Everybody knows that there are $D_n=n! \left( 1-\frac1{2!}+\frac1{3!}-\cdots+(-1)^{n}\frac1{n!} \right)$ derangements of $\{1,2,\dots,n\}$ and that there are $D_n(q)=(n)_q! \left( 1-\frac{1}{(1)_q!}+\frac1{(2)_q!}-\frac1{(3)_q!}+\cdots+(-1)^{n}\frac1{(n)_q!} \right)$ elements in $\mathrm{GL}(q,n)$ which do not have $1$ as an eigenvalue; here $q$ is a prime-power, $(k)_q!=(1)_q(2)_q\cdots(k)_q$ are the $q$-factorials, and $(k)_q=1+q+q^2+\cdots+q^{k-1}$ are the $q$-numbers.

Now, there are $D_n^+=\tfrac12\bigl(|D_n|-(-1)^n(n-1)\bigr)$ and $D_n^-=\tfrac12\bigl(|D_n|+(-1)^n(n-1))\bigr)$ even and odd derangements of $\{1,2,\dots,n\}$, as one can see, for example, by computing the determinant $\left| \begin{array}{cccccc} 0 & 1 & 1 & \cdots & 1 & 1 \\ 1 & 0 & 1 & \cdots & 1 & 1 \\ 1 & 1 & 0 & \cdots & 1 & 1 \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ 1 & 1 & 1 & \cdots & 0 & 1 \\ 1 & 1 & 1 & \cdots & 1 & 0 \end{array} \right|$ and looking at the result.

How should one define $D_n^+(q)$ and $D_n^-(q)$?