Prove in full detail that the set is a vector space So I'm doing a review test and I have this problem:
Prove in full detail, with the standard operations in R2, that the set
{(x,2x): x is a real number}

is a vector space.
Attempt:
Given:
$(x_1, 2x_1) \in \mathbb{R}^2$
and
$(x_2, 2x_2) \in \mathbb{R}^2$
Addition:
$(x_1, 2x_1) + (x_2, 2x_2) = (x_1 + x_2, 2x_1 + 2x_2) \in \mathbb{R}^2$
$ = (x_1 + x_2, 2(x_1 + x_2)) \in \mathbb{R}^2$
$ ≃ (x, 2x) \in \mathbb{R}^2$
Thus the set is closed under addition
Scalar multiplication:
$c(x_1, 2x_1) = (cx_1, 2(cx_1)) \in \mathbb{R}^2$
$ ≃ (x, 2x) \in \mathbb{R}^2$
Thus the set is closed under scalar multiplication

Are these operations enough to prove that the set is a vector space? Or do I have to go through each of the following (or in other words do I have to to the same thing for each property in the definition):


Write
$$V=\{(x,2x)\mid x\in\mathbf{R}\}.$$
Here the main thing is to note that $(x_1,2x_1)+(x_2,2x_2)$ and $c(x_1,2x_1)$ [edit]a bad typo was here - it used to read c(x_1,2x_2)[/edit] can both be written in the form $(y,2y)$ for some $y\in\mathbf{R}$, in other words that they are elements of $V$. You have not shown that $V$ is closed under addition and scalar multiplication unless you do that.
As Vhailor pointed out, once you do this, you get the vector space axioms for free, because the set $V$ inherits them from $\mathbf{R}^2$, which is (hopefully) already known to you to be a vector space with respect to these very operations. So, to fix your proof, show that
1) $(x_1,2x_1)+(x_2,2x_2)\in V$ for all $x_1,x_2\in\mathbf{R}$.
2) $c(x,2x)\in V$ for all $x\in\mathbf{R}$.
Look up subspace criteria or something like that.
