Subspace intersecting many other subspaces V is a vector space of dimension 7. There are 5 subspaces of dimension four. I want to find a two dimensional subspace such that it intersects at least once with all the 5 subspaces.
Edit: All the 5 given subspaces are chosen randomly (with a very high probability, the intersection is a line).
If i take any two of the 5 subspaces and find the intersection it results in a line. Similarly, we can take another two planes and find another line. From these two lines we can form a 2 dimensional subspace which intersect 4 of the 5 subspaces. But can some one tell me how we can find a two dimensional subspace which intersect all the 5 subspace.
It would be very useful if you can tell what kind of concepts in mathematics can i look for to solve problems like this?
Thanks in advance.
Edit: the second paragraph is one way in which i tried the problem. But taking the intersection of the subspace puts more constraint on the problem and the solution becomes infeasible.
 A: Assuming your vector space is over $\mathbb R$, it looks to me like "generically" there should be a finite number of solutions, but I can't prove that this finite number is positive, nor do I have a counterexample.  We can suppose your two-dimensional subspace $S$ has an orthonormal basis $\{ u, v \}$ where  $u \cdot e_1 = 0$ (where $e_1$ is a fixed nonzero vector).   There are 10 degrees of freedom for choosing $u$ and $v$.  The five subspaces are
the kernels of five linear operators $F_j$ of rank 3; for $S$ to have nonzero intersection with ${\rm ker} F_j$ you need scalars $a_j$ and $b_j$ with $a_j^2 + b_j^2  = 1$ and $F_j (a_j u + b_j v) = 0$.  This gives 5 more degrees of freedom for choosing points $(a_j, b_j)$ on the unit circle, minus 15 for the equations $F_j (a_j u + b_j v) = 0$, for a net of 0 degrees of freedom, and thus a discrete set of solutions (finite because the equations are polynomial).
For actually finding solutions in particular cases, I found Maple's numerical solver fsolve worked pretty well - the system seems too complicated for the symbolic solvers.
A: By "intersects" do you mean that the intersection is nonzero? Then in general there won't be a solution: the two-dimensional subspace spanned by the intersection of subspaces 1 and 2 and the intersection of subspaces 3 and 4 will intersect subspace 5 only in the zero vector.
For example, if $e_1, \ldots, e_7$ form a basis of $V$ take $U_1$ spanned by $e_1, e_2, e_3, e_4$, $U_2$ spanned by $e_1, e_5, e_6, e_7$, $U_3 = U_1$, $U_4$ spanned by $e_2, e_5, e_6, e_7$, and $U_5$ spanned by $e_3, e_4, e_5, e_6$.    
Edit:  Oops, this example is bogus, please ignore.
A: Here comes my ridiculously late answer: if you want to answer questions like this, the right methods are Schubert calculus. Ledoux and Malham give a good survey of this topic.
Specifically, there's a way of translating each subspace into a partition, and a way of multiplying linear combinations of partitions together. If the 5 partitions multiply to a linear combination containing at least one partition with at most 7 - 2 = 5 columns and 2 rows, then there is a 2 dimensional subspace that intersects all 5 of them. 
Here's the translation recipe: the requirement that the $2$-dimensional subspace $W$ intersect a $4$-dimensional subspace $V$ is the same as asking that $W$ is in the Schubert variety indexed by $\{4,7\}$ for any flag that has $V$ as its 4th element. Using the bijection "sh" in Section 4.3 of the survey, this turns into the partition $\{2,0\}$. Now we need to actually figure out how to compute these products. 
Anders Buch has written some code to multiply partitions in this way; see this Sage interface. I find it a little hard to use because it doesn't natively allow you to iteratively multiply partitions; as far as I can tell you have to add this functionality yourself.
I think in this situation you can proceed by hand using Pieri's rule, also mentioned in the survey. This is made easier because when you multiply partitions by hand, the number of boxes add, so the only admissible partition is $\{5,5\}$. You just have to figure out the number of ways to obtain $\{5,5\}$ using Pieri's rule. I got 6 ways (I could have messed up for sure). It's certainly at least 1, which answers your question in the affirmative. If the answer is really 6, then there should be 6 such subspaces.
