The Two Subspace Problem, miniaturised

It turns out that $(V,W)$ is isomorphic to $(V',W')$ if and only if $\dim V=\dim V'$ and $\dim (V\cap W)=\dim (V' \cap W')$ . Assume that $\dim V=\dim V'$ and $\dim (V\cap W)=\dim (V' \cap W')$ . We can construct a basis of $\mathbb{R}^4$ of the form $\mathscr{B}=(u_1,..,u_p,v_1,..,v_q, w_1,..,w_q,z_1,..,z_r)$ such that $(u_1,..,u_p)$ is a basis of $V \cap W$ , $(u_1,..,u_p,v_1,..,v_q)$ is a basis of $V$ , and $(u_1,..,u_p,w_1,..,w_q)$ is a basis of $W$ . We can construct an analogous basis $\mathscr{B'}$ for $V'$ and $W'$ and then find an invertible matrix $A$ that maps $\mathscr{B}$ to $\mathscr{B'}$ , defining an isomorphism from $(V,W)$ to $(V',W')$ .

Let's count the cases, for all possible dimensions of $\dim V$ and $\dim (V\cap W)$ . We have five trivial cases where $V=W$ , with $0 \leq \dim V \leq 4$ , and, in addition, for the pair $(\dim V, \dim (V\cap W))$ we have the possible values $(1,0), (2,0), (2,1)$ and $(3,2)$ , for a total of $\boxed{9}$ . Note that $\dim(V \cap W)\geq 2$ for two three-dimensional subspaces $V,W$ of $\mathbb{R}^4$ since $\dim V + \dim W = \dim(V \cap W)+\dim (V+W)=2p+2q$ , referring to the basis $\mathscr{B}$ constructed above.