When solving a (linear) recurrence relation using the characteristic polynomial method sometimes there are roots with multiplicity. Trying to find the solution coefficients (using the initial conditions) using a matrix, a special form arises (Vandermonde like matrix): Assume $ r_1,\dots,r_m$ are the roots of the characteristic polynomial with multiplicities $ k_1,\dots, k_m$ (respectively) s.t. $ \sum_{i=1} ^m k_i =n$ (where $ n$ is the degree of the characteristic polynomial). Then the solution coefficients can be obtained using the following matrix
$$ \begin{pmatrix} 1 & 0 & \cdots & 0 & 1 & 0 & \cdots & 0 & \cdots & 1 & 0 & \cdots & 0 \newline r_1 & r_1 & \cdots & r_1 & r_2 & r_2 & \cdots & r_2 & \cdots & r_m & r_m & \cdots & r_m \newline r_1 ^2 & 2r_1 ^2 & \cdots & 2^{k_1 -1}r_1 ^2 & r_2 ^2 & 2r_2 ^2 & \cdots & 2^{k_2-1}r_2 ^2 & \cdots & r_m ^2 & 2r_m ^2 & \cdots & 2^{k_m -1}r_m ^2 \newline \vdots & \vdots & \cdots & \vdots & \vdots & \vdots & \cdots & \vdots & \cdots & \vdots & \vdots & \cdots & \vdots \newline r_1 ^{n-1} & (n-1)r_1 ^{n-1} & \cdots & (n-1)^{k_1 -1}r_1 ^{n-1} & r_2 ^{n-1} & (n-1)r_2 ^{n-1} & \cdots & (n-1)^{k_2-1}r_2 ^{n-1} & \cdots & r_m ^{n-1} & (n-1)r_m ^{n-1} & \cdots & (n-1)^{k_m -1}r_m ^{n-1} \newline \end{pmatrix} $$ I saw somewhere that the determinant is $ (\prod_{i=1} ^{m} r_i ^{k_i \choose 2})(\prod_{1\leq j<i\leq m} (r_{i}-r_{j})^{k_j k_i})$ without an explanation. I would be glad to know how to get to this result, I tried to use the methods of the proof of Vandermonde matrix but couldn't reach that result (or any other result) and I found the proof when there is one root with multiplicity $ n$ but couldn't generalize it to the general case. Thanks!
