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I was studying for some quals and I remember running into this problem last year and I couldn't get anywhere with it. Even now I'm kind of stumped. I was wondering if you guys had any ideas. Here's the problem:

Let $ V $ be a vector space and let $ 1\leq n< \operatorname{dim}(V) $ be an integer. Let $ \{V_i\} $ be a collection of $ n $-dimensional subspaces of $ V $ with the property that $$ \operatorname{dim}(V_i\cap V_j) = n-1 $$ for every $ i\neq j $. Show that at least one of the following holds:

(i) All $ V_i $ share a common $ (n-1) $-dimensional subspace.

(ii) There is an $ (n+1) $-dimensional subspace of $ V $ containing all $ V_i $.

poopstraw
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1 Answers1

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First let us show the following:

For any $i,j,k$, if we have $V_k\cap V_i\neq V_i\cap V_j,$ then $V_i\subseteq V_k+V_j$.

Note that $\dim(V_k+V_j)=\dim(V_k)+\dim(V_j)-\dim(V_k\cap V_j)=n+1$, so this is our unique candidate for a common $n+1$-dimensional superspace.

Now, if $V_k\cap V_i\neq V_i\cap V_j$, we know that there exists a vector $\nu_{k,i}\in V_k\cap V_i$ which is linearly independent of $V_j$. But since $V_i\cap V_j$ is an $n-1$ dimensional subspace of the $n$-dimensional space $V_i$, we get that $$ V_i=\textrm{span} \{\nu_{k,i}\}+V_i\cap V_j\subseteq V_k+V_j $$

Now, all we have to argue is that if $V_k\cap V_{i_0}\neq V_{i_0}\cap V_j$ for some $i_0$, then for any $i$ such that $V_k\cap V_i=V_i\cap V_j$, we must still have $V_i\subseteq V_k+V_j$.

This follows since $V_i\cap V_{i_0}\neq V_k\cap V_i$. Otherwise, $V_{i_0}$ would contain the $n-1$ dimensional subspace $V_i\cap V_j=V_i\cap V_k$ and thus, counting dimensions, we would have $$ V_{i_0}\cap V_k=V_i\cap V_k=V_i\cap V_j=V_{i_0}\cap V_j $$ Hence, applying the previous, we have $$ V_i\subseteq V_{i_0}+V_k\subseteq V_j+V_k $$

All in all, we've established the desired.

WoolierThanThou
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