Prove the following set of vectors is a subspace Let $\,S\,$ be the set of vectors in $\,\mathbb R^3$ that are of the form $\;\begin{bmatrix} a \\ b  \\ a-2b  \\ \end{bmatrix} \;$ where $a$ and $b$ are real numbers. 

Prove $\,S\,$ is a subspace of $\,\mathbb R^3$.



So I know in order to show that a set of vectors is a subspace the $0$ vector must be an element of the set, vector addition must be closed, and scalar multiplication must be closed. 

I started off showing the $0$ vector is an element of the set by having: 
$\;0\cdot a+0\cdot b+0\cdot \left(a-2b\right)=0\,$ which it does so that part is proven. 

Now how do I show the closure? 
For addition I wrote: 

\begin{align}
a &amp;= c_1a+c_2b+c_3\left(a-2b\right) \\
b &amp;= d_1a+d_2b+d_3\left(a-2b\right)
\end{align}

and then combined $\,a+b\,$ to be $\,\left(c_1+d_1\right)a + \left(c_2+d_2\right)b + \left(c_3+d_3\right)\left(a-2b\right)\,$ and since these are linear combinations of one another it is closed under addition? Is this correct? And how do I show closed under scalar multiplication?
Let $\vec{v}_1, \vec{v}_2$ be two arbitrary vectors of this form:

$$
\vec{v}_1=(a_1,b_1,a_1-2b_1)^T,\\
\vec{v}_2=(a_2,b_2,a_2-2b_2)^T.
$$

What we need to show is that their arbitrary linear combination $x\vec{v}_1+y\vec{v}_1$ is again of such form:
$$
x\vec{v}_1+y\vec{v}_1=(xa_1+ya_2,xb_1+yb_2,x(a_1-2b_1)+y(a_2-2b_2))^T=(xa_1+ya_2,xb_1+yb_2,xa_1+ya_2-2(xb_1+yb_2))^T
$$
You seem to be a little confused. You need to show that $$S=\{\begin{pmatrix} a\\ b\\a-2b \end{pmatrix}:a,b\in \mathbb R\}$$ is a subspace. 

The zero vector $\begin{pmatrix} 0\\ 0\\0 \end{pmatrix}$ is in $S$ since you may let $a=b=0$ to obtain it. 

Closure law holds as $$\begin{pmatrix} a_1\\ b_1\\a_1-2b_1 \end{pmatrix}+\begin{pmatrix} a_2\\ b_2\\a_2-2b_2 \end{pmatrix}=\begin{pmatrix} a_1+a_2\\ b_1+b_2\\a_1+a_2-2(b_1+b_2) \end{pmatrix}\in S$$ by letting $a=a_1+a_2$ and $b=b_1+b_2$.

Similarly for scalar multiplication you may select $a$ to be $\alpha a$ and $b$ to be $\alpha b$.