Vector Spaces: Theorems Part 7 - Example Solved Problems

Vector Spaces - Theorems Part 7

Example Problems

 

Example

Verify that the following sets are bases for R³ or not.

(i) {(1, 0, −1), (2, 5, 1), (0, −4, 3)}

(ii) { 1, − 3, − 2), (− 3, 1, 3), (− 2, − 10, − 2) }

Solution:

(i) Let v₁ = (1,0,− 1), v₂ = (2, 5, 1) and v₃ = (0,−4, 3)

To verify that the set is bases for R³,

Let us verify the linear independence.

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

a₁ (1, 0, −1) + a₂ (2, 5, 1) + a₃ (0,− 4, 3) = (0, 0, 0)

(a₁, 0, −a₁)+(2α₂, 5a₂, α₂) + (0, −4a₃, 3α₃) = (0, 0, 0)

a₁+2a₂ = 0             …... (1)

5a₂−4a₃ = 0       ……. (2)

−α₁+ a₂+3a₃ = 0         ... (3)

By solving these three equations, we get

 a₁ =0, a₂ =0 and a₃=0

The given set is linearly independent.

We know that any linearly independent subset of V that contains exactly n vectors is a basis for V.

 Hence the set is a basis for R³.

(ii) Let v₁ = (1,− 3,−2), v₂ = (− 3, 1, 3) and v₃=(−2, −10, −2).

Let a₁v₁ + a₂v₂ + a₃v₃ = 0

 a₁(1, −3, −2) + a₂(−3, 1, 3) + a₃(−2,−10, −2) = (0, 0, 0)

 (a₁, −3a₁, − 2a₁) + (−3a₂, α₂, 3a₂) + (− 2a₃, − 10a₃, − 2a₃) = (0, 0, 0)

a₁−3a₂−2a₃=0      ………..(1)

−3a₁+ a₂ − 10a₃ = 0      ………..(2)

− 2α₁ +3α₂−2a₃ = 0      ………..(3)

Now the equation (1)×3; 3a₁−9a₂−6a₃ = 0

From (4) and (5); a₂+2a₃ =0

⇒ a₂=2a₃

 a₂ / ‒2 = a₃ / 1

 a₂=−2 and a₃ = 1. Then a₁ = −4.

Hence the given sets are linearly dependent.

The given set is not a basis for R³.

 

Example 6

Let S = {(2,−3, 1), (1, 4, −2), (−8, 12,−4), (1, 37, −17), (−3, −5, 8)} Show that S is a basis for R³.

Solution:

Let v₁ = (2,3, 1), v₂ = (1,4,−2)

v₃ = (−8, 12, 4), v₄ = (1, 37, −17) and v₅ = (−3,−5, 8)

The given vectors of triplets of real numbers. They generate R³ if they are linearly independent.

But the dimension of R³ is only 3.

Randomly, we can select three vectors and then we can show that they are linearly independent degenerate R³.

Let us consider S = { v₁, v₃, v₅ }

Then a₁v₁+ a₂v₃ + a₃v₅ = 0

 a₁(2, −3, 1) + a₂(−8, 12,−4)+ a₃ (−3, −5, 8) = (0, 0, 0)

(2a₁, 3a₁, a₁) + (−8a₂, 12a₂, −4a₂) + (−3a₃, −5a₃, 8a₃) = 0

2a₁−8α₂−3a₃ = 0            .. (1)

− 3a₁+12a₂−5a₃=0           …….(2)

 a₁ −4a₂+8a₃ = 0            …….(3)

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

.S= {v₁, v₃, v₅ } is linearly independent and forms a basis.

 

Example 7

Let S={(2, 3, 5), (8, 12, 20), (1, 0, −2), (0, 2, 1), (7, 2, 0) }.

Show that S is a basis for R³.

Solution:

It is given that v₁ = (2,−3, 5), v₂= (8,−12, 20), v₃ = (1,0,−2), v₄= (0,2−1) and v₅ = (7, 2, 0).

Let us select a basis for R³ that is a subset of S.

Let us select v₁ = (2, 3, 5) to be a vector in the basis.

