The number of subspaces in a finite field

Question

In my lecture note, there is the following statement:

The number of 2-dimnesional subspaces in $F_2^3$ equal to the number of 1-dimnesional subspaces in $F_2^3$

Here, $F_2^3=F_2 \times F_2 \times F_2$ where $F_2=\{0,1\}$.

I am wondering, why this should be true? is there some theorems explain this? or this is just special case?

Answer 1

This is just an instance of a general result:

If $F$ is a finite field, $V$ an $F$-vector space of finite dimension $n$, and $0\leq k\leq n$, then there are as many $k$-dimensional subspaces in $V$ as there are $(n-k)$-dimensional subspaces.

To see this, you can use the following to facts:

• First, show that if two vector spaces over $F$ have the same dimension, then they have the same number of subspaces of each dimension.

• Second, if $V$ is a vector space of dimension $n$, the rule which assigns to each subspace $S$ of $V$ its annihilator $S^\perp=\{\phi\in V^*:\phi(v)=0\;\forall v\in S\}$ which is a subspace of the dual space $V^*$, is a bijection between the set of subspaces of $V$ and the set of the subspaces of $V^*$, and $\dim S+\dim S^\perp=n$ for all subspaces $S\subseteq V$.

Answer 2

If V is a vector space of dimension n and F ve a finite field with q elements then number of subspace of dim k is ($q^n-1$)($q^n-q$).....($q^n-q^{k-1}$)/($q^k-1$)($q^k-q$).....($q^k-q^{k-1}$). In your case n=3 and q=2.Using this verify that in both the cases the number of subspace is 7.