Find the number of $n-1$ dimensional subspaces $V_2$ such that $V_1+V_2=V$ where $\dim V_1=1$

Question

Let $V$ be an $n$ dimensional vector space over a finite field $F$ of order $p$ where $p=q^t$ and $q$ is a prime.

Let $V_1$ be a $1$ dimensional subspace of $V$. Show that number of $n-1$ dimensional subspaces $V_2$ of $V$ such that $V_1+V_2=V$ is $p^{n-1}$.

My try:

I took $n=3,p=2$. Then $V$ has $7$ one-dimensional and $7$ two-dimensional subspaces.

Listing the $1$ dimensional subspaces:

$\langle (1,0,0)\rangle ,\langle (0,1,0)\rangle ,\langle (0,0,1)\rangle ,\langle (1,1,0)\rangle ,\langle (0,1,1)\rangle ,\langle (1,0,1)\rangle ,\langle (1,1,1)\rangle $

Listing the $2$ dimensional subspaces:

$\langle [1 0 0], [0 1 0] \rangle, \langle [1 0 0], [0 0 1]\rangle,\langle [1 0 0] ,[0 1 1]\rangle ,\langle [0 1 0], [0 0 1]\rangle ,\langle [1 1 0], [0 0 1]\rangle ,\langle [1 0 1], [0 1 0]\rangle ,\langle [1 0 1] ,[0 1 1]\rangle $

If we take $V_1=(1,0,0)$ then we have $4$ choices for $V_2$ namely $\langle [0 1 0], [0 0 1]\rangle ,\langle [1 1 0], [0 0 1]\rangle ,\langle [1 0 1], [0 1 0]\rangle ,\langle [1 0 1] ,[0 1 1]\rangle $

So the answer is in this case $2^2=4$ and is true for any subspace $V_1$ of $V$

But how to frame the argument in general?I am unable to proceed in this case.Please help.

Answer 1

I'm going to reverse your notation to correspond to usual conventions. So let $p$ be a prime, let $q$ be a power of $p$.

Let's also establish the following notation: $[k] = [k]_q = \frac{q^k-1}{q-1}$, $[k]! = [k][k-1]\cdots [2][1]$, and $\genfrac [ ] {0pt}{} {n} {k} = \frac{[n]!}{[k]! [n-k]!}$.

Let $\def\F{\mathbb F} \F$ be a field of size $q$, and let $V$ be an $n$-dimensional vector space over $\F$. We fix a subspace $V_1$ of dimension $1$.

In general, the number of subspaces of $V$ of dimension $k$ is $\genfrac [ ] {0pt}{} {n} {k}$. In particular the number of subspaces of dimension $n-1$ is $\genfrac [ ] {0pt}{} {n} {n-1} = [n]$.

Note that subspaces of dimension $n-1$ which contain $V_1$ correspond one to one with subspaces of $V/V_1$ of dimension $n-2$, so using the previous paragraph, the number of these is $[n-1]$.

We want to count the number of subspaces of $V$ of dimension $n-1$ which do not contain $V_1$, thus the number is $[n] - [n-1] = q^{n-1}$.

Answer 2

I am sure that @fredgoodman 's approach is the right way to go about these questions in general. I only offer this to show that this particular question does not need the whole $q$-apparatus.

Let $V$ be $n$-dimensional over the finite field $\mathbb{F}_q$, and let $U$ be a fixed $r$-dimensional subspace. How many subspaces are there such that $V=U\oplus W$?

Let $e_1,e_2,\dots, e_r$ be a basis of $U$, and extend this to a basis $e_1,e_2,\dots, e_n$ of $V$.

The requirement $V=U+W$ implies that $W$ has a basis $e_{r+1}+u_{r+1}, \dots, e_n+u_n$ with the $u_i\in U.$

Moreover the requirement $U\cap W=0$ implies that different choices of the $u_i$ yield different spaces $W$.

So the number of $W$ is just the number of ways of choosing $(n-r)$ not necessarily different vectors in $U$; that is $(q^{r})^{n-r}$.

(The amended question asks only for the case $r=1$.)