[Math] Prove in full detail that the set is a vector space

linear algebravector-spaces

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):

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Best Answer

Since you are working in a subspace of $\mathbb{R}^2$, which you already know is a vector space, you get quite a few of these axioms for free. Namely, commutativity, associativity and distributivity.

With the properties that you have shown to be true you can deduce the zero vector since $0 v=0$ and your subspace is closed under scalar multiplication, and same thing for the inverse, $-1 v=-v$.

You seemed to have skipped a few steps in your reasoning though, after doing the addition of two vectors from your subspace, in order to show that the resulting vector actually is in that same subspace you should show explicitly that it is of the form $(x,2x)$ (you're almost there).

For scalar multiplication, you seem to have taken a generic vector of $\mathbb{R}^2$ instead of a vector belonging to your subset, so it needs a bit of correction as well.

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