3
$\begingroup$

Let $V$ be a vector space of dimension $n$ over $\mathbb{F}_q$, and let $U$ be a subspace of dimension $k$. I want to compute the number of subspaces $W$ of $V$ of dimension $m$ such that $W\cap U=0$.

I know why the number of subspaces of $V$ that contain $U$ and have dimension $m$ is $\binom{n-k}{m-k}_q$, but I don't understand why $q^{km}\binom{n-k}{m}_q$ is number of these subspaces?

  • 0
    yes, this is Gaussian integer2012-05-04

1 Answers 1

3

Hint

Consider the standard projection $\pi : V \to V/U,\quad v \mapsto v + U.$ For a subspace $W$ of $V$, $W\cap U = \{0\}$ is equivalent to $\dim(\pi(W)) = \dim(W)$.

Using this characterization, you have to count the possible $\pi$-images of $W$ in $V/U$, and for a fixed such $\pi$-image $M$ the suitable $\pi$-preimages of $M$ in $V$.

Let me remark that in the spacial case $m = n - k$, you get the number of complements of $U$ in $V$ as $q^{k(n-k)}$.