\documentclass[ALCO,Unicode,published]{cedram} \usepackage{amsmath,amsxtra,amsthm,amssymb,graphics,hyperref,graphicx} \usepackage{latexsym,color} \newcommand{\ncom}{\newcommand} \newcommand{\cA}{{\mathcal M}} \newcommand{\cB}{{\mathcal B}} \newcommand{\cD}{{\mathcal Z}} \newcommand{\cE}{{\mathcal E}} \newcommand{\cF}{{\mathcal F}} \newcommand{\cI}{{\mathcal I}} \newcommand{\cM}{{\mathcal M}} \newcommand{\cN}{{\mathcal N}} \newcommand{\C}{{\mathbb C}} \newcommand{\Fq}{{{\mathbb F}_q}} \newcommand{\F}{{{\mathbb F}}} \newcommand{\kq}[1]{{{ {[#1]_q} }}} \newcommand{\qb}[2]{{{ {\binom{#1}{#2}}_q }}} \newcommand{\bin}[2]{{{ \binom{#1}{#2}}} } \newcommand{\oc}[1]{{{ \overline{#1} }}} \ncom{\noi}{\noindent} \ncom{\bq}{\begin{equation}} \ncom{\eq}{\end{equation}} \ncom{\beqn}{\begin{eqnarray*}} \ncom{\eeqn}{\end{eqnarray*}} \ncom{\ba}{\begin{array}} \ncom{\ul}{\underline} \ncom{\ea}{\end{array}} \ncom{\beq}{\begin{eqnarray}} \ncom{\eeq}{\end{eqnarray}} \ncom{\rar}{\rightarrow} \ncom{\lrar}{\longrightarrow} \ncom{\bc}{\begin{center}} \ncom{\ec}{\end{center}} \ncom{\bt}{\begin{theo}} \ncom{\bcon}{\begin{conj}} \ncom{\et}{\end{theo}} \ncom{\econ}{\end{conj}} \ncom{\bl}{\begin{lemm}} \ncom{\el}{\end{lemm}} \ncom{\bco}{\begin{coro}} \ncom{\ds}{\displaystyle} \ncom{\eco}{\end{coro}} \ncom{\bp}{\begin{prop}} \ncom{\ep}{\end{prop}} \ncom{\bex}{\begin{exam}} \ncom{\eex}{\end{exam}} \ncom{\bexr}{\begin{exer}} \ncom{\eexr}{\end{exer}} \ncom{\bprob}{\begin{problems}} \ncom{\eprob}{\end{problems}} \ncom{\bd}{\begin{defi}} \ncom{\ed}{\end{defi}} \ncom{\brm}{\begin{rema}} \ncom{\erm}{\end{rema}} \ncom{\ol}{\overline} \ncom{\wh}{\widehat} \ncom{\pf}{\begin{proof}} \ncom{\eprf}{\end{proof}} \ncom{\be}{\begin{enumerate}} \ncom{\ee}{\end{enumerate}} \ncom{\seq}{\subseteq} \newcommand{\inp}[2]{\langle {#1} ,\,{#2} \rangle} \title[A $q$-analog of the adjacency matrix of the $n$-cube]{A $q$-analog of the adjacency matrix of the $n$-cube} \author[\initial{S.} Ghosh]{\firstname{Subhajit} \lastname{Ghosh}} \address{Bar-Ilan University\\ Department of Mathematics\\ Ramat-Gan\\ 5290002 (Israel)} \email{gsubhajit@alum.iisc.ac.in} \author[\initial{M.} \middlename{K.} Srinivasan]{\firstname{Murali} \middlename{K.} \lastname{Srinivasan}} \address{Indian Institute of Technology, Bombay\\ Department of Mathematics\\ Powai, Mumbai\\ 400076 (India)} \email{murali.k.srinivasan@gmail.com} \thanks{The first named author is partially supported by the Israel science foundation grant $\#957/20$.} \keywords{$n$-cube, $q$-analog} \subjclass{05E18, 05C81, 20C30} \begin{abstract} Let $q$ be a prime power and define $(n)_q = 1+q+q^2+\cdots +q^{n-1}$, for a nonnegative integer $n$. Let $B_q(n)$ denote the set of all subspaces of $\mathbb F_q^n$, the $n$-dimensional $\mathbb Fq$-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 \[ M_q(n)(X,Y) = \begin{cases} 1 & \text{if }Y\subseteq X, \dim(Y)= \dim(X)-1,\\ q^k & \text{if }X\subseteq Y, \dim(Y)=k+1, \dim(X)=k,\\ 0 & \text{otherwise.