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Let $V_2=\{(x,y)\in \mathbb{Q}^2 \mid x^2=-y^2\}$ with the addition and the scalar multiplication of $\mathbb{Q}^2$.

To check if $V_2$ is a $\mathbb{Q}$-vector space, we must check the axioms, right?

We have the following: $(x_1, y_1), (x_2, y_2)\in V_2 : x_1^2=-y_1^2, x_2^2=-y_2^2$

$x_1^2x_2^2=y_1^2y_2^2 \Rightarrow x_1x_2=\pm y_1y_2$

$(x_1,y_1)+(x_2,y_2)=(x_1+x_2,y_1+y_2) : (x_1+x_2)^2=x_1^2+2x_1x_2+x_2^2=-y_1\pm 2y_1y_2-y_2^2$ it doesn't imply that $(x_1+x_2)^2=-(y_1+y_2)^2$, right?

Does it follow from that that $V_2$ is not closed under the addition, and so it is not a $\mathbb{Q}$-vector space?

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    $\begingroup$ if x^2=-y^2, then they are both zero $\endgroup$
    – Ben
    Dec 3, 2016 at 22:33
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    $\begingroup$ @Ben Hah, I didn't even notice that! Nice. $\endgroup$ Dec 3, 2016 at 22:34
  • $\begingroup$ So, is the ony element of $V_2$ the $(0,0)$ ? @Ben $\endgroup$
    – Mary Star
    Dec 3, 2016 at 22:35

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Ben correctly pointed out that if $x^2=-y^2$ and $x,y\in\mathbb{Q}$, then we must have $x=y=0$ (why?). So $V_2$ is the trivial vector space.

To make the question less trivial, let's look at $\mathbb{C}$ instead of $\mathbb{Q}$. Now $V_2$ has lots of elements, e.g. $(1, i)$. Is it a vector space?

Here your argument is relevant, but not complete. You're right that this version of $V_2$ is not a vector space, since it's not closed under addition; however, you haven't proved that. You need to exhibit a counterexample: find an example of vectors $(x_1, y_1)$ and $(x_2, y_2)$ in $V_2$ such that $(x_1+x_1, y_1+y_1)$ is not in $V_2$.

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You are correct that $x_1^2=-y_1^2$ and $x_2^2=-y_2^2$ does not imply $(x_1+x_2)^2=-(y_1+y_2)^2$ when you manipulate them as formal polynomials. However, remember that $x_1,y_1,x_2$, and $y_2$ must actually be rational numbers here, and there might not actually exist a counterexample in the rational numbers even if the formal manipulation does not allow you to simplify $(x_1+x_2)^2$ to $-(y_1^2+y_2)^2$. What rational numbers $x$ and $y$ are there such that $x^2=-y^2$? (Hint: There aren't very many!)

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    $\begingroup$ Does only the $(x,y)=(0,0)$ satisfy that restriction? $\endgroup$
    – Mary Star
    Dec 3, 2016 at 22:37
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    $\begingroup$ Yep. Try and prove it! $\endgroup$ Dec 3, 2016 at 22:37
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    $\begingroup$ The same way you would check that any set is closed under addition? Let $v$ and $w$ be elements of $V_2$, and check whether $v+w$ must also be an element of $V_2$. In this case there aren't many possibilities for $v$ and $w$, and you can just check all the possibilities individually. $\endgroup$ Dec 3, 2016 at 22:46
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    $\begingroup$ @MaryStar You can take a look at this question math.stackexchange.com/questions/2041775/… $\endgroup$
    – Arnaud D.
    Dec 3, 2016 at 22:48
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    $\begingroup$ @MaryStar Yes. Is $(0, 0)+(0, 0)\in \{(0, 0)\}$? $\endgroup$ Dec 3, 2016 at 22:58

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