Given a prime power, $q=p^e$, is it known how many elements of $GL(n, q)$ have 1 as an unrepeated eigenvalue? By unrepeated eigenvalue, I mean that the dimensions of the eigenspace is 1. Or, how many elements fix all the elements of a one dimensional subspace of $k^n$, but fixes no other elements.
I have tried some counting arguments, starting with counting subspaces to fix and then working out how many bijections there can be to a subspace of dimension $n-1$ from remaining fixed basis vectors but I am stuck on this step.
Fix $q$ and let $N_n$ be the number you’re searching for. Choose, for each line of $L$ of $\mathbb{F}_q^n$, a subspace $S_L$ such that $S_L \oplus L=\mathbb{F}_q^n$.
A matrix $M \in GL_n(q)$ fixes the elements of $L$ and no other element iff it’s of the form $M(s+l)=l+u(s)+v(s)l$, for all $s \in S,l \in L$, and $u \in GL(S), v \in S^*$ and ($u$ is invertible without one as an eigenvalue or the eigenspace of $u$ with eigenvalue $1$ has rank one and $(u-id,v)$ has rank $n-1$.
It follows that $N_n=L_n \times (q^{n-1}U_{n-1}+ N_{n-1}D_{n-1})$, where: $U_n$ is the number of matrices $A\in GL_n(\mathbb{F}_q)$ such that $A-I$ is invertible, $D_n$ is the number of lines $L \in \mathbb{F}_q^n$ such that for a given matrix $S \in \mathcal{M}_n(\mathbb{F}_q)$ with rank $n-1$, $(S,L)$ has rank $n$ (and it doesn’t depend on $S$), and $L_n$ is the number of lines in $\mathbb{F}_q^n$.
It’s easy to check that $D_n=q^n-q^{n-1}, L_n=\frac{q^n-1}{q-1}$. It remains to find $U_n$ to obtain a recurrence relation on $N_n$, since $N_1=1$ – we have $N_n=\frac{q^n-1}{q-1}(q^{n-1}U_{n-1}+(q^{n-1}-q^{n-2})N_{n-1})$.
Note that since $U_1=q-2$, we find $N_2=(q+1)(q(q-2)+(q-1))=(q+1)(q^2-q-1)=q^3-2q-1$, as predicted by Jyrki Lahtonen.
I’m not sure how to find an exact formula $U_n$ exactly though. It’s easy to see that $U_n=q^{n^2}+O(q^{n^2-1})$, and we can give another recurrence formula:
We have $|GL_n(\mathbb{F}_q)|=\sum{|U(V)||\mathcal{L}(S_V,V)|U_{r}}$, where the sum is over the subspaces $V$ of $\mathbb{F}_q^n$, with $S_V$ a fixed subspace such that $\mathbb{F}_q^n=S_V \oplus V$, and $r=\dim\,S_v$, and $U(V)$ is the set of unipotent endomorphisms on $V$.
In other words, $|GL_n(\mathbb{F}_q)|=\sum_{r =0}^n{q^{r(n-r)}U_rO_{n-r}NS_{n-r}}$, where $O_t$ is the number of nilpotent matrices $t \times t$ in $\mathbb{F}_q$, and $NS_t$ is the number of subspaces of dimension $t$, ie $NS_t=\frac{(q^n-1)\ldots (q^n-q^{t-1})}{|GL_t(\mathbb{F}_q)|}=\prod{r=0}^{t-1}{\frac{q^n-q^r}{q^t-q^r}}$.
By the Fine-Herstein theorem (1958) $O_t=q^{t^2-t}$, so that $|GL_n(\mathbb{F}_q)|=\sum_{r=0}^n{q^{r(n-r)+r^2-r}U_{n-r}\prod_{t=0}^{r-1}{\frac{q^n-q^t}{q^r-q^t}}}$, ie
$$\prod_{t=0}^{n-1}{(q^n-q^t)}=\sum_{r=0}^n{U_{n-r}q^{r(n-1)} \prod_{t=0}^{r-1}{\frac{q^n-q^t}{q^r-q^t}}}.$$
This gives a self-contained recursive formula for $U_n$, and thus we have expressed $N_n$ in a (theoretically) completely computable way by saying $N_1=1$ and $$N_n=\frac{q^n-1}{q-1}(q^{n-1}U_{n-1}+(q^{n-1}-q^{n-2})N_{n-1}).$$