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?
 A: 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
A: Too long for a comment. I solved your question in a different way.
...
As I said in the comments, the first condition is not necessary and the second and the third conditions can be combined to make only one condition. See: https://en.wikipedia.org/wiki/Linear_subspace
SUB-SPACE CONDITION: A nonempty subset $W$ is a subspace of the vector space $V$ over the field $K$, if, whenever $w_1$, $w_2$ are elements of $W$ and $c_1$, $c_2$ are elements of $K$, it follows that $c_1 w_1 + c_2 w_2$ is in $W$.
Note that:
i) Sub-space condition $\implies$ 1. Condition: Let $c_1=c_2=0\in K$, $w_1=w_2=w\in W$. There is such an element $w$ since $W\neq\emptyset.$
ii) Sub-space condition $\implies$ 2. Condition: Let $c_1=c_2=1\in K$, $w_1,w_2\in W$ any.
iii) Sub-space condition $\implies$ 3. Condition: Let $c_1\in K$ any, $c_2=0\in K$, $w_1=w\in K$ any, $w_2=0\in W$.
Now, let me solve your question in this way by checking only one condition:
Let $w_1=(r_1-s_1,r_1+2s_1,-s_1)$ and  $w_2=(r_2-s_2,r_2+2s_2,-s_2)$ be two vectors in the sub-space $W$ you wrote. And let $c_1,c_2\in K=\Bbb{R}$. Then, here is kinda mind confusing, but we can show that $c_1w_1+c_2w_2=(r-s,r+2s,-s)$ where $r=c_1r_1+c_2r_2$ and $s=c_1s_1+c_2s_2$. So, Sub-space condition is satisfied and $W$ is a sub-space.
I think they separate this condition into three sub-conditions  to give quick answers to sub-space questions. Let me give an example: Let $A=\{(s,s+1)|s\in\Bbb{R}\}$. By your condition 1, for some $s$, we must have $(s,s+1)=(0,0)$. Then we have $s=s+1=0$ but then $s=s+1$ gives $0=1$. Contradiction. $A$ is not a sub-space.
