Let $V$ be vector space of dimension 6 over field $Z/7Z$. Let $W_1$ and $W_2$ be two subspaces of $V$. Let $\dim(W_1)=4$, $\dim(W_2)=3$. what about dimension of $W_1 \cap W_2 $? I know the formula $\dim(W_1 + W_2) = \dim(W_1) + \dim(W_2) - \dim(W_1 \cap W_2)$. I tried with this formula but could not find the proper answer. I am confused with, is dimension ($W_1 \cap W_2$) one? or greater than equal to one or anything else?
Note that $W_1 \subseteq W_1 + W_2 \subseteq V$. So $4 = \dim(W_1) \leq \dim(W_1 + W_2) \leq \dim(V) = 6$. Putting in the information we know into your formula gives that:
$$\dim(W_1 + W_2) = 7 - \dim(W_1 \cap W_2).$$
Using our inequality with this gives:
$$4 \leq 7 - \dim(W_1\cap W_2) \leq 6.$$
This inequality simplifies to:
$$ 1 \leq \dim(W_1 \cap W_2) \leq 3.$$
Without more information, we can't say more.