Get from a transformation matrix to the resultant span of the solution set Guess that's a very basic question, but anyway:
I have the following transformation matrix:
$$\begin{bmatrix}1 & 0 & 0 & -\tfrac{1}{3}\\ 0 & -6 & -3 & -1\end{bmatrix}$$
And I know that the span of the solution set is $Sp\{(2, -1, 0, 6), (0, -1, 2, 0)\}$. But how do I get it?
I always thought that solving such matrix requires using parameters, so the solution would always be a general one (i.e., parameterized). But I guess that's not the case.
(If there's some missing part here, please let me know)
 A: To find the null space of a matrix, one wants to solve the equation $Ax=0$, where $A$ is the matrix of interest.  Let's do this for your case.
We want to find all the $x=(x_1,x_2,x_3,x_4)$ such that
$$\underbrace{\begin{bmatrix}1 & 0 & 0 & -\tfrac{1}{3}\\ 0 & -6 & -3 & -1\end{bmatrix}}_{A} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{bmatrix} = \begin{bmatrix}0\\ 0\end{bmatrix}.$$
Therefore we know
\begin{aligned}
  x_1 - (1/3) x_4 &= 0 \\
  -6 x_2 - 3 x_3 - x_4 &= 0
\end{aligned}
We have two equations, but we have four variables.  Therefore, if the equations are not redundant, we'll end up with a two dimensional vector space as the null set.  Let's find this vector space.
Let $x_4=s$  Then we know that $x_1=(1/3)s$.  Let $x_3=t$  Then we know that 
$$-6x_2 -3t -s=0.$$
Equivalently, this shows us that $x_2 = t/2 - s/6$.  This shows us that any vector in the null space of A is of the form
$$\begin{bmatrix}
  1/3 s \\ t/2 - s/6 \\ t \\ s
\end{bmatrix} = 
s\begin{bmatrix}1/3 \\ -1/6 \\ 0 \\ 1\end{bmatrix} + t\begin{bmatrix}0 \\ 1/2 \\ 1 \\ 0\end{bmatrix},$$
where $s$ and $t$ are real numbers.  Another way to state that is
$$\text{null}(A) = \text{span}\left(\begin{bmatrix}1/3 \\ -1/6 \\ 0 \\ 1\end{bmatrix},\begin{bmatrix}0\\ 1/2 \\ 1 \\ 0\end{bmatrix}\right)= \text{span}\left(\begin{bmatrix}2 \\ -1 \\ 0 \\ 6\end{bmatrix},
\begin{bmatrix}0\\ 1 \\ 2 \\ 0\end{bmatrix}\right).$$
