Number of $m$-dimensional subspaces of $V$ is same as number of $(n-m)$-dimensional subspaces. 
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?
 A: 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 $\dim U^{\bot} = n- \dim U$.
A: No this is not correct. Not only there is more than one subspace complementary to a given subspace $U$, but choosing one such complementary subspace $W$, it is also complementary to many other subspaces $U'$ of the same dimension as $U$. So one cannot exclude the possibility that the same subspace $W$ is chosen as the complementary subspace many times, and this means one does not get any inequality between their numbers.
What one would need for this kind of argument is a standardised method to make a choice of complementary subspace that would guarantee that for $U'\neq U$ the complementary subspaces chosen will be different. However there is no easy such method.
A: Problem: How do you choose the complementary subspace? Maybe you choose the same complement for different subspaces. Then your estimate does not work...
Example: $\mathbb F_2^2$, the subspace $\langle (1,1)\rangle$ is complementary to $\langle (0,1)\rangle$ and $\langle (1,0)\rangle$. Why should you use a different one?
Solution: Using a nondegenerate bilinear form $b$, you can find $b$-orthogonal complements. This way your choice of complement becomes unique.
In the example above with the product $(a,b)\cdot(c,d)=ac+bd$, the space $\langle (0,1)\rangle$ is orthogonal to $\langle (1,0)\rangle$ and $\langle (1,1)\rangle$ to itself. Thus there are as many 1-dimensional subspaces as (2-1)-dimensional subspaces. (ok, the statement is stupid but the correspondence works)
