Proving The (Kind-Of) Inclusion-Exclusion Principle

Question

Let $V$ be an $F$-vector space and let $U$ and $W$ be finite dimensional subspaces of $V$.

Show that $U+W$ is finite dimensional, and moreover that

$$\dim_F(U+W)=\dim_F(U)+\dim_F(W)-\dim_F(U\cap W)$$

Conceptually, this makes complete sense. I can draw it as a venn-diagram and it's logically correct. I just have no idea how to prove it. I understand that some things in $U$ might be in $W$ (vice versa) and so we need to remove what they have in common, but in the best case, they share nothing in common so $\dim_F(U\cap W)=0$. Unfortunately, this isn't always the case.

Answer 1

Let $n$ and $m$ be the dimensions for U and V repsectively. Let $B:=\{v_1, ..v_r\}$ be a basis for $U \cap V$.

We need $n-r$ extra vectors to complete a basis for $U$. Similarly, we need $m-r$ extra vectors to complete a basis for $V$. Put those two basis together and you get a basis for $U \cup V$ after removing the $r$ repeated vectors (can you prove this fact?), so the dimension of $U \cup V$ is $m+n-r$

Answer 2

Well, I'd consider the linear mapping $\phi:U\times W\rightarrow U+W:(u,w)\mapsto u+w$.

The kernel is $U\cap W$ and so there is a linear isomorphism $\psi: (U\times W)/(U\cap W)\rightarrow U+W: (u,w)+ (U\cap W)\mapsto \phi(u,w)=u+w$.

From here the dimension formula follows:

$\dim (U\times W)/(U\cap W) = \dim U+W$ by the isomorphism,

$\dim (U\times W)/(U\cap W) =\dim U\times W - \dim U\cap W$ by the quotient, and

$\dim U\times W = \dim U + \dim W$.