Since 4v₁ = 4(2,−3, 5) = (8, 12, 20) = v₂,

v₁ and v₂ are linearly dependent.

 We should not consider v₂ = (8, 12, 20) as a set. Hence v₁, v₃, v₄ and v₅ are the remaining set of vectors. Since this set belongs to R³, we can consider any three among v₁, v₃, v₄ and v₅.

As a random, let us consider S = { v₁, v₃, v₄ }

Then

 a₁v₁ + a₂ v₃ + a₃v₄ = 0

a₁ (2,−3, 5)+a₂ (1, 0, −2)+ a₃ (0,2,−1)=(0, 0, 0)

 (2a₁,−3a₁, 5a₁) + (a₂, 0, − 2a₂) + (0, 2a₃, ‒ a₃) = (0, 0, 0)

 2a₁ + a₂ = 0       …….(1)

−3α₁+2a₃ = 0      …….(2)

5α₁−2a₂−a₃=0      …….(3)

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

S = {v₁, v₃, v₄ } is a subset and linearly independent.

The set S forms a basis for R³

 

Example 8

Verify whether S={(1, 4, 6), (1, 5, 8), (2, 1, 1), (0, 1, 0)} is linearly independent or not on R³.

Solution:

Let S = {(1, 4, 6), (1, 5, 8), (2, 1, 1), (0, 1, 0)}

Let v₁ = (1,4,−6), v₂ = (1, 5, 8), v₃ = (2, 1, 1) and v₄ = (0, 1, 0)

Then a₁v₁ + a₂v₂ + a₃v₃ + a₄v₄ = 0

  a₁(1, 4, − 6) + a₂(1, 5, 8) + a₃(2, 1, 1) + a₄(0, 1, 0) = (0, 0, 0)

 (a₁, 4a₁, − 6a₁) + (a₂, 5a₂, 8a₂) + (2a₃, a₃, a₃) + (0, a₄, 0) = (0, 0, 0)

 a₁+ a₂+2a₃=0      ... (1)

4a₁ + 5a₂ + a₃ +4 =0      ... (2)

 −6α₁+8a₂+ a₃ = 0      ... (3)

Let us put a₄ = 1 in equation (2). Then

 a₁+ a₂+2a₃ = 0          .. (1)

 4a₁ +5α₂+a₃ = −1          .. (2)

 −6a₁+8a₂+ a₃ = 0          .. (3)

Solving these equations, we get

a₁ = ‒ 5/37

a₁ = − 13/111

a₁ = 14/111

and a₄ = 1.

Hence the given sets are linear dependent on R³.

 

Example 9

Verify, whether the given polynomials

{x³−2x²+1, 4x²−x+3, 3x−2} generate P₃(R) or not.

Solution:

Given that the sets are x³ − 2x²+1, 4x²−x+3 and 3x−2.

All the three polynomials have degree less than or equal to 3.

The given sets of polynomials belong to P₃ (R).

But, generally the given set of vector in P_n(R) has the dimension n+1.

.. The above given three polynomials must have the dimension 4.

. The given set of polynomials cannot span P₃(R).

Thus the given polynomials cannot generate P₃(R).

 

Example 10

Verify the following set of vectors in R³ forms a basis or not? S={v₁, v₂, v₃} where v₁ = (1, 0, − 1), v₂ = (2, 5, 1) & v₃ = (0, − 4, 3).

Solution:

To verify these vectors are linearly independent choose a, b, c are scalars such that.

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

(a, 0,− a)+(2b, 5b, b) + (0, −4c, 3c) = (0, 0, 0)

⇒ a+2b+0=0     ...(1)

 0+5b−4c=0      ...(2)

− a+b+3c=0      ...(3)

By solving these three equations (1) (2) & (3) we get

  a=0, b=0 & c = 0.

  The vectors are linearly independent.

Let (x, y, z) ∈ R³

(x, y, z) = a (1, 0, −1) + b (2, 5, 1) + c (0, −4, 3)

(x, y, z) = (a, 0,− a) + (2b, 5b, b) + (0, −4c, 3c)

a+2b+0c = x      …….(4)

0a+5b−4c=y      …….(5)

− a+b+3c=z      …….(6)

From equations (4), (5) & (6) we notice that L(s) = V.

