Existence of a certain subspace in a vector space 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.
 A: 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.  
A: There is (up to a certain point) a certain analogy between
vector spaces and elementary set theory, where disjoint union (of sets) is replaced
by direct sum (of vector spaces).
In the “disjunctive normal form” of a Boolean algebra, one decomposes 
every “case” as a list of mutually incompatible “basis cases” (that
may be seen as “atoms”).
Similarly, for any family of vector spaces, we may find a family of 
“atoms” such that every space in the original family can be decomposed as
a direct sum of “atoms”.
The main difference between elementary set theory and linear algebra
on this point is that the decomposition is not unique, because 
a subspace always has several complements (a set on the other hand has
always a unique complement in a superset).
Let us show how it works, by pushing your analysis further. There are spaces $A_1,A_2,A_3$ such that
$$
U_1=U \oplus A_1, U_2=U \oplus A_2, U_3=U \oplus A_3
$$
For any $i\neq j$, we have $A_i \cap A_j \subseteq U_i \cap U_j=U$, so
$A_i \cap A_j =\lbrace 0 \rbrace$.
There are spaces $B_1,B_2,B_3$ such that
$$
W_1=(U\oplus A_2 \oplus A_3) \oplus B_1,
W_2=(U\oplus A_1 \oplus A_3) \oplus B_2,
W_3=(U\oplus A_1 \oplus A_2) \oplus B_3
$$
Putting $u={\sf dim}(U),a_i={\sf dim}(A_i),b_j={\sf dim}(B_j)$, you have the
equations
$$
\begin{array}{lcl}
k &=& u+a_1+a_2+b_3=u+a_1+a_3+b_2=u+a_2+a_3+b_1,\\ 
k-1 &=& u+a_1=u+a_2=u+a_3 
\end{array}
$$
So $a_1=a_2=a_3=(k-1)-u$, and we deduce $b_1=b_2=b_3=u-k+2$. Since dimensions
are nonnegative, we must have $k\geq u$ and $u\geq k-2$, so $u$ is one of $k-2$ or
$k-1$.
If $u=k-2$, then all the $b_i$ are zero, and all the $a_i$ are equal to $1$. Then,
you have  a $Z$ : take $Z=U\oplus A_1 \oplus A_2 \oplus A_3$.
If $u=k-1$, then  you obviously have a $U$.
I leave to you the pleasure of working out the general case form this, feel free
to ask more questions if you need to.
