Dimension of $(U+V+W)$ Let $U, V, W$ be subspaces of a vector space. I know that in general, the equation
$\dim(U+V+W)=\dim U+\dim V+\dim W-\dim(U \cap V)-\dim (U\cap W)-\dim (V \cap W)+\dim (U \cap V \cap W)$
doesn't hold. However, in the examples I've looked at, it always seems to be the case that
$\dim(U+V+W) \leq \dim U+\dim V+\dim W-\dim(U \cap V)-\dim (U\cap W)-\dim (V \cap W)+\dim (U \cap V \cap W)$.
My question is: is it possible to have
$\dim(U+V+W)>\dim U+\dim V+\dim W-\dim(U \cap V)-\dim (U\cap W)-\dim (V \cap W)+\dim (U \cap V \cap W)?$
I tried writing $\dim(U+V+W)$ as $\dim((U+V)+W)$ and using the formula for the dimension of the sum of 2 vector spaces, which reduces the problem to proving
$\dim(U \cap W)+\dim (V \cap W) \leq \dim(U \cap V \cap W)+\dim ((U+V) \cap W)$, but I'm not sure how to proceed further.
Thanks for the help.
 A: Yes. This can be explained using quivers in representation theory.
Consider the graph with one central vertex connected to three other vertices, none of which are connected to each other. This is a specific "quiver." We can associate the central vertex with an overarching vector space and the three outer vertices with three subspaces $U$, $V$, $W$. This is now a "quiver representation." It is possible to take the direct sum of two quiver representations (of the same quiver), by taking the direct sum of pairs of spaces in corresponding vertices. Any quiver representation will be a direct sum of "indecomposable" subrepresentations, those that are not direct sums of subreps. We can classify quiver representations up to isomorphism (an isomorphism is an invertible linear transformation which restricted to subspaces of one quiver rep map onto the subspaces of the other quiver rep), then, by classifying indecomposable reps.
The indecomposables are mentioned in John Baez's blog post here, in his first comment. For our purposes we can ignore the first five and focus on the last eight. Label their multiplicities $a,b,c,y,x,z,u,v$ respectively in the quiver. Then from these we can calculate the dimensions:

*

*$\dim(U+V+W)=a+b+c+x+y+z+u+2v$

*$\dim U=a+y+z+u+v$

*$\dim V=b+x+z+u+v$

*$\dim W=c+x+y+u+v$

*$\dim(U\cap V)=z+u$

*$\dim(V\cap W)=x+u$

*$\dim(W\cap U)=y+u$

*$\dim(U\cap V\cap W)=u$
Therefore in the naive inclusion-exclusion equation with $\dim(U+V+W)$ on the LHS,
$$ {\rm LHS}\le{\rm RHS} \iff 0\le v. $$
So your inequality is true. And, indeed, equality holds iff the exceptional quiver (with three lines intersecting trivially in 2D) is not present (note this is not possible for sets with union and intersection: it is not possible for three disjoint singleton sets to union to a set of cardinality two).
A: You should note that $(U\cap W)+(V\cap W)\subset(U+V)\cap W$. Indeed, if $x\in(U\cap W)+(V\cap W)$ then there exists $a\in U\cap W$ and $b\in V\cap W$ such that $x=a+b$. Since $a\in U$ and $b\in V$, we have $x\in U+V$ and since $a\in W$ and $b\in W$, we have $x\in W$, hence $x\in (U+V)\cap W$.
Now $U+V+W=(U+V)+W$ and by Grassman's formula :
$\dim(U+V+W)=\dim(U+V)+\dim(W)-\dim((U+V)\cap W)$
$\leqslant\dim(U)+\dim(V)-\dim(U\cap V)+\dim(W)-\dim((U\cap W)+(V\cap W))$
$=\dim(U)+\dim(V)+\dim(W)-\dim(U\cap V)-\left[\dim(U\cap W)+\dim(V\cap W)-\dim((U\cap W)\cap(V\cap W)))\right]$
$=\dim(U)+\dim(V)+\dim(W)-\dim(U\cap V)-\dim(U\cap W)-\dim(V\cap W)+\dim(U\cap V\cap W)$