S generates V⇒S is a basis of V.

Hence the given set is basis for R³.

 

Example 11

Let S = {v₁, v₂, v₃} where v₁ = (1, −3, −2), v₂ = (−3, 1, 3), v₂ = (−2, −10, −2). Verify whether S forms a basis or not?

Solution:

To verify these vectors are linearly independent or not choose a, b, c as scalars such that,

 a(1,−3,−2) + b(−3, 1, 3) + c(−2, −10,−2) = (0, 0, 0)

 (a,− 3a, −2a)+(−3b, b, 3b) + (−2c, 10c, −2c) = (0, 0, 0)

Then a−3b−2c=0       …….(1)

−3a+b−10c = 0       …….(2)

−2a+3b−2c=0       …….(3)

 =  1 [1−2+30] + 3[6−20] − 2[−9+2]

= 1[28] + 3[14] −2[7]

= 28−42+ 14

= 42−42=0

Since  = 0 the given vectors are linearly dependent.

Hence, the given set of vectors does not form a basis.

The set of vectors is not a basis for R³.

 

Example 12

Verify the following set forms a basis or not for P₂(R) given that {−1−x+2x², 2+x−2x², 1−2x+4x²}

Solution:

To verify these vector are linearly independent or not let us choose a, b, c as scalars such that

 a(−1−x+2x²) + b(2+x−2x²) + c(1−2x+4x²) = 0

(−a−ax +2ax²) + (2b+ bx − 2bx²) + (c−2cx+4cx²)=0

Then −a+2b+c=0         ...(1)

−a+b−2c=0         ...(2)

2a−2b+4c=0         ...(3)

Let 

= −1[4 −4] − 2[− 4 + 4] + 1[2 − 2] = 0

 The given set of vectors are linearly dependent.

Hence these set of vectors does not form a basis for P₂(R).

 

Example 13

Verify the following set of vectors forms a basis or not for P₂(R) given that {−1 +2x+4x², 3−4x−10x², −2−5x−6x² }

Solution:

To verify these vectors are linearly independent or not let us choose a, b, c as scalars such that

 a (− 1 + 2x + 4x²) + b (3 − 4x − 10x²) + c (− 2 − 5x − 6x²) = 0

(−a+2ax+4ax²)+(3b − 4bx − 10bx²) + (− 2c −5cx − 6cx²) = 0

Then

−a+3b−2c=0         ...(1)

2a−4b−5c=0         ...(2)

4a−10b−6c=0         ...(3)

= −1[24−50] −3[−12+20] − 2[−20+16]

= −1[−26] −3[8] −2[4]

= 26−24+8

= 10

Hence the determinant of the matrix is 10≠0 the given set of vectors are linearly independent.

Hence the given set of vectors forms a basis for P₂(R).

 

Example 14

Let u, v and w be three distinct vectors of a vector space V. If {u, v, w} is a basis for V, then show that {u+v+w, v+w, w} also a basis for V.

Solution:

It is given that, u, v, w be the distinct vectors and { u, v, w} is a basis for V.

Now to prove {u+v+w, v+w, w} is a basis for V.

We have to prove that it is linearly independent.

Let us consider a₁v₁ + a₂v₂ + a₃v₃ ... + a_nv_n = 0

That means a₁ = 0, a₂ = 0, a₃ = 0... a_n = 0.

 a₁(u +v+w) + a₂(v + w) + a₃(w) = (0, 0, 0)

 a₁u + a₁v + a₁w + a₂v + a₂w + a₃w = (0, 0, 0)

 а₁(u) + (α₁ + a₂) v + (α₁ + a₂+ а₃) w = 0

Since it is given that { u, v, w} is a basis for V, it is linearly independent.

(ie) a₁ =0, a₁+α₂ = 0, a₁ + a₂+ a₃=0

 a₁ =0, a₂ = 0 and a₃ = 0.

 { u + v + w, v+w, w} is linearly independent.

Hence it is a basis for the vector space V.

 

Example 15

Let u and v be two vectors of a vector space V. If {u, v} is a basis for V and c₁ and c₂ are non−zero scalars, then prove that

{ u + v, c₁u } and { c₁u, c₂v} are also the bases for V.

