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The answer with the most of upvotes on MO is this answer on $\dim(U+V+W)$.

Question:

1. Is it nonetheless true that every three vector subspaces $U$, $V$ and $W$ of a vector space $M$ satisfy $$ \dim(U +V + W) \le $$ $$ \dim U + \dim V + \dim W - \dim (U \cap V) - \dim (U \cap W) - \dim (V \cap W) + \dim(U \cap V \cap W) $$ ?

2. And, more generally, that $$\dim(\sum_{i = 1}^{n} U_i) \le \sum_{r=1}^{n} (-1)^{r+1} \sum_{i_1 < i_2 < \dots < i_r} \dim(\bigcap_{s=1}^{r}U_{i_s}) ? $$

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Sebastien Palcoux
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2 Answers2

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Yes, for $n=3$.

I assume known the equality in the case $n=2$, so $\dim (U+V)= \dim(U) +\dim(V) - \dim (U\cap V)$.

Using this twice, we have $\dim(U +V + W) = \dim(U +V) + \dim (W) - \dim((U +V) \cap W) = \dim(U ) + \dim(V) + \dim (W) - \dim(U \cap V) - \dim((U +V) \cap W) $.

As $(U+V) \cap W \supset (U \cap W) + (V \cap W) $ we have

$\dim((U +V) \cap W) \ge \dim ( (U \cap W) + (V \cap W))$

Then $ \dim ( (U \cap W) + (V \cap W)) = \dim(U \cap W) + \dim(V \cap W) - \dim((U \cap W) \cap (V \cap W))$.

As $(U \cap W) \cap (V \cap W)= U \cap V \cap W$, the claim follows.

darij grinberg
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quid
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  • Is it clear that the general case can be proved idem by induction? – Sebastien Palcoux Jul 27 '15 at 14:14
  • Originally I thought so (and even wrote it). But I redacted it in the grace period as some doubts emerged. As I should be doing something else right now I will not check right now. – quid Jul 27 '15 at 14:16
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The answer to question 2. is No for every $n\geq4$.

Here is a counterexample for $n=4$:

Let $\mathbf{k}$ be a field, and let $M=\mathbf{k}^{6}$.

Let $U_{1}=\left\{ \left( a,b,c,d,e,f\right) \in M\ \mid\ a=b\text{ and }d=e\right\} $.

Let $U_{2}=\left\{ \left( a,b,c,d,e,f\right) \in M\ \mid\ b=c\text{ and }e=f\right\} $.

Let $U_{3}=\left\{ \left( a,b,c,d,e,f\right) \in M\ \mid\ c=d\text{ and }f=a\right\} $.

Let $U_{4}=\left\{ \left( a,b,c,d,e,f\right) \in M\ \mid \ a+c+e=b+d+f\right\} $.

Then, your inequality becomes

$6\leq4+4+4+5-2-2-2-3-3-3+1+1+1+1-1$,

which is false.

To get counterexamples for any $n\geq4$, extend this by setting $U_{5} =U_{6}=\cdots=U_{n}=0$.

darij grinberg
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