Showing that a transformation $T:\mathbb R^3 \to \mathbb R^2$ is linear OK, I am trying to prove the following transformation is linear, and find the basis for $\ker(T)$ and Im$(T)$ (also denoted in our textbook by $N(T)$ and $R(T)$ ). Then we're suposed to find the nullity and rank of $T$. 
$T: \Bbb{R}^3 \rightarrow \Bbb{R}^2$ defined by  $T(a_1, a_2, a_3) = (a_1-a_2, 2a_3)$
We want to see that the transformation preserved addition and scalar multiplication. So I define vector $a$ as $(a_1, a_2, a_3)$ and $b$ as $(b_1, b_2, b_3)$. So the first question is whether $T((a_1+b_1, a_2+b_2, a_3+b_3)$ = $T((a_1-a_2), 2a_3) + T(b_1-b_2, 2b_3)$
and when I plug in vectors $a+b$ to the transformation I get: 
$((a_1+b_1)-(a_2+b_2), 2(a_3+b_3))$ which works. So addition is preserved. 
The next question is whether it preserves scalar multiplication, or if $T(ca+b) = cT(a) + T(b)$ and as it happens:
 $T(ca_1+b_1, ca_2+b_2, ca_3+b_3) = ((ca_1+b_1-ca_2+b_2), 2(ca_3+b_3))$
and then if we break up the vectors we find that we get $(ca_1-ca_2, 2ca_3)+(b_1-b_2, 2ba_3)$ so the transformation is linear. 
To find the kernel we look for the set of vectors for which $T(a_1,a_2,a_3) = 0$. 
That happens whenever $a_1 = a_2$ and $a_3 = 0$ 
But that is where I get stuck because the definition of a kernel doesn't seem to fit. What is the basis for the kernel in this case? If a kernel is a set of vectors then this is making little or no sense to me from the get-go. Because I am not sure what the basis would be if the set of vectors are all those where $a_1 = a_2$ unless it's something like $(a_1, a_2, 0)$. And the dimension of the kernel is 2, I wold think intuitively, but I want to better understand why that is so I can get through the rest of the problem. 
 A: "... the definition of a kernel doesn't seem to fit." What do you mean?
You're looking at vectors of the form $(\lambda,\lambda,0)=\lambda(1,1,0)$, $\lambda\in\Bbb R$. This is just $\langle (1,1,0)\rangle$. This has dimension $1$.
A: First, because of paragraph 6, paragraphs 3-5 are not necessary. plugging in $ca+b$ is enough to demonstrate linearity.
Next, the kernel consists of all vectors of the form $(t,t,0)$, $t\in \mathbb{R}$. This means that $\{(1,1,0)\}$ is a basis for the kernel of this linear transformation.
A: A few things:
Your proof that $T$ is linear works.  However, you could save time by doing this in one step, showing that $T(\vec a+ c \vec b)=T(\vec a)+c\,T(\vec b)$; plugging in $c=1$ to this equality gives you addition, and plugging in $\vec a=\vec 0$ gives you scalar multiplication.
As for the second part of the question: the kernel is indeed the set of vectors such that $a_1=a_2$ and $a_3=0$.  That is, we have a system of two equations on a 3-dimensional system which means that our solution should be one-dimensional.  In particular, we can state that the vector $\vec v=(1,1,0)$ forms a basis of the kernel, since any vector satisfying these equations will be a multiple of $\vec v$
A: Your verification that $T$ is a linear transformation could use some work.  Try expanding out the first expression to show explicitly that $T(x + y) = T(x) + T(y)$.  Also, check your (linear) algebra in your proof that $T$ preserves scalar multiplication.
As for the kernel, you are correct $T(a_1, a_2, a_3) = 0$ when $a_1 = a_2$ and $a_3 = 0$. So $$N(T) = \{(a, a, 0) : a \in \mathbb{R}\}$$  Can you see why $\{(1, 1, 0)\}$ is a basis for $N(T)$?
It's pretty clear that $T$ is onto.  For any $(x,y) \in \mathbb{R}^2$, $(x, 0, y/2) \mapsto (x, y)$  so $R(T) = \mathbb{R}^2$.  You can also use the rank-nullity theorem, which gives us
$$3 = \dim \mathbb{R}^3 = \dim N(T) + \dim R(T) = 1 + \dim R(T)$$
