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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.

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.

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 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.

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.

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 and thus they are simultaneously diagonalisable and their eigenvectors do not depend on $l$.

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 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.