Non-distributivity of subspaces I believe that for subspaces $X,Y,Z$, distributivity does not apply, but could someone give an example to illustrate this: $(X\cap Y)+(X\cap Z)\neq X\cap(Y+Z)$? And perhaps suggest the criteria for this inequality to hold? Thanks.
 A: Let the vector space be $\mathbb R^2$, $Y$ and $Z$ the $x$- and $y$-axes, respectively, and $X=\{(x,x):x\in\mathbb R\}$.
Assuming you want conditions for the equality to hold: You will find in the (great!) book Quadratic Algebras by Alexander Polishchuk and Leonid Positselski the proof of the following statement

Let $V$ be a vector space and let $W_1$, $\dots$, $W_n$ be subspaces of $V$.  Then $W_i$ generate a distributive sublattice in the lattice of all subspaces of $V$ if and only if there exists a basis $\mathcal B$ of $V$ such that each $W_i$ is generated by a subset of $\mathcal B$.

From this, one gets a condition on three subspaces $X$, $Y$, $Z$ of a vector space to generate a distributive lattice. This is probably stronger than what you want, of course. (But it is in fact a theorem that if the equality in your question holds, then it also holds for all permutations of $X$, $Y$, $Z$)
A: I just happened to come to this page, and I thought the following example may still be of interest to you, even though you posed this question quite a while ago.
Consider the vector space $R^2$.  Let $V_1$ and $V_2$ be the subspaces spanned by the standard vectors $\delta_1$ and $\delta_2$ respectively.  Then $V_1 + V_2 = R^2$.  Next, let $W$ denote the subspace spanned by $\delta_1 + \delta_2$.  Then
$$W \cap (V_1 + V_2) = W \cap R^2 = W$$
But $W \cap V_1 = W \cap V_2 = \{0\}$, the zero subspace of $R^2$, so that 
$$(W \cap V_1) +  (W \cap V_2) = \{0\}.$$ 
That is 
$$W \cap (V_1 + V_2) \neq  (W \cap V_1) + (W \cap V_2)$$.
Thus dsitributivity is violated.  This is an example from Joseph Jauch, Foundations of Quantum Mechanics, Addison Wesley, 1968.  Also see Steven Roman,  Advanced Linear Algebra, Springer, 1992
