Vector Spaces: Theorems Part 1: Example Solved Problems

Vector Spaces: Theorems Part 1

Example Problems

 

Example 1: Write the zero vector of M_3×4 (F).

Solution:

The given matrix M_3×4 is a matrix of order (3 × 4). (ie) Three rows and 4 columns.

 The zero vector of M_3×4 = 

 

Example 2

If M =  what are M₁₃, M₂₁ and M₂₂?

Solution:

Given that M is a non−square matrix with 2 rows and 3 column (ie) M_2x3 (F).

In general M_2x3= 

 M₁₃=3, M₂₁ =4 and M₂₂ = 5.

 

Example 3

Perform the indicated operations for the following

(v) (2x⁴ − 7x³+4x+3) + (8x³ + 2x² − 6x + 7)

(vi) (−3x³ +7x² + 8x − 6) + (2x³ − 8x + 10)

 (vii) 5(2x⁷ – 6x⁴ + 8x² ‒ 3x)

 (viii) 3(x⁵ – 2x³ + 4x + 2)

Solution:

(v) (2x⁴ − 7x³+4x+3) + (8x³ + 2x² − 6x + 7)

 = (2 + 0)x⁴ + (− 7 + 8) x³ + (0 + 2)x² + (4 − 6)x + (7 + 3)

= 2x⁴ + x³ + 2x² − 2x + 10

(vi) (−3x³ +7x² + 8x − 6) + (2x³ − 8x + 10)

= (− 3 + 2)x³ + (7 + 0)x² + (8 − 8)x + (− 6 + 10)

= −x³ + 7x² + 0x + 4

(vii) 5(2x⁷ – 6x⁴ + 8x² ‒ 3x) = 10x⁷ − 30x⁴ + 40x² − 15x

(viii) 3(x⁵ – 2x³ + 4x + 2) = 3x⁵ − 6x³ + 12x+6

Example 4

Let S={0,1} and F= R. In F (S, R) show that f=g and f+g=h where f(t)=2t+1, g (t)=1+4t−2t² and h(t)=5^t+1.

Solution:

It is given that S={0, 1} and F= R.

 To prove f=g we have to show that this statement is valid for t=0 and t = 1.

 f=g  ⇒ 2t+1=1+4t−2t²

when t=0; 1= 1 and when t=1; 3 = 3

Since 1= 1 and 3=3 the statement f= g is valid.

To prove f+g=h; then

 (2t+1)+(1+4t−2t²) = 5^t+1

when t=0;  1+1=1+1  ⇒   2=2

when t = 1;       3+ (3)=5+1  ⇒   6=6

 The statement f+g = h is verified.

 

Example 5

In any vector space V, show that (a+b) (x+y)= ax+ay + bx+by for any x,y ∈ V and any a, b ∈ F.

Solution:

We know that for each element a in F and each pair of elements x, y in V, then

 a (x + y) = ax + ay and also for each pair of elements (a, b) in F and each element x in V then (a+b) x = ax + bx.

Based on these two conditions, (a+b) (x + y) can be written as

 (a+b) (x + y) = a (x + y) + b (x+y) (or) (a+b)x + (a + b)y

 = ax + ay + bx+ by (or) ax + bx + ay + by

 

Example 6

Let V={(a₁, a₂); a₁, a₂ = F} where F is a field. Define addition of elements of V and for C∈F and (a₁, a₂) ∈ V, define c (a₁, a₂) = (a₁, 0). Is V is a vector space over F with these operations.

Solution:

Given that V = {(a₁, a₂); a₁, a₂ ∈ F}

 c (a₁, a₂) = (a₁, 0) for c ∈ F, (a₁, a₂) ∈ V.

 0 (a₁, a₂) = (a₁, 0) is the zero vector which is a contradiction the rule 3, that the zero vector to be unique.

  V is not a vector space over F with these operations.

 

Example 7

Let V = {(a₁, a₂); a₁, a₂ ∈ R}. For (a₁, a₂), (b₁, b₂) ∈ V and c ∈ R, define (a₁, a₂) + (b₁, b₂) = (a₁ + 2b₁, a₂+3b₂) and c (a₁, a₂ = (ca₁, ca₂). Verify whether V is a vector space over R with these operations?

