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?
HINT: Denote by $\operatorname{Sub}(V)$ the set of subspaces of the vector space $V$. Use the map
$$\begin{matrix} \operatorname{Sub}(V) &amp;\longrightarrow &amp;\operatorname{Sub}(V^*)\\
U &amp; \longmapsto &amp; U^{\bot}
\end{matrix}$$
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 $\dim U^{\bot} = n- \dim U$.
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)