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?

  • 2
    $\begingroup$ Hint: $\dim(V_1 \cap V_2) = 1$, what does this tell you about the possible values of $\dim(V_1 \cap V_2 \cap V_3)$? $\endgroup$
    – wythagoras
    Jun 30, 2018 at 20:59

2 Answers 2


$\dim (V_1\cap V_2\cap V_3)$ can only be $0$ or $1$ since $\dim (V_1\cap V_2)=1$. The question amounts to prove

$$\dim (V_1\cap V_2\cap V_3)=0\Longrightarrow \dim (V_1+ V_2+ V_3)=3$$

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

$$\dim (V_1+ V_2+ V_3)\leq 3.$$

But we also have

$$\dim (V_1+ V_2+ V_3)\geq 3$$

since $V_2\not \subseteq V_1$.

  • $\begingroup$ I don't understand how you got the last two inequalities, even though you indicated what they follow from. I would think one needs some kind of inclusion-exclusion formula to deduce them. $\endgroup$
    – user557
    Jul 1, 2018 at 0:43
  • $\begingroup$ @user538518 No inclusion-exclusion. For the first, $V_1+V_2+V_3$ is a sum of three subspaces of dimension $1$, therefore its dimension is at most $3$. For the last one, $V_2$ has dimension $2$ and $V_1$ is not contained in $V_2$ so the sum has dimension at least one more than $V_2$ $\endgroup$ Jul 1, 2018 at 0:48
  • $\begingroup$ Why do we need that the sum is direct? The equalities like $V_1=(V_1\cap V_2)+(V_1\cap V_3)$ hold without assumptions on the triple intersection, and don't they imply $V_1+V_2+V_3=\sum V_i\cap V_j$? $\endgroup$
    – user557
    Jul 1, 2018 at 17:16
  • $\begingroup$ @user538518 No, the equality that you mention doesn't hold without the triple intersection assumption. If the triple intersection is not trivial, then $(V_1\cap V_2)+(V_1\cap V_3)=(V_1\cap V_2)$ has dimension $1$ and certainly cannot be equal to $V_1$. $\endgroup$ Jul 1, 2018 at 19:54
  • $\begingroup$ Why is $(V_1\cap V_2)+(V_1\cap V_3)=V_1\cap V_2$ if the triple intersection is not trivial? I thought you used that the triple intersection is trivial to conclude that $(V_1\cap V_2)\cap(V_1\cap V_3)$ is trivial, which along with $V_1=(V_1\cap V_2)+(V_1\cap V_3)$ gives $V_1=(V_1\cap V_2)\oplus (V_1\cap V_3)$. $\endgroup$
    – user557
    Jul 1, 2018 at 20:21

Hint: Since the $V_j$'s are all 2 dimensional and you have $\dim(V_1 \cap V_2)=1,$ and $V_1 \cap V_2\cap V_3$ is a subspace of $V_1 \cap V_2$, then it must be $V_1 \cap V_2\cap V_3$ has dimension $1$, or $V_1 \cap V_2 \cap V_3 =0$, this last case tells us $$(V_1 \cap V_2 )+V_3=(V_1 \cap V_2) \oplus V_3$$


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