linear transformation Let $T : R^4 → R^2$ be the linear transformation given by
$$T(x_1; x_2; x_3; x_4) = (2x_1 + 2x_2 + 4x_3 + 6x_4; x_1 + 2x_2 + 2x_3 + 4x_4):$$
i. Find a basis for nullspace(T) and hence:
$\hspace{0.1in}$A. Find nullity(T).
$\hspace{0.1in}$B. Find nullspace(T), that is, describe it explicitly.
ii. Find a basis for range(T) and hence:
$\hspace{0.1in}$A. Find rank(T).
$\hspace{0.1in}$B. Find range(T), that is, describe it explicitly.
 A: I'll try to give you a hint as to the methods via a different example. 
If I had a similar system such as 
$$(x,y)\mapsto(2x+y,4x+2y),$$
I can write as a matrix: 
$$\begin{pmatrix}
2&1\\4&2
\end{pmatrix},$$
which in rref becomes: 
$$\begin{pmatrix}
1&0\\0&0
\end{pmatrix}.$$
This tells us the null space, that is, the solution set to the equation $Ax=0$, is 
$$x=0, y=t, $$
that is, $y$ is a free variable (null space is the $x$-axis). In other words, a basis for the null space is 
$$\left\{
\begin{pmatrix}
0\\1
\end{pmatrix}
\right\}.$$
The rref also tells us a way to represent the range, that is, by observing the pivot columns of the rref, we select those columns from the original matrix to describe the range. In this case a basis for the range, or column space, is 
$$\left\{
\begin{pmatrix}
2\\4
\end{pmatrix}
\right\}.$$
(Why does it work this way? That is, why does rref tell us the null space, and it's pivot columns tell us a basis for the range? Here's how I like to think of it: the row operations can be thought of as matrix multiplication of $Ax$ by an elementary matrix on the left, that is, $E_1Ax$, and this continues, $E_2E_1Ax$, etc. Each one is invertible, so a vector is a solution to $Ax=0$ if and only if it's a solution to the rref. So the null space of $A$ is equal to the null space of the rref. This doesn't quite hold for the range, since we are in a sense continually acting on the codomain. However, any linear relations that hold in the matrix $A$ are preserved by the invertible row operations, so linear dependence and independence is preserved. Thus, since the pivot columns form a basis for the (Edit: range of) rref, so do the corresponding columns of the original matrix $A$ (Edit: form a basis for the range of $A$).) 
