The title pretty much describes it. To be proven: for every subspace $U \subseteq \mathbb R^2$ there exists a linear map $\Phi$ with $\rm null(\Phi) = U$.
What would be the general approach to the problem? I've already proved it for $U = \{0\}$ and $U = \mathbb R^2$, but I can't think of an example of a map matrix in the case $U = \{tx |~ t \in \mathbb R\}$ (in other words the subspace of all lines through the origin). Any suggestions? Thanks in advance!
In the case $U=\{tx: t\in \mathbb R\}$ there exists $y$ such that $\{x,y\}$ is a basis for $\mathbb R^{2}$. Define $\Phi (ax+by)=(0,b)$. Then $\Phi$ is linear and its kernel is $U$.