Answer to Question #154705 in Linear Algebra for Sourav Mondal

Question

Let : 

V = R3, 

W = {(X1, X2, X3) ! x1 - x2 = .X3}. 

Show that W is a subspace of V. Further, 

find a basis for W, and hence, find the 

dimension of W.

Answer 1

Let : 

V = R3, 

W = {(X1, X2, X3) ! x1 - x2 = .X3}. 

Show that W is a subspace of V. Further, 

find a basis for W, and hence, find the 

dimension of W.

For W to be a subspace of V

W must be empty

$\forall$ x,y $\in$ W (x + y) $\in$ W

also, ∀ x $\in$ W and c $\in$ F cx $\in$ W

let (0, 0, 0) $\in$ W

then 0 - 0 = 0

$\implies$ (0, 0, 0) $\in$ W

$\implies$ W is not empty

let v=(v₁, v₂, v₃) and r= (r₁, r₂, r₃) $\in$ W

since v and r $\in$ W

then, v₁ - v₂ = v₃ and r₁ - r₂ = r₃

v + r = (v₁ + r₁) , (v₂ + r₂), (v₃ + r₃)

$\because$ (v₁ + r₁) - (v₂ + r₂) = (v₃ + r₃)

(v₁ - v₂) +( r₁ - r₂) = (v₃ + r₃)

v₃ + r₃ = (v₃ + r₃)

LHS = RHS

Also

r= (r₁, r₂, r₃) $\in$ W and c $\in$ F

rc = (cr₁,c r₂, cr₃)

= c r₁ - c r₂ = cr₃

$\because$ W is a subspace of V

consider the set

B = {(2, 1, 1), (4, 2, 2)}

we will prove that the set is linearly independent

(0, 0, 0) = a(2, 3, -1) + b(4, 2, 2)

solving this a=b= 0

$\because$ the set the set is linearly independent

Also,

(r₁, r₂, r₃) = a(2, 3, -1) + b(4, 2, 2)

then r₁ = -4a + 2r₂

r₂ = 4 + r₃

r₃ = r₂ - 4

$\because$ the set B span W

$\implies$ B is a Basis of W

since the set B contains 2 elements, the Dimension of W is 2