Proving a subset is a subspace of $\mathbb{R}^3$ & Writing basis/dimensions I need help showing that 
$W = 
\begin{bmatrix}
    3s - 2t \\
    s + 2t \\  
    2s + 3t \\ 
\end{bmatrix}$ such that $s$ and $t$ are real numbers is a subspace of $\mathbb{R}^3$.
I also need to write a basis of $W$ and state the dimension of $W$. 
My ideas:
I know that the conditions for a subset being a subspace is that it most include the zero vector, and be closed under addition and multiplication. However, I do not know how to do this.
I am not so sure about writing a basis of $W$, or stating the dimension of $W$, either.
 A: It's better to write the subset as:
$$W=\left\{\begin{bmatrix}
    3s - 2t \\
    s + 2t \\  
    2s + 3t \\ 
\end{bmatrix}:s,t \in \mathbb{R}\right\}.$$

I know that the conditions for a subset being a subspace is that it most include the zero vector, and be closed under addition and multiplication. However, I do not know how to do this.



*

*To show the zero vector belongs to the set, we find values of $s \in \mathbb{R}$ and $t \in \mathbb{R}$ such that
$$\begin{bmatrix}
    3s - 2t \\
    s + 2t \\  
    2s + 3t \\ 
\end{bmatrix}=\begin{bmatrix}
    0 \\
    0 \\  
    0 \\ 
\end{bmatrix}.$$

*To show that it's closed under addition, we take two arbitrary elements from $W$, add them together, and show that the result is in $W$.  I.e. show
$$
\begin{bmatrix}
    3s - 2t \\
    s + 2t \\  
    2s + 3t \\ 
\end{bmatrix}
+
\begin{bmatrix}
    3s' - 2t' \\
    s' + 2t' \\  
    2s' + 3t' \\ 
\end{bmatrix}
=
\begin{bmatrix}
    3(s+s') - 2(t+t') \\
    (s+s') + 2(t+t') \\  
    2(s+s') + 3(t+t') \\ 
\end{bmatrix}
$$
belongs to $W$.

*To show that it's closed under scalar multiplication, we take an arbitrary element from $W$, and multiply it by an arbitrary scalar $\alpha \in \mathbb{R}$, and show that the result is in $W$.  I.e., we show
$$
\alpha
\begin{bmatrix}
    3s - 2t \\
    s + 2t \\  
    2s + 3t \\ 
\end{bmatrix}
=
\begin{bmatrix}
    3\alpha s - 2\alpha t \\
    \alpha s + 2\alpha t \\  
    2\alpha s + 3\alpha t \\ 
\end{bmatrix}
$$
is in $W$.
To identify a basis (a spanning, linearly independent subset), the co-efficients of $s$ and the coefficients of $t$ should give hint.  We can rewrite $W$ as:
$$W=\left\{s\begin{bmatrix}
    3 \\
    1 \\  
    2 \\ 
\end{bmatrix}+t\begin{bmatrix}
    - 2 \\
    2 \\  
    3 \\ 
\end{bmatrix}:s,t \in \mathbb{R}\right\}.$$  Note: we need to check that the basis vectors are indeed linearly independent.
The dimension of $W$ is the size of any of it's bases, by the Dimension Theorem, so once you've found a basis, the answer to this part is "what's it's size?".
