I read this problem,
If $V$ and $W$ are 2-dimensional subspaces of $\mathbb{R}^{4}$ what are the possible dimensions of $V\cap W$?
I know the answer is: 0, 1, or 2. But I would like more insight about this if possible. For instance, in $\mathbb{R}^{3}$ an intersection of two-dimensional subspaces could never form a 0-dimensional subspace--though a plane and line could do so. I feel like the Rank-Nullity Theorem is relevant here but I'm struggling to see just how it bears on this. If anyone has something that could help to explain just when certain subspaces are possible I would appreciate it.
The dimension of a subspace $V$ of $\Bbb R^n$ is determined by the dimension of its orthogonal complement $V^\perp$. $\operatorname{dim}V=n-\operatorname{dim}\ V^\perp$ where $$V^\perp=\{x\in \Bbb R^n| \langle x,y\rangle=0,\text{for all}\ y\in V\} $$ $\langle,\rangle$ is the usual dot product. $(V\cap Y)^\perp=V^\perp+Y^\perp$, so the question becomes equivalent to looking at possible dimensions for sums of vector spaces. For example if $V$ and $Y$ are 2-dimensional in $\Bbb R^4$, then $\operatorname{dim}(V^\perp+Y^\perp)=4,3,$ or $2$ by an easy argument. For $\Bbb R^3$, we have $\operatorname{dim}(V^\perp+Y^\perp)$ can only be $3$ or $2$, so we can conclude your cases.