Solution:

Given that V = {(a₁, a₂): a₁, a₂ ∈ R}

Here (a₁, a₂), (b₁, b₂) ∈ V and c ∈ R.

(a₁, a₂)+(b₁, b₂) = (a₁+2b₁, a₂ + 3b₂) and

 c (a₁, a₂)=(ca₁, ca₂)

(a₁, a₂) + (b₁, b₂) = (a₁ +2b₁, a₂ +3b₂)

(b₁, b₂) + (a₁, a₂) = (b₁ + 2a₁, b₂ +3α₂)

 (a₁, a₂) + (b₁, b₂) ≠ (b₁, b₂) + (a₁, a₂)

This is the contradiction to the rule 1.

 V is not a vector space over R, with these operations.

 

Example 8

Let V={(a₁, a₂, a₃... a_n); a_i ∈ R for i = 1, 2, 3 ... n }.. V is a vector space over R.

Define u = (a₁, a₂, а₃ ... a_n) ∈ Fⁿ

v = (b₁, b₂, b₃ ... b_n) ∈ Fⁿ

and c ∈ F. Then u + v = { a₁+b₁, a₂+b₂, a₃ +b₃, ... a_n+b_n}

and cu= (ca₁, ca₂, ca₃, ca₄... ca_n).

Is V a vector space over the field of complex numbers with the operations of coordinate wise addition and multiplication.

Solution:

No. A real valued vector scalar multiply with a complex number will not be real valued vector.

 

Example 9

Let V={(a₁, a₂, а₃ ... а_n) ; α_i ∈ С for i = 1, 2, 3 ... n} and V is a vector space over C by u= (a₁, a₂, a₃... a_n) ∈ Fⁿ and

v = (b₁, b₂, b₃ ... b_n) ∈ Fⁿ. c ∈ F then,

u+v= (a₁+b₁, a₂ +b₂, a₃ + b₃ + ...)

cu = (ca₁, ca₂, ca₃... ca_n)

Is V a vector space?

Solution:

All the eight conditions of vector space are satisfied by the above operations. V is a vector space over the filed of real numbers, with the operations addition and multiplication.

[This example has been already explained in Example. 1 given in examples for vector space].

 

Example 10

Verify that the set V of all ordered triples of real numbers of the form (x, y, 0) and defined the operations + and • by

 (i) (x, y, 0) + (x', y', 0) = (x + x', y+y', 0); (ii) c . (x, y, 0) = (cx, cy, 0) is a vector space or not.

Solution:

Let us consider u = (x, y, 0) and v = (x′, y′, 0) and w = (x", y", 0) ∈ R³.

(i) Closure axiom

 u+v= (x, y, 0)+(x', y', 0) = (x+x', y+y', 0) ∈ R³

(ii) Associative property

Let u+(v+w) = (x, y, 0) + [(x', y', 0)+(x", y", 0)]

= (x, y, 0)+(x′+x", y′+y", 0]

= [x+x+x", y+y′+y", 0]          …………(1)

Similarly

(u + v) + w = [(x, y, 0) + (x', y', 0)] + (x", y", 0)

 = [x + x', y + y', 0] + (x'', y'', 0)

= [x + x' + x'', y+y'+y", 0]

From (1) and (2) Associativity holds.

(iii) Existence of Identity

Let e = (e₁, e₂, 0) and u= (x, y, 0).

 u+e=(x, y, 0)+(e₁, e₂, 0) = (x + e₁, y + e₂, 0) = u.

  x+e₁ =x, y+e₂ = y, 0=0.

 e₁=0, e₂=0, 0=0.

⇒ e=(0, 0, 0) ∈ R³.

(iv) Existence of inverse

Let u= (x, y, 0), u' = (x₁,y₁, 0)

u+u' = e

⇒ (x, y, 0) + (x₁, y₁, 0) = (e₁, e₂, 0)

⇒ x+x₁=e₁, y+y₁=e₂, 0+0=0

  u' = (x₁,y₁, 0) exists.

(v) c(x, y, 0) = (cx, cy, 0) for all real numbers c and all u and v in V.

