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](https://mathoverflow.net/questions/255869/eigenvectors-of-a-matrix-with-entries-involving-combinatorics) for a set of eigenvectors of $M(l, n)$. A side question is how one can obtain the eigenvectors using elementary methods. 



Eigenvalues of a matrix with entries involving combinatorics
One can prove this statement along the following lines.


1. Prove that ${\rm Trace}(M(l,n)) = 1+l +\cdots + l^{n-1}$.
2. Prove that $M(l,n)^p = M(l^p,n)$.


Clearly, these statements 1 and 2 together imply that the eigenvalues 
are $1,l,\ldots, l^{n-1}$.


We will assume throughout that $l\geq 2$, otherwise the statement is obvious.


The proof of the first statement is by induction on $n$. The case $n=1$ is obvious. We are now looking for the cardinalities of sets of $k\_1,\ldots, k\_n\in [0,l-1]$ whose sum is divisible by $l-1$. If this number is $A\_{n,l}$, then we have 
$$
A\_{n,l} = 2 A\_{n-1,l} + (l^{n-1} - A\_{n-1,l})
$$
by splitting these sets into those with $k\_n = 0\mod (l-1)$ and 
$k\_n\neq 0\mod (l-1)$.


Clearly, this recursion proves statement 1.


Let me prove the statement 2 for p=2. Let me ignore the $(-1)^{i+j}$ signs in the definition of the matrix, since these can be removed by conjugating by a ${\rm diag}((-1)^i)$ matrix.


We are looking for a number of ways of writing 
$$
l^2\, j - i = a\_1 + \ldots + a\_n
$$
with $a\_r \in [0,l^2-1]$. We can write for each $r$
$$
a\_r = l \,b\_r + c\_r
$$
with $b\_r,c\_r\in [0,l-1]$.


We then have 
$$
l^2\,j - i = l\, \sum\_r b\_r + \sum\_r c\_r.
$$


This shows that $\sum\_r c\_r = -i \mod l$, so 
$$
\sum\_r c\_r = l j\_1 - i
$$
for some $j\_1$. Clearly, $j\_1\in [1,\ldots,n]$.


Then we have 
$$
\sum\_r b\_r = l j - j\_1.
$$
Therefore, 
$$
M(l^2,n)\_{i,j} = \sum\_{j\_1} M(l,n)\_{i,j\_1}M(l,n)\_{j\_1,j}.
$$
This proves the desired matrix identity. 


The argument for $p>2$ is completely analogous (or one can prove
$M(l^p,n) = M(l^{p-1},n)M(l,n)$). So we are done.


