Let $V_1,V_2,V_3$ be 2-dimensional subspaces of a vector space $V$ with $\dim(V_i\cap V_j)=1$ for $i\ne j$. Prove that either $\dim(V_1\cap V_2\cap V_3)=1$ or $\dim (V_1+V_2+V_3)=3$.
Originally I thought I can apply some kind of inclusion-exclusion principle $$\dim(V_1+V_2+V_3)=\dim V_1+\dim V_2+\dim V_3-\sum_{i\ne j} \dim(V_i\cap V_j)+\dim (V_1\cap V_2\cap V_3)$$
But then I realized that this formula is not true when the $V_i$ are distinct lines in $\mathbb R^2$. So how do I prove what is required then?
Best Answer
$\dim (V_1\cap V_2\cap V_3)$ can only be $0$ or $1$ since $\dim (V_1\cap V_2)=1$. The question amounts to prove
So assume that $\dim (V_1\cap V_2\cap V_3)=0$. Since $\dim(V_i\cap V_j)=1$ iff $i\neq j$, we derive
$$V_1=(V_1\cap V_2)\oplus(V_1\cap V_3)$$ and similarly for $V_2$ and $V_3$.
Therefore $$V_1+V_2+V_3=(V_1\cap V_2)+(V_1\cap V_3)+(V_2\cap V_3)$$ which implies that
But we also have
since $V_2\not \subseteq V_1$.