TITLE: Number of $m$-dimensional subspaces of $V$ is same as number of $(n-m)$-dimensional subspaces.
QUESTION [7 upvotes]: Let $V$ be an $n$-dimensional vector space over a finite field $F$. For $ 0\le m \le n$, the number of $m$-dimensional subspaces of $V$ is same as the number of $(n-m)$-dimensional subspaces.

I tried using the following argument: 
Corresponding to every $m$ dimensional subspace, we can find a complementary subspace of dimension $(n-m)$.
So,
$$
\text{number of subspaces of dimension }m\le \text{number of subspaces of dimension }(n-m)
$$
Reversing the argument, we get they are equal.
Is this correct?

REPLY [4 votes]: HINT: Denote by $\operatorname{Sub}(V)$ the set of subspaces of the vector space $V$. Use the map
$$\begin{matrix} \operatorname{Sub}(V) &\longrightarrow &\operatorname{Sub}(V^*)\\
U & \longmapsto & U^{\bot}
\end{matrix}$$
Where $U^{\bot} = \{ \phi \in V^* : \phi (U) = \{ 0 \} \}$. Since $V$ and $V^{**}$ are canonically isomorphic, you can consider an analogous map
$$\operatorname{Sub}(V^*) \longrightarrow \operatorname{Sub}(V^{**}) \ "=" \ \operatorname{Sub}(V)$$
And note that these two are inverse each other.
To conclude, you should know that $\dim U^{\bot} = n- \dim U$.