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.