The Three Subspace Problem, miniaturised

The two isomorphism classes consist of triples whose intersection is the origin, and triples whose intersection is a line through the origin.

Let $\ell_{12}, \ell_{13}, \ell_{23}$ be the three lines of pairwise intersection of the planes ( $\ell_{ij} = P_i \cap P_j$ ). If they are distinct, then I claim that vectors $v_{12}, v_{13}, v_{23}$ along these three lines are linearly independent. This is because each plane $P_i$ equals the span of the two (linearly independent) vectors whose index includes $i.$ But if the vectors were linearly dependent, that span would equal the span of all three vectors (again, because the corresponding lines are distinct, a dependence relation between the vectors would have to have all three coefficients be nonzero, so each vector would be a linear combination of the other two). So all three of the planes would be the same, which is not allowed.

So now I can find an invertible matrix $B$ sending $v_{12}, v_{13}, v_{23}$ to the three unit vectors along the $x,y,z$ -axes, respectively. Then the images of the planes under multiplication by $B$ are forced to be the $xy$ -plane, the $xz$ -plane, and the $yz$ -plane. So any ordered triple of planes whose intersection is the origin is isomorphic to this ordered triple of coordinate planes, and isomorphism is an equivalence relation, so any two triples whose intersection is the origin are isomorphic to each other. Note also that multiplication by an invertible matrix preserves intersection dimension, so this is a complete isomorphism class.

The other case is that the planes intersect in a common line through the origin. The only thing left to show is that any two such triples are isomorphic. The proof is similar: By pre-multiplying each triple by an appropriate invertible matrix, we may assume that triple $i$ ( $i=1,2$ ) consists of: the $xy$ -plane, the $xz$ -plane, and another plane $P_i,$ which by the initial assumption must be spanned by $\begin{pmatrix} 1\\0\\0\end{pmatrix}$ and $\begin{pmatrix}0\\a_i\\ b_i\end{pmatrix}$ where $a_1, b_1, a_2, b_2$ are nonzero. In this case, the matrix $\begin{pmatrix} 1 & & \\ & a_2/a_1 & \\ & & b_2/b_1 \end{pmatrix}$ transforms triple 1 into triple 2.

My solution is very similar to Patrick's. We consider Case 1 where the intersection of the three planes is just the origin and Case 2 where the intersection is a line $L$ .

In Case 1, we can choose nonzero vectors $v_{ij}\in V_i\cap V_j$ . The vectors $v_{12},v_{13},v_{23}$ will be independent; for example, $v_{23}$ fails to be in $V_1$ , by construction, and therefore fails to be a linear combination of $v_{12}$ and $v_{13}$ . We can construct $v_{ij}'\in V_i'\cap V_j'$ analogously, and we can find a unique, invertible matrix $A$ with $Av_{ij}=v_{ij}'$ and therefore $AV_k=V_k'$ , showing that any two triples in this case are isomorphic.

In Case 2, we can choose nonzero vectors $u\in L, v_1 \in V_1\setminus L, v_2 \in V_2\setminus L$ ; the vectors $u,v_1,v_2$ will be linearly independent, by construction. Any vectors $v_3 \in V_3\setminus L$ can be expressed as $v_3=au+bv_1+cv_2$ . Let $w_1=bv_1,w_2=cv_2,w_3=v_3-au \in V_3$ , so that $w_3=w_1+w_2$ . Define $u',w_1',w_2',w_3'$ analogously. There exists a unique, invertible matrix $A$ such that $Au=u',Aw_1=w_1',Aw_2=w_2'$ . Now we have $Aw_3=Aw_1+Aw_2=Aw_1'+Aw_2'=w_3'$ as well, so that $AV_k=V_k'$ since $V_k=span(u,w_k)$ and $V_k'=span(u',w_k')$ .

Thus there are $\boxed{2}$ isomorphism classes, corresponding to Cases 1 and 2.