} \end{cases} \] 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 \[ (n-k)_q - (k)_q \text{ with multiplicity } \binom{n}{k}_q,\;\;k=0,1,\ldots ,n, \] 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. \end{abstract} \dedicatory{To the memory of Chandan and beloved Lawson} \datepublished{2023-06-19} \begin{document} \maketitle \section{{Introduction}} \label{sec:intro} One aspect of algebraic combinatorics is the study of eigenvalues and eigenvectors of certain matrices associated with posets and graphs. Among the most basic such examples is the adjacency matrix of the $n$-cube, which has an elegant spectral theory and arises in a variety of applications (see ~{\cite{cst,st}}). This paper defines a $q$-analog of this matrix, studies its spectral theory, and gives an application to weighted counting of rooted spanning trees in the $q$-analog of the $n$-cube. Let $q$ be a prime power and define $(n)_q = 1+q+q^2+\cdots +q^{n-1}$, for a nonnegative integer $n$. Let $B_q(n)$ denote the set of all subspaces of $\F_q^n$, the $n$-dimensional $\Fq$-vector space of all column vectors with $n$ components. The set of $k$-dimensional subspaces in~$B_q(n)$ is denoted $B_q(n,k)$ and its cardinality is the {\em $q$-binomial coefficient $\qb{n}{k}$}. The {\em Galois number} $$G_q(n)=\sum_{k=0}^n \qb{n}{k}$$ is the total number of subspaces in $B_q(n)$. The set $B_q(n)$ has the structure of a graded poset of rank $n$, under inclusion. Recall the definition of the adjacency matrix $M(n)$ of the $n$-cube: let $B(n)$ denote the set of all subsets of $\{1,2,\ldots ,n\}$. The rows and columns of $M(n)$ are indexed by elements of $B(n)$, with the entry in row $S$, column $T$ equal to 1 if $|(S\setminus T)\cup (T\setminus S)|=1$ and equal to 0 otherwise. Define a $B_q(n)\times B_q(n)$ complex matrix $M_q(n)$ with entries given by \begin{eqnarray}\label{eq:mt1} M_q(n)(X,Y) &=& \begin{cases} 1 & \mbox{if }Y\seq X, \dim(Y)= \dim(X)-1,\\ q^k & \mbox{if }X\seq Y, \dim(Y)=k+1, \dim(X)=k,\\ 0 & \mbox{otherwise.} \end{cases} \end{eqnarray} We think of $M_q(n)$ as {\em a $q$-analog of the adjacency matrix of the $n$-cube}. Note that $M(n)$ is symmetric, has entries in $\{0,1\}$, and has all row sums equal to $n$. The significance of the definition above for the $q$-analog comes from the fact that though $M_q(n)$ lacks the first two properties it does have all row sums equal. Indeed, let $X\in B_q(n,k)$. Then the number of subspaces covering $X$ is $(n-k)_q$ and the number of subspaces covered by $X$ is $(k)_q$ and so the sum of the entries of row $X$ of $M_q(n)$ is $q^k(n-k)_q + (k)_q = (n)_q$. A scaling (i.e., a diagonal similarity) of $M_q(n)$ is symmetric. Let $D_q(n)$ be the~$B_q(n)\times B_q(n)$ diagonal matrix with diagonal entry in row $X$, column $X$ given by~$\sqrt{q^{\bin{k}{2}}}$, where $k=\dim(X)$. Then for $X\in B_q(n,k)$, $Y\in B_q(n,r)$ the entry in row~$X$, column $Y$ of $D_q(n)M_q(n)D_q(n)^{-1}$ is given by \beqn \lefteqn{\sqrt{q^{\bin{k}{2}}}\, M_q(n)(X,Y)\, \sqrt{q^{-\bin{r}{2}}}}\\ &=&\begin{cases} \sqrt{q^{\bin{k}{2}}}\, q^k\, \sqrt{q^{-\bin{k+1}{2}}} = \sqrt{q^k} & \mbox{if $X\seq Y$ and $r=k+1$,}\\ \sqrt{q^{\bin{k}{2}}} \sqrt{q^{-\bin{k-1}{2}}} = \sqrt{q^{r}} & \mbox{if $Y\seq X$ and $r=k-1$,}\\ 0 & \mbox{otherwise,} \end{cases}\\ &=&\begin{cases} \sqrt{q^{\min\{\dim(X),\dim(Y)\}}}& \mbox{if $X\seq Y$ or $Y\seq X$, and $|\dim(X)-\dim(Y)|=1$,} \\ 0 & \mbox{otherwise,} \end{cases} \eeqn yielding a symmetric matrix. It follows that $M_q(n)$ is diagonalizable and that its eigenvalues are real. In fact they are integral and the (eigenvalue, multiplicity) pairs of $M_q(n)$ are a $q$-analog of those for $M(n)$; the eigenvalues of $M(n)$ are $n-2k=(n-k)-(k)$ with multiplicity $\binom{n}{k}$, $k=0,1,\ldots ,n$ (see~{\cite{cst,st}}). \bt \label{mt1} The eigenvalues of the matrix $M_q(n)$ are $$(n-k)_q - (k)_q \mbox{ with multiplicity } \qb{n}{k},\;\;k=0,1,\ldots ,n.$$ \et We give two proofs of Theorem~\ref{mt1} in this paper. The first proof, given in Section~\ref{sec:eigenvals}, is based on two basic results of Terwilliger ~{\cite{t1,t2,t4}}. A result from ~{\cite{t1}} on the existence of a symmetric Jordan basis with respect to the up operator, together with formulas for the action of the up and down operators on this basis, reduces Theorem~\ref{mt1} to showing that the eigenvalues of $K_q(n)$, a certain $(n+1)\times (n+1)$ tridiagonal matrix, are $(n-k)_q - (k)_q,\;0\leq k \leq n$. The matrix $K_q(n)$ is a $q$-analog of the famous tridiagonal matrix $K(n)$ of Mark Kac ~{\cite{a,k,tt}} (both these matrices are defined in Section~\ref{sec:eigenvals}). Now $K_q(n)$ occurs in Terwilliger's classification of Leonard pairs ({\cite{t2, t4}}) and as such its eigenvalues/eigenvectors were known (see, for example, Lemma~4.20 in ~{\cite{t3}}), completing the first proof of Theorem~\ref{mt1}. Another natural problem is to write down eigenvectors of $M_q(n)$. For $X\in B_q(n)$ with $\dim(X)=k$ define $$\pi(X) = \frac{q^{\bin{k}{2}}}{P_q(n)}, $$ where $P_q(n) =\prod_{k=0}^{n-1} (1+q^k)$. We have $$\sum_{X\in B_q(n)}\pi(X)\;=\;\frac{\sum_{k=0}^n q^{\bin{k}{2}}\qb{n}{k}} {P_q(n)}\;=\;1$$ where the second equality follows by the $q$-binomial theorem (so $\pi$ is a probability vector on $B_q(n)$). Define an inner product on the (complex) vector space of column vectors with components indexed by $B_q(n)$ as follows: given vectors $u,v$ define \beq \label{inp} \inp{u}{v}_\pi =\sum_{X\in B_q(n)}\ol{u(X)}v(X)\pi(X). \eeq Since $P_q(n)$ is independent of $k$, the argument showing that $D_q(n)M_q(n)D_q(n)^{-1}$ is symmetric shows that $M_q(n)$ is self-adjoint with respect to the inner product~\eqref{inp}. Recall that a classical result exhibits an explicit orthogonal eigenbasis of $M(n)$ (under the standard inner product), see ~{\cite{cst,st}}. Up to scalars, this basis is canonical in the sense that no choices are involved in writing it down. Moreover, the product $M(n)v$, for $v$ in the eigenbasis, can be evaluated easily, determining the eigenvalues of $M(n)$. In Section~\ref{sec:eigenvecs} we extend this method to $M_q(n)$. \bt \label{evectors} There is an inductive procedure to write down a canonical eigenbasis of $M_q(n)$, orthogonal with respect to the inner product~\eqref{inp}. \et In the course of proving Theorem~\ref{evectors} we also evaluate the products $M_q(n)v$, for $v$ in the eigenbasis, thereby giving an