How can you find a matrix given you know its kernel/nullspace? Suppose we are given that $\phi : \mathbb{R}^4 \rightarrow \mathbb{R}^3$, and also that $\ker\phi$ is the span of 
$\{\begin{pmatrix} 1 \\ 0 \\ 1 \\ 1 \end{pmatrix}, \begin{pmatrix} 2 \\ 1 \\ 0 \\ 1 \end{pmatrix}\}$
How can we find a matrix which corresponds to the linear map $\phi$?
Edit: I'm not looking for the unique matrix corresponding to $\phi$, merely any matrix which satisfies the given conditions.
 A: Call your vectors $v_1,v_2$. Pick two vectors $v_3,v_4$ such that $\{ v_1,\dots,v_4 \}$ is a linearly independent set. Pick linearly independent images $w_1,w_2$ for them. (The linear independence ensures that the kernel contains only your given vectors.) Then you want $A$ such that
$$A \begin{bmatrix} v_1 & v_2 & v_3 & v_4 \end{bmatrix} = \begin{bmatrix} 0 & 0 & w_1 & w_2 \end{bmatrix}$$
which you might write as $AV=W$. So $A=W V^{-1}$.
A: Short answer: Knowing the kernel is not enough to determine the whole matrix (unless the kernel is not the whole domain).
Explanation: To determine a matrix uniquely, you have to know the image of each basis vector of the domain (after you fix a certain basis in the domain). So you can complete the basis of the kernel to a basis of the whole domain. Here you have to add two vectors $v_1$ and $v_2$. For them you can set a different image in $\mathbb R^3$ (the image shall not be zero and $\phi(v_1)$ must be linearly independent with $\phi(v_2)$, because you do not want to make the kernel bigger). For nearly each chose of $\phi(v_1)$ and $\phi(v_2)$ you will end up with a different linear map $\phi$. 
