TITLE: Eigenvalues of a matrix with entries involving combinatorics
QUESTION [14 upvotes]: Let $F(n, l, i, j)$ be the cardinality of the set
\begin{eqnarray*}
\{(k_1, \cdots, k_n)\in\mathbb{Z}^{\oplus n}|0\leq k_r\leq l-1\text{ for }1\leq r\leq n\text{, }k_1+\cdots+k_n=lj-i\}.
\end{eqnarray*}
Define an $n\times n$ matrix $M(l, n)$ by 
\begin{eqnarray*}
M_{ij}(l, n)=(-1)^{i+j}F(n, l, i, j).
\end{eqnarray*}
In fact $M(l, n)$ is related to the Adams operations on $U(n)$, and I can show using algebraic topology that the eigenvalues are $1, l, l^2, \cdots, l^{n-1}$. Note that the last column vector of $M(l, n)$ is $(0, 0, \cdots, 0, 1)$ and so it is an eigenvector corresponding to the eigenvalue 1. When $n=2$, $M(l, 2)$ is $\begin{pmatrix}l&0\\ 1-l& 1\end{pmatrix}$.
Question: Are there more elementary ways to show this?
Added: For a fixed $n$ and different $l$, the matrices $M(l, n)$ commute with each other (more precisely, $M(l_1, n)M(l_2, n)=M(l_1l_2, n)$) and thus they are simultaneously diagonalisable and their eigenvectors do not depend on $l$. See the answer to this question for a set of eigenvectors of $M(l, n)$. A side question is how one can obtain the eigenvectors using elementary methods.

REPLY [2 votes]: As I mentioned in the question, the matrix $M(l, n)$ is related to the Adams operations on $U(n)$. Let me elaborate further on this. 
Recall that for a finite CW complex $X$, $K^{-1}(X)$ is $[X, U(\infty)]$, the group of homotopy classes of maps from $X$ to $U(\infty)$. Let $G$ be a compact Lie group and $\delta: R(G)\to K^{-1}(G)$ be the map which sends a $G$ representation $\rho$ to the homotopy class of it viewed the map $G\stackrel{\rho}{\rightarrow}U(n)\hookrightarrow U(\infty)$. If $G=U(n)$, then by a theorem of L. Hodgkin, its $K$-theory is the exterior algebra
\begin{eqnarray}
K^*(U(n))=\bigwedge\nolimits^*_\mathbb{Z}(\delta(\sigma_n), \delta(\wedge^2\sigma_n), \cdots, \delta(\wedge^n\sigma_n)),
\end{eqnarray}
where $\sigma_n$ is the standard representation of $U(n)$. The Adams operation $\psi^l$ on the primitive $\mathbb{Z}$-module of this exterior algebra has $lM(l, n)$ as the matrix representation with respect to the basis $\delta(\sigma_n), \cdots, \delta(\wedge^n\sigma_n)$. The definition of Adams operations implies that they commute with each other, so do $M(l, n)$ for different positive $l$ and fixed $n$, and their eigenvectors are independent of $l$ due to simultaneous diagonazability.
Using the definition of Chern character and Adams operation, one can easily get the following statement: 

Let $X$ be a finite CW-complex and $\alpha\in K^{-1}(X)$. Then we have 
  \begin{eqnarray}
\text{ch}(\psi^l(\alpha))=\sum_il^i\text{ch}_{2i-1}(\alpha)
\end{eqnarray}
  where $\text{ch}_k(\alpha)$ is the degree $k$ term of $\text{ch}(\alpha)$.

Returning to the case $X=U(n)$, we know that $H^*(U(n), \mathbb{Q})$ is also an exterior algebra on $n$ primitive generators of degrees $1, 3, 5, \cdots, 2n-1$. So if $\alpha$ is a (rational) primitive element of $K^*(U(n))\otimes\mathbb{Q}$ which is also an eigenvector of $\psi^l$, then there exists $1\leq j\leq n$ such that $\text{ch}_{2i-1}(\alpha)$ is zero for $i\neq j$ but nonzero for $i=j$, and the corresponding eigenvalue is $l^j$. So the eigenvalues of $lM(l, n)$ are $l, l^2, \cdots, l^n$. This recent paper has more details of the above discussion.