1
$\begingroup$

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?

|cite|improve this question|||||
$\endgroup$
  • 5
    $\begingroup$ if x^2=-y^2, then they are both zero $\endgroup$ – Ben Dec 3 '16 at 22:33
  • 1
    $\begingroup$ @Ben Hah, I didn't even notice that! Nice. $\endgroup$ – Noah Schweber Dec 3 '16 at 22:34
  • $\begingroup$ So, is the ony element of $V_2$ the $(0,0)$ ? @Ben $\endgroup$ – Mary Star Dec 3 '16 at 22:35
3
$\begingroup$

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 $V_2$.

|cite|improve this answer|||||
$\endgroup$
2
$\begingroup$

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!)

|cite|improve this answer|||||
$\endgroup$
  • 2
    $\begingroup$ Does only the $(x,y)=(0,0)$ satisfy that restriction? $\endgroup$ – Mary Star Dec 3 '16 at 22:37
  • 2
    $\begingroup$ Yep. Try and prove it! $\endgroup$ – Eric Wofsey Dec 3 '16 at 22:37
  • 1
    $\begingroup$ 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. $\endgroup$ – Eric Wofsey Dec 3 '16 at 22:46
  • 1
    $\begingroup$ @MaryStar You can take a look at this question math.stackexchange.com/questions/2041775/… $\endgroup$ – Arnaud D. Dec 3 '16 at 22:48
  • 2
    $\begingroup$ @MaryStar Yes. Is $(0, 0)+(0, 0)\in \{(0, 0)\}$? $\endgroup$ – Noah Schweber Dec 3 '16 at 22:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.