Let $U,\ W \leq V$ subspaces of a vector space V. What could the dimension of $U \cap W$ be, if $\dim{U} = 4,\ \dim{W} = 5,\ \dim{V} = 7$?

Now, given that $\dim{U + W} = \dim{U} + \dim{W} - \dim{U \cap W}$, and since U and V are subspaces, I assumed the best case scenario, where $U + W = V$, in which case, solving the previous equation, it gives a dimension of 2 for the intersection.

I feel this method somewhat lacking formality, but it's the best thing I can think of right now. Any feedback is appreciated deeply.

Solution 1:

The question suggests that you should determine all the possibilities. Your idea is good: since both spaces are contained in $V$, we have $\dim (U+W) \leq 7$. Also, since $W\leq U+W$, we get $\dim(U+W)\geq 5$. Hence $\dim(U\cap W)$ is between $9-7=2$ and $9-5 =4$.