# 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?

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.

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Do you want a constructive method or a mathematical proof of existence? If you want the latter and you are sure that such a subspace exists the pigeonhole principle would be something I would keep in the back of my mind. – Listing Jun 17 '11 at 14:41
Thanks for the fast reply. I am looking for a constructive method. i checked for the existence of the solution by comparing the number of variables and constraints, i think the solution exists. Thank you for the term "pigeonhole princple", i didnt know this before. i will look into it. – karthik Jun 17 '11 at 15:09
Is the intersection of two subspaces given to be a line? Certainly it is at least that, but it could be a plane. Was that part of the problem statement, or is it an assertion of yours? – Ross Millikan Jun 17 '11 at 23:06
The second paragraph is an assertion of my own. The five subspaces are chosen randomly. So with a very high probability the intersection between two subspaces is a line. The intersection to be a plane has very low probability. – karthik Jun 18 '11 at 7:42

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.

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Thank you very much for the answer. The vector space is over C. But i hope this will not affect the solution. – karthik Jun 21 '11 at 11:16
Over $\mathbb C$ it becomes even more likely that solutions will exist, because a system of $n$ polynomials in $n$ variables will have solutions (except in degenerate cases). – Robert Israel Jun 22 '11 at 21:52

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.

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a) The two-dimensional subspace spanned by $e_4$ and $e_5$ intersects each of these subspaces at least in a line. b) The conditions say that the intersection of any two subspaces results in a line; that would seem to exclude $U_3=U_1$. – joriki Jun 17 '11 at 20:47
Thanks for your answer. Yes i mean nonzero intersection. One thing is not clear for me. we are looking for a 2 dimensional subspace which intersect all the five subspace. I think, it is not necessary that these 5 subspaces intersect. For example, $e_1$ and $e_2$ be the basis of two dimensional subspace we are looking for. It can happen that neither of these two basis are in $U_1$. But ($e_1 + e_2$) is in $U_1$. It is possible that $U_1$ and $U_2$ intersect in someother direction $e_3$. And $e_3$ is not in our final subspace at all. Please correct me if i misunderstood something. – karthik Jun 18 '11 at 14:10