TITLE: Existence of a certain subspace in a vector space
QUESTION [9 upvotes]: Let $V$ be a vector space, and $W_1, W_2, \ldots, W_m \subset V$ its $k$-dimensional subspaces. Every their pairwise intersection is $k-1$-dimensional (for all $i \neq j$, $\dim (W_i \cap W_j) = k - 1$). Show that there is either:

a $k-1$-dimensional subspace $U \subset W_i$ in all $W_i$, or
a $k+1$-dimensional subspace $Z \supset W_i$ containing all $W_i$.


My try:
First, notice that $\dim W_i < \dim V$, since otherwise $W_i = V$. Thus there is always room for a $Z$.
$m=1$ is trivial.
For $m=2$ there is definitely a set $U = W_1 \cap W_2$, $\dim U = k - 1$. There is also a set $Z = W_1 + W_2$, since we can consider the basis $u_1, u_2, \ldots, u_{k-1}$ in $U$ add to it two vectors (one to complement the basis in $W_1$ and the other for $W_2$) and obtain a basis of $Z$. So for $m=2$ there are both $U$ and $Z$.
For $m=3$ things get interesting. We have k-dimensional $W_1, W_2, W_3$, and their pairwise intersections are all $k-1$-dimensional.

$U_1 = W_2 \cap W_3$, $\dim U_1 = k - 1$
$U_2 = W_3 \cap W_1$, $\dim U_2 = k - 1$
$U_3 = W_1 \cap W_2$, $\dim U_3 = k - 1$
$U = W_1 \cap W_2 \cap W_3$, $\dim U = k - 1 - x$

The same can be said about the sums instead of intersections:

$Z_1 = W_2 + W_3$, $\dim Z_1 = k + 1$
$Z_2 = W_3 + W_1$, $\dim Z_2 = k + 1$
$Z_3 = W_1 + W_2$, $\dim Z_3 = k + 1$
$Z = W_1 + W_2 + W_3$, $\dim Z = k + 1 + y$

The problem is, essentially, to show that either $x=0$ or $y=0$. It is probably also true that $x \leq 1$ and $y \leq 1$, but I have no idea how to prove that either.
So, that's where I got stuck. The picture I have in mind is the sequence of increasing subspaces $(U, U_i, W_i, Z_i, Z)$ with expanding basises, but their basises expand in some complicated manner I do not fully comprehend.
I suppose that the method for solving the case $m=3$ could be applied to all other $m$s, probably resulting in a proof by induction.
Any help would be appreciated.

REPLY [2 votes]: As you suspected, an induction argument on $m$ seems to be possible, along the following lines:
Assume $m > 2$.  Suppose the set of subspaces $\{W_1, \ldots, W_{m-1}\}$ has the required property.  There are two cases to consider:
(a) There exists a $(k-1)$-dimensional subspace $U$ contained in each $W_i$ for $1 \leq i \leq m-1$.  
In this case, it is not too hard to see that $U = W_1 \cap \cdots \cap W_{m-1}$.  If $U \subseteq W_m$, then $U$ is a $(k-1)$-dimensional subspace contained in all the $W_i$, and so we are done.  If $U \not\subseteq W_m$, then there exists $u \in U$ such that $u \notin W_m$, i.e., $u \in W_i$ for each $1 \leq i \leq m-1$ but $u \notin W_m$.  Hence $W_i = (W_i \cap W_m) \oplus \langle u \rangle$.  This implies $\sum_{i=1}^m W_i \subseteq W_m \oplus \langle u \rangle$, i.e., all the subspaces $W_i$ are contained in the $(k+1)$-dimensional subspace $W_m \oplus \langle u \rangle$.  
(b) There exists a $(k+1)$-dimensional subspace $Z$ containing each $W_i$ for $1 \leq i \leq m-1$.  
Then we have $Z = W_1 + \cdots + W_{m-1}$.  If $W_m \subseteq Z$, then $Z$ is a $(k+1)$-dimensional subspace containing all the $W_i$, and we are done.  Otherwise, $W_m \not\subseteq Z$, and there exists $w \in W_m$ such that $w \notin W_i$ for all $1 \leq i \leq m-1$.  So $W_m = (W_i \cap W_m) \oplus \langle w \rangle$ for each $1 \leq i \leq m-1$.  Taking the intersection, we have $W_m = (\bigcap_{i=1}^m W_i) \oplus \langle w \rangle$, so we can take $\bigcap_{i=1}^m W_i$ as the $(k-1)$-dimensional subspace contained in every $W_i$.  
I think this works, but please let me know if you find any problems.