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Given a square matrix $A$ and a delay matrix $D(\lambda) = \textrm{diag}(\lambda^{m_1},\dots,\lambda^{m_N})$, where all $m_i$'s are integers.

I'm interested in the generalised characteristic polynomial of $A$:

$\tilde p_A(\lambda) = \det\left(D(\lambda) - A\right)$

For all $m_1 = \dots = m_N = 1$, $\tilde p_A$ is clearly the standard characteristic polynomial $p_A$. The order of $\tilde p_A$ is $\tilde N = \sum_{i=1}^N m_i$.

Is this a well-known structure, because I cannot find anything about it?

It is well known, that:

$p_A(\lambda) = \sum_{k = 0}^N (-1)^{N-k} \textrm{ tr}_k(A) \lambda^k$

with the $k$-th trace, $\textrm{ tr}_k(A)$, of a square matrix $A$ being the sum of all $k \times k$ sub-determinants got from $A$ be deleting corresponding rows and columns (so that the diagonal of each sub-determinant coincides with the diagonal of $A$). We define $\textrm{tr}_0(A) = 1$.

For the generalised case I conjecture:

$\tilde p_A(\lambda) = \sum_{k = 0}^{\tilde N} \lambda^k \sum_{I \in I_k} (-1)^{\tilde N-k} \det(A_{I}) $

where, $I_k = \{ I \subset \left\{ 1, \dots, N \} \middle| \sum_{i \in I} m_i = k \right\}$ is the set of index sets which sum up to $k$, $A_I$ is the sub-matrix of $A$ with the columns and rows with index $I$ deleted.

I'm looking for a proof of the equation.

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