Check whether the set of all pairs of real numbers of the form (1 , x) with operations (1 , y) + (1 , y’) = (1 , y + y’) and k(1 , y) = (1 , ky) is a vector space.

Subject: Applied Mathematics 4

Topic: Vector Spaces

Difficulty: Medium

Let V = {(1,x) | x $ \epsilon $ IR} and define (1,y) + (1,y') = (1,y + y') and k(1,y) = (1,ky)

(i) Let u = (1,x); v = (1,y); w = (1,z) $ \epsilon $ V

(u + v) = (1,x) + (1,y) = (1,x + y)

Since x,y $ \epsilon $ IR; x + y $ \epsilon $ IR

Therefore, (u + v) = (1, x+ y) $ \epsilon $ V and u,v $ \epsilon $ V $ \implies $ u + v $ \epsilon $ V

Therefore, V is closed under addition

(ii) ku = k(1,x) = (1,kx)

Because x $ \epsilon $ IR and k $ \epsilon $ IR $ \implies $ kx $ \epsilon $ IR

Therefore, ku $ \epsilon $ V u $ \epsilon $ V $ \implies $ ku $ \epsilon $ V

Therefore, V is closed under scalar multiplication.

(iii) u + v = (1,x) + (1,y)

= (1,x + y)

= (1, y + x)

= (1,y) + (1,x) = v + u

Therefore, '+' is commutative in V

(iv) (u + v) + w

= [(1,x) + (1,y)] + (1,z)

= [(1,x + y)] + (1,z)

= (1,(x+y) + z)

= (1,x + (y + z))

= (1,x) + (1, y + z)

= (1,x) + [(1,y) + (1,z)]

= u + (v + w)

Therefore, '+' is associative in V

(v) Consider I = (1,0)

u + I = (1,x) + (1,0) = (1,(x + 0)) = (1,x)

Therefore, I = (1,0) is additive identity in V

Therefore, V has additive identity.

(vi) Consider, -u = (1,-x)

u + (-u) = (1,x) + (1,-x)

= (1,x- x) = (1,0) = I (Additive identity)

Therefore, (-u) is inverse of u in V

Therefore, (-u) is additive identity

(vii) k(u + v) = k[(1,x) + (1,y)]

= k(1, x+y)

= (1,k(x+y)

= (1, kx + ky)

= (1,kx) + (1,ky)

= k(1,x) + k(1,y)

= du + dv

Therefore, Distributive of scalar multiplication is satisfied.

(viii) (k + l)u = (k + l)(1,x)

= (1, (k+l)x)

= (1,(kx + lx))

= (1,kx) + (1,lx)

= k(1,x) + l(1,x)

= ku + lu

Therefore, distributivity of scalar is satisfied in V

(ix) kl(u) = (kl)(1,x) = (1,(kl)x)

= (1,k(lx))

= k(1,lx)

= k(l(1,x))

= k(lu)

Therefore, Associative law of scalar is satisfied in V

(x) Consider k = 1

1.u = 1(1,x) = (1,x) = u

1 is the multiplication identity in V

Therefore, all the ten axioms are satisfied.

Therefore, V is a vector space.