Linear algebra - Dimension theorem. Suppose we have a vector space $V$, and $U$, $W$ subspaces of  $V$.
Dimension theorem states:
$$ \dim(U+W)=\dim U+ \dim W - \dim (U\cap W).$$
My question is:
Why is $U \cap W$ necessary in this theorem?

Maybe you are looking for a more intuitive answer.
The sum $U+W$ of subspaces $U$ and $W$ is the smallest subspace of $V$ that contains $U$ and $W$. If $U$ and $W$ are not so independent then $U+W$ is very close to both subspaces. This happens just when the intersection $U\cap W$ is big. The extreme case is when $U=W$. In this case $U+W,\  U\cap W,\  U$ and $W$ are all equal and we have:
$$dim(U+W)=dim(U)=dim(W)=dim(U\cap W)$$
On the other hand, in this case: $$dim(U)+dim(W)=2dim(U+W)$$
I hope this helps you to understand why in the Dimension Theorem to obtain $dim(U+W)$ we have to substract $dim(U\cap W)$ to the sum of the dimensions of both spaces.