(vi) c [d(x, y, 0) = c [dx, dy, 0]

= [cdx, cdy, 0]

= d [cx, cy, 0]

= d[c (x, y, 0)]

(vii) c [u+v] = c [(x, y, 0) + (x′, y', 0)]

= [c(x, y, 0) + c(x', y', 0)]

(viii) (c+d) u = (c+ d) (x, y, 0)

= c(x, y, 0) + d (x, y, 0)

 Since the given functions satisfy all the 8 conditions for vector space, it is a vector space.

 

Example 11

Consider the set V of all ordered triples of real numbers (x, y, z) and define the operations + and • by

(i) (x+y+z) + (x', y', z') = (x + x', y+y', z+z') (ii) c. (x, y, z) = (cx, y, z) Verify the given function is a vector space or not under the operations + and •.

Solution:

It is easy to verify that all the conditions satisfy for the vector space except the 8th condition. Here 0) = (0, 0, 0) and the negative of the vector (x, y, z) is the vector (‒x, ‒y, ‒z).

Now let us verify the last 8th condition.

(ie) (a + b) (x) = ax + bx

Here (a+b) (x, y, z) = [(a + b) x, y, z]

Then ax + bx becomes

 a (x, y, z) + b (x, y, z) = (ax, y, z) + (bx, y, z)

= [(ax + bx), 2y, 2z]

Hence (a + b) (x) ≠ ax + bx.

The above condition is not satisfied by the given function.

V is not a vector space under the prescribed operations.

 

Example 12

Let V be the set of all positive real numbers. Define the vector addition and scalar multiplication as follows. x+y=xy and kx = x^k. Determine whether or not V is a vector space over R with respect to the above operations.

Solution:

Let V={(x)/x+y=xy & kx = x^k, x, y ∈ R⁺ = (0, ∞) }

(i) Let x₁ ∈ V & y₁ ∈ V  ⇒ x₁+y₁ = x₁у₁ ∈ V

Closure axion is satisfied.

(ii) For all x, y ∈ V, x+y=y+x ( commutative law).

Let x₁, y₁ ∈ V

Now x₁+y₁ = x₁y₁= y₁x₁= y₁+x₁

(iii) For all x, y, z ∈ V, (x+y) + z = x + (y + z) (Associative law).

Let x₁, y₁, z₁ ∈ V then

Now (x₁+y₁) + z₁ = (x₁ y₁) +z₁ = x₁ y₁ z₁ ∈ V          ...(1)

Then x₁ + (y₁ + z₁) = x₁ + (y₁z₁) = x₁ y₁ z₁ ∈ V          ...(2)

From (1) & (2) we have (x+y) + z = x+(y+z)

 (iv) For x ∈ V, there exists 0 ∈ V such that x+0=x

Let x₁ = V & e ∈ V

 x₁+e=x₁e=ex₁ = e +x₁

 ⇒ e= l ∈ V

(v) For all x ∈ V, ⇒ 1•x=x

Let x₁ ∈ V ⇒ 1•x₁=x=x₁ ∈ V

(vi) For any x,y ∈ V ⇒ x+y=0

Let x₁ ∈ V

x₁+(− x₁)=(− x₁) + x₁ = − x₁ x₁ = 1/x₁ ∈ R⁺ = V

(vii) For all a, b ∈ F and x ∈ V, (ab) x = a (bx)

Let k, l ∈ F and x ∈ V

 (kl) x = k(lx) = k(x^l) = x^lk = (kl) (x) = k (lx)

(viii) For a ∈ F, x, y ∈ V ⇒ a (x + y) = ax + ay

Let k ∈ F and x₁,y₁ ∈ V

 k (x₁+y₁) = k (x₁ y₁) = (x₁ y₁)^k = x₁^k y₁^k = x₁^k + y₁^k

 = kx₁+ky₁

 (ix) For a, b ∈ F, and x ∈ V then (a + b) x = ax + bx

Let k, l ∈ F & x₁ ∈ V

  (k + 1) x₁ = x₁^k+1 = x₁^k . x₁^l = x₁^k + x₁^l = kx₁ + lx₁

Since all the properties are satisfied, V is a vector space.

 

Example 13

Determine whether the set of all 2×2 matrix of the form  , a, b ∈ R with respect to standard matrix addition and scalar multiplication is a vector space or not? If not, list all the axioms that fail to hold.

