Fix arbitrary real numbers $x_1,\ldots ,x_n$ which are pairwise distinct, i.e. so that $x_i \neq x_j$ for any pair $i \neq j$. Let $A = (a_{ij})$ be the following $n \times n$ matrix: Its diagonal entries are given by the equation,
$$a_{ii}=\sum_{j\neq i}\frac{x_i}{x_i-x_j},$$
while its off-diagonal entries given by the equation,
$$a_{ij}=\frac{x_i}{x_i-x_j}$$,
for $i\neq j$. For instance when n=2, the matrix looks like:
$$A=\begin{pmatrix}
\frac{x_1}{x_1-x_2} & \frac{x_1}{x_1-x_2}\\
\frac{x_2}{x_2-x_1} & \frac{x_2}{x_2-x_1}
\end{pmatrix}$$
Prove that the set of eigenvalues for the matrix A is of the form $\left \{0,1,\ldots,n-1\right \}$.
I'm completely lost as to how to continue. I've tried to work on the determinant of $A-\lambda I$ for $2 \times 2$ and $3 \times 3$ matrices $A$ but I haven't managed to find anything helpful towards the proof.
Update 1
I'm not entirely sure how to write this formula nicely as a mathematical expression, but as code in python I have that the $k$th element of $v_0$ is
p = product([L[i-1] - L[j-1] for i in [1..n] for j in [i+1..n] if i != k and j != k])
v[k-1] = p if k % 2 == 1 else -p
where L refers to the list [x_1,...,x_n], and I want for $1 \leq i < j \leq n$ and for $i \neq j \neq k$
I also have that $v_i$ is equal to $diag(x_1,…,x_n)^{i}v_0$
Update 2
the eigenvectors for the case where $n=4$ are,
$$ v_0 = \begin{pmatrix}
1\\
-\frac{x_1^2 – x_1x_3 – (x_1 – x_3)x_4}{x_2^2 – x_2x_3 – (x_2 – x_3)x_4}\\
\frac{x_1^2 – x_1x_2 – (x_1 – x_2)x_4}{x_2x_3 – x_3^2 – (x_2 – x_3)x_4}\\
-\frac{x_1^2 – x_1x_2 – (x_1 – x_2)x_3}{x_2x_3 – (x_2 + x_3)x_4 + x_4^2}
\end{pmatrix},$$
$$v_1 = \begin{pmatrix}
1\\
-\frac{x_1^2x_2 – x_1x_2x_3 – (x_1x_2 – x_2x_3)*x_4}{x_1x_2^2 – x_1x_2x_3 – (x_1x_2 – x_1x_3)x_4}\\
-\frac{(x_1 – x_2)x_3x_4 – (x_1^2 – x_1x_2)x_3}{x_1x_2x_3 – x_1x_3^2 – (x_1x_2 – x_1x_3)x_4}\\
-\frac{(x_1^2 – x_1x_2 – (x_1 – x_2)x_3)x_4}{x_1x_2x_3 + x_1x_4^2 – (x_1x_2 + x_1x_3)x_4}
\end{pmatrix},$$
$$v_2 = \begin{pmatrix}
1\\
-\frac{x_1^2x_2^2 – x_1x_2^2x_3 – (x_1x_2^2 – x_2^2x_3)x_4}{x_1^2x_2^2 – x_1^2x_2x_3 – (x_1^2x_2 – x_1^2x_3)x_4}\\
-\frac{(x_1 – x_2)x_3^2x_4 – (x_1^2 – x_1x_2)x_3^2}{x_1^2x_2x_3 – x_1^2x_3^2 – (x_1^2x_2 – x_1^2x_3)x_4}\\
-\frac{x_1^2 – x_1x_2 – (x_1 – x_2)x_3)x_4^2}{x_1^2x_2x_3 + x_1^2x_4^2 – (x_1^2x_2 + x_1^2x_3)x_4}
\end{pmatrix},
$$
$$v_3 = \begin{pmatrix}
1\\
-\frac{x_1^2x_2^3 – x_1x_2^3x_3 – (x_1x_2^3 – x_2^3x_3)x_4}{x_1^3x_2^2 – x_1^3x_2x_3 – (x_1^3x_2 – x_1^3x_3)x_4}\\
-\frac{(x_1 – x_2)x_3^3x_4 – (x_1^2 – x_1x_2)x_3^3}{x_1^3x_2x_3 – x_1^3x_3^2 – (x_1^3x_2 – x_1^3x_3)x_4}\\
-\frac{x_1^2 – x_1x_2 – (x_1 – x_2)x_3)x_4^3}{x_1^3x_2x_3 + x_1^3x_4^2 – (x_1^3x_2 + x_1^3x_3)x_4}
\end{pmatrix}$$
Update 3
I managed to work out a formula for the $j$th element of the eigenvector with $eigenvalue=\lambda$ and $size=n$,
$$(-1)^{j+1}\frac{x_j^\lambda}{x_1^\lambda} \prod_{k\neq 1,k\neq j}^n \frac{x_1-x_k}{x_j-x_k}
$$
I'm just not sure now as to how to use the formula for one entry of an eigenvector to prove the set of eigenvalues
Best Answer
For $ k = 0, 1, \ldots, n-1$, consider the (horizontal) vector $v_k$ with $i$th coordinate $$ \sum \prod_{j=1, a_j \neq i}^{k} {x_{a_j}}.$$
For example, with $n = 3$, we have
$v_0 = (1, 1, 1)$,
$v_1 = (x_2 + x_3, x_3 + x_1, x_1 + x_2)$,
$v_2 = ( x_2x_3, x_3x_1, x_1x_2)$.
With $n = 4$, we have
$v_0 = (1, 1, 1, 1)$,
$v_1 = (x_2 + x_3 + x_4, x_3 + x_4 + x_1, x_4 + x_1 + x_2, x_1 + x_2 + x_3)$,
$v_2 = ( x_2x_3+x_3x_4+x_4x_2, x_3x_4+x_4x_1+x_1x_3, x_4x_1+x_1x_2+x_2x_4, x_1x_2 + x_2x_3 + x_3x_2)$,
$v_3 = (x_2x_3x_4, x_3x_4x_1, x_4x_1x_2, x_1x_2x_3)$.
Claim: $v_kA = (n-1-k) v_k$.
Proof: Expand it. A lot of the cross terms cancel out.
For example, with $v_0$, the column sum is $n-1$, so $v_0 A = (n-1) v_0$.
For example, with $v_{k-1}$, the numerators are all $\prod x_i$, and by looking at the denomninators, they cancel out to 0, so $v_{k-1} A = 0 $.
Do you see how we get $v_k A = (n-1-k)v_k$?
Corollary: The eigenvalues are $0, 1, 2, \ldots, n-1 $.