Vector Spaces: Theorems Part 3 - Example Solved Problems

Vector Spaces: Theorems Part 3

Example Problems

Example 15

Prove that a subset W of a vector space V is a subspace of V iff L(W) = W.

Solution:

Given that W is a subspace of V.

To prove L (W) = W, let x ∈ L(W), such that x_i ∈ W, α_i ∈ F.

 x=Σα_ix_i for  i = 1, 2, 3 ... n

 x=Σα_ix_i ∈ W.

Since W is a subspace of V, it is closed under addition and multiplication.

 L(W)  W and W  L (W)

  L (W) = W.

Hence the proof.

Example 16

Examine the following vectors are in the span S.

(i) (2,−1, 1) and S={(1, 0, 2), (− 1, 1, 1) }

 (ii) 2x³−x²+x+3 and S={x³+x²+x+1,x²+x+1,x + 1 }

Solution:

(i) Let consider (x, y) ∈ V.

 x = α (1, 0, 2) + β(‒1, 1, 1)

where x = (2, ‒1, 1), v₁ = (1, 0, 2) and v₂ = (− 1, 1, 1)

 (2, 1, 1) = α(1, 0, 2) + β(− 1, 1, 1)

⇒ (2,−1,1) = (α, 0, 2α) + (−β, β, β)

  α−β=2; 0+β=−1 and 2α + β=1

  α − β =2          ... (1)

 β = −1         ... (2)

 2a+β=1         ... (3)

From (2): β=−1

Then from (1): α+1=2 ⇒ α = 1

From (3): 2a+β=1  ⇒ 2(1)+(−1) = 1

  Equations (1), (2) and (3) are satisfied.

  The given vector x is in the span of S.

 (ii) Let x=2x³− x²+x+3

 v₁ = x³ + x² + x + 1; v₂ = x²+x+1 and v₃ = x+1

  x = αv₁ + βv₂ + γv₃

 2x³− x² + x + 3 = α (x³ + x² + x + 1) + β(x² + x + 1) + γ(x + 1)

 2x³−x²+x+3 = αx³ + αx² + αx + α + βx² + βx + β + γx + γ

 = x³(α) + x²(α + β) + x (α + β + γ) + (α+β+γ)

Equating the coefficients on both sides we get

 α=2        ………(1)

 α+β = −1        ………(2)

  α+β+γ=1        ………(3)

  a+β+γ=3        ………(4)

From (1) we have α=2.

Put α=2 in equation (2). We get β =−3

Substitute α = 2 and β=−3 in equation (3),

 a+β+γ=1

⇒ 2−3+γ=1 ⇒ γ =2.

Now substitute α=2, β=−3 and γ=2 in equation (4).

  α+β+γ=3

⇒ 2−3+2 ≠ 3.

  The given vector is not in span of S.

α+γ= 1          ……….(1)

 β+γ=2          ……….(2)

 −α=−3           ……….(3)

 β=4          ……….(4)

From (3) and (4), α=3 and β=4.

From equation (1), α+γ='1 ⇒ γ=−2

From equation (2), β+γ=2   ⇒ 4−2=2

 All the equations are satisfied.

Hence the given vector is in span of S.

(iv) 

 α+γ= 1          ……….(1)

 β+γ=0          ……….(2)

 −α=0           ……….(3)

 β=1          ……….(4)

From equations (3) and (4), we have α=0 and β=1

From equation (1): α+γ=1  ⇒  0+γ=1  ⇒ γ=1

From equation (2): β+γ=0    ⇒   1+1 ≠ 0

The given vector is not in span of S.

Example 17

Determine whether the set of vectors X₁ = (1, 1, 2), X₂ = (1, 0, 1), X₃ = (2,1,3) span R³.

Solution:

Let us consider (x, y, z) ∈ V.

 (x, y, z) = α(1, 1, 2) + β(1, 0, 1) +  γ(2, 1, 3).

  We must determine whether an arbitrary vector (x, y z) in R³ can be expressed as a linear combination (x, y, z) = αX₁+βX₂+γX₃ of the vectors X₁, X₂ & X₃.

 (x, y, z) = (α, α, 2α) + (β, 0, β) + (2γ, γ, 3γ)

⇒ a+β+2γ=x

  α+0β+γ=y

 2α+β+3γ=z

This problem is reduced to determining whether this system of equations is consistent for all values of x, y & z.

The system will be consistent for all x, y, z the matrix of coefficient,

But A =  is invertible.

But A =  =1[0−1] −1[3−2] + 2[1−0]

= 1[− 1] − 1[1] + 2[1]

= −1−1+2=0

So the matrix A is not invertible.

Hence, X₁, X₂ & X₃ do not span R³.

Example 18

Verify whether the vectors u = (1, 2, 3) ; v = (0, 1, 2); w = (0, 0, 1) generates R³ or not.

Solution:

 Let (x, y, z) ∈ R³, and a, b, c ∈ F.

 (x, y, z) = au + by + cw

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

 = (a, 2a, 3a) + (0, b, 2b) + (0, 0, c)

(x, y, z) = (a, 2a + b, 3a+2b+c)

Comparing the corresponding terms on both sides, we get

a = x;

2a + b = y;

b=y−2x

3a+2b+c=z

c=z−2(y−2x)−3x.

c=z−2y+x

Hence any vector in R³ can be generated by u, v, w.

Example 19

Verify whether the matrices  generates M_2×2 (R) or not.

Solution:

Where a₁₁, a₁₂, a₂₁, a₂₂ ∈ F = R

Comparing on both sides

 a₁₁ = a+b+c; a₂₁ = c

 a₁₂=b; a₂₂=a+b+c

⇒ a₁₁=a₂₂

 The given matrices do not generate M_2×2(R) because each of these matrices has equal diagonal entires. So any linear combination of these matrices has equal diagonal entries.

Hence every 2×2 matrix is not a linear combination of these three matrices.