Solution:

(i) For all x, y ∈ V, then x+y=y+x (commutative law).

From equations (1) & (2) we have u+v = v+ u

(ii) For all x, y, z ∈ V, (x + y) + z = x + (y + z) (Associative law)

From equations (3) and (4) (u+v) + w = u + (v+w)

 (iii) For x ∈ V, there exists 0 ∈ V such that x+0 = x.

 (iv) For all x,y ∈ V ⇒ x+y=0.

 (v) For all x ∈ V  ⇒  1.x=x.

 (vi) For all a, b ∈ F and x ∈ V then (ab) x = a (bx).

 (vii) For a ∈ F and x, y ∈ V then a (x+y)=ax+ay, u, v ∈ V, k ∈ F.

 (viii) For a, b = F & x ∈ V then (a + b) x = ax + bx

Let k, l ∈ F & u ∈ V

 = ku + lu

Since all the above 8 results are satisfied V is a vector space.

 

Example 14

Let U and V be the vector spaces over a field F. Let Z = {(u, v) | u ∈ U and v ∈ V}. Prove that Z is a vector space over F under the following operations. (u₁, v₁) + (u₂, v₂) = (u₁+u₂, v₁+v₂) and c (u₁, v₁) = (cu₁, cv₁)

Solution:

Given that Z= {(u,v); u ∈ U and v ∈ V}

Also (u₁, v₁) + (u₂, v₂) = (u₁ + u₂, v₁+v₂)        ...(1)

 c (u₁, v₁) = (cu₁, cv₁)        ...(2)

(i) For all x,y ∈ V, then x+y=y+x (commutative law).

Let (u₁, v₁), (u₂, v₂) ∈ z

Now (u₁, v₁)+(u₂, v₂)=(u₁ + u₂, v₁+v₂)         ...(3)

and (u₂, v₂)+(u₁, v₁)=(u₂+u₁, v₂+v₁)          ...(4)

From equations (3) & (4), we have

(u₁, v₁) + (u₂, v₂) = (u₂, v₂) + (u₁, v₁)

(ii) For all x, y, z ∈ V, then (x + y) + z = x + (y + z) (Associative law)

Let (u₁, v₁) (u₂, v₂), (u₃, v₃) ∈ Z.

Now [(u₁, v₁) + (u₂, v₂)] + (u₃, v₃)

= (u₁ + u₂, v₁+v₂) + (u₃, u₃)

= (u₁+u₂+u₃, v₁+v₂ + v₃)          ...(5)

and (u₁, v₁) + [(u₂, v₂) + (u₃, v₃)]

= (u₁, v₁) + (u₂+ u₃, v₂ + v₃)

= (u₁ + u₂ + u₃, v₁ + v₂+ v₃)       ...(6)

From equations (5) & (6) we have

  [(u₁, v₁) + (u₂, v₂)] + (u₃, v₃) = (u₁, v₁) + [(u₂, v₂) + (u₃, V₃)]

(iii) For x ∈ V, there exists 0 ∈ V such that x+0=x.

Lt (0u, 0v) ∈ Z

 (u, v) + (0u, 0v) = (u + 0u, v + 0v) = (u, v)

 (0u, 0v) is the zero vector in Z.

 (iv) For all x,y ∈ V ⇒ x+y=0.

Let (u, v) ∈ Z then there exists (‒u, ‒v) ∈ Z

 (u, v) + (−u, −v) = (u−u, v − v) = (0, 0) = (0u, 0v) = 0z

(v) For all x ∈ V ⇒ 1.x=x.

Let (u, v) ∈ Z ⇒ 1 (u, v) = (lu, 1v) = (u, v)

(vi) For all a, b ∈ F and x ∈ V, (ab) x = a (bx).

Let a, b ∈ F and (u, v) ∈ Z

 (ab) (u, v) = (abu, abv) = a (bu, bv) = a [b (u, v)]

(vii) For a ∈ F, x, y ∈ V, then a (x+y) = ax+ay

Let c ∈ F & (u₁, v₁), (u₂, v₂) ∈ Z

 c[ (u₁, v₁) + (u₂, v₂) ] = c[ u₁ + u₂, v₁ + v₂ ]

= [cu₁ + cu₂, cv₁ + cv₂ ]

= c(u₁, v₁)+c (u₂, v₂)

(viii) For a, b ∈ F and x ∈ V, (a + b) x = ax + bx.

