Question #154705

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.


1
Expert's answer
2021-01-11T14:50:33-0500

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



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