# 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

(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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