alternate proof of Theorem~\ref{mt1}. Let us make a few informal remarks about the proof of Theorem~\ref{evectors}. For the matrix $M(n)$ one writes down an orthogonal eigenbasis, directly and explicitly for each $n$. It is not clear how to extend this direct approach to $M_q(n)$. On the other hand, we can inductively understand the eigenbasis of $M(n)$ and then try to extend that approach to $M_q(n)$. Roughly speaking, $M(n+1)$ can be built using two copies of $M(n)$ and this allows us to write down an eigenbasis of $M(n+1)$ given an eigenbasis of $M(n)$. The Goldman--Rota recurrence for the Galois numbers ({\cite{gr,kc,ku}}) \beq \label{gri} G_q(n+1) &=& 2G_q(n) + (q^n - 1)G_q(n-1),\;n\geq 1,\;\; G_q(0)=1,\;G_q(1)=2,\eeq suggests the possibility of building $M_q(n+1)$ from two copies of $M_q(n)$ and $(q^n-1)$ copies of $M_q(n-1)$ and using this to write down an eigenbasis of $M_q(n+1)$ given eigenbases of $M_q(n)$ and $M_q(n-1)$. This is implemented in Section~\ref{sec:eigenvecs} using a linear algebraic interpretation of the Goldman--Rota recurrence that was worked out in ~{\cite{s3}} (and summarized in Section~\ref{sec:decomp} of the present paper). The two copies of $M_q(n)$ occurring in $M_q(n+1)$ is along the same lines as the $q=1$ case (although we need some powers of~$q$ not visible in the $q=1$ case, see~\eqref{ind} and cases (a), (b) in the proof of Theorem~\ref{mt}). The extra feature here is the $q^n-1$ copies of $M_q(n-1)$ in $M_q(n+1)$. Here a central role is played by the characters of the additive abelian group $\F^n_q$. There is a copy of $M_q(n-1)$ inside $M_q(n+1)$ for every nontrivial irreducible character of~$\F^n_q$, so $q^n-1$ copies in all. These introductory remarks on Theorem~\ref{evectors} are continued at the beginning of Section~\ref{sec:eigenvecs}. We originally arrived at the matrix $M_q(n)$ through a reversible Markov chain with state space $B_q(n)$, transition matrix $\frac{1}{(n)_q}M_q(n)$, and stationary distribution $\pi$ (see ~{\cite{gs}}). Since the spectral theory of $M_q(n)$, as a $q$-analog of the spectral theory of $M(n)$, is of independent interest we are presenting it separately in this paper. Another application concerns weighted enumeration of rooted spanning trees in the $q$-analog of the $n$-cube. Let us first recall the remarkable product formula (see Example 9.12 in ~{\cite{st}}) for the number of rooted spanning trees in the $n$-cube. Let $C(n)$ denote the $n$-cube (i.e., the Hasse diagram of the poset $B(n)$ treated as a graph) and let $\cF(n)$ denote the set of all rooted spanning trees of $C(n)$. The Laplacian eigenvalues of $C(n)$ are well known to be $2k,\;k=0,1,\ldots ,n$ with multiplicity $\bin{n}{k}$. It thus follows from the Matrix-Tree theorem that the number of rooted spanning trees of $C(n)$ is given by \beq \label{hc} \sum_{F\in \cF(n)} 1&=&\prod_{k=1}^n (2k)^{\bin{n}{k}}. \eeq The {\em $q$-analog $C_q(n)$} of $C(n)$ is defined to be the Hasse diagram of $B_q(n)$ treated as a graph (note that $M_q(n)$ is not the adjacency matrix of the graph $C_q(n)$). The eigenvalues of the Laplacian of $C_q(n)$ are not known (note that $C_q(n)$ is not regular) but it was shown in ~{\cite{s2}} that the product of the nonzero eigenvalues of the Laplacian is more tractable and led to a product formula for the number of spanning trees of $C_q(n)$, although the individual terms in this product are not explicitly given but only as a positive combinatorial sum. Here we obtain an explicit $q$-analog of~\eqref{hc} by a weighted count of the rooted spanning trees of $C_q(n)$. Let $F\in \cF_q(n)$, the set of all rooted spanning trees of $C_q(n)$. Orient every edge of~$F$ by pointing it towards the root. Let $e=(X,Y)$ be an oriented edge of $F$. We say $e$ is {\em{spin up}} if $\dim(Y)=\dim(X)+1$ and is {\em{spin down}} if $\dim(Y)=\dim(X)-1$. The {\em{weight}} of $F$ is defined by \beqn w(F)&=&\sum_{(X,Y)}\dim(X), \eeqn where the sum is over all spin up oriented edges of $F$. In Section~\ref{sec:trees} we prove the following generalization of~\eqref{hc} (the proof can be read at this point, assuming Theorem~\ref{mt1}). \bt \label{qct} We have $$\sum_{F\in\cF_q(n)}q^{w(F)} = \prod_{k=1}^n ((1+q^{n-k})(k)_q)^{\qb{n}{k}}. $$ \et \section{{Eigenvalues of \texorpdfstring{$M_q(n)$}{Mq(n)} }} \label{sec:eigenvals} The spectral theory of $M_q(n)$ goes hand in hand with that of a $q$-analog of the Kac matrix. Recall that the Kac matrix is a $(n+1)\times (n+1)$ tridiagonal matrix $K(n)$ with diagonal $(0,0,\ldots ,0)$, subdiagonal $(1,2,\ldots ,n)$ and superdiagonal $(n,n-1,\ldots ,1)$: \[K(n) = \left[\ba{cccccc} 0&n&&&&\\ 1&0&n-1&&&\\ &2&0&n-2&&\\ &&\ddots&\ddots&\ddots&\\ &&&n-1&0&1\\ &&&&n&0 \ea\right].\] The eigenvalues of $K(n)$ are $n-2k, k=0,1,\ldots ,n$, and its eigenvectors have been written down (see ~{\cite{a,cst,ek,k,tt}}). We define the $q$-analog of the Kac matrix to be the $(n+1)\times (n+1)$ tridiagonal matrix $K_q(n)$ with diagonal $(0,0,\ldots ,0)$, subdiagonal $((1)_q,(2)_q,\ldots ,(n)_q)$ and superdiagonal $((n)_q,q(n-1)_q,\ldots ,q^{n-1}(1)_q)$: \[K_q(n) = \left[\ba{cccccc} 0&(n)_q&&&&\\ (1)_q&0&q(n-1)_q&&&\\ &(2)_q&0&q^2(n-2)_q&&\\ &&\ddots&\ddots&\ddots&\\ &&&(n-1)_q&0&q^{n-1}(1)_q\\ &&&&(n)_q&0 \ea\right].\] More formally, let us index the rows and columns of $K_q(n)$ by the set $\{0,1,2,\ldots ,n\}$. If $c_0,c_1,\ldots ,c_n$ denote column vectors in $\F_q^{n+1}$ with $c_i$ having a 1 in the component indexed by $i$ and 0's elsewhere (and we set $c_{-1}=c_{n+1}=0$) then, for $0\leq k \leq n$, column $k$ of $K_q(n)$ is \beq \label{ctm} &(k+1)_q\;c_{k+1} + q^{k-1} (n-k+1)_q\;c_{k-1}.& \eeq We now discuss the significance of $K_q(n)$ for studying $M_q(n)$. The proper framework for studying this are the up (and down) operators on the poset of subspaces. For a finite set $S$, we denote the complex vector space with $S$ as basis by $\C[S]$. We denote by $r$ the rank function (given by dimension) of the graded poset $B_q(n)$. Then we have (vector space direct sum) $$\C[B_q(n)]=\C[B_q(n,0)]\oplus \C[B_q(n,1)]\oplus \cdots \oplus\C[B_q(n,n)].