How to prove the existence of non-intersecting subspaces? Let $V=\Bbb R^n$, $P,P'$ two $k$-dim subspaces, there must exist a $(n-k)$-dim subspace $Q$ whose intersection with both $P$ and $P'$ is zero. 
Although I have no idea how to prove this result, I feel it might be somehow connected to this problem.
Thanks for any help.
EDIT I tried arguing analogously to Steve's answer to the linked problem, but I was just not sure how to make a proper analogy, or these two problems are not actually related at all. By the way, geometric intuition tells me the result is still true for any finite collection of $k$-dim subspaces.
 A: Rather than look at a set of subspaces, you might look at the space $\def\R{\Bbb R}(\R^n)^{n-k}$ of $n-k$-tuples of vectors, each of which span a subspace that is a candidate for$~Q$ (some of them are linearly dependent, in which case $Q$ would not have the right dimension, but these $n-k$-tuples are among the ones that we are going to exclude). Fixing some basis of$~P$, we can complete those fixed $k$ vectors with our varying $n-k$-tuple and compute the determinant, giving a polynomial function $f:(\R^n)^{n-k}\to\R$ (in fact an $n-k$-linear alternating one) that is not identically zero (since $P$ certainly does have complementary subspaces). Now the "good" $n-k$-tuples are those for which $f$ does not vanish, and these form an open dense subset of $(\R^n)^{n-k}$. A similar construction for $P'$ gives another open dense subset, and the intersection of the two is still open and dense. In particular it is not empty, and the span of any $n-k$-tuple taken from it gives a$~Q$.
This argument immediately generalises to any finite set of $k$-dimensional subspaces to which one seeks a common complementary subspace.
A: The underlying fact here is that subspaces $X$ of codimension $k$ in $V={\mathbb R}^n$ form a manifold called the Grassmannian $Gr(V,k)$, whose dimension is easily computed.  The condition that a given $X\in Gr(V,k)$ should have nontrivial intersection with $P$ is a positive-codimension condition (also easily computed). Thus you have two positive-codimension subvarieties in $Gr$ defined by $P$ and $P'$ and it suffices to choose a point not in one of those subvarieties (this would work for any finite number of $P$'s and in fact even for a countable number of them).
