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I have a set $U=\{(x,y,z)\in\Bbb{R^3} |x^2+y^2+z^2=0\}$. I believe that it isn't a subspace of $\Bbb{R^3}$ because it isn't closed under vector addition. Here is my reasoning:

Let $$Let \,\vec{v}=(v_1,v_2,v_3) \, and \, \vec{u}=(u_1,u_2,u_3)\, where \, \vec{v}\in U$$

$$Thus \,\, \vec{v}+\vec{u}=(u_1+v_1,u_2+v_2,u_3+v_3)$$

$$Now \, \, x+y+z=(u_1+v_1)^2+(u_2+v_2)^2+(u_3+v_3)^2$$

$$=u_1^2+2u_1v_1+v_1^2 + u_2^2+2u_2v_2+v_2^2 + u_3^2+2u_3v_3+v_3^2$$ $$=(u_1^2+u_2^2+u_3^2)+(v_1^2+v_2^2+v_3^2)+2(u_1v_1+u_2v_2+u_3v_3)$$ $$=2(u_1v_1+u_2v_2+u_3v_3)\neq0$$

Thus U is not closed under vector addition.

Is this reasoning sound?

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  • $\begingroup$ The zero vector is the only point in that set, and it is indeed a subspace. $\endgroup$ – Adam Hughes Apr 4 '17 at 3:15
  • $\begingroup$ Okay I understand then the only point is (0,0,0). I'm just not sure how to show that algebraically that the expression equals zero. $\endgroup$ – Patrick Robertson Apr 4 '17 at 3:18
  • $\begingroup$ $x^2 \ge 0$ for all $x,$ and $x^2=0$ if and only if $x = 0, x^2+y^2 + z^2 = 0 \iff x = y = z = 0$ $\endgroup$ – Doug M Apr 4 '17 at 3:26
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It's not at all clear that $2(u_1v_1 + u_2v_2 + u_3v_3)$ is nonzero. It sort of looks nonzero, but if we don't know what sort of constraints are placed on the $u_i$ and $v_i$. In this case, it's helpful to first ask what those constraints are.

For $(x,y,z) \in U$, we have that $x^2 + y^2 + z^2 = 0$. But $x^2$, $y^2$, and $z^2$ are all non-negative numbers; if they add up to zero, they must all be zero. So the only point in $U$ is $(0,0,0)$. That single point is certainly closed under addition; $(0,0,0) + (0,0,0) = (0,0,0)$, and that's already an exhaustive list of all possible sums of members of $U$.

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