Proving that the range of linear transformation is a linear subspace I need some help figuring out how to prove this question. 
True or false, the set $S = \left \{ A\mathbf{y}: \mathbf{y} \in \mathbb{R}^4\right \}$ is a subspace of $\mathbb{R}^3$ where A is a fixed $3\times4$ real matrix. 
Well I will need to show that the zero vector is in the set S. Then show the closure axioms hold. Or I can show for $A\mathbf{u}, A\mathbf{v}$ and some scalar $c$ in our field, $A\mathbf{u}+cA\mathbf{v} \in S$
What I have so far:
For $\mathbf{y} = \begin{pmatrix}
0\\ 
0\\ 
0\\
0 \end{pmatrix} \in \mathbb{R}^4$. It is clear that, $A\mathbf{y} = \begin{pmatrix}
0\\ 
0\\ 
0
\end{pmatrix}   \in S $. (Hopefully this is right so far)
Now, we need to prove the closure axioms.
Suppose $A\mathbf{u}, A\mathbf{v} \in S$ where $\mathbf{u} = \begin{pmatrix}
u_{1}\\ 
u_{2}\\ 
u_{3}\\
u_{4} \end{pmatrix},  \mathbf{v} = \begin{pmatrix}
v_{1}\\ 
v_{2}\\ 
v_{3}\\
v_{4} \end{pmatrix} \in \mathbb{R}^4 $
Here is where I'm stuck, I know I am suppose to show $A\mathbf{u} +  cA\mathbf{v} \in S$
Any clues or hints would be appreciated. Thanks. Please try to only post partial solutions or hints to get me going.

I think I figured it on, now I just need some confirmation so continuing from what I have above, we have $A\mathbf{u}+cA\mathbf{v} = A\begin{pmatrix}
u_{1}\\ 
u_{2}\\ 
u_{3}\\
u_{4} \end{pmatrix} + cA\begin{pmatrix}
v_{1}\\ 
v_{2}\\ 
v_{3}\\
v_{4} \end{pmatrix} $
$
= A\begin{pmatrix}u_{1}\\ 
u_{2}\\ 
u_{3}\\
u_{4} \end{pmatrix} + A\begin{pmatrix}
cv_{1}\\ 
cv_{2}\\ 
cv_{3}\\
cv_{4} \end{pmatrix} $
$
= A\begin{pmatrix}
u_{1}+cv_{1}\\ 
u_{2}+cv_{2}\\ 
u_{3}+cv_{3}\\
u_{4}+cv_{4} \end{pmatrix} \in S$ as $\begin{pmatrix}
u_{1}+cv_{1}\\ 
u_{2}+cv_{2}\\ 
u_{3}+cv_{3}\\
u_{4}+cv_{4} \end{pmatrix} \in \mathbb{R}^4$
