Bounds on dimension of intersection of three subspaces Let three distinct  subspaces of $\mathbb{R}^{10}$ be $W_1,W_2,W_3$ have dimensions $9$ each. Then, what are the upper and lower bounds on the dimension of $W=\displaystyle\cap_{i=1}^3W_i$? Can we say that $dim (W)=7$? I know  that the standard formula for sets does not apply here because the subspaces need not be independent. Any ideas. Thanks beforehand.
 A: Fred gave you an upper bound for the dimension of $W$, if all spaces are the same, so let's work on a lower bound: In $\mathbb{R}^{10}$, we have the orthogonal complement of every space $U$, that satisfies $\dim(U) + \dim(U^{\perp}) = 10$. Now we can compute that 
$$W^{\perp} = W_1^{\perp} + W_2^{\perp} + W_3^{\perp},$$
as the orthogonal complement always exchanges intersection and sum. By the dimension formula, we know that $\dim(W_i^{\perp}) = 1$ for all $i$, and hence we get
$$\dim(W^{\perp}) \leq 3.$$
This allows us to compute the desired lower bound as
$$\dim(W) = 10 - \dim(W^{\perp}) \geq 10-3 = 7.$$
Of course you should show that this bound is tight, so you should give spaces $W_i$ that really achieve this bound. Given the construction above, you only need three linearly independent vectors to generate the one-dimensional spaces $W_i^{\perp}$, so it's not that hard to find the example achieving this bound.
edit: Regarding the question on how to show that the orthogonal complement exchanges sum and intersection:
We need to show for all subspaces $U,V$ that 
$$(U\cap V)^{\perp} = U^{\perp} + V^{\perp}.$$
Using that $(U^{\perp})^{\perp} = U$, we can equivalently show
$$U^{\perp} \cap V^{\perp} = (U + V)^{\perp}.$$
They can both be done element-wise, but I personally prefer the second one:
Let $x \in U^{\perp} \cap V^{\perp}$. Then $<x,u> = <x,v> = 0$ for all $u \in U$ and all $v \in V$. Thus also $<x,u+v> = 0$. But every element of $U+V$ is of the form $u+v$ for $u \in U$ and $v \in V$ and hence we have shown that $x \in (U + V)^{\perp}$. I'm sure you can now do the other direction on your own.
A: Note that if $W_i \neq W_j$ then $(W_i + W_j) = W$. Now use the fact that dim $(U+V) =$ dim $U$ + dim $V$ - dim $U \cap V$ to find dim $W_i \cap W_j$. Finally if $W_i \cap W_j \not \subset W_k$ for $k \neq i,j$ then $W = W_k + W_i \cap W_j$, so apply the dimension formula again.   
A: If you choose a basis for $R^{10}$, each of the spaces $W_i$ is represented by a linear equation with 10 unknowns; their intersection is the solution of the system of these three equations, an by Rouché-Capelli theorem the solution has dimension 10 minus the rank of the matrix of the coefficients of the system, and this rank is less or equal than 3 (the number of equations). Thus your conjecture is correct: $dim\ W \geq 7$.
