Eigenvalues of a matrix with entries involving combinatorics
One can prove this statement along the following lines.
- Prove that ${\rm Trace}(M(l,n)) = 1+l +\cdots + l^{n-1}$.
- 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.
Let $F(n,\ell)$ be the matrix with coefficients $$F_{i,j}(n,\ell)=[t^{\ell j-i}] \left(\frac{1-t^\ell}{1-t}\right)^n,\,\;\;\;1\leq i,j \leq n$$ Above Pat Devlin pointed out that it suffices to show that the $(n-1)\times (n-1)$ submatrix $L(n,\ell)$ with coefficients $$L_{i,j}(n,\ell)=[t^{\ell j-i}] \left(\frac{1-t^\ell}{1-t}\right)^n,\,\;\;\;1\leq i,j \leq n-1$$ has eigenvalues $\ell,\ell^2,\ldots,\ell^{n-1}$.
In fact, for positive integer $\ell\geq 2$ the matrices $P(n,\ell)$ with coefficients $$P_{i,j}(n,\ell)=\frac{1}{\ell^n} [t^{(j+1)\ell-i-1}] \left(\frac{1-t^\ell}{1-t}\right)^{n+1},\,\;\;\; 0\leq i,j \leq n-1$$ are known. They are the transition matrices for the Markov chains describing the propagation of carries when $n$ integers which have independent uniform $\ell$-ary ''digits'' are added (clearly $\ell^n\cdot P(n,\ell)=L(n+1,\ell)$).
These matrices are subject of the fascinating article Carries, Combinatorics and an Amazing Matrix by John Holte (American Mathematical Monthly, 104 (2), 1997)).
Holte proved that $P(n,\ell)$ has eigenvalues $1,\ell^{-1},\ldots,\ell^{-(n-1)}$, that the eigenvectors do not depend on the base $\ell$, and described the left and right eigenvectors explicitly. He also showed that $P(n,a)\cdot P(n,b)=P(n,ab)$.
The $P(n,b)$ also appear in the probability of card shuffling. They are the transition matrices for the Markov chains describing the descents in the permutations generated by shuffling a deck of $n$ cards with successive $b$-shuffles. (Persi Diaconis and Jason Fulman, Carries, shuffling and an amazing matrix, AMM November 2009 (arXiv preprint)).
Here's part way to an answer. I reduce the problem to one that sounds simpler, and I also argue why $l^{n-1}$ is an eigenvalue.
Let $f_{n, l} (k)$ denote the number of ways to write $k$ as an ordered sum $k = a_1 + \cdots + a_{n}$ where $a_{i} \in \{0, 1, \ldots, l-1\}$ [so your function is $F(n, l, i, j) = f_{n,l} (lj - i)$]. The matrix you considered is
$$ M = \Big( (-1)^{i+j} f_{n,l} (lj - i) \Big)_{i,j}. $$ Since you only want the eigenvalues, we can ignore the $(-1)^{i+j}$ term since doing so will give us a matrix similar to $M$. Moreover, since (as noted in original post) the last column of this matrix is $(0, 0, \ldots , 0, 1)$, we can ignore the last row and last column of $M$ when finding the other eigenvalues. Thus, your claim is equivalent to the following:
Equivalent problem: Consider the $(n-1) \times (n-1)$ matrix $A$ whose $(i,j)$-entry is given by $f_{n,l} (lj - i) = F(n, l, i,j)$. Show that the eigenvalues of $A$ are $l, l^2, \ldots, l^{n-1}$.
We first claim that the row sums of $A$ are constant and equal to $l^{n-1}$ [this shows one of the eigenvalues we need]. To see this, note the sum along row $i$ is equal to $\sum_{j=1} ^{n-1} f_{n,l} (lj - i)$, which is equal to the number of $n$-tuples in $\{0, 1, \ldots, l-1\}^n$ whose sum is $-i$ modulo $l$. This is clearly $l^{n-1}$ since the first $n-1$ values are free choices, and the last is then determined by the others.
Some other facts: Let $R$ be the $(n-1) \times (n-1)$ matrix with $1$'s along the diagonal $i+j=n$ and $0$'s elsewhere. Then because $f_{n,l} (k) = f_{n,l} (n(l-1)-k)$, this implies $AR = RA$ (i.e., $a_{i,j} = a_{n-i, n-j}$). (This fact looks very useful for finding eigenvectors, and it's why I think one should ignore the last row and column of $M$.)
It's also not difficult to see (if one's familiar with generating functions in enumeration problems) that $$\sum_{k=0}^{\infty} x^{k} f_{n,l} (k) = \left( 1 + x + x^2 + \cdots + x^{l-1} \right)^n = \left( \frac{1-x^{l}}{1-x} \right)^n.$$