$$ An element $v\in \C[B_q(n)]$ is {\em homogeneous} if $v\in \C[B_q(n,i)]$ for some $i$, and if~$v\not= 0$, we extend the notion of rank to nonzero homogeneous elements by writing $r(v)=i$. For $0\leq k \leq n$, the {\em $k^{th}$ up operator} $U_{n,k}:\C[B_q(n)]\rar \C[B_q(n)]$ is defined, for $X\in B_q(n)$, by $U_{n,k}(X)=0$ if $\dim(X)\not=k$ and $U_{n,k}(X)= \sum_{Y} Y$, where the sum is over all $Y\in B_q(n)$ covering $X$, if $\dim(X)=k$. Similarly we define the {\em $k^{th}$ down operator} $D_{n,k}:\C[B_q(n)]\rar \C[B_q(n)]$ (we have $U_{n,n}=D_{n,0}=0$). Set $U_n=U_{n,0}+U_{n,1}+\cdots +U_{n,n}$ and $D_n=D_{n,0}+D_{n,1}+\cdots +D_{n,n}$, called, respectively, the {\em up and down} operators on $\C[B_q(n)]$. If we think of the elements of $\C[B_q(n)]$ as column vectors with components indexed by the {\em standard basis} elements $B_q(n)$ then $M_q(n)$ is the matrix of the operator $$\cA_q(n) = U_n +\sum_{k=0}^n q^{k-1}D_{n,k}$$ with respect to the basis $B_q(n)$. For $0\leq k \leq n$, define $s_k\in \C[B_q(n,k)]$ by $$ s_k = \sum_{X\in B_q(n,k)} X,$$ and define $R_q(n)$ to be the subspace of $\C[B_q(n)]$ spanned by $s_0,s_1,\ldots , s_n$. Elements of $R_q(n)$ are called {\em radial vectors}. Clearly, $R_q(n)$ is closed under $\cA_q(n)$ and $\dim(R_q(n))=n+1$. We have \beqn \cA_q(n)(s_k)= (k+1)_q\;s_{k+1} + q^{k-1}(n-k+1)_q\;s_{k-1},\;\;0\leq k \leq n. \eeqn It follows from~\eqref{ctm} that the matrix of $\cA_q(n): R_q(n)\rar R_q(n)$ with respect to the basis $\{s_0,\ldots ,s_n\}$ is $K_q(n)$. Thus, knowing the eigenvalues of $K_q(n)$ would at least tell us some of the eigenvalues of $M_q(n)$. The eigenvalues and eigenvectors of $K_q(n)$ have been determined by Terwilliger ~{\cite{t3}} (we shall not use the eigenvector information given in the result below in this paper). \bt \label{mt2} (i) The eigenvalues of $K_q(n)$ are $$(n-k)_q - (k)_q,\;k=0,1, \ldots ,n.$$ (ii) For $0\leq k \leq n$, there is a right eigenvector of $K_q(n)$ corresponding to the eigenvalue $(n-k)_q - (k)_q$ whose component $i,\;0\leq i \leq n$ is given by $$ _3\phi_2 \left( \ba{c} q^{-i}, q^{-k}, -q^{k-n}\\0, q^{-n} \ea \; \vline \; q, q \right)$$ where the basic hypergeometric series notation is from ~{\cite{gr1}}. \et \pf This is Lemma 4.20 in ~{\cite{t3}} (replace $b$ by $q$, $d$ by $n$, and $j$ by $n-k$). Part (i) also follows from Theorem 2 in Johnson ~{\cite{j}} (by taking $a=z=1$ and $h=k=0$). \eprf To show that there are no other eigenvalues of $M_q(n)$ and that the eigenvalue multiplicities are as claimed in Theorem~\ref{mt1} we shall use another result of Terwilliger ~{\cite{t1}}. A {\em symmetric chain} in $\C[B_q(n)]$ is a sequence \beq \label{gjc} &s=(v_k,\ldots ,v_{n-k}),\;\;\;k\leq n/2,& \eeq of nonzero homogeneous elements of $\C[B_q(n)]$ such that \begin{itemize} \item $r(v_i)=i$ for $i=k,\ldots ,n-k$. \item $U_n(v_i)$ is a nonzero scalar multiple of $v_{i+1}$, for $i=k,\ldots , n-k-1$ and $U_n(v_{n-k})=0$. \item $D_n(v_{i+1})$ is a nonzero scalar multiple of $v_i$ for $i=k,\ldots , n-k-1$ and $D_n(v_k)=0$. \end{itemize} Note that the elements of the sequence $s$ are linearly independent, being nonzero and of different ranks. We say that $s$ {\em starts} at rank $k$ and {\em ends} at rank $n-k$. Note that the subspace spanned by the elements