Prove that $ A = \{ \left (t, 2t, 3t \right) :t \in \mathbb R \} $ is a subspace of $\mathbb R^3$.
$Proof$:
If $t = 0$, then $(t, 2t, 3t) = (0, 0, 0)$. So, $(0, 0, 0) \in A$.
Suppose $(t, 2t, 3t) \in A$. Then $t \in \mathbb R$.
Suppose $(v, 2v, 3v) \in A$. Then $v \in \mathbb R$
So, $(t, 2t, 3t) + (v, 2v, 3v) = ((t + v), 2(t + v), 3(t + v))$. Since both $t \in \mathbb R$ and $v \in \mathbb R$, $t + v \in \mathbb R$.
Thus, $((t + v), 2(t + v), 3(t + v)) \in \mathbb A$.
If $(t, 2t, 3t) \in A$ and $t \in \mathbb R$, then $c((t, 2t, 3t) = (ct, 2ct, 3ct)$. Since both $c \in \mathbb R$ and $t \in \mathbb R$, $ct \in \mathbb R$.
Thus $(ct, 2ct, 3ct) \in \mathbb A$.
Therefore,$ A = \{ \left (t, 2t, 3t \right) :t \in \mathbb R \} $ is a subspace of $\mathbb R^3$.
Does the proof work?
Thanks.
edit: I thank you people for confirming.