A proof that the number of 𝑚-dimensional subspaces of 𝑉 is same as number of (𝑛−𝑚)-dimensional subspaces.

by A_P   Last Updated July 11, 2019 19:20 PM - source

(This question has been asked before here. Based on the accepted answer (which is a hint), I fleshed out a more complete answer which I'd love feedback on. I hope this is an appropriate question to be asking here.)

Halmos 17.7: If $V$ is an $n$-dimensional vector space over a finite field, and if $0 \le m \le n$, then the number of $m$-dimensional subspaces of $V$ is the same as the number of $(n — m)$-dimensional subspaces.

Solution: Let $\operatorname{Sub}(V)$ denote all subspaces of $V$.

Let $A = \{a_1, ..., a_n\}$ be a basis for $V$, and $\{a_1, ..., a_m\}$ be a basis for subspace $W$.

Let $B = \{b_1, ..., b_n\}$ be the dual basis of $A$.

Let $C = \{c_1, ..., c_n\}$ be the dual basis of $B$. Via the canonical isomorphism, we can identify $a_k$ with $c_k$.

We have that the annihilator $\operatorname{ann}(W) = \operatorname{span}\{b_{m+1}, ..., b_n\}$, and $\operatorname{ann(ann}(W)) = \operatorname{span}\{c_1, ..., c_m\} = W$. This means that the annihilator gives an isomorphism between $\operatorname{Sub}(V)$ and $\operatorname{Sub}(V^*)$.

$\operatorname{ann}$ maps each $m$-dim subspace of $V$ to an $(n-m)$-dim subspace of $V^*$. Because there are equally many $m$-dim subspaces of $V$ and $V^*$, there are as many $m$-dim subspaces of $V$ as $(n-m)$-dim subspaces.

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Updated July 18, 2018 12:20 PM