of $s$ is closed under $U_n,D_n$ and also $\cA_q(n)$. The following result was proved (in an equivalent form) by Terwilliger ~{\cite{t1}} (see Item~5 of Theorem~3.3 on top of page 208). For a proof using Proctor's $\mathfrak{sl}(2,\C)$ technique ~{\cite{p}} see Theorem 2.1 in ~{\cite{s2}} (where also the result is stated differently but in an equivalent form to that given below). \bt \label{sjb} There exists a basis $T_q(n)$ of $\C[B_q(n)]$ such that \begin{enumerate} \item $T_q(n)$ is a disjoint union of symmetric chains in $\C[B_q(n)]$. \item Let $0\leq k \leq n/2$ and let $(v_k,\ldots ,v_{n-k})$ be any symmetric chain in $T_q(n)$ starting at rank $k$ and ending at rank $n-k$. Then \beqn U_n(v_u)&=& q^k(u+1-k)_q\;v_{u+1},\;k\leq u < n-k.\\ D_n(v_{u+1})&=& (n-k-u)_q\;v_u,\;k\leq u < n-k. \eeqn \end{enumerate} \et We now give the \pf[Proof of Theorem~\ref{mt1}] Observe the following. (i) The number of symmetric chains in $T_q(n)$ starting at rank $k$ and ending at rank $n-k$, for $0\leq k \leq n/2$, is $\qb{n}{k} - \qb{n}{k-1}$. (ii) Let $s=(v_k,\ldots ,v_{n-k})$ be a symmetric chain in $T_q(n)$ starting at rank $k$, where $0\leq k \leq n/2$. Then the subspace spanned by $\{v_k,\ldots ,v_{n-k}\}$ is closed under $\cA_q(n)$ and, by Theorem~\ref{sjb}, the matrix of $\cA_q(n)$ with respect to the basis $s$ is $q^k K_q(n-2k)$. (iii) By Theorem~\ref{mt2} the eigenvalues of $q^k K_q(n-2k)$ are \beqn \lefteqn{ q^k((n-2k-i)_q - (i)_q),\;i=0,1,\ldots ,n-2k}\\ &=& ((n-i)_q - (i)_q),\;i=k,\ldots ,n-k. \eeqn (iv) It now follows from items (i), (ii), (iii) above that the eigenvalues of $\cA_q(n)$ are \beqn (n-j)_q - (j)_q,\;j=0,\ldots ,n, \eeqn with respective multiplicities \beqn \sum_{i=0}^{\min\{j,n-j\}} \qb{n}{i} - \qb{n}{i-1} = \qb{n}{j}. \eeqn That completes the proof. \eprf \section{{A decomposition of \texorpdfstring{$\C[B_q(n)]$} {CBq(n)} }} \label{sec:decomp} In this and the next section we give proofs of Theorems~\ref{mt1} and~\ref{evectors} by inductively writing down an eigenbasis for the operator $\cA_q(n)$. This is based on a direct sum decomposition of the vector space $\C[B_q(n)]$ that was given in the paper ~{\cite{s3}}. This decomposition yields a linear algebraic interpretation of the Goldman--Rota recurrence (see equations~\eqref{d1},~\eqref{d2},~\eqref{grd} and Remark~\ref{remark1}) and is of independent interest. In ~{\cite{s3}} it was used to inductively write down an explicit eigenbasis for the Bose--Mesner algebra of the Grassmann scheme. Here we recall the relevant definitions and results from ~{\cite{s3}}. All the omitted proofs may be found in Section~2 of ~{\cite{s3}}. We can write down the Goldman--Rota identity in terms of the $q$-binomial coefficients, \beq \label{nsw} \qb{n+1}{k} &=& \qb{n}{k} + \qb{n}{k-1} + (q^n - 1)\qb{n-1}{k-1},\;n,k\geq 1,\eeq with $\qb{0}{k}=\delta(0,k)$ and $\qb{n}{0}=1$. Note that~\eqref{gri} follows by summing~\eqref{nsw} over $k$. We shall now give a linear algebraic interpretation to~\eqref{gri} and \eqref{nsw}. Denote the standard basis vectors of $\F^n_q$ by the column vectors $e_1,\ldots ,e_n$. We identify $\F_q^k$, for $k