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.

  • $\begingroup$ For the $m = 3$ case we have $\dim(W_1 + W_2 + W_3) = \sum_{i = 1}^3 \dim(W_i) - \sum_{1 \leq i \leq j \leq 3} \dim(W_i \cap W_j) + \dim(W_1 \cap W_2 \cap W_3) \implies k + 1 + y = 3k - 3(k - 1) + k - 1 - x \implies x + y = 1$. Since $x, y$ are non-negative integers, we must have $x = 0$ or $y = 0$. I'm not sure if this helps with the general case though. $\endgroup$ – Pratyush Sarkar Dec 15 '13 at 5:47
  • $\begingroup$ @Pratyush: I don't think your formula is correct $\endgroup$ – Seub Dec 15 '13 at 13:56
  • $\begingroup$ @Seub Why not? You can easily prove $\dim(W_1 + W_2) = \dim(W_1) + \dim(W_2) - \dim(W_1 \cap W_2)$ by starting with a basis for $W_1 \cap W_2$ and then extending it. Similarly, you can prove the formula I used. Its just like the formula for cardinality or probability of unions of sets. $\endgroup$ – Pratyush Sarkar Dec 15 '13 at 15:36
  • $\begingroup$ @Pratyush: why don't you try to prove it, maybe you'll see where it goes wrong (the sum of vector subspaces does not have as nice properties as union of sets). For a counter-example, consider three distinct lines (through the origin) in the plane. $\endgroup$ – Seub Dec 15 '13 at 17:29
  • $\begingroup$ @Seub You're right. Sorry about that. So its only true for sum of two subspaces. I guess for larger sums the formula will be not as nice. You can use the formula for two sums repeatedly but it will contain expressions like $\dim((W_1 + W_2)\cap W_3)$ which cannot be simplified since $(W_1 + W_2)\cap W_3 \neq (W_1 \cap W_3) + (W_2 \cap W_3)$ (an equality that we do have for unions). Thanks for correcting me. I learned something today. $\endgroup$ – Pratyush Sarkar Dec 15 '13 at 19:31

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.

  • $\begingroup$ Seems to work. This implicitly relies upon the distributive property of vector space addition and intersection ($A \cap (B \oplus C) = A \cap B \oplus A \cap C$, in the taking the intersection part in (b)) which is not completely obvious. Thank you. $\endgroup$ – Pastafarianist Dec 15 '13 at 12:58
  • 3
    $\begingroup$ @Pastafarianist: what you wrote in your comment is not true. What is true is that $(B \oplus A) \cap (C \oplus A) = (B \cap C) \oplus A$. $\endgroup$ – Seub Dec 15 '13 at 14:02

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.


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