Kernel/Image expression 
Express the kernel of the 1 × 4 matrix A = \begin{bmatrix}1&2&3&4 \end{bmatrix}
  as the
  image of a 4 × 3 matrix B.

I understand that the kernel of a matrix is solving the system for A$\vec{x}$ = 0, but I have no idea what this question is asking nor how to do it. Are there any kind souls who can walk me through it?
 A: $Ax=0\Rightarrow [1 \ 2 \  3 \  4][x \  y \  z \  v ]^t=0\Rightarrow x+2y+3z+4v=0\Rightarrow x=-2y-3z-4v$
$\Rightarrow \begin{pmatrix} 
x  \\
y \\
z \\
v 
\end{pmatrix}=\begin{pmatrix} 
-2y  \\
y \\
0 \\
0 
\end{pmatrix}+\begin{pmatrix} 
-3z  \\
0 \\
z \\
0 
\end{pmatrix}+\begin{pmatrix} 
-4v  \\
0 \\
0 \\
v 
\end{pmatrix}\Rightarrow ker(A)=<\begin{pmatrix} 
-2  \\
1 \\
0 \\
0 
\end{pmatrix},\begin{pmatrix} 
-3  \\
0 \\
1 \\
0 
\end{pmatrix},\begin{pmatrix} 
-4  \\
0 \\
0 \\
1 
\end{pmatrix}>
$
Can you continue now?
A: Do you know the Rank-nullity Theorem? It states that, if $L: E \rightarrow F$ is a linear transformation, then $dim(E) = dim(im(L)) + dim(ker(L))$. Let $A:\mathbb{R}^{4} \rightarrow \mathbb{R}$ be the linear transformation such that its matrix in the canonical basis is your matrix $\mathbf{a}$. The image of $A$ is one dimensional, so from the Rank-nullity Theorem we have that $dim(ker(A)) = 3$. So, if you can find three linear independent vectors $\vec{x}_1, \vec{x}_2, \vec{x}_3$ in $\mathbb{R}^{4}$ such that $A\vec{x}_i = 0$, you will have found a basis of $ker(A)$. One such basis can be $\vec{x}_1 = (2, -1, 0, 0)$, $\vec{x}_2 = (3, 0, -1, 0)$, $\vec{x}_3 = (4, 0, 0, -1)$. Then, to construct a matrix $\mathbf{b}$ such that the image of its correspondent linear tranformation (in the canonical basis) $B: \mathbb{R}^{3} \rightarrow \mathbb{R}^{4}$ will be equal to $ker(A)$, it is enough to demand that $\vec{x}_1, \vec{x}_2, \vec{x}_3 \in im(B)$ (can you see why this is true?); we can set, for example, $B\vec{e}_i = \vec{x}_i$, and we will have
$ \mathbf{b} = 
\begin{bmatrix}
    2  & 3  & 4 \\
    -1 & 0  & 0 \\
    0  & -1 & 0 \\
    0  & 0  & -1  
\end{bmatrix}$
