Extending the Intersection of Subspace For two subspace, one can express the dimension of the sum as 
$$ \dim(U_1 + U_2) = \dim U_1 + \dim U_2 - \dim (U_1 \cap U_2).$$ 
However, the obvious extension to three subspacess fails, in the sense that 
$$\dim (U_1 + U_2 + U_3) \neq \dim U_1 + \dim U_2 + \dim U_3 - \dim (U_1 \cap U_2) - \dim (U_1 \cap U_3) \\ - \dim (U_2 \cap U_3) + \dim (U_1 \cap U_2 \cap U_3).$$
I showed the above with a counter example but can't seem to understand it intuitively. Why does it fail?  In general, is the "obvious extension" true for $\dim(U_1+....+U_n)$ when $n$ is even and false when $n$ is odd? Is it just that two a special case?
 A: Using the original formula we get this:
$$
\dim(U_1 + U_2 + U_3) = \dim(U_1+U_2) + \dim(U_3) - \dim((U_1+U_2)\cap U_3)\\
= \dim(U_1) + \dim(U_2) - \dim(U_1 \cap U_2) + \dim(U_3) - \dim( (U_1 + U_2) \cap U_3)\\
= \dim(U_1) + \dim(U_2) + \dim(U_3) - \dim(U_1 \cap U_2) - \dim( (U_1+U_2) \cap U_3)
$$
The issue is how the "+" operation interacts with the "$\cap$" operation for subspaces, which is not very obvious. 
The "obvious extension" more or less assumes that $(U_1 + U_2)\cap U_3 = (U_1 \cap U_3) +  (U_2 \cap U_3) $, and this is not true in general. Now it is true that, $(U_1 + U_2)\cap U_3$ contains $(U_1 \cap U_3) +  (U_2 \cap U_3)$, so if your formula doesn't work it will overestimate the dimension.
A: 2 is a special case. The counterexample you have for 3 works for any $n\ge3$; take $n$ distinct one-dimensional subspaces of ${\bf R}^2$. Their sum is ${\bf R}^2$, which has dimension 2; the sum of the dimensions is $n$; the intersections all have dimension zero, so the "equation" gives $2=n$. 