Let (a, b) ∈ F, (u, v) ∈ Z then

 (a + b) (u, v) = (au + bu, av + bv)

= (au, av) + (bu, by)

= a(u, v) + b(u, v)

From the above 8 results, Z is a vector space over F.

 

Example 15

Let V be the set of all m×n matrices with real entries. Let F be the field of rational numbers. Let us define (A+B)_ij=A_ij+B_ij for all A, B ∈ M_m×n (F) & C ∈ F, also (CA)_ij=CA_ij, Prove that V is a vector space over the field F.

Solution:

Let us consider A=[a_ij]_m×n , B = [b_ij] _m×n , C = [c_ij]_m×n and a_ij, b_ij, c_ij ∈ F.

(i) For all x,y ∈ V, x+y=y+x (commutative law.)

A + B = [a_ij]_m×n + [b_ij]_m×n = [α_ij + b_ij]_m×n

= [b_ij + a_ij]_m×n = [b_ij]_m×n +[a_ij]_{m× n} = B+A

A + B = B+A

(ii) For all x, y, z ∈ V, (x + y) + z = x + (y + z).

 [A + B] + C = [ [a_ij]_m×n + [b_ij]_m×n ] + [C_ij]_m×n

= [α_ij + b_ij]_m×n + [C_ij]_m×n

= [a_ij + b_ij + C_ij]_m×n

= [a_ij + (b_ij + C_ij)]_m×n

 = [a_ij]_m×n + [ [b_ij]_m×n + [C_ij]_m×n ]

=A+ (B+C)

  (A + B) +C = A+(B+C)

(iii) For x ∈ V, there exists 0 ∈ V such that x+0=x.

 A+0 = [a_ij]_m×n + [0]_m×n  = [a_ij+0]_m×n = [α_ij]_m×n = A

Similarly, 0+ A = A ⇒ 0=[0]_m×n is the identity matrix.

(iv) For all x, y ∈ V, x+y=0

 A+ (−A) = [a_ij]_m×n  + [−a_ij]_m×n = [a_ij − a_ij]_m×n = [0]_m×n

  Also (−A) + A = [−a_ij]_m×n + [a_ij]_m×n = [− a_ij + a_ij]_m×n = [0]_m×n

 −A is the additive inverse of the matrix A.

 (v) For all x ∈ V, 1⋅ x = x.

  1 • A = 1 [a_ij]_m×n = [1 ⋅ α_ij]_m×n = [a_ij]_m×n =A.

(vi) For all a, b ∈ F and x ∈ V then (ab) x = a (bx)

 (ab) A = (ab) [a_ij]_m×n = [(ab) a_ij]_m×n = [ a (ba_ij) ]_m×n

= a[ba_ij]_m×n = a [b [a_ij]_m×n] = a(bA)

 (ab) A = a (bA)

(vii) For a ∈ F and x,y ∈ V then a (x + y) = ax+ay

 a [A + B] = a [[a_ij]_m×n + [b_ij]_m×n] = a [a_ij+b_ij]_m×n

= [a (a_ij + b_ij)]_m×n

= [aa_ij + ab_ij]_m×n

= [aa_ij]_m×n + [ab_ij]_mxn = a [a_ij]_mxn = a [b_ij]_m×n

= aA + aB

  a [A+B] = aA + aB.

(viii) For a, b ∈ F and x ∈ V then (a+b) x = ax + bx

 (a+b) A = (a+b) [a_ij]_m×n

= a[α_ij]_m×n + b[a_ij]_m×n = aA +bA

 (a + b)A = aA + bA

From the above 8 results, the set of all m×n matrices with real entries is a vector space over F.

 

Example 16

Given that V = {(u₁, u₂) / u₁, u₂ ∈ R}. Also (u₁, u₂), (v₁, v₂) ∈ V and C ∈ R. Define (u₁, u₂) + (v₁, v₂) = (u₁+2v₁, u₂+3v₂) and c (u₁, u₂) = (cu₁, cu₂). Prove that V is not a vector space over R for the above operations.

