Linear algebra set-theoretic intersection Today,I took a linear algebra test and I had the following question in the test:
$W,U,V$ are all vector spaces. 
Prove the statement is true :
$W \cap [(W+V) \cap U + (U+V) \cap W]=(U+V)\cap W$
Can anyone help me? I want to know if what I did was right or wrong.
Thank you all.
 A: This is actually pretty tricky. One approach (perhaps the easiest) is to use the modular law.
Define $$\gamma = (U+V) \cap W$$
We're trying to show:
$$W \cap [(W+V) \cap U + \gamma]=\gamma$$
Since $\gamma \subseteq W,$ hence by the modular law, we have:
$$W \cap [(W+V) \cap U + \gamma] = [W \cap (W + V) \cap U]+\gamma= W \cap U+\gamma$$
But since $W \cap U \subseteq W \cap (U+V) = \gamma,$
Hence $$W \cap U + \gamma = \gamma.$$
So we're done. In fact, this shows that the identity of interest holds in any modular lattice. I wonder if the converse holds?
A: Because of distributive property
$ W∩\left[\left(W+V\right)∩U+\left(U+V\right)∩W\right]$
$ W∩\left[W∩U+V∩U+U∩W+V∩W\right]=$
$W∩W∩U+W∩V∩U+W∩U∩W+W∩V∩W$
Now $W∩W$ is same as just $W$ so it becomes
$W∩U+W∩V∩U+U∩W+V∩W $
Since $W∩U$ and $U∩W$ are the same, having that term more than once makes no difference so we can leave only one. And you'll get
$ W∩V∩U+U∩W+V∩W $
And also the term $  W∩V∩U $ is included in $U∩W+V∩W$ so it makes no difference if we remove it too. And by the distributive property you can get the final answer.
