Let $q$ be a prime power and define $(n)_q:=1+q+q^2+\cdots+q^{n-1}$, for a non-negative integer $n$. Let $B_q(n)$ denote the set of all subspaces of $\mathbb{F}_q^n$, the $n$-dimensional $\mathbb{F}_q$-vector space of all column vectors with $n$ components.

Define a $B_q(n)\times B_q(n)$ complex matrix $M_{q,n}$ with entries given by \begin{equation} M_{q,n}(X,Y):= \begin{cases} 1&\text{ if }Y\subseteq X, \dim(Y)=\dim(X)-1,\\ q^{\dim(X)}&\text{ if }X\subseteq Y, \dim(Y)=\dim(X)+1,\\ 0&\text{ otherwise.} \end{cases} \end{equation} We think of $M_{q,n}$ as a $q$-analog of the adjacency matrix of the $n$-cube. We show that the eigenvalues of $M_{q,n}$ are \begin{equation} (n-k)_q - (k)_q\text{ with multiplicity }\binom{n}{k}_q,\quad k=0,1,\dots,n, \end{equation} and we write down an explicit canonical eigenbasis of $M_{q,n}$. We give a weighted count of the number of rooted spanning trees in the $q$-analog of the $n$-cube.

This talk is based on a joint work with M. K. Srinivasan.