Solution:

It is given that { u, v} is a basis for V.

(i) To prove { u + v, c₁u }, { c₁u, c₂v} are bases for V,

Then a₁ (u+v) + a₂ (c₁u) = 0

 . a₁u + a₁v + а₂c₁u = 0

 (a₁ + c₁α₂)u + (a₁)v=0.

Since {u, v} is basis, it is linearly independent.

  (a₁ + a₂c₁) = 0 and a₁=0. Since a₁ = 0, a₂ = 0

 a₁(u + v) + a₂ (c₁u) = 0 we get a₁ =0, a₂ = 0.

Therefore { u+v, c₁u } is linearly independent.

(ii) Now to prove { c₁u, c₂v} is a basis for V

Let a₁(c₁u) + a₂(c₂v) = 0

(a₁ c₁) u + (a₂ c₂) v=0

Since {u, v} is linearly independent,

a₁c₁=0 and a₂ c₂ = 0

 a₁=0 and a₂ =0

a₁ (c₁u) + a₂ (c₂v)=0 we get a₁ = 0 and a₂ = 0.

{c₁u, c₂v} is linearly independent.

Hence it forms a basis for V.

 

Example 16

Find the dimension of W; where

W = {(a₁, a₂, a₃, a₄, a₅) ∈ F⁵/a₁+a₃+α₅=0, a₂ = α₄}

Solution:

 a₁, a₂, a₃ are choosen by us to be they are independent.

  (a₁, a₂, a₃, а₂, − а₁ − а₃) = α₁ (1, 0, 0, 0, − 1) + a₂ (0, 1, 0, 1, 0) + a₃(0, 0, 1, 0, −1)

{(1, 0, 0, 0, −1), (0, 1, 0, 1, 0), (0, 0, 1, 0, −1)) is a basis of W.

  dim (W) = 3.

 

Example 17

Find the dimensions of W, where

W= {(x₁,x₂, x₃) | x₁+x₂+x₃=0}.

Solution:

  (x₁, x₂, x₁−x₂) = x₁(1, 0, − 1) + x₂(0, 1, − 1)

  B = {(1, 0, −1), (0, 1, −1)} is a basis of W

 dim (W) = 2.

 

Example 18

Find the dimensions of W, where W is given by

W = { (a₁, a₂, a₃, a₄, а₅) ∈ F⁵/a₁−a₃−a₄=0}.

Solution:

 (a₁, a₂, a₃, а₁ ‒ a₃, a₅) = a₁(1, 0, 0, 1, 0) + a₂(0, 1, 0, 0, 0) + a₃(0, 0, 1, − 1, 0) + a₅(0, 0, 0, 0, 1).

 B = { (1, 0, 0, 1, 0), (0, 1, 0, 0, 0), (0, 0, 1, − 1, 0), (0, 0, 0, 0, 1)} is a basis of W

 . dim (W) = 4.

 

Example 19

Find the dimensions of W, where W is given by

W = { (a₁, a₂, a₃, a₄, a₅) ∈ F⁵/a₂=a₃=a₄ and a₁+a₅=0}

Solution:

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

  B = {(1, 0, 0, 0, ‒1), (0, 1, 1, 1, 0)}

  dim (W) = 2.

 

Example 20

Find the dimension of W where W is given by W= { a, b, c); 2a+3b= c; 7c+9b = a }.

Solution:

W = { a, b, c)/2a+3b−c=0, a−7c−9b=0}

2a+3b−c=0

a−9b−7c=0

AX= B

 p(2) < 3 = number of unknown.

 Infinite number of solution exists.

 Leta = k, where k is arbitrary

2k+3b−c=0

c = 2k+ 3b

k−9b−7(2k + 3b) = 0

k−9b−14k – 21b=0

− 30b = 13k

b = − 13/30 k

dim (W) = 1.

 

Example 21

Determine the basis and dimension of the solution space of the linear homogeneous system x+y−z=0; −2x−y + 2z = 0 ; −x +z = 0.

Solution:

From the given system of linear equations

The Augmented matrix is

⇒ x−z=0 & y=0

x=z & y=0

Put x = t; x=t, y = 0, z=t.

