Question

Can someone help me with this question:

Let W be the set of all vectors v = (v1,v2,v3) in R3 such that v1-2v2+v3 =0

Prove that W is a subspace of R3

Answer 1

Assume v satisfies the conditions you listed above and let c be a scalar. Does cv still satisfy the conditions?

Now assume there is a second vector w that also satisfies the conditions. Does v+w
satisfy the conditions?

If the answer to both questions is yes, then you have a subspace.

Answer 2

Definition: \[v_1 - 2v_2 = v_3 = 0: v_3 = 0, v_1 = 2v_2\]\[w -> (2v_2,v_2,0)\]
Associative Condition:
\[w + w' = (2v_2,v_2,0) + (2v'_2,v'_2,0) = (2(v_2+v'_2),(v_2+v'_2),0) = (2v''_2,v''_2,0)\] where \[v''_2 = v_2 + v'_2\]
Existence of Identity:
\[w + e = w\]
\[(2v,v,0) + (2x,x,0) = (2v,v,0)\]\[=> (2(v + x),(v + x),0) = (2v,v,0)\]\[=> v + x = v\]\[=> x = 0\]\[e -> (0,0,0)\]
Existence of Identity:
\[(2v,v,0) + (2v',v',0) = (0,0,0)\]\[(2(v + v'),(v + v'),0) = (0,0,0)\]\[v + v' = 0\]\[v' = -v\]\[(2v,v,0)^{-1} = (-2v,-v,0)\]
Therefore W is a group and since it is within V, it is a subgroup of V

Answer 3

Thanks everyone for your help. It is appreciated.

Answer 4

check 3 conditions
1> (0,0,0) is there in subspace
2> scalar multiplication is there in subspace
3> addition of any arbitrary vectors must be there in subspace