verify $R(T)=R^2$(Let $T:R^3 \rightarrow R^2$ be the linear mapping defined by $T(a_1,a_2,a_3)=(a_1-a_2,2a_3)$. 
Let $T:R^3 \rightarrow R^2$ be the linear mapping defined by $T(a_1,a_2,a_3)=(a_1-a_2,2a_3)$.

I want to verify $R(T)=R^2$
So $R(T)=T(x)=T(a_1,a_2,a_3)=(a_1-a_2,2a_3)$
Since it is closed under scalar multiplication and addition, $(a_1-a_2,2a_3)$ is indeed in $R^2$, so $R(T)=R^2$. Is that the right way of saying?
 A: If I'm understanding the question correctly, then we are given that $T:\Bbb{R}^3\to\Bbb{R}^2$ is defined by letting
$$T(x_1,x_2,x_3)=(x_1-x_2,2x_3)$$
for all $x_1,x_2,x_3\in\Bbb{R}$, and we need to show that $\text{Range}(T)=\Bbb{R}^2$.
Note that $T(1,0,0)=(1,0)$ and $T\left(0,0,\frac{1}{2}\right)=(0,1)$. Since $(1,0),(0,1)\in\text{Range}(T)$, we have that
$$\text{Range}(T)\supseteq\text{Span}\left\{(1,0),(0,1)\right\}=\Bbb{R}^2.$$
Hence $\text{Range}(T)=\Bbb{R}^2$.
Note: the above argument will work as long as the scalars don't come from a field of characteristic $2$.
A: It is easy to see that for any $(x,y)\in R^2$ the triple $(x,0,{y\over 2})\in R^3$ maps to $(x,y)$ and thus it is surjective i.e. $R(T) = R^2$.
A: It seems to me that, in order to complete your proof, you need to show that
$\dim R(T) = 2; \tag 1$
you have already shown that
$R(T) \subset \Bbb R^2 \tag 2$
is a subspace; if (1) is established, then we have
$R(T) = \Bbb R^2, \tag 3$
since a two-dimensional subspace of $\Bbb R^2$ must be the whole thing.  
This can simply be done by setting
$a_1 = 1, \; a_2 = 0, \; a_3 = 0, \tag 4$
so that
$(a_1 - a_2, 2a_3) = (1, 0) \in R(T), \tag 5$
and then by taking
$a_1 = a_2, \; a_3 = \dfrac{1}{2}, \tag 6$
whence
$(a_1 - a_2, 2a_3) = (0, 1) \in R(T), \tag 7$
so that $R(T)$ contains the basis
$\{(1, 0), (0, 1) \}, \tag 8$
which shows that (1), and hence (3), binds. 
On the other hand we may directly take, for any
$(b_1, b_2) \in \Bbb R^2, \tag 9$
$a_1 = a_2 + b_1, \; a_3 =  \dfrac{b_2}{2};  \tag{10}$
thus
$(a_1 - a_2, 2a_3) = (b_1, b_2) \in \Bbb R^2, \tag {11}$
and we are done.