A Solution:

Given that V={(u₁, u₂) / u₁, u₂ ∈ R }

(u₁, u₂), (v₁, v₂) ∈ V and c∈R

Define that (u₁, u₂) + (v₁, v₂) = (u₁ + 2v₁, u₂ + 3v₂)             ...(1)

and c (u₁, u₂)=(cu₁, cu₂)             ...(2)

(i) For x,y ∈ V, x+y=y+x (commutative law).

(u₁, u₂) + (v₁, v₂) = (u₁ + 2v₁, u₂+3v₂)     ...(3)

(v₁, v₂) + (u₁, u₂) = (V₁+2u₁, v₂+3u₂)     ...(4)

From (3) & (4)

(u₁, u₂) + (v₁, v₂) ≠ (v₁, v₂) + (u₁, u₂)

which is a contraction to the first rule.

  V is not a vector space over R with the above operations.

 

Example 17

Prove that the set of all polynomials over the field F is a vector space V.

Solution:

Let us consider the set of all polynomials be

 (a) u(x)=a₀+a₁x+a₂x² + a₃x³ + a₄x² + ……. + a_nxⁿ

where a₀, a₁, a₂, a₃... a_n ∈ F

(a) v(x)=b₀+ b₁x+ b₂x² + b₃x³ + b₄x² + ……. + b_nxⁿ

where b₀, b₁, b₂, b₃... b_n ∈ F

(c) u(x)+v(x) = (a₀ + b₀) + (a₁ + b₁)x + (a₂ + b₂)x² + ... + (a_n+b_n) xⁿ

(d) cu (x) = ca₀ + ca₁x + ca₂x² + ……. + ca_nxⁿ     where  c ∈ F.

(i) For x, y ∈ V, then x+y=y+x (commutative law).

 u(x) + v(x)=(a₀+b₀) + (a₁ + b₁) x + (α₂+ b₂) x² + ... + (a_n+b_n) xⁿ

 v(x) + u(x) = (b₀ + a₀) + (b₁ + a₁)x + (b₂+ a₂)x² + …. + (b_n+a_n)xⁿ.

 u(x) + v(x) = v(x) + u(x)

(ii) For x, y, z ∈ V, (x + y) + z = x + (y + z) (Associative law).

Here u (x), v (x), w (x) ∈ V.

we can easily prove that

[u (x) + v(x)] + w(x) = u(x) + [v (x)+w (x)]

(iii) For x ∈ V there exists 0 ∈ V such that x+0=x.

 0(x) + u(x) = (0 +0x + 0x² + 0x² + ... + 0xⁿ) + (a₀ + a₁x + a₂x² + ... + a_nxⁿ)

= (0+a₀) + (0+a₁)x+(0 + a₂)x² + ... + (0+ a₁) xⁿ

= a₀ + a₁x + a₂x² + a₃x³ + ... +a_nxⁿ

= u(x)

Similarly we can prove that u (x) + 0 (x) = u(x)

From the above two relations 0(x) is the additive identity in V.

(iv) For x,y ∈ V such that x+y=0.

Here u(x) ∈ V  ⇒  u (x) ∈ V

 u(x) + [−u (x)] = 0(x) and also [−u (x)] + u(x) = 0 (x)

 − u (x) is the additive inverse of u (x).

(v) For x ∈ V, 1.x=x

Here u(x) ∈ V and 1 ∈ F

 1 ⋅ u (x) = 1 (a₀ + a₁x + a₂x² + a₃x³ + ... + a_nxⁿ)

= (1a₀) (1.a₁)x+(1.a₂)x² + (1.a₃)x³ + ... +(1.a_n)xⁿ

= a₀ + a₁x+a₂x² + α₃x³ + ... + a_nxⁿ = u(x)

 1 • u (x) = u(x)

(vi) For a, b ∈ F and x ∈ V, (ab) x = a (bx)

Here u(x) ∈ V and a, b ∈  F.

