Is there always a bijection between the subspaces of dimension m and codimension m in a finite dimensional vector space? Say $U$ is a vector space of dimension $n$ over $\mathbb{F}$, if $\mathbb{F}$ is finite we know there exists a bijection between the subspaces of dimension $n-m$ and $m$. Can this be generalized to any $\mathbb{F}$(keeping $dim\ U$ finite)?
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(Answer after help from comments)
Yes, there is.
We can map any $M$ with $dim\ M = m$ to its annihilator which has dimension $n-m$ (note that it is an injective map), do the same in $U'$ for its subspaces with dimension $n-m$ and use the fact that isomorphic finite dimensional vector spaces have the same number of $m$-dimensional subspaces(which is an easy exercise).
 A: For finite dimensional vector spaces $V$ there is a natural bijection between subspaces of dimension $m$ of $V$ and subspaces of dimension $n-m$ of the dual of $V$. Given a subspace $A$ of dimension $m$, the corresponding subspace is the set of all linear functionals that vanish on $A$.
Edit: to incorporate Slade's comment and finish the story, if we have a nondegenerate bilinear form on $V$, not necessarily an inner product because this works over a completely general field, this gives us an isomorphism with $V^{\ast}$ and hence the bijection we seek.
A: If the vector space has an inner product, then every subspace has a unique orthogonal complement, and that gives you the bijection you seek.
So the problem is: for which fields can one define an inner product for vector spaces of that field?
The easy answer is $\mathbb R$ or $\mathbb C$.
In general, you need an ordered field, such as a subfield of $\mathbb R$.
I don't know what would be the analogue of $\mathbb C$ for a general ordered field.
