# Number of Subspaces of a Vector Space

If $V$ is an $n(<\infty)$ dimensional vector space over a finite prime field $\mathbb{F}_p$, then it is well known that the number of $1$-dimensional subspaces of $V$ is equal to the number of $n-1$ dimensional subspaces.

Question: Can one give a "natural" map (bijection) $f$ which associates to a $1$-dimensional subspace, an $n-1$ dimensional subspace?

(Here "natural bijection" can be understood as in case of conjugacy classes of $S_n$: writing each permutation of $S_n$ as a product of disjoint cycles, with monotonic (fixed) order, we have a natural bijection between conjugacy classes of $S_n$ and partitions of $n$. In case of finite dimensional vector spaces over $\mathbb{R}$, $\mathbb{C}$, "orthogonal complement'' with respect to an inner product gives a natural bijection between $1$-dimensional and $n-1$ dimensional subspaces; is there such type of bijection in above question?)

## 1 Answer

You could of course associate to the one-dimensional subspace $\langle (a_{1}, \dots, a_{n}) \rangle$ (so I am taking $(a_{1}, \dots, a_{n}) \ne 0$) the $(n-1)$-dimensional subspace $$\{ (x_1, \dots, x_n) : a_1 x_1 + \dots + a_n x_n = 0 \}.$$ But this is not natural, as it depends on the choice of a basis on $V$. And I don't think that your orthogonal complement is natural either, at least not in the sense I take as usual, as it depends on the choice of an inner product.