(ab) u (x) = (ab) [a₀ + a₁x + a₂x² + a₃x³ + ... + a_nxⁿ]

= ab a₀ + ab a₁x + ab a₂x² + aba₃x³ + ... + ab a_nxⁿ

= a [ba₀ + ba₁x + ba₂x² + ba₃x³ + ... + ba_nxⁿ]

= a [bu (x)]

⇒ (ab) u (x) = a [bu (x)]

 (vii) For a ∈ F and x, y = V, a (x + y) = ax+ay

 a [u(x) + v(x)] = a [(a₀+b₀) + (a₁ + b₁) x + (a₂ + b₂) x² + ... + (a_n + b_n) xⁿ]

= aa₀ + ab₀ + aa₁x + ab₁x + aa₂x² + ab₂x² +….. + aa_nxⁿ + ab_nxⁿ

= a [a₀ + a₁x + a₂x² + a₃x³ + ... + a_nxⁿ] + a [b₀ + b₁x + b₂x² + ... + b_nxⁿ]

= a u(x) + b v (x)

  a [u (x) + v (x)] = a u (x) + b v (x).

(viii) For a, b ∈ F and x ∈ V, (a + b) x = ax + bx.

  (a + b) u (x) = (a + b) [a₀ + a₁x + a₂x² + a₃x³ + ... + a_nxⁿ]

= [aa₀+aa₁x+aa₂x² + aa₃x³ + ... + aa_nxⁿ] + [ba₀+ba₁x+ba₂x² + ba₃x³ + ... + ba,xⁿ]

= a [a₀ + a₁x + ... a_nxⁿ] + b [a₀ + a₁x + a₂x² + ... + a_nxⁿ]

= a u(x) + b u(x)

From the above 8 results it is proved that the set of all polynomials over a field F is a vector space V.

 

Example 18

Let V be the set of sequences {a_n} of real numbers. Define: {a_n}+{b_n}={a_n+b_n} and t {a_n}={ta_n}. Prove that V is a vector space over R with the above operations.

Solution:

Given that

{a_n},{b_n} ∈ V

{a_n} + {b_n} = {a_n+b_n}           .......(1)

 t{a_n} = {ta_n}              .......(2)

(i) For all x,y ∈ V, x+y=y+x (commutative law)

Here {a_n}, {b_n} ∈ V

Then {a_n}+{b_n}={a_n+b_n}        ...(3)

also {b_n}+{a_n}={b_n + a_n}        ...(4)

From (3) & (4) {a_n}+{b_n} = {b_n} + {a_n}

(ii) For all x, y, z ∈ V, (x + y)+z = x+(y+z) (Associative law)

[{a_n} + { b_n } ] + { c_n } ] = {a_n+b_n } + { c_n } = { a_n+b_n+c_n}        …..(5)

{a_n } + [ { b_n } + { c_n } ] = {a_n} + { b_n + c_n } = {a_n + b_n + C_n }         …..(5)

From (5) & (6) [{a_n}+{b_n} ] + { c_n} = {a_n}+[ {b_n}+{ c_n}]

(iii) For x ∈ V, there exists 0 ∈ V such that x+0=x

 { a_n } + {0} = {a_n + 0 } = { a_n }

(iv) For all x, y ∈ V, x+y=0

 {a_n} + {−a_n} = {a_n + (−a_n) } = { 0 }

(v) For x ∈ V, 1.x=x

 1.{a_n} = { 1a_n } = {a_n}

(vi) For all a, b ∈ F and x ∈ V, then (ab)x = a(bx)

(ab) {x_n} = { ab x_n}    by (2).

 a{bx_n}={ab x_n}         by (2).

  ab {x_n} = a {bx_n}

(vii) For a ∈ F and x,y ∈ V, a (x+y)=ax+ay.

 a[{x_n} + {y_n } ] = a { x_n + y_n } = {a (x_n + y_n) } = { ax_n+ay_n }

         ……….(7)

 a{x_n}+a{y_n} = {ax_n} + { ay_n } = { ax_n + ay_n }

        ………..(8)

From (7) and (8) a[{x_n}+{ y_n } ] = a { x_n } + α { y_n }

(viii) For a, b ∈ F and x ∈ V then (a + b) x = ax + bx.

 (a + b) { x_n} = {(a + b) x_n } = { ax_n + bx_n}       ……….(9)

 a{x_n}+b{x_n}={ax_n} + { bx_n } = { ax_n + bx_n}       ……….(10)

From (9) & (10), (a+b) {x_n} = a{x_n} + b{x_n}

From the above 8 results, V is a vector space over R.