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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.

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  • $\begingroup$ Solid! Now, you should try to generalize your proof to show that if $\mathbf v\in\mathbb R^n$, then $\{t\mathbf v : t\in\mathbb R\}$ is a subspace of $\mathbb R^n$. $\endgroup$
    – MPW
    Apr 17, 2014 at 2:33

2 Answers 2

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This looks fine. Alternatively you could note that $$ A=\operatorname{Span}\{(1,2,3)\} $$

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Yes. This is correct. If the zero vector is in $A$ and $A$ is closed under vector addition and scalar multiplication, then you have shown that $A$ is a subspace of $\mathbb{R^3}$. Nicely done!

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