This shows the vector v₁=  span the solution space, since it is linearly independent. It is a basis.

The solution space is 1−dimensional

 

Example 22

Let V = M_2×2 (F) . W₁ = { ∈ V/a,b,c ∈F}. W₂={∈ V/a,b ∈F}.

(i) Prove that W₁ is a subspace of V and find the dimension of W₁.

 (ii) Prove that W₂ is a subspace of V and find the dimension of W₂.

 (iii) Find the dimensions of W₂+ W₂ and W₁ ∩ W₂

Solution:

Let V=M_2×2 (F)

(i) Given that 

are linearly independent.

These vectors span W₁.

Hence W₁ is a subspace of V

Here dimension of W₁ = 3.

(ii) 

are linearly independent.

 These vectors span W₂.

Hence W₂ is a subspace of V.

Here the dimension of W₂ =2.

(iii) From (i) & (ii) Combining all the bases vectors of W₁ and W₂.

We have

But these vectors

 can span the vector 

  is in both W₁ and W₂.

Hence the dimension of W₁ ∩ W₂ = 1.

Now dim (W₁+ W₂) = dim (W₁) + dim (W₂) − dim (W₁ ∩ W₂)

dim (W₁ + W₂)=3+2−1=4

 dim (W₁ + W₂)=4

 

Example 23

Let W₁ and W₂ be the subspaces of a finite dimensional vector space V. Write down the necessary and sufficient conditions on W₁ and W₂ so that dim (W₁ ∩ W₂) = dim (W₁)

Solution:

Given that W₁ and W₂ be two subspaces at V.

Then W₁<W₂. Let a₁ and a₂ be the basis for W₁ and W₂ and they are finite.

Let us prove this by the method of contradiction.

Suppose W₁ is not a subspace of W₂, then there will be no common element. Since thee is no common element, the vector v ∈ V doesnot generate the basis for W₂.

The set has dimension of W₁ ∩ W₂ is a subset of dim (W₁).

 (ie) dim (W ∩ W₂)  dim (W₁)

Now conversely, let W₁ is a subspace of W₂ then

 (W ∩ W₂)=W₁ and

W₂ is the subspace of W₁, then (W₁ ∩ W₂) = W₂

W₁ ∩ W₂ has same vectors as W₁.

  dim (W₁ ∩ W₂) = dim (W₁)

Thus both W₁ and W₂ have same dimensions.

 

Example 24

Let W₁ and W₂ be the subspaces of a vector space V having dimensions, m and n respectively where m≥n.

(i) Prove that dim (W ∩ W₂) ≤ n

(ii) Prove that dim (W₁+ W₂) ≤m+n.

Solution:

Let W₁ and W₂ be two subspaces of V, and it is given that dim (W₁) = m and dim (W₂) = n.

(i) Let (W₁ ∩ W₂)  W₂

Then dim (W₁ ∩ W₂) ≤ dim (W₂)

dim (W₁ ∩ W₂) ≤ n

Hence dim (W₁ ∩ W₂) ≤n.

 (ii) dim (W₁+ W₂) = dim (W₁) + dim (W₂) − dim (W₁ ∩ W₂)

= m+n−dim (W₁ ∩ W₂)

But dim (W₁ ∩ W₂) is non negative, then

dim (W₁ ∩ W₂) ≤ m + n

 .dim (W₁+ W₂) ≤ m+n

 

Example 25

Use Lagrange interpolation formula to find the polynomial for the following points (−2, −6), (1, 5), (1, 3)

Solution:

Let us consider

Here x₀ = −2, x₁ = − 1, x₂ = 1, y₀ = −6, y₁ = 5 and y₂ = 3.

The Lagrange's interpolation formula is

f (x) =  −2(x² − 1) – 5/2 (x² +x−2)+ 1/2(x²+3x+2)

f (x) = ½ [ − 4 (x² − 1) − 5 (x² +x−2)+(x²+3x+2)]

= ½[−4x²+4−5x²−5x+10+x²+3x+2]

= 1/2 [−8x² −2x+16]

f(x) = −4x²−x+8