Determining a matrix with given image and kernel. When constructing matrices, I'm not understanding how to have a given column space and null space. For example, if the column space is $x_1 = -2x_2$ then it can be written as $(-2 ,1)$ as a column, but I don't understand why. What if it were $x_2 = -2x_1$, is it $(1,-2)$? And how do I represent $x_2=3x_1+2x_3$ or $3x_1+2x_2+4x_3 = 0$ as a column space? 
For the null space I have the same problem. To check if the given null space is that of the matrix, I would multiply each element in the null space vector by the matrix and see if each row equals $0$. This works if I am given a line, like $x_1=4x_2$, but how can I do it if it is a plane like $x_1+2x_2+3x_2 = 0$? Do I need to put this in parametric form?
 A: Part I. Let's start from how to pass from cartesian equations to parametric equations of a vector subspace. Take $W=\{(x_1,x_2)\in \mathbb{R}^2 \,:\, x_2=-2x_1\}$: this is a vector subspace of $\mathbb{R}^2$ and its dimension is $1$. To find a generator of this subspace we need to find a non-zero vector such that $2x_1+x_2=0$, for instance $(1,-2)$. Thus we can write
$$
W = \langle \begin{pmatrix}1\\-2\end{pmatrix}\rangle = \left\{\lambda\begin{pmatrix}1\\-2\end{pmatrix} \,:\, \lambda\in \mathbb{R}\right\}.
$$
If you have the equation of a plane in $\mathbb{R}^3$ you need to find two independent generators of this subspace to write its parametric equations. Taking the first of your examples, consider $U=\{ (x_1,x_2,x_3)\in \mathbb{R}^3 \,:\, x_2=3x_1+2x_3\}$: this is a vector subspace of $\mathbb{R}^3$ of dimension $2$. To find two independent generators, the faster way I know is the following: let $x_1=1$ and $x_2=0$, then $x_3=-\frac 32$; let $x_1=0$ and $x_2=1$, then $x_3=\frac 12$. Thus $\left(1,0,-\frac 32\right)$ and $\left(0,1,\frac 12\right)$ generates $U$. You can also remove denominators by multiplying the two vector by $2$, and they still remain generators. So
$$
U=\langle \begin{pmatrix}2\\0\\-3\end{pmatrix}, \begin{pmatrix}0\\2\\1\end{pmatrix}\rangle = \left\{\lambda\begin{pmatrix}2\\0\\-3\end{pmatrix} + \mu\begin{pmatrix}0\\2\\1\end{pmatrix} \,:\, \lambda\in \mathbb R,\,\mu\in \mathbb{R}\right\}.
$$

Part II. Suppose you are asked to construct a $n\times n$ matrix $A$ with assigned image (or column space) $\text{Imm}\,A$ and kernel (or null space) $\text{Ker}\,A$. Note that, from the dimension theorem, we must have $\dim \text{Imm}\,A + \dim \text{Ker}\,A = n$. Furthermore, note that 

$$
 \mathbf{v}\in \text{Ker}\,A \iff \sum_{i=1}^n v_iA^i = \mathbf{0},
$$

where $A^1,\,\ldots,\,A^n$ are the columns of $A$. In other words, the coordinates of a vector in the kernel of $A$ are the coefficients of a null linear combination of the columns of $A$.
Suppose $n=2$, $\text{Imm}\,A = \{(x_1,x_2)\in \mathbb{R}^2 \,:\, x_2=-2x_1\}$ and $\text{Ker}\,A = \{(x_1,x_2)\in \mathbb{R}^2 \,:\, x_1=x_2\}$, both of dimension $1$. If you follow part I you can write
$$
\text{Imm}\,A = \langle \begin{pmatrix}1\\-2\end{pmatrix}\rangle
\quad\text{and}\quad 
\text{Ker}\,A = \langle \begin{pmatrix}1\\1\end{pmatrix}\rangle
$$
Thus a matrix $A$ satisfying these conditions is
$$
A=\begin{pmatrix} 1&-1\\-2&2\end{pmatrix},
$$
where the first column derives from the knowledge of $\text{Imm}\,A$ and the second follows from the condition above in the yellow box.
Suppose $n=3$, $\text{Imm}\,A = \{(x_1,x_2,x_3)\in \mathbb{R}^3 \,:\, x_2=3x_1+2x_3\}$ and $\text{Ker}\,A = \{(x_1,x_2,x_3)\in \mathbb{R}^3 \,:\, x_1=x_2,\ x_2=x_3\}$. Thus
$$
\text{Imm}\,A = \langle \begin{pmatrix}2\\0\\-3\end{pmatrix}, \begin{pmatrix}0\\2\\1\end{pmatrix}\rangle
\quad\text{and}\quad 
\text{Ker}\,A = \langle \begin{pmatrix}1\\1\\1\end{pmatrix}\rangle
$$
and then a matrix $A$ satisfying these conditions is
$$
A=\begin{pmatrix} 2&0&-2\\0&2&-2\\-3&1&2\end{pmatrix}.
$$
