Proving subspaces I want to prove that for all $r,s\in\mathbb{R}$
\begin{pmatrix}
r-s\\
r+2s\\
-s\\
\end{pmatrix}
is a subspace of $\mathbb{R}^3$.
I started with:
1. Contains origin) It contains the origin when $r=s=0$
2. Closed under addition) $$\begin{pmatrix}
r-s\\
r+2s\\
-s\\
\end{pmatrix}+\begin{pmatrix}
r-s\\
r+2s\\
-s\\
\end{pmatrix}=2\begin{pmatrix}
r-s\\
r+2s\\
-s\\
\end{pmatrix}\in\mathbb{R^3}\text{ (I'm not sure about this step)}$$
3. Closed under multiplication)$$c\begin{pmatrix}
r-s\\
r+2s\\
-s\\
\end{pmatrix}\in\mathbb{R^3}\text{ (also unsure about what should be stated here?)}$$
What else can I add or change to 2. and 3. that would make this a proof?

For 2, you have to show that 2 elements of the given form when added gets you another element of the given form.  In other words, you can't use the same $r,s$ for each.   so let $r,s,x,y\in \mathbb{R}$, you need to show that
$$\begin{pmatrix}
r-s\\
r+2s\\
-s\\
\end{pmatrix}+\begin{pmatrix}
x-y\\
x+2y\\
-y\\
\end{pmatrix}=\begin{pmatrix}
r-s+x-y\\
r+2s+x+2y\\
-s-y\\
\end{pmatrix}$$
is of that form,  i.e. you can find real numbers $a,b$ such that $$r-s+x-y=a-b$$
$$r+2s+x+2y=a+2b$$
$$-s-y=-b$$
Obviously $a$ and $b$ will be in terms of $r,s,x,y$,  it should be pretty obvious what the values have to be.
Similar for #3
