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)
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 $$\text{null}(A) = \text{span}\left(\begin{bmatrix}1/3 \\ -1/6 \\ 0 \\ 1\end{bmatrix},\begin{bmatrix}0\\ 1/2 \\ 1 \\ 0\end{bmatrix}\right)= \text{span}\left(\begin{bmatrix}2 \\ -1 \\ 0 \\ 6\end{bmatrix}, \begin{bmatrix}0\\ 1 \\ 2 \\ 0\end{bmatrix}\right).$$