Show that the range of a linear transformation is a subspace Let $T : \mathbb V \to \mathbb W$ be a linear transformation from a vector space $\mathbb V$ into a vector space $\mathbb W$. Prove that the range of $T$ is a subspace of $\mathbb W$.
OK here is my attempt...
If we let $x$ and $y$ be vectors in $\mathbb V$, then the transformation of these vectors will look like this... $T(x)$ and $T(y)$.
If we let $\mathbb V$ be a vector space in $\mathbb R^3$ and $\mathbb W$ be a vector space in $\mathbb R^2$, then
$$
T \begin{pmatrix} x_1\\ x_2 \\ x_3 
\end{pmatrix} = T\begin{pmatrix} x_1 + 2x_2 \\ 3x_3 + 4 \end{pmatrix}.
$$ 
Now if we tried to row reduce the matrix $\begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}$
we would get $\begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix} .$
SO the range of $T$ are the linear combinations of the pivot colums of the matrix above.
This is as much as I can do by myself. But now that i think about it, I believe this is wrong because the linear combinations of the pivot columns will give out any vector in $\mathbb R^2$, and not in the subspace of $\mathbb W$.
Any help will be appreciated.
 A: Given any two vector spaces $\mathbb{V, W}$ over $F$, and $T:\mathbb{V}\to\mathbb{W}$ is a linear transformation.


*

*$0_\mathbb{V}\in \mathbb{V}$. Since $T$ is linear, $T(0_\mathbb{V})=0_\mathbb{W},$ which implies $0_\mathbb{W}\in R(T).$

*Given any $x,y\in R(T),$ which implies there exist $v_x,v_y\in \mathbb {V}$ s.t. $T(v_x)=x, T(v_y)=y,$ but this implies for any $c\in F, T(cv_x+v_y)=^\dagger cT(v_x)+T(v_y)=cx+y,$ so $cx+y\in R(T).$
The 1. and 2. complete the proof.
$\dagger:$ linearity of $T:\mathbb{V}\to\mathbb{W}$.
A: Hint: Let $V$ and $W$ be vector spaces over a fixed field $k$ and $T : V \to W$ a morphism. What is the range of $T$? We have that $$T(V) = \{T(v) : v \in V\}.$$ Thus we have to show that $T(V) \subseteq W$ is a subspace. Do you know what one has to show for a set being a subspace?
Comment. It is a very good idea to get used to a concept by explicitely considering an example like you did and your idea is right.
A: The range of $T$ is a set inside the vector space $W$. To prove that a subset of a vector space is a subspace, you need to show that the subset is non-empty and closed under addition and scalar multiplication. Can you see why the range of $T$ is not empty? Now, pick two arbitrary members of the range of $T$. What can you say about them? How can you prove that their sum (well-defined in $W$) is also in the range of $T$? And similarly for the product with a scalar.
