Let $$V=\{(x, y, z) \in \mathbb{R}^3\,|\,x + y + z = 0\}$$ and $$W=\{(x, y, z) \in \mathbb{R}^3\,|\,x^2 + y^2 + z^2 = 0\}$$

I know I have to prove that the subsets are not empty (I am sure that both subsets are non-empty) and that they are also closed under both vector addition and scalar multiplication but I am not sure how to do so. Any help would be appreciated, thank you!

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    $\begingroup$ $W=\{(0,0,0)\}$ is a trivial subspace of $ \mathbb{R}^3$ . $\endgroup$ – Behrouz Maleki Sep 3 '16 at 7:36
  • $\begingroup$ what happens to if $v_1,v_2 \in V$, and then you do $v_1+v_2$? what happens to the coordinates of $v_1+v_2$ $\endgroup$ – An old man in the sea. Sep 3 '16 at 7:40

Let's start with $V$. You have to show the following:

  • $0 \in V$.
  • If $u \in V$ and $v \in V$, then $u+v \in V$.
  • If $u \in V$ and $c \in \mathbb{R}$, then $cv \in V$.

The first item is clear.

The second one, let $u = [u_1,u_2,u_3] \in V$ and same for $v \in V$, Then $u_1 + u_2 + u_3 = 0$ and $v_1 + v_2 + v_3 =0$. Add both equations you get $(u_1 + v_1) + (u_2 + v_2) + (u_3 + v_3) = 0$. Let $w = u+ v$. Therefore $w_1 + w_2 + w_3 = 0$. So $w \in V$.

The last one, if $u = [u_1,u_2,u_3] \in V$ Then $u_1 + u_2 + u_3 = 0$, multiple $c \in \mathbb{R}$ on both sides, and you get $cu_1 + cu_2 + cu_3 = 0$, so $cu \in V$.

As for $W$, it is a trivial subspace, meaning that only the $0$ vector is an element of $W$.

  • $\begingroup$ thank you! very clear answer $\endgroup$ – Ashley Sep 3 '16 at 7:46

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