Show that vectors [ x y z] where $x+y+z=0$ is a subspace $V$ of $\mathbb{R}^3$. Show that vectors [x y z] where $x+y+z=0$ is a subspace $V$ of $\mathbb{R}^3$.
I am having trouble understanding how to properly set this problem up. I know the properties required to be met to be considered a subspace, I just am unsure how [x y z] gets set up in this problem. It says these are vectors, plural, but there are only 3 variables so this is throwing me off. Thanks for any help!
 A: The set that we are considering here is
$$W:=\{(x,y,z)\in \mathbb{R}^3\;|\;x+y+z=0\}$$
that is the set of ALL the vectors $(x,y,z)$ such that $x+y+z=0$.
In order to show that $W$ is a subspace of $\mathbb{R}^3$, you should show that it is closed with respect to linear combinations, that is
if $(x_{1},y_{1},z_{1})$ and $w(x_{2},y_{2},z_{2})$ are in $W$ then
also
$$a(x_{1},y_{1},z_{1})+b(x_{2},y_{2},z_{2})=(ax_1+b x_2,ay_1+b y_2,az_1+b z_2)\in W$$ 
for all $a,b\in\mathbb{R}.$
This is equivalent to say that if $x_1+y_1+z_1=0$ and  $x_2+y_2+z_2=0$ then 
$$(ax_1+b x_2)+(ay_1+b y_2)+(az_1+b z_2)=0$$
which can be easily verified.
A: Observe that the vectors $s:=\langle 1,0,-1\rangle$ and $t:=\langle 1,-1,0\rangle$ are linearly independent vectors in $W:= \{\textbf{x} = \langle x,y,z \rangle: x+y+z = 0\}$ i.e $W= \textbf{span}\{s,t\}\Rightarrow W$ is a vector space. 
A: One way to prove that a given nonempty subset $V$ of a vector space is a subspace is to use the one-step vector subspace test. It says that $V$ is a subspace if and only if $u+\lambda v\in V$ whenever $u, v\in V$ and $\lambda\in\Bbb R$.
Note that your $V$ is a nonempty subset of $\Bbb R^3$ since $(0,0,0)\in V$. To use the one-step vector subspace test, suppose $(a,b,c),(p,q,r)\in V$ and let $\lambda\in\Bbb R$. Then
$$
(a,b,c)+\lambda(p,q,r)=(
\underbrace{a+\lambda p}_{=x}, 
\underbrace{b+\lambda q}_{=y},
\underbrace{c+\lambda r)}_{=z}
$$
But this vector $(x,y,z)$ satisfies
\begin{align*}
x+y+z
&= (a+\lambda p)+(b+\lambda q)+(c+\lambda r) \\
&= \underbrace{a+b+c}_{=0}+\lambda\cdot\underbrace{(p+q+r)}_{=0} \\
&= 0
\end{align*}
That is, $(a,b,c)+\lambda(p,q,r)\in V$. Hence $V$ is a subspace.
A: Obviously $f\colon\mathbb{R}^3\to \mathbb{R}$, $f(x,y,z):=x+y+z$ is linear and $W$ is as the kernel of $f$ a subspace of $\mathbb{R}^3$.
A: If I understand correctly these are just 3D vectors where the sum of their coordinates is 0.
e.g. (0,0,0), (1,-1,0),(2,-1,-1)
You can then check if the subspace properties hold for such vectors.
