+2a₃=0             ……..(3)

  4a₂+6a₃=0             ……..(4)

From (1) and (3): a₁−2a₃=0 ⇒  a₁=2a₃

From (2) and (4): 2a₂+3a₃ = 0  ⇒   2a₂ = −3a₃

     a₁ = 2a₃ and 2a₂=−3a₃

If we put a₃ = 2:

Then a₁ = 4, a₂ = −3 and a₃=2.

 a₁v₁ + a₂v₂ + a₃v₃ = 0

⇒  4v₁−3v₂+2v₃ = 0

Hence, the set V₁, V₂, V₃ are linearly dependent.

Example 18

Verify whether the given sets in R³ are linearly dependent or linearly independent.

(i) {(1, −1, 2), (1, −2, 1), (1, 1, 4)}

(ii) {(1,−1,2), (2, 0, 1), (− 1, 2, −1)}

Solution:

(i) Let v₁ = (1, 1, 2), v₂ = (1, − 2, 1), v₃ = (1, 1, 4)

We know that a₁v₁ + a₂v₂ + a₃ν₃=0

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

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

 a₁ + a₂ + a₃=0              ………(1)

 −a₁−2a₂+a₃=0              ………(2)

 2α₁ + a₂ + 4a₃ = 0              ………(3)

From (1) and (2) ⇒ (1) + (2)

⇒ − a₂+2a₃ = 0

⇒   a₂=2a₃              ………(4)

Substitute equation (4) in (3)

  2a₁+2a₃+4a₃=0

  a₁=−3a₃               ……….(5)

Put a₃ = 1; then a₁ = −3, a₂ = 2

⇒  −3v₁ + 2v₂ + v₃ = 0

Hence v₁, v₂, v₃ linearly dependent.

(ii) Let v₁ = (1,−1, 2), v₂ = (2, 0, 1) and v₃ = (−1,2,−1)

We know that a₁v₁ + a₂v₂ + a₃v₃ = 0

 a₁(1, − 1, 2) + a₂(2, 0, 1) + a₃(− 1, 2, − 1) = 0

  (a₁,−a₁, 2a₁) + (2a₂, 0, a₂) + (−a₃, 2a₃, −a₃) = 0

  a₁+2a₂−a₃=0             ... (1)

 −a₁+2a₃=0             ... (1)

 2α₁+a₂−a₃=0         ….(3)

By solving these 3 equations, we get a₁ =0, a₂ =0 and a₃ = 0.

Hence the sets v₁, v₂, v₃ are linearly independent.

Example 19

Determine the given set in P₄(R) is linearly dependent or linearly independent for x⁴−x³+5x²−8x+6, −x⁴+x³−5x²+5x−3, x²+3x² − 3x+5 and 2x⁴+x³+4x² + 8.x.

Solution:

Let us consider

v₁ = x² − x³ +5x² − 8x+6

v₂ = −x⁴ +x³ − 5x² + 5x−3

v₃ = x²+3x²−3x+5

v₄ = 2x²+x³ + 4x² + 8x

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

 a₁v₁ + a₂v₂ + a₃v₃ + a₄v₄ = 0

⇒ a₁ (x⁴ − x³ +5x² − 8x+6) + a₂ (− x² + x³− 5x² + 5x − 3) + a₃(x⁴ + 3x² − 3x+5) + a₄(2x⁴+x³ + 4x² + 8x) = 0

 ⇒ a₁x⁴ − a₁x³ + 5a₁x² − 8a₁x + 6α₁ − a₂x⁴ + a₂x³−5a₂x² + 5a₂x − 3α₂+ a₃x⁴ + 3a₃x² − 3a₃x² + 5a₃ +2a₄x⁴ +α₄x³ +4α₄x² + 8а₄x = 0

 x⁴ [a₁ − a₂+ a₃ +2a₄] + x³[− a₁ + a₂+a₄] + x² [5a₁ − 5a₂+ 3a₃ + 4a₄] + x[− 8a₁ +5a₂ − 3a₃ + 8a₄] + [6α₁−3α₂+5α₃] = 0

Comparing the coefficients on both sides we get

a₁−a₂+ α₃+2a₄ = 0      ...(1)

−α₁+ a₂+ 0а₃ + a₄ = 0      ...(2)

5a₁−5α₂+3α₃ + 4a₄ =0      ...(3)

−8a₁+5a₂−3a₃ + 8a₄ =0      ...(4)

6a₁−3a₂+5a₃+0α₄=0      ...(5)

From (1) + (2): a₃+3a₄=0

⇒ a₃ = −3a₄                 ...(6)

From (3) + (4): −3a₁+12a₄ = 0

⇒ a₁ = 4a₄                 ...(7)

From (2): −4a⁴ + a₂+a₄ = 0.

 а₂=3а₄                 ...(8)

By solving these equations and putting these values

 a₁=4, a₂=3, a₃=−3 & a₄ = 1.

Hence, the given set in P₄(